Determining the amperage mat a main disconnect fuse block

[Scott Bray]

I inspected an apartment in a building with 200 apartment units, The main disconnect for the sub panel of the apartment is located at the electrical meter in the basement of the building. There are two 60 amp cartridge type fuses in a fuse block as the disconnect. I stated that it was a 60 amp service and the building supervisor stated no it is a 120 amp service because you combine the two legs of the service. I am confusing my self, I always thought you rated the disconnect by the weakest link, which would be the 60 amp fuse. Which is correct, is it a 60 amp service or 120 amp service ? Thanks in advance.:confused:

[Jim Port]

That would be a 60 amp panel, not 120 amps. You do not add the leg ratings together when describing the service or feeder.

[Scott Bray]

That would be a 60 amp panel, not 120 amps. You do not add the leg ratings together when describing the service or feeder.

Thanks,That's what I reported and the building supervisor and there electrician are stating that I am wrong,I started second guessing my self.

[Bruce Low]

Think of it as a "2 pole" breaker that has fuses instead..... A 2 pole fuse block. As with breakers, the ratings are not additive. So, 60 amp is correct IF you confirmed that the service feeders to the fuse block can support 60 amps (# 4 Al or # 6 Cu).

Bruce Low
Bottom Line Home Inspection
Northeast Wisconsin
Go Pack Go

[Roland Miller]

Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder.

At 120 volts you still only have 60 amps available.:)

[Jim Port]

At 120 volts you still only have 60 amps available.:)

Fixed this for you. At 120 volts you still only have 60 amps available per leg.

[Roland Miller]

Fixed this for you. At 120 volts you still only have 60 amps available per leg.


I guess you will have to show me where you could measure 120 amps on a 3-wire, 120/240, single phase transformer with the 120 volt phase/circuit each pulling 60 amps;)

[Jim Port]

Think about it, you could try to pull 75 amps on one leg of the 2 pole breaker while the other leg is at 0. Depending on the trip curve, the breaker will trip. You could also have each of the 2 hots pulling 60 amps and the breaker does not trip. 60 + 60=120.

[Vern Heiler]

Think about it, you could try to pull 75 amps on one leg of the 2 pole breaker while the other leg is at 0. Depending on the trip curve, the breaker will trip. You could also have each of the 2 hots pulling 60 amps and the breaker does not trip. 60 + 60=120.

Jim, take a look at "Code Check Electrical" pg. 14.

[Jim Port]

I don't have a copy Vern, can you post the page?

[Vern Heiler]

I don't have a copy Vern, can you post the page?
Having trouble up loading scanned file:(

[Roland Miller]

Think about it, you could try to pull 75 amps on one leg of the 2 pole breaker while the other leg is at 0. Depending on the trip curve, the breaker will trip. You could also have each of the 2 hots pulling 60 amps and the breaker does not trip. 60 + 60=120.

Lets use instantaneous fuses at 60 amps for fun... show me where you could put an amp meter and actually measure more then 60 amps. the math is more like +60 +(-60) = 0. 60 amps is still all you can draw with the fuses at 120 volts.. the 0 would be on the neutral with 60 respectively on each phase or leg..never more..

Breaker trip curves make this an unrealistic problem.

[Vern Heiler]

I don't have a copy Vern, can you post the page?

I've tried several times and IN does not tell me why it failed, just that it failed.

What the page shows is that the neutral leg has the difference current of the two 240 legs, thus you can only have the max current of either of the circuit breakers not the sum of the two.

[Jim Port]

The load on the neutral doesn't matter in this example. Leg A carries 60 amps, leg B carries 60 amps equaling 120 amps of power. The 120 power is not returning on the other leg. It is on the neutral.

[Vern Heiler]

I think I have done it:D Yay

[Jim Port]

Ok, I understand that. Leg A carries 40 amps, Leg B carries 20 amps. The neutral load will be the difference of between A and B or 20 amps. This is why a MWBC requires the hots to be on opposite legs.

If both hots were on the same leg of the panel, the neutral now carries the additive of the two hots or 60 amps. This would create a fire hazard on the now undersized conductor.

Each leg can carry up to the limit of the breaker. It does not matter how much the other leg carries.

[Vern Heiler]

Ok, I understand that. Leg A carries 40 amps, Leg B carries 20 amps. The neutral load will be the difference of between A and B or 20 amps. This is why a MWBC requires the hots to be on opposite legs.

If both hots were on the same leg of the panel, the neutral now carries the additive of the two hots or 60 amps. This would create a fire hazard on the now undersized conductor.

Each leg can carry up to the limit of the breaker. It does not matter how much the other leg carries.

I think where you are getting tripped up is that as soon as you put a load on the other leg, then they are shared, and the neutral only carries the difference.

[Jim Port]

A load on Leg A does not affect anything on Leg B. The only thing that changes is the load on the neutral.

[Jim Port]

Think of this, a 20 amp breaker is installed in a panel. What capacity does the circuit have? 20 amps.

Now add another 20 amp breaker for a second circuit. Again the capacity is 20 amps. It has not affected the original circuit capacity.

The only thing is that you can now pull 40 amps of power from the panel.

[Roland Miller]

The load on the neutral doesn't matter in this example. Leg A carries 60 amps, leg B carries 60 amps equaling 120 amps of power. The 120 power is not returning on the other leg. It is on the neutral.

Jim, Do you want to talk about amps or power. Cause if you are talking about power you have just changed the subject.

No where, No how will you ever be able to measure more then 60 amps with an amp meter. Even if you stand on your head and hold your tongue right....:confused:

- - - Updated - - -


Think of this, a 20 amp breaker is installed in a panel. What capacity does the circuit have? 20 amps.

Now add another 20 amp breaker for a second circuit. Again the capacity is 20 amps. It has not affected the original circuit capacity.

The only thing is that you can now pull 40 amps of power from the panel.


Never heard of 40 Amps of power. What is it??

[Vern Heiler]

A load on Leg A does not affect anything on Leg B. The only thing that changes is the load on the neutral.

If leg A is carrying 60a and leg B is carrying 60a, and the neutral only carries the difference (0), then the single wire that makes up the secondary of the xfmr 240v leg to leg, now has 120a of current with 60a breakers.

[Roland Miller]

If leg A is carrying 60a and leg B is carrying 60a, and the neutral only carries the difference (0), then the single wire that makes up the secondary of the xfmr 240v leg to leg, now has 120a of current with 60a breakers.


No, no and no not ever....never..Please read the other posts..:(

[Roland Miller]

http://www.mikeholt.com/instructor2/img/product/pdf/1302643809-sample.pdf

Maybe this will help, hopefully.:thumb:

[Jim Port]

Each leg can carry 60 amps. You have two legs available at 60 amps. 60 x 2 = 120.

[Vern Heiler]

No, no and no not ever....never..Please read the other posts..:(

I don't think I made myself clear. One piece of wire with a 60a breaker at each end will only carry 60a without tripping the breaker.

[Jim Port]

Never heard of amps of power? Power = amps x voltage. Can also be expressed in watts. The only difference is that amps ate dependent on voltage while watts are constant, regardless of voltage.

20 amps x 120v = 2400 watts. 20 amps x 240v = 4800 watts.
2400 watts/120 = 20 amps
2400 watts/240 = 10 amps.

- - - Updated - - -

The Op stated the fuse block had 2 60 amp fuses in it. Each leg can carry 60 amps. Two legs at 60 amps is 120 amps.

This is not a transformer discussion.

[Vern Heiler]

Never heard of amps of power? Power = amps x voltage. Can also be expressed in watts. The only difference is that amps ate dependent on voltage while watts are constant, regardless of voltage.

20 amps x 120v = 2400 watts. 20 amps x 240v = 4800 watts.
2400 watts/120 = 20 amps
2400 watts/240 = 10 amps.

- - - Updated - - -

The Op stated the fuse block had 2 60 amp fuses in it. Each leg can carry 60 amps. Two legs at 60 amps is 120 amps.

This is not a transformer discussion.

He also stated it is the disconnect for the service.

[Jim Port]

And each leg of the disconnect can support 60 amps. There are 120 amps of power available at 120 volts or 60 amps at 240 volts.

- - - Updated - - -

Repeated again from post #4.



Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder.

[Vern Heiler]

And each leg of the disconnect can support 60 amps. There are 120 amps of power available at 120 volts or 60 amps at 240 volts.

- - - Updated - - -

Repeated again from post #4.
Take the diagram I posted for you and draw out where the 120a comes from and goes to.

[Jim Port]

With 60 amps on each leg there would be 0 Amps on the neutral. The 120 power is not returned on the opposite leg, but the neutral. A 240 volt circuit would return the power on the other leg.

[Vern Heiler]

With 60 amps on each leg there would be 0 Amps on the neutral. The 120 power is not returned on the opposite leg, but the neutral. A 240 volt circuit would return the power on the other leg.

Draw it out. And remember; " Kirchhoff's current law (KCL) (http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws#Kirchhoff.27s_current_law _.28KCL.29) The current entering any junction is equal to the current leaving that junction.

[Jim Port]

Think of this, a 20 amp breaker is installed in a panel. What capacity does the circuit have? 20 amps.

Now add another 20 amp breaker for a second circuit. Again the capacity is 20 amps. It has not affected the original circuit capacity.

The only thing is that you can now pull 40 amps from the panel.

I really don't know how to make it any simpler than this.

[Vern Heiler]

I really don't know how to make it any simpler than this.
Then you really didn't like MI5:). Draw it out if you can. Also remember that current flows only in one direction at a time in the secondary coil or any wire for that matter.

[Vern Heiler]

This should help. Note that all of the current goes through one of the legs.

[wayne soper]

it doesn't really matter what the fuse or breaker size is, what matters is the size of the main wire and what IT is rated for.We all have seen 100 amp breakers on wires rated for 60 AMPS. So what size was the main wire coming into the panel? You guys always get so caught up in who's got the biggest code book that you overlook the question or best advice and start running around in circles. The first question to Scott , In my opinion, should have been, What size are the main wires?

[Jim Port]

Wayne, as described in the OP, there would be a descrepency as to the correct sizing of the wires. One person is saying the service is twice as large as it actually is. If you were to believe that person the wiring is too small when in actuality it could be correct. The wiring does not need to be sized for 120 amps, but you do have 120 amps between the two legs available for usage.

[Vern Heiler]

Wayne, as described in the OP, there would be a descrepency as to the correct sizing of the wires. One person is saying the service is twice as large as it actually is. If you were to believe that person the wiring is too small when in actuality it could be correct. The wiring does not need to be sized for 120 amps, but you do have 120 amps between the two legs available for usage.

Jim please show me how you can have 120 amps with 60 amp breakers.

[wayne soper]

The question what is the size of the service 60 or 120 amps. That was answered in the first few posts. I don't see how that's overlooking the question as you've suggested.

Since Scott never mentioned the conductor size we can only assume that it was at least a 60 amp conductor but as you've stated maybe someone should have asked. :)

Yes, but you know we can never assume! That's why I asked the question. and, if the OP was answered in the first few posts, why are there 40 odd posts after the fact hemming and hawing about which end of a candlestick will light. I just think that sometimes a simple answer or question is enough. we don't need to turn a post into a $hit fight,

[Jim Port]

Again, each leg can carry 60 amps. There are 2 legs, 60 + 60 =120.

Assume that you could get a single pole 60. Install it on leg A and load it up. You can flow 60 amps, ignoring trip curves. Now install another single pole 60 on leg B. Again you can pull 60 amps, even if you turned off the first breaker. With both breakers on you have 120 amps available.

[Vern Heiler]

Again, each leg can carry 60 amps. There are 2 legs, 60 + 60 =120.

Assume that you could get a single pole 60. Install it on leg A and load it up. You can flow 60 amps, ignoring trip curves. Now install another single pole 60 on leg B. Again you can pull 60 amps, even if you turned off the first breaker. With both breakers on you have 120 amps available.

If you can't diagram a simple parallel circuit, don't try to explain it to one who can!
( I was only trying to help but you refuse to open your eyes!)

[Mike Borchardt]

It is not possible to exceed the amperage the fuse is rated for (unless its a slow blow fuse and the amperage is exceeded for a brief period of time such as starting a motor), but that's not the type of fuse were talking about. The only reason I mention slow blow is because I assume feed back on it. Keep in mind were talking AC current and the two legs are 180 deg out of phase. Leg A is positive 120 volts at the same time leg B is -120 volts. To have 120 volts you need one leg A OR B to neutral or ground. To have 240 volts you need legs A AND B with no neutral. Leg A OR B to neutral is (120 + 0 =120 volts). Legs A AND B is (Positive 120 + minus 120 = 240 volts). Remember were talking about vectors not sums. A positive 120 volts when added to negative 120 volts does not equal 0 as one would expect. What is being measured is the difference between the two peaks of a sign wave. The sign wave swings from positive 120 volts to negative 120 volts and the difference between the two peaks is 240 volts. The two peaks being 180 degrees out of phase means there is a voltage - not an amperage - differential between leg A and B. Regardless which leg is positive or negative at any given time each leg will only allow a load up to the fuse size which in this case is 60 amps. :) This is a 60 amp service. Jim, your logic isn't totally correct. You are saying that each leg has 60 amps available so this equates to 120 total amps. You are correct in saying each leg can supply 60 amps for a total of 120 amps, And, if each leg had a load of 60 amps at 120 volts, you would consume 120 amps and the most the neutral would carry is 60 amps because the legs are phased 180 degrees apart. So far so good. Here's where the formula fails; If you insert one double pole 60 amp breaker and put a 60 amp load on it, you've maxed out the available amount of current on both legs and any more breakers whether it be single or double pole will blow a fuse. The amperage available, determined by the amperage of the fuses, is based on 240 volts, not 120 volts. If you exceed 60 amps at 120 volts on either leg, you will blow a fuse, If you exceed 60 amps at 240 volts, you will blow fuses. this is why its a 60 amp panel.

[Mike Borchardt]

You cannot state the amperage without referencing the voltage.


The load on a 2-pole, 20 amp CB supplying a 240 volt, 20 amp load is 20 amps. The same 2 pole 20 amp CB with a neutral (MWBC) can supply 40 amp (20*2) @ 120 volts each.
Almost correct. A two pole 20 amp CB with a neutral (MWBC) will supply two 20 amp 120 volt circuits or one 20 amp 240 volt circuit.

[Mike Borchardt]

Again, each leg can carry 60 amps. There are 2 legs, 60 + 60 =120.

Assume that you could get a single pole 60. Install it on leg A and load it up. You can flow 60 amps, ignoring trip curves. Now install another single pole 60 on leg B. Again you can pull 60 amps, even if you turned off the first breaker. With both breakers on you have 120 amps available.
Jim, you are absolutely correct. And the neutral would be zero volt.

[Mike Borchardt]

This should help. Note that all of the current goes through one of the legs.
In regards to the picture, I'm assuming due to the size of the light bulbs, this is an unbalanced circuit. Show a balanced circuit and the current path would be quite different. Jim is correct in stating a single pole 60 amp breaker on each leg could give you 120 amps total consumption. Think of it this way; If I'm running a 60 amp 120 volt motor connected to a circuit breaker off of leg A, and at the same time I'm running a 60 amp 120 volt motor connected to a circuit breaker off of leg B, even though each leg is maxed at 60 amps, I'm still consuming 120 amps.

[Vern Heiler]

In regards to the picture, I'm assuming due to the size of the light bulbs, this is an unbalanced circuit. Show a balanced circuit and the current path would be quite different. Jim is correct in stating a single pole 60 amp breaker on each leg could give you 120 amps total consumption. Think of it this way; If I'm running a 60 amp 120 volt motor connected to a circuit breaker off of leg A, and at the same time I'm running a 60 amp 120 volt motor connected to a circuit breaker off of leg B, even though each leg is maxed at 60 amps, I'm still consuming 120 amps.

You are 100% wrong! If you would look at the circuit diagram you would see that ALL of the total current is passing through one of the legs of the transformer and any additional current on either side of the neutral will be added to that current, thus the fuse or breaker in that leg will trip.

And by the way you don't have a phase difference across the coil, it is a single phase. What you are describing is a multi-phase where one leg is lagging the other by some number of degrees.

[Jim Port]

This is not about a transformer. This is about how much power can be supplied through a 2 pole 60 amp fused disconnect as asked about in the OP.

[Vern Heiler]

This is not about a transformer. This is about how much power can be supplied through a 2 pole 60 amp fused disconnect as asked about in the OP.

And what do you think is on the other ends of the SEC conductor wires?

[Jim Port]

And what do you think is on the other ends of the SEC conductor wires?

That is upstream of anything germane to this discussion. It will have no bearing, unless it is too small, on the ability to pull up to 120 amps through the disconnect. Since this is an apartment complex I seriously doubt the transformer size is an issue.

[Vern Heiler]

That is upstream of anything germane to this discussion. It will have no bearing, unless it is too small, on the ability to pull up to 120 amps through the disconnect. Since this is an apartment complex I seriously doubt the transformer size is an issue.

Read what is written in post #4, under "3-Wire Edison Circuits". The issue is, and has been from since the statement that you gave, saying 120 amps could be drawn through a 60 amp service if it was all 120v loads. I put arrows showing the current paths on the fig. 42, which I stole from Code Check (hope I don't hear from there lawyer). The whole page 14 is to help understand where the currents are going and the role of the neutral. All of the current must pass through one of the legs and when the current exceeds the OPD, it trips or blows. So if you have 60 amps all on one side of the buss (transformer) and you add additional amps on the other side of the buss, they add and trip the CB.

[Jim Port]

If you look at the current flow in this diagram you will see that all the current does not flow through one leg. http://www.ibiblio.org/kuphaldt/electricCircuits/AC/02168.png

If your statement about all the current through one leg were true, turning off or losing one fuse would cause everything not to work. That would not be the case as the 120 loads from the other leg would still function. Think about when one leg is lost at the meter or main breaker.

[Vern Heiler]

If you look at the current flow in this diagram you will see that all the current does not flow through one leg. http://www.ibiblio.org/kuphaldt/electricCircuits/AC/02168.png

If your statement about all the current through one leg were true, turning off or losing one fuse would cause everything not to work. That would not be the case as the 120 loads from the other leg would still function. Think about when one leg is lost at the meter or main breaker.

83.33 amps IS the total current and the total current IS going through one of the legs, just because it is going through both of the legs is not the issue. What is diagrammed is a balanced load so none of the TOTAL current goes through the neutral.

[Jim Port]

Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder.

Just posting this one more time to show that I am not the only one that is saying you have 120 amp available between the two hot legs. Vern, I don't know why you cannot focus on this as two 120 volt 60 amp circuits. This has nothing to do with the transformer upstream from the disconnect.

[Vern Heiler]

Just posting this one more time to show that I am not the only one that is saying you have 120 amp available between the two hot legs. Vern, I don't know why you cannot focus on this as two 120 volt 60 amp circuits. This has nothing to do with the transformer upstream from the disconnect.

Jim, it does not have to be a transformer, even though it is. It could be a series of resistors, a series of batteries (if you look at it as a snap shot in time) or any other means of creating a difference in potential +,-,+,-, at the three conductors, the result would be the same. All of the current would pass through one of the legs regardless and some of the total current will pass through the neutral as well in an unbalance load. I did not design this or discover this, it is just the rules of electron flow. Talk to Tesla or Kirchhoff or Edison if you want to change it:D. I have done all I can to show you where and why the current flows. Code Check has done all it can to ex plane the current flowing in the 3-wire circuit. You keep insisting 120 amps can flow through the 3-wire circuit with 60 amp fuses but have not produced a single bit of evidence this can happen. Your best bet is to not share this belief with any electrician to avoid embarrassment.

[Vern Heiler]

It's surprising that such a simple concept has turned this thread into such a train wreck. Let's try one more example based on the OP.

The 2-60 amp fuses feed a service panel, there are only two loads, 2-120 volt 60 amp water heaters, one connected on ØA/N the other on ØB/N. The load on ØA is 60 amps, the load on ØB is 60 amps. Total 120 volt load is 60+60=120 amps

This is the same thing stated back in post #4.

First show me the TWO phases. Second show me the current flow (it has to come from somewhere!) It's a closed circuit, which all working circuits are, you can't just add more from nowhere. Draw the diagram and scan it. I and the world wait with baited breath.

[Jim Port]

You have ignored my examples, posts from Robert M and also the recent additions from Mike B as evidence of the capacity of the service. You have not explained how the circuit could continue to work with the one breaker turned off even though it will.

I have talked to other electricians and they agree with myself and Robert. The capacity is 60 amps per leg. Two legs at 60 gives you 120 amps at 120 volts or 60 amps at 240 volts. Mike gives a good example of how this can work that echoed the example I gave. Leg A is powering a 120 volt 60 amp load. Add another 60 amp 120 volt load to Leg B. You are now running 120 amps of load through the panel, 60 on each leg. Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly.

[Vern Heiler]

You have ignored my examples, posts from Robert M and also the recent additions from Mike B as evidence of the capacity of the service. You have not explained how the circuit could continue to work with the one breaker turned off even though it will.

I have talked to other electricians and they agree with myself and Robert. The capacity is 60 amps per leg. Two legs at 60 gives you 120 amps at 120 volts or 60 amps at 240 volts. Mike gives a good example of how this can work that echoed the example I gave. Leg A is powering a 120 volt 60 amp load. Add another 60 amp 120 volt load to Leg B. You are now running 120 amps of load through the panel, 60 on each leg. Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly.

What examples? Anyone can say something is so, proving it is another story. I continue to wait for your or Roberts diagram showing where the 120 amps comes from and goes to. (I would not use your electricians;)).

In the Mike example (if that is the diagram you are talking about), the current on leg A is the same current that came from or goes to leg B not additive. You need to read things a little closer!

"Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly." The amps will drop to the load on the not tripped side of the neutral and is shifted to the neutral that was carrying 0 amps.

[Mike Borchardt]

You are 100% wrong! If you would look at the circuit diagram you would see that ALL of the total current is passing through one of the legs of the transformer and any additional current on either side of the neutral will be added to that current, thus the fuse or breaker in that leg will trip.

And by the way you don't have a phase difference across the coil, it is a single phase. What you are describing is a multi-phase where one leg is lagging the other by some number of degrees.



The sine wave or sinusoid is a mathematical curve (http://en.wikipedia.org/wiki/Curve) that describes a smooth repetitive oscillation (http://en.wikipedia.org/wiki/Oscillation). It is named after the function sine (http://en.wikipedia.org/wiki/Sine), of which it is the graph (http://en.wikipedia.org/wiki/Graph_of_a_function). It occurs often in pure and applied mathematics (http://en.wikipedia.org/wiki/Mathematics), as well as physics (http://en.wikipedia.org/wiki/Physics), engineering (http://en.wikipedia.org/wiki/Engineering), signal processing (http://en.wikipedia.org/wiki/Signal_processing) and many other fields. Its most basic form as a function of time (t) is:
http://upload.wikimedia.org/math/5/c/6/5c6f2c4463feb5d9e20ddb438af87e83.pngwhere:


A, the amplitude (http://en.wikipedia.org/wiki/Amplitude), is the peak deviation of the function from zero.
f, the ordinary frequency (http://en.wikipedia.org/wiki/Frequency), is the number (http://en.wikipedia.org/wiki/Real_number) of oscillations (cycles) that occur each second of time.
ω = 2πf, the angular frequency (http://en.wikipedia.org/wiki/Angular_frequency), is the rate of change of the function argument in units of radians (http://en.wikipedia.org/wiki/Radian) per second
φ, the phase (http://en.wikipedia.org/wiki/Phase_(waves)), specifies (in radians) where in its cycle the oscillation is at t = 0.

When φ is non-zero, the entire waveform appears to be shifted in time by the amount φ/ω seconds. A negative value represents a delay, and a positive value represents an advance. And by multi phase your probably referring to 3 phase which represents three wave forms. Your confused with the meaning of phase. Phase does not represent the time difference between two wave forms. It represents a point in time within a wave form. One wave is voltage raising above 0 volts to dropping through 0 volts to a negative voltage and rising to 0 volts. In a wave form representing any given AC voltage, the peak positive voltage with respect to the peak negative voltage is represented in degrees of the total wave. The two peaks are 180 degrees apart. Therefor the peak positive voltage and the peak negative voltage do not happen at the same time and in fact happen 180 degrees out of phase.

[Vern Heiler]

The sine wave or sinusoid is a mathematical curve (http://en.wikipedia.org/wiki/Curve) that describes a smooth repetitive oscillation (http://en.wikipedia.org/wiki/Oscillation). It is named after the function sine (http://en.wikipedia.org/wiki/Sine), of which it is the graph (http://en.wikipedia.org/wiki/Graph_of_a_function). It occurs often in pure and applied mathematics (http://en.wikipedia.org/wiki/Mathematics), as well as physics (http://en.wikipedia.org/wiki/Physics), engineering (http://en.wikipedia.org/wiki/Engineering), signal processing (http://en.wikipedia.org/wiki/Signal_processing) and many other fields. Its most basic form as a function of time (t) is:
http://upload.wikimedia.org/math/5/c/6/5c6f2c4463feb5d9e20ddb438af87e83.pngwhere:


A, the amplitude (http://en.wikipedia.org/wiki/Amplitude), is the peak deviation of the function from zero.
f, the ordinary frequency (http://en.wikipedia.org/wiki/Frequency), is the number (http://en.wikipedia.org/wiki/Real_number) of oscillations (cycles) that occur each second of time.
ω = 2πf, the angular frequency (http://en.wikipedia.org/wiki/Angular_frequency), is the rate of change of the function argument in units of radians (http://en.wikipedia.org/wiki/Radian) per second
φ, the phase (http://en.wikipedia.org/wiki/Phase_(waves)), specifies (in radians) where in its cycle the oscillation is at t = 0.

When φ is non-zero, the entire waveform appears to be shifted in time by the amount φ/ω seconds. A negative value represents a delay, and a positive value represents an advance. And by multi phase your probably referring to 3 phase which each wave form lags 120 degrees.



And if you were to look at any point on the secondary with an o'scope, you would see the sine wave. A representation of time on the X and amplitude on the Y axis. There is a negative high and a positive high developed on the single wire secondary at the same point in time. The voltages are positive and negative in respect to the neutral which is a chosen reference point. If you could see current as a larger wire being more current and a smaller wire being less current, you would see the wire (coil in this example) grow and shrink for its entire length at the same time. It would be like watching a window parting sill move up and down.

[Mike Borchardt]

Uh how is that almost correct, haven't you've said the same thing. :confused: The same 2 pole 20 amp CB with a neutral (MWBC) can supply 40 amp (20*2) @ 120 volts each.

We probably did say the same thing. Yes you can supply a total of 40 amps but the amperage would be two circuits of 120 volt at 20 amps each. Maybe its the way you worded it that confused me.

- - - Updated - - -


Uh how is that almost correct, haven't you've said the same thing. :confused: The same 2 pole 20 amp CB with a neutral (MWBC) can supply 40 amp (20*2) @ 120 volts each.

We probably did say the same thing. Yes you can supply a total of 40 amps but the amperage would be two circuits of 120 volt at 20 amps each. Maybe its the way you worded it that confused me.

[Vern Heiler]

No diagram needed. This is a simple concept that requires no more information or examples beyond the ones that have already been provided in this thread. If you want to continue to complicate this with notions of sine waves go right ahead.

As for Jim's electrician's, I would use them any day of the week because they actually know what they're talking about. :D

What you mean is "I can't draw it!"

"As for Jim's electrician's, I would use them any day of the week because they actually know what they're talking about."

And I would use Roland Miller:D

[Vern Heiler]

There is nothing to draw, so call it what you will. The question has been answered many times in this thread by several different posters, providing numerous different examples and anything further would simply be a waste of time. :)

Aw go ahead and draw something, you've wasted this much time, and you can embarrass me:D.

[Mike Borchardt]

And if you were to look at any point on the secondary with an o'scope, you would see the sine wave. A representation of time on the X and amplitude on the Y axis. There is a negative high and a positive high developed on the single wire secondary at the same point in time. The voltages are positive and negative in respect to the neutral which is a chosen reference point. If you could see current as a larger wire being more current and a smaller wire being less current, you would see the wire (coil in this example) grow and shrink for its entire length at the same time. It would be like watching a window parting sill move up and down.

OMG, If there was a negative high and a positive high developed on the single wire at the same point in time, the negative high and the positive high would be self canceling. This is not possible within a single wave form. It sounds to me your o'scope is not in sync with the wave form. I don't doubt what you're seeing, I doubt the settings on the scope.

[Vern Heiler]

OMG, If there was a negative high and a positive high developed on the single wire at the same point in time, the negative high and the positive high would be self canceling. This is not possible within a single wave form. It sounds to me your o'scope is not in sync with the wave form. I don't doubt what you're seeing, I doubt the settings on the scope.
Learn to read Mike, "The voltages are positive and negative in respect to the neutral which is a chosen reference point."

[JOHN PAVAN]

What examples? Anyone can say something is so, proving it is another story. I continue to wait for your or Roberts diagram showing where the 120 amps comes from and goes to. (I would not use your electricians;)).

In the Mike example (if that is the diagram you are talking about), the current on leg A is the same current that came from or goes to leg B not additive. You need to read things a little closer!

"Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly." The amps will drop to the load on the not tripped side of the neutral and is shifted to the neutral that was carrying 0 amps.


It really seems that people are talking past each other. I wonder if the following isn't a better way to look at it. As far as asking electricians, I don't know how many of them haven't insisted to me, when I was supplying pool pumps, that it is cheaper to run things on 240 than 120! When I did the watt calculation for them, they scratch their heads, and look dubious.

The service is 60 Amps @ 240 Volts- which would allow a maximum of 14,400 watts - A * V = W. If you pulled 60 amps at 120- you get 7,200 watts, two such circuits would yield the same maximum Watts- 14400. Watts is the total measurement of electricity used- Power- after all Watts as in Kilowatts is what you get billed for.

I think it is customary to list the rating of the service at 240 Volts, so we say 60 Amps. One could say it is 120 Amps @ 120 volts, but we all know that is not the custom. If we insist that it is 60 Amps @ 120 Volts, that can't be right, because we know that we can pull 14400 watts with two 60 Amp 120 volt circuits. The gentleman who said the wire size should be checked is of course correct, because either the wire size or the breaker could be the limiting factor, and of course the breaker should never exceed the wire capacity

[Vern Heiler]

It really seems that people are talking past each other. I wonder if the following isn't a better way to look at it. As far as asking electricians, I don't know how many of them haven't insisted to me, when I was supplying pool pumps, that it is cheaper to run things on 240 than 120! When I did the watt calculation for them, they scratch their heads, and look dubious.

The service is 60 Amps @ 240 Volts- which would allow a maximum of 14,400 watts - A * V = W. If you pulled 60 amps at 120- you get 7,200 watts, two such circuits would yield the same maximum Watts- 14400. Watts is the total measurement of electricity used- Power- after all Watts as in Kilowatts is what you get billed for.

I think it is customary to list the rating of the service at 240 Volts, so we say 60 Amps. One could say it is 120 Amps @ 120 volts, but we all know that is not the custom. If we insist that it is 60 Amps @ 120 Volts, that can't be right, because we know that we can pull 14400 watts with two 60 Amp 120 volt circuits. The gentleman who said the wire size should be checked is of course correct, because either the wire size or the breaker could be the limiting factor, and of course the breaker should never exceed the wire capacity
And as Roland said, " Do you want to talk about amps or power. Cause if you are talking about power you have just changed the subject."

My challenge stands; someone, anyone, draw a circuit diagram showing where the current goes that is different than the one I posted! Because with mine, there can not be 120 amps on a 60 amp service with 60 amp breakers!

[Jim Port]

It really seems that people are talking past each other. I wonder if the following isn't a better way to look at it. As far as asking electricians, I don't know how many of them haven't insisted to me, when I was supplying pool pumps, that it is cheaper to run things on 240 than 120! When I did the watt calculation for them, they scratch their heads, and look dubious.

The service is 60 Amps @ 240 Volts- which would allow a maximum of 14,400 watts - A * V = W. If you pulled 60 amps at 120- you get 7,200 watts, two such circuits would yield the same maximum Watts- 14400. Watts is the total measurement of electricity used- Power- after all Watts as in Kilowatts is what you get billed for.

I think it is customary to list the rating of the service at 240 Volts, so we say 60 Amps. One could say it is 120 Amps @ 120 volts, but we all know that is not the custom. If we insist that it is 60 Amps @ 120 Volts, that can't be right, because we know that we can pull 14400 watts with two 60 Amp 120 volt circuits. The gentleman who said the wire size should be checked is of course correct, because either the wire size or the breaker could be the limiting factor, and of course the breaker should never exceed the wire capacity

Thank you John for pointing out the math aspect of this issue and showing that two legs of 120 is the same capacity as one leg at 240.

Vern, look at the following link. Insert a 60 amp 120 volt load on both legs A and B of the panel and tell me where the current is flowing to. If you said the neutral you would be correct. Now how much is flowing through the main breaker. How much is flowing through breaker A, how much through breaker B? Answer 60 and 60.

http://www.samlexamerica.com/support/documents/WhitePaper-120240VACSingleSplitPhaseandMultiWireBranchCircuit s.pdf

[JOHN PAVAN]

And as Roland said, " Do you want to talk about amps or power. Cause if you are talking about power you have just changed the subject."

My challenge stands; someone, anyone, draw a circuit diagram showing where the current goes that is different than the one I posted! Because with mine, there can not be 120 amps on a 60 amp service with 60 amp breakers!

So your contention is that one could only get a maximum of 7200 Watts if everything is on single pole circuits?

[Vern Heiler]

So your contention is that one could only get a maximum of 7200 Watts if everything is on single pole circuits?


Just draw the circuit would you! Then you can put me out of my misery:(.

I was too quick to answer that! 120 volt loads on opposite legs will react as though they were 240 volt loads as far as current is concerned. So if the loads are balanced you will have the full wattage available, still only 60 amps total.

[H.G. Watson, Sr.]

I inspected an apartment in a building with 200 apartment units, The main disconnect for the sub panel of the apartment is located at the electrical meter in the basement of the building. There are two 60 amp cartridge type fuses in a fuse block as the disconnect. I stated that it was a 60 amp service and the building supervisor stated no it is a 120 amp service because you combine the two legs of the service. I am confusing my self, I always thought you rated the disconnect by the weakest link, which would be the 60 amp fuse. Which is correct, is it a 60 amp service or 120 amp service ? Thanks in advance.:confused:

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y.

Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Vern, you're getting yourself stuck on trying to apply DC electrical theory and a simple circuit for same and trying to apply (your "directional arrows on your 'diagram'") which doesn't "fly" with AC. Unlike DC current on a simple circuit with a purely resistive load, AC current doesn't flow "one way" from a positive battery terminal with a negative for reference - each conductor pulses the exchange in a "pull me/push me" fashion - on your scope you'll see a flat line for each half cycle for each conductor with center tap reference. Not all loads are purely resistive, in fact in todays "usual" residence with other than coil/resistive electric range, other than electric WH and other than electric resistive heat, few loads are "purely resistive". Modern electronics, including the electronics involved in compact fluo., televisions, etc. motor loads refrigeration, etc. multitude of rectiifiers & power supplies, none of which are purely resistive loads (most inductive, and PF increasingly "of issue") for large multi-unit buildings.

Likely the main power feeder is 60 amps 208/120. IIRC the OP is in CT. Although early utility in No. NY and CT may have been DC "back in the day" the entire US has been on a 60 Hz AC standard for over a century (for other than RR, aircraft & marine) and Con Ed does not supply DC power to residential occupancies.

Be that as it may, there is no way the load side feeder from the fuse block to the residential unit can be called a "service". Presuming it is a dedicated feeder and not a polyphase riser tapped or fed-through, at best it could be called a "main power feeder" for the occupancy.

I don't recall the OP having indicated the "meters" were "utility owned and/or controlled" for that matter either, not that it matters.

60 amp fuse on each of two split half-phase legs is still a 60 amp supply (and likely each are split half phases from poly-phase supply) The "Service point" before the feeder to the unit not in the unit.

[Jim Port]

If it was drawn it would not make it true. Each breaker or fuse will add capacity. Since there are two legs able to supply 7200 watts each, the total capacity of the service is 14,400 watts. (60*120)=(7200) *2 = 14,400 watts.

Vern, pull one of the fuses out of the block and tell me how many amps can flow through the remaining leg. How many legs can support that current? How many amps can flow through both legs?

[Vern Heiler]

Thank you John for pointing out the math aspect of this issue and showing that two legs of 120 is the same capacity as one leg at 240.

Vern, look at the following link. Insert a 60 amp 120 volt load on both legs A and B of the panel and tell me where the current is flowing to. If you said the neutral you would be correct. Now how much is flowing through the main breaker. How much is flowing through breaker A, how much through breaker B? Answer 60 and 60.

http://www.samlexamerica.com/support/documents/WhitePaper-120240VACSingleSplitPhaseandMultiWireBranchCircuit s.pdf

Did you read this part? "The maximum current flowing in the common Neutralwill be limited to the breaker capacity (Maximum current will flow in the common Neutral when one of the split phase branch circuitsis not loaded
and the loaded split phase branch circuit is drawing itsfull rated capacity)."

[Jim Port]

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y..


Who cares what the voltage is? The question was how much power could be supplied through a 2 fuse 60 amp pullout disconnect. Stick to the subject and stop speculating.


Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Why do you think this has anything to do with the original question?

- - - Updated - - -


Did you read this part? "The maximum current flowing in the common Neutralwill be limited to the breaker capacity (Maximum current will flow in the common Neutral when one of the split phase branch circuitsis not loaded
and the loaded split phase branch circuit is drawing itsfull rated capacity)."

Correct, the neutral current will never be more than the ungrounded current in a MWBC. If only one leg is flowing the maximum current and the other leg is flowing 0, the current on the neutral is still limited to the breaker maximum.

Where do you think the extra current would be coming from?

[Vern Heiler]

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y.

Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Vern, you're getting yourself stuck on trying to apply DC electrical theory and a simple circuit for same and trying to apply (your "directional arrows on your 'diagram'") which doesn't "fly" with AC. Unlike DC current on a simple circuit with a purely resistive load, AC current doesn't flow "one way" from a positive battery terminal with a negative for reference - each conductor pulses the exchange in a "pull me/push me" fashion - on your scope you'll see a flat line for each half cycle for each conductor with center tap reference. Not all loads are purely resistive, in fact in todays "usual" residence with other than coil/resistive electric range, other than electric WH and other than electric resistive heat, few loads are "purely resistive". Modern electronics, including the electronics involved in compact fluo., televisions, etc. motor loads refrigeration, etc. multitude of rectiifiers & power supplies, none of which are purely resistive loads (most inductive, and PF increasingly "of issue") for large multi-unit buildings.

Likely the main power feeder is 60 amps 208/120. IIRC the OP is in CT. Although early utility in No. NY and CT may have been DC "back in the day" the entire US has been on a 60 Hz AC standard for over a century (for other than RR, aircraft & marine) and Con Ed does not supply DC power to residential occupancies.

Be that as it may, there is no way the load side feeder from the fuse block to the residential unit can be called a "service". Presuming it is a dedicated feeder and not a polyphase riser tapped or fed-through, at best it could be called a "main power feeder" for the occupancy.

I don't recall the OP having indicated the "meters" were "utility owned and/or controlled" for that matter either, not that it matters.

60 amp fuse on each of two split half-phase legs is still a 60 amp supply (and likely each are split half phases from poly-phase supply) The "Service point" before the feeder to the unit not in the unit.

H.G. If you can get current to flow in two directions at the same point in time in the same conductor you will be the first to do so. The potential difference along the secondary at any point in time is +-+-+- and it could be a series of batteries for that point in time.

[Vern Heiler]

"Where do you think the extra current would be coming from?"

That was my question to you, I've always said 60 amps is the maximum. Where do you think the sixty first amp comes from? (And I diagrammed my current flow)

[Jim Port]

Sixty amps is able to flow through each fuse. You have two fuses. How much is 2 x 60?

Lets try this another way again. Say you have 1500w toaster oven on leg A and are warming dinner. Now you also want a cup of coffee and start your 1500w coffee maker that is on leg B. How many amps are being used, 12.5 or 25 amps?

[Vern Heiler]

Sixty amps is able to flow through each fuse. You have two fuses. How much is 2 x 60?

Lets try this another way again. Say you have 1500w toaster oven on leg A and are warming dinner. Now you also want a cup of coffee and start your 1500w coffee maker that is on leg B. How many amps are being used, 12.5 or 25 amps?

The 60 amps on leg A is the same 60 amps that is on leg B in a closed circuit that is a total of 60 amps.

[Vern Heiler]

The 60 amps on leg A is the same 60 amps that is on leg B in a closed circuit that is a total of 60 amps.

That is 60 amps producing 3000 watts of power. Good night gentlemen....I gotta get up in the morning.

[Jim Port]

You didn't answer the question. How many amps are being used by the two 1500 watt appliances?

[Vern Heiler]

You didn't answer the question. How many amps are being used by the two 1500 watt appliances?
Oh sorry:o...12.5. And the reason this works is that the two appliances are treated as one 240 volt load with none of the current on the neutral. I'm really off to bed this time.

[Jim Port]

Oh sorry:o...12.5

WRONG. 12.5*2 = 25, not 12.5.

I will gladly stick with my electricians.

[Mike Borchardt]

Here's an attempt to explain why you can get 120 amps from two 60 amp fuses. To explain this I will use as an example a double pole 60 amp circuit breaker. We know this breaker is connected to both legs in the panel. We also know the voltage will be 240 volts AC across the legs. Now we put a 60 amp load on the breaker. Follow me Vern? Good! If you calculate the voltage by the amperage, (240 volts x 60 amps = 14400 watts). The panel doesn't give a hoot that I'm using a two pole breaker that's connected to both legs. The fact is I'm consuming 14400 watts. Am I still good Vern. OK. Now, we know we can only get 60 amps from one leg and we know this leg gives us 120 volts in respect to neutral. this calculated out (60 amps x 120 volts = 7200 watts). Say what? 7200 isn't very good, we need 14400 watts. So where do we get the other 7200 watts to make 14400 watts? Lets use both legs. LEG A (60 amps x 120 volts = 7200 watts + LEG B (60 amps x 120 volts = 7200 watts)). Fact, we have 7200 watts available on each leg. Fact, if we use both legs to feed the double pole circuit breaker we will have 14400 watts total. Here is the proof your looking for. The proof is we can add together the current on each leg to achieve the 14400 watts we need. there is no dispute about this. Its real, its fact, its true. But, if this is possible, that means each leg must be compatible to be use this way. It must also mean each leg is independent of each other. So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect. This works and I proved it. Now, its fact we have 7200 watts per leg. Calculating this out works like this, (7200 watts / 120 volts = 60 amps). If we can add leg A to leg B for a sum of 120 amps, there MUST be 60 independent amps per leg. They MUST be compatible to work together. And in fact, if we use both legs regardless of anything else, there is useable, consumable, and undisputable 120 amps available. A sixty amp, 240 volt panel has 120 amps to be used as we see fit which is the same as saying a sixty amp 240 volt circuit breaker can supply 14400 watts. Each leg supplies 60 amps, independent of each other, without interfering with each other, with out relying on each other. I proved it earlier in this post. A drawing or diagram isn't necessary. Its simple math. Please except this as a vial, friendly explanation to help you understand.

- - - Updated - - -

[Vern Heiler]

Here's an attempt to explain why you can get 120 amps from two 60 amp fuses. To explain this I will use as an example a double pole 60 amp circuit breaker. We know this breaker is connected to both legs in the panel. We also know the voltage will be 240 volts AC across the legs. Now we put a 60 amp load on the breaker. Follow me Vern? Good! If you calculate the voltage by the amperage, (240 volts x 60 amps = 14400 watts). The panel doesn't give a hoot that I'm using a two pole breaker that's connected to both legs. The fact is I'm consuming 14400 watts. Am I still good Vern. OK. Now, we know we can only get 60 amps from one leg and we know this leg gives us 120 volts in respect to neutral. this calculated out (60 amps x 120 volts = 7200 watts). Say what? 7200 isn't very good, we need 14400 watts. So where do we get the other 7200 watts to make 14400 watts? Lets use both legs. LEG A (60 amps x 120 volts = 7200 watts + LEG B (60 amps x 120 volts = 7200 watts)). Fact, we have 7200 watts available on each leg. Fact, if we use both legs to feed the double pole circuit breaker we will have 14400 watts total. Here is the proof your looking for. The proof is we can add together the current on each leg to achieve the 14400 watts we need. there is no dispute about this. Its real, its fact, its true. But, if this is possible, that means each leg must be compatible to be use this way. It must also mean each leg is independent of each other. So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect. This works and I proved it. Now, its fact we have 7200 watts per leg. Calculating this out works like this, (7200 watts / 120 volts = 60 amps). If we can add leg A to leg B for a sum of 120 amps, there MUST be 60 independent amps per leg. They MUST be compatible to work together. And in fact, if we use both legs regardless of anything else, there is useable, consumable, and undisputable 120 amps available. A sixty amp, 240 volt panel has 120 amps to be used as we see fit which is the same as saying a sixty amp 240 volt circuit breaker can supply 14400 watts. Each leg supplies 60 amps, independent of each other, without interfering with each other, with out relying on each other. I proved it earlier in this post. A drawing or diagram isn't necessary. Its simple math. Please except this as a vial, friendly explanation to help you understand.

- - - Updated - - -
Mike thank you for your attempt to get 120 amps through a 60 amp wire. Please accept my picture of reality.

[Vern Heiler]

"So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect."

Mike, what you are describing is the voltage in reference to neutral, not the current. If you have a 240 volt panel, no neutral is required and the voltage across the two legs is in phase, do you think you reverse half of the current just because you put a neutral in? Doesn't happen!

[Jim Port]

Took awhile to find this from a member from here. Haven't seen Roger here in a few years.

http://www.diychatroom.com/attachments/f18/22595d1279219372-multiwire-diagram-multi2.jpg

[Vern Heiler]

Took awhile to find this from a member from here. Haven't seen Roger here in a few years.

http://www.diychatroom.com/attachments/f18/22595d1279219372-multiwire-diagram-multi2.jpg
Well Roger is probably collecting his Nobel Prize for getting current to flow in two directions at the same time. (and what happens to the neutral when it is carrying double the amps of either leg?)

[Jim Port]

(and what happens to the neutral when it is carrying double the amps of either leg?)

If you notice the two hots are fed from opposite legs of the panels so it will not carry 2x the current.

Do you feel that a 15 amp breaker will affect a 20 amp circuit on the other leg of the panel?

[Vern Heiler]

It shows 10a on one leg and 5a on the other leg with 15a on the neutral. Just bump the numbers up to 60a on each leg and tell me what is on the neutral. And they can't go in both directions at the same time on the neutral.

[Jim Port]

The diagram clearly states in the box that the neutral is carrying 5 amps, the difference between the 10 on A and the 5 on B.

Simplify this, draw two independent 60 amp 120 volt circuits, each with a neutral. You now have 2 hots (ungrounded) and two grounded conductors. How much current can flow through one hot? How much can flow through the other hot? What is the combined current through both hots at 120 volts? Can you turn off either hot without affecting the operation of the other circuit?

[Jim Port]

If both legs are carrying 60 amps or any other balanced value and are on opposing legs, the neutral is carrying 0. The opposing currents offset each other. This is why the neutral does not need to be twice as large as the ungrounded conductors.

[Vern Heiler]

Um, no! It shows 10a going from right to left and 5a going from left to right on the same wire.

[Vern Heiler]

If both legs are carrying 60 amps or any other balanced value and are on opposing legs, the neutral is carrying 0. The opposing currents offset each other. This is why the neutral does not need to be twice as large as the ungrounded conductors.

Jim, can you dispute anything on the last drawings I posted. Does the current go in the same direction on the wires at any given point in time. Does the current going in equal the current coming out.

Your biggest problem is mixing current and voltage. The 180 phase shift in your diagram is voltage. The voltage is more positive than the neutral on top and less positive on the bottom, but all one single current has developed the potentials. Its just that we are measuring from the middle to the ends.

[Jim Port]

Those are the opposing amplitudes and are subtractive, not additive.

Please answer the questions I posted above. If you do the math correctly you should come up with the same answers as myself, Robert M, John P and Mike.

I really want to hear you explain what the affect of turning off one leg has on the other.

If you had a 2" hose connected to Hoover Dam and added another 2" hose would you get the same amount as the single hose? Same concept.

[Vern Heiler]

Those are the opposing amplitudes and are subtractive, not additive.

Please answer the questions I posted above. If you do the math correctly you should come up with the same answers as myself, Robert M, John P and Mike.

I really want to hear you explain what the affect of turning off one leg has on the other.

If you had a 2" hose connected to Hoover Dam and added another 2" hose would you get the same amount as the single hose? Same concept.

Jim, in my last attachment, fig 1 is one leg turned off. Don't let facts get in the way of your reality.

I am done, no hard feelings, just done!

[Rick Cantrell]

I'm not about to choose a side, but you guys are to be commended.
Nearly 100 post on this and no name calling, slanders, or put downs.

[Jim Port]

Vern, sorry to see that you could not see the facts regardless of the examples presented or by whom. If you had looked at this as 2 parallel 120 circuits it would have been easy to see how to have 120 amps available through the disconnect. The math was shown several times.

[Vern Heiler]

Vern, sorry to see that you could not see the facts regardless of the examples presented or by whom. If you had looked at this as 2 parallel 120 circuits it would have been easy to see how to have 120 amps available through the disconnect. The math was shown several times.
I understand the math, I have forgotten more formulas than most if not all electricians have ever seen. Microwave radio technician and main frame computer repair for over 20 years.

[Jim Port]

Ignore the transformer and focus on 2 60 amp single pole circuits. Each can flow 60 amps. You say you can understand the math, but cannot accept that 60 x 2 = 120?

[Vern Heiler]

Ignore the transformer and focus on 2 60 amp single pole circuits. Each can flow 60 amps. You say you can understand the math, but cannot accept that 60 x 2 = 120?
First do you agree the two diagrams are 60 amp circuits?

[Vern Heiler]

Your diagrams do not actually depict what were talking about which is a 3-wire circuit.
We are going to get there! Bear with me.

[Mike Borchardt]

This is as simple as it gets. I'm having a real hard time you don't see this. Each load can take 60 amps. I'm going to word this very carefully. In reference to the second drawing - This is what exists in a main pannel less the transformer. The secondary of the transformer with a center tap makes two 120 volt individual circuits both providing 60 amps in reference to the center tap / neutral or ground. The center tap is essential for creating two seperate 60 amp supplies. Please note the word seperate. Seperate means not of the same body, individual, not together. Each leg in the main panel is powered by is own power source in reference to neutral which means they are individual sources seperate from each other. Not of the same body, individual, not together. Each source supplies 60 amps. There are two seperate sources. 60 amps on leg A and 60 amps on leg B. I can put a load on leg A of 60 amps. I can also put a load on leg B of 60 amps. The total amount of amperage I can load from the panel is 120 amps. Please read and think about the last statement. The total amount of amperage I can load from the panel is 120 amps. Not from each leg, The panel. The total load of both legs. Your drawings are nonsense. You show series resistance and label each one 60 amps. Your diagrams relate more to DC then AC. The arrows show DC paths. Its sad that you are so obstinate in trying to prove something that is wrong instead of listening to what others are telling you, or you just don't understand whats being said. This is not criticism but If you truly believe everyone who posted there response is wrong, I suggest you consult a professor or engineer in the electrical field (not an electrican) with your drawings and get a second opinion before producing any more arguments to your beliefs.

[Garry Sorrells]

I'm not about to choose a side, but you guys are to be commended.
Nearly 100 post on this and no name calling, slanders, or put downs.


I was thinking the same thing. Very glad for the forum that the discussion could continue without degrading due to frustration. Good demonstration that it is possible. Great example for others to follow.

[Garry Sorrells]

Have enjoyed the ping-pong match.

Have a few questions for you guys.

1) How do you typically name the service cable AWG 4/3 coming from the triplex connection to the meter ? ____________________.

2) How would you describe a service cable that was AWG 4/2 ? ______________

3) Using the answer from question # 1 . What is the total number of Amps that the service can supply? ___________

4) Using the answer from question #2 . What is the total number of Amps that the service can supply? ___________

5) If you had two separate service panels and each was serviced by only one leg of a 4/3 service cable how many Amps would each box be able to supply ? ___________________

6) If a property needed 128 Amps to satisfy the the load.
__A) What would you expect to see as a service cable ?___________
__B) What would you expect to see as a main service panel ? ___________________
__C) How many and what size breaker would you to find ? _____________

[Vern Heiler]

I understand your frustration:).

[Vern Heiler]

I was thinking the same thing. Very glad for the forum that the discussion could continue without degrading due to frustration. Good demonstration that it is possible. Great example for others to follow.

Ah but so few "Likes" from the gallery:(. Nothing scheduled for the afternoon, so going sailing. Race start is 6:30...

[Roland Miller]

What examples? Anyone can say something is so, proving it is another story. I continue to wait for your or Roberts diagram showing where the 120 amps comes from and goes to. (I would not use your electricians;)).

In the Mike example (if that is the diagram you are talking about), the current on leg A is the same current that came from or goes to leg B not additive. You need to read things a little closer!

"Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly." The amps will drop to the load on the not tripped side of the neutral and is shifted to the neutral that was carrying 0 amps.

I will have to say you have a lot of staying power. No one will ever produce a diagram or the math that will show the 120 amps exists because there isn't any such animal. I wonder if they balance their bank account using the same math:confused: Probably why this countries' banking system is in such a mess..

[Jim Port]

Roland the math and how you could get 120 amps from that panel was described numerous times in this thread. That panel can supply 120 amps at 120 volts or 60 amps at 240 volts. Use Ohms Law and you will see how it is the same amount of watts.

One circuits ability to flow current does not affect another circuit. If it did you would be limited to whatever the smallest circuit breaker installed was. I have asked VH to explain how he thinks it is possible, but so far he has not answered the question. He also could not add up the amps flowing on 2 single pole circuits and get the correct answer.

[Roland Miller]

I agree and you could say it 100 more times and still not convince those who cannot grasp the simple concept. There are numerous examples that already in this thread.http://i15.photobucket.com/albums/a381/Ladyofthelost/smileys/deadhorse-1.gif

I know what you mean!! This discussion was never about power or watts or volt-amps. It was about whether you can get 120 amps from 2 60 amp breakers on opposite phases. The answer is no and will remain no even to "those who cannot grasp the simple concept". Whan the direct question was asked about whether or not everyone wanted to talk about watts, etc. There was no clear answer. Only repeating that the 120 amps was valid. No one here can draw a circuit, calculate or measure 120 amps from a 60 amp service..

[Jim Port]

I agree and you could say it 100 more times and still not convince those who cannot grasp the simple concept. There are numerous examples that already in this thread.http://i15.photobucket.com/albums/a381/Ladyofthelost/smileys/deadhorse-1.gif

I apologize for continuing this so long. I was hoping to use this as a teaching moment and that the students would finally grasp the subject and understand what they were missing. Oh well, you can't reach everyone. Sad.

I should have realized that some were not getting it when the load of 2 12.5 amp circuits still equaled 12.5 amps.

[Roland Miller]

Let me put into dollars since everyone should under stand money.. You put $60 into your account (or pocket) you immediately take $60 dollars out for something. Some of you would reach in your pocket and expect to find 120 dollars:confused:. The net result is 0 dollars. Same math with this circuit..no 120 amps anywhere...because it is additive using "absolute" values. Try putting it on a number line and move your finger back and forth, it will never spike up to 120..

And some are trying to confuse the issue by saying this is not a DC circuit and its not but it doesn't make any difference. It would work with AC or DC.

[Jim Port]

I would ask those that believe that there is only 60 amps available from that panel to post the same question asked the same way on an electrical forum and post a link to that discussion.

[Mike Borchardt]

Let me put into dollars since everyone should under stand money.. You put $60 into your account (or pocket) you immediately take $60 dollars out for something. Some of you would reach in your pocket and expect to find 120 dollars:confused:. The net result is 0 dollars. Same math with this circuit..no 120 amps anywhere...because it is additive using "absolute" values. Try putting it on a number line and move your finger back and forth, it will never spike up to 120..

And some are trying to confuse the issue by saying this is not a DC circuit and its not but it doesn't make any difference. It would work with AC or DC.Ok, lets use a pair of pants that has two pockets that represent the two legs in a panel. Put 60 dollars in each pocket. How much money do you have on your person? Oh wow, must be 120 bucks.

[Jim Port]

Try this: Typical 1Ø, 100 amp service, 120/240 volts, has only two fully loaded 20 amp circuits both on the same leg how many total amps do those two circuits draw?


Almost 4 hours later and still no answer. Looks like a stumper. ;)

[Roland Miller]

Almost 4 hours later and still no answer. Looks like a stumper. ;)
And my answer is: The problem is about apples. Why is someone introducing a question about oranges? The question Jim is referring to is about oranges. The source and the circuit was sufficiently defined early on so lets try to answer that question before moving on. The question (as a reminder) is about a single phase, 120/240 volt service with 60 main. How many amps is the maximum at any time? Not watts, not VA not all on the same phase.....or for those of you who are stuck, draw, measure or calculate it to prove it will be 120 amps. (you can't BTW)

Oh and BTW---the answer is +60 and (-60). Because when one phase is positive the other is negative. It does not add up to 120 amps. Never more than 60..

The wattage (VA) is positive in both cases because VA=E x I and VA= (-E) x (-I) both positive answers. Basic 5th grade math. but we are not talking about this..

[Jim Port]

Again the concept that each leg at 120 volts can supply 60 amps eludes.

Thankfully the electricians understand this.

Could you answer Roberts question above above the two 20 amp breakers on the same leg?

[Jim Port]

From post #6, where the debate started.


At 120 volts you still only have 60 amps available.:)

After that is was people trying to show the math errors.

[Vern Heiler]

I will have to say you have a lot of staying power. No one will ever produce a diagram or the math that will show the 120 amps exists because there isn't any such animal. I wonder if they balance their bank account using the same math:confused: Probably why this countries' banking system is in such a mess..

There diagram will work just fine! Just needs a little help.....

[Jim Port]

Congratulations, you finally see using Diagram 1 that it is possible to flow 120 amps @ 120 volts.

[Vern Heiler]

Congratulations, you finally see using Diagram 1 that it is possible to flow 120 amps @ 120 volts.

I don't know if I should call the fire department or the scientific community and tell them to hold there nominations, we have a winner of the Nobel prize without a doubt;).

[Jim Port]

Not exactly breaking news or a fire hazard, single pole circuits have operated safely like that for years.

[Vern Heiler]

Not exactly breaking news or a fire hazard, single pole circuits have operated safely like that for years.

How exactly do you explain getting twice as much current through the single neutral/ground wire in the second diagram without it glowing red?

[Jim Port]

Because the hots are from opposing legs of the panel they would only carry the difference between the two hots, not the sum of the hots. Every single phase 120/240 panel operates the same way. The service is one large multi-wire circuit.

[Roland Miller]

There diagram will work just fine! Just needs a little help.....
This diagram ignores basic transformer theory and continues to compare apples and oranges..

- - - Updated - - -


How exactly do you explain getting twice as much current through the single neutral/ground wire in the second diagram without it glowing red?
He can't explain it because the current will never exceed 60 amps on the neutral.

[Roland Miller]

One of the principles I have learned over the years in Math, Electrical Engineering and teaching is you must keep track and do the math operations on the unit labels all the way through the problem or the answer will be meaningless- as it this example. I have also learned the you can't mix apples and oranges to come up with something of meaning such as in a lot peoples thinking here:hand:. And I have also learned that a politician is always right even when they are wrong . They just have to convince a lot of people they are right...(doesn't make them correct)...

[Vern Heiler]

One of the principles I have learned over the years in Math, Electrical Engineering and teaching is you must keep track and do the math operations on the unit labels all the way through the problem or the answer will be meaningless- as it this example. I have also learned the you can't mix apples and oranges to come up with something of meaning such as in a lot peoples thinking here:hand:. And I have also learned that a politician is always right even when they are wrong . They just have to convince a lot of people they are right...(doesn't make them correct)...

I agree. The biggest problem some are having is trying to apply ohms law before understanding the circuit. You must understand the circuit before applying the math.

[Roland Miller]

Because the hots are from opposing legs of the panel they would only carry the difference between the two hots, not the sum of the hots. Every single phase 120/240 panel operates the same way. The service is one large multi-wire circuit.


I think it is interesting how you try to convince everyone there is 120 amps somewhere and then you post this in complete disagreement with yourself???? I think you have confused yourself.

- - - Updated - - -


This is as simple as it gets. I'm having a real hard time you don't see this. Each load can take 60 amps. I'm going to word this very carefully. In reference to the second drawing - This is what exists in a main pannel less the transformer. The secondary of the transformer with a center tap makes two 120 volt individual circuits both providing 60 amps in reference to the center tap / neutral or ground. The center tap is essential for creating two seperate 60 amp supplies. Please note the word seperate. Seperate means not of the same body, individual, not together. Each leg in the main panel is powered by is own power source in reference to neutral which means they are individual sources seperate from each other. Not of the same body, individual, not together. Each source supplies 60 amps. There are two seperate sources. 60 amps on leg A and 60 amps on leg B. I can put a load on leg A of 60 amps. I can also put a load on leg B of 60 amps. The total amount of amperage I can load from the panel is 120 amps. Please read and think about the last statement. The total amount of amperage I can load from the panel is 120 amps. Not from each leg, The panel. The total load of both legs. Your drawings are nonsense. You show series resistance and label each one 60 amps. Your diagrams relate more to DC then AC. The arrows show DC paths. Its sad that you are so obstinate in trying to prove something that is wrong instead of listening to what others are telling you, or you just don't understand whats being said. This is not criticism but If you truly believe everyone who posted there response is wrong, I suggest you consult a professor or engineer in the electrical field (not an electrican) with your drawings and get a second opinion before producing any more arguments to your beliefs.

If the diagram with this post was mathematically correct (valid) you should be able to hook this circuit up in the lab and measure 120 amps on the neutral. And that won't happen cause it will be 0 amps when they are combined.

- - - Updated - - -


I agree. The biggest problem some are having is trying to apply ohms law before understanding the circuit. You must understand the circuit before applying the math.


Yes, You have it nailed. When you don't understand the whole package it is impossible to analyze it..

[Roland Miller]

Again the concept that each leg at 120 volts can supply 60 amps eludes.

Thankfully the electricians understand this.

Could you answer Roberts question above above the two 20 amp breakers on the same leg?

I know the answer to this question, but the question and the answer have nothing to do with the discussion.. yah know--the apples and oranges thing again:)

[Jim Port]

In the 120 volt circuits there are two ungrounded conductors, one for each circuit carrying 60 amps.

[Vern Heiler]

In the 120 volt circuits there are two ungrounded conductors, one for each circuit carrying 60 amps.
​SHOW ME THE MONEY!

[Jim Port]

​SHOW ME THE MONEY!

This is the same thing I have been saying since the start of this thread. Two 120 volt legs at 60 amps equals the ability of that panel to supply 120 amps worth of load. Nothing has changed, including two peoples lack of understanding.

[Jim Port]

I know the answer to this question, but the question and the answer have nothing to do with the discussion.. yah know--the apples and oranges thing again:)

The answer is at the heart of the question. Post your answer please.

I also have not seen the link posted to any electrical forum.

[Roland Miller]

There diagram will work just fine! Just needs a little help.....
http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29204d1375848868-determining-amperage-mat-main-disconnect-fuse-block-img.jpg

Grounded or ungrounded it doesn't matter --answer still the same. If these two transformers were wound as drawn there is no difference between them. You need to understand basic transformer theory. Since Jim and Robert don't, how can they understand the circuit and math? Smoke (lots of smoke) with a trace of bad math and mirrors = wrong answer.. If what you say is so then you could prove it in the lab. and you would find that one was a +60 amps and the other (at the same instant) would be a -60 amps. Besides circuit #1 is not the circuit we are talking about. Just more confusion from a couple of politicians that are weak in math..:confused:

[Jim Port]

Circuit 1 is exactly the situation that was described when the capacity of the panel was described at 120 volts. It was also stated that the capacity was 60 amps at 240 volts.

Stop deflecting and answer the questions. This has never been about transformers.

[Vern Heiler]

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29204d1375848868-determining-amperage-mat-main-disconnect-fuse-block-img.jpg

Grounded or ungrounded it doesn't matter --answer still the same. If these two transformers were wound as drawn there is no difference between them. You need to understand basic transformer theory. Since Jim and Robert don't, how can they understand the circuit and math? Smoke (lots of smoke) with a trace of bad math and mirrors = wrong answer.. If what you say is so then you could prove it in the lab. and you would find that one was a +60 amps and the other (at the same instant) would be a -60 amps. Besides circuit #1 is not the circuit we are talking about. Just more confusion from a couple of politicians that are weak in math..:confused:
I understand what you are saying Roland but + and - amps might just be a bit of a misnomer. Current flows in a direction due to + and - potentials. What is important is that if you push one electron in one end of a wire you only get one out the other end of the wire, not two.

[Roland Miller]

I understand what you are saying Roland but + and - amps might just be a bit of a misnomer. Current flows in a direction due to + and - potentials. What is important is that if you push one electron in one end of a wire you only get one out the other end of the wire, not two.

It is all about which way the electrons are flowing at any instant in an AC circuit. DC is a little more straight forward.

Sorry Jim and Robert you are using oranges to prove an apple theory and you have failed at a very basic level.. 5th grade math. Catch you on the next one if you don't overload in theory;)

[Jim Port]

Sorry Jim and Robert you are using oranges to prove an apple theory and you have failed at a very basic level.. 5th grade math. Catch you on the next one if you don't overload in theory;)

So you are leaving this discussion without proving anything except that you do not understand a simple 120 circuit? If you were so sure of yourself why did you not ask electricians on one of their forums?

[Roland Miller]

So you are leaving this discussion without proving anything except that you do not understand a simple 120 circuit? If you were so sure of yourself why did you not ask electricians on one of their forums?


Apparently you think the tail should wag the dog. I have nothing to prove. The math and science of electricity is all that is needed to prove my position.. I have posted enough information that anyone of average intelligence can work through it and come to the same conclusion I have posted. They couldn't do the same for your position. So you can continue you struggles with the math and science of electricity alone....some nuts are too tough to crack:D

[Jim Port]

It is just a shame the wrong conclusion would be reached. Vern even drew the diagram showing the 2 legs carrying 60 amps at 120 volts and some still don't see it.

Why will you not answer Robert's question? How hard is it to add 20 + 20?

[Jim Port]

It is just a shame that homeowners will have to pay to have erroneous information like this refuted simply because of a lack of understanding of the subject matter on the part of the inspector.

[Vern Heiler]

It is just a shame the wrong conclusion would be reached. Vern even drew the diagram showing the 2 legs carrying 60 amps at 120 volts and some still don't see it.

Why will you not answer Robert's question? How hard is it to add 20 + 20?
Jim, I was wanting to just let this go until you tried to put words in my mouth. You really are a spin master. Vern might have shown that each leg has 60 amps, but Vern also said they are the SAME 60 amps.

[Jim Port]

Diagram 1 shows 2 independent 120 circuits. They cannot be the same 60 amps being measured. Just because the values are the same does not mean they are the same amps. Vern is wrong about that.

[Vern Heiler]

Still no takers on the simple question in post #122. :(
Oh hell ya!

It's 40 amps on the same leg, and if you were to add a 10 amp load on the other leg it would still be 40 amps. If you were to look at the series of diagrams I posted (Fig 1 through Fig 5), and were to be able to comprehend, you would see how this is. Please feel free to make the arrows double ended (pointing both directions) to satisfy your distaste for a single point in time (DC representation). And yes I do know that I left out a load by mistake and drew an extra wire that would cause a short, but if you can't fix that you should not get anywhere near electricity!

[Vern Heiler]

So you're saying if I install a third circuit breaker on the other leg with a 10 amp load the total 120 volt connected load is still 40 amps?
You are still mixing AMPS AND POWER. We are talking AMPS and have been since the time it was stated that you could have 120 amps on a 3-wire 60 amp circuit. Ten of the same 40 amps that is on the first leg now goes through the second leg. The other 30 amps still returns through the neutral. Again if you want to talk amps, lets talk amps. If you want to talk power, lets talk power.

[Jim Port]

No, you would have a panel with a 50 amp load, 40 on one leg plus 10 on other other. The neutral load would be 30 amps. The current is not crossing to the other hot as they are 120 volt and the return is on the grounded conductor, not the other hot leg.

[Vern Heiler]

No, you would have a panel with a 50 amp load, 40 on one leg plus 10 on other other. The neutral load would be 30 amps. The current is not crossing to the other hot as they are 120 volt and the return is on the grounded conductor, not the other hot leg.

First give a definition of "amp load".

Second tell us how the current got there.

If you don't get this right, I'm joining Roland on the side-line. I mean after all most people don't really need to have a picture drawn for them, its just an expression!

[Jim Port]

Do you really need someone to tell you that running equipment from a panel places a load on it? Amps are a measure of the load. A demand load calculation for a service gives you the expected load in amps.

The current got there by the equipment being used. If it were turned off there would be no load.

Go ahead and sit out. Robert and myself are sure of our answers and have given numerous examples to prove ourselves correct.

[Vern Heiler]

Do you really need someone to tell you that running equipment from a panel places a load on it? Amps are a measure of the load. A demand load calculation for a service gives you the expected load in amps.

The current got there by the equipment being used. If it were turned off there would be no load.

Go ahead and sit out. Robert and myself are sure of our answers and have given numerous examples to prove ourselves correct.

A load is work being performed, work requires power...do you see where this is going?

The current got there by the equipment being used!!!! Roland... slide over and hand me a beer, I'm done!

[Roland Miller]

The math and science of electricity is all that is needed to prove my position.. I have posted enough information that anyone of average intelligence can work through it and come to the same conclusion I have posted. They couldn't do the same for your position. So you can continue you struggles with the math and science of electricity alone.

Electricity is exactly predictable with math. Once the math is done you could build the circuit in a lab and measures every value (and verify) you calculated. You couldn't do that with your fuzzy math problem..

[Jim Port]

What the circuit will support is right on the breaker handle or fuse label. Why you and Vern will not believe this is beyond me.

I see that you are still not sure enough of your position to ask the same question on and electrical foruma and post a link here so others can see the correct answer.

You said the math is easy to prove and it is. Watch this: 60 amps x 120 volts = 7200 watts, 7200 x 2 for the number of legs in the panel = 14400 watts.

60 amps x 240 volts = 14400 watts.

Look they match.

[Roland Miller]

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29204d1375848868-determining-amperage-mat-main-disconnect-fuse-block-img.jpg

There is not a single point in either circuit that will see more than 60 amps. In fact the second circuit would work without the grounded (center) conductor..

- - - Updated - - -


What the circuit will support is right on the breaker handle or fuse label. Why you and Vern will not believe this is beyond me.

I see that you are still not sure enough of your position to ask the same question on and electrical foruma and post a link here so others can see the correct answer.

You said the math is easy to prove and it is. Watch this: 60 amps x 120 volts = 7200 watts, 7200 x 2 for the number of legs in the panel = 14400 watts.

60 amps x 240 volts = 14400 watts.

Look they match.
Here's the correct math

29207

You are still mixing apples and oranges. you are taking a 240 volt calculation and using 120 volts to prove you point.

- - - Updated - - -

I don't need to check with anyone.. I just use math and science. You use smoke and mirrors..:p

[Roland Miller]

What the circuit will support is right on the breaker handle or fuse label. Why you and Vern will not believe this is beyond me.



I see you have shifted your position again. So you agree that a 2 pole-60 amp CB or 2 60 fuses will only support 60 amps now. Amazing. Never the 120 amps you have been spouting about...This is more than funny..

[Jim Port]

My position has NEVER changed. I have said since the beginning of this thread that the panel in question can support 120 amps at 120 volts or 60 amps at 240 volt. Go back and check.

Again you confused how many legs in the panel there are. Myself and several others have tried to point out your confusion and math errors. I see that you are going to continue to twist this however you want instead of admitting your blindness.

Good Day.

[Roland Miller]

My position has NEVER changed. I have said since the beginning of this thread that the panel in question can support 120 amps at 120 volts or 60 amps at 240 volt. Go back and check.

Again you confused how many legs in the panel there are. Myself and several others have tried to point out your confusion and math errors. I see that you are going to continue to twist this however you want instead of admitting your blindness.

Good Day.

Thanks for saying you agreed with me from the very beginning. If you would drop the 120 amp thing you would get a passing grade. I am sorry I am still laughing... oh you have tried to confuse amps and watts, 240 volt calculations with 120 volt calculations and you are still loosing. Do you still have that roofers card? I hear it covers everything and what you are sitting on badly needs covered.

Math teaches a life skill called adaptive reasoning. Something you apparently don't have.:deadhorse:

[Jim Port]

I do not agree with you, you simply do not understand that there are TWO INDEPEDENT legs in the panel, EACH capable of allowing 60 amps to flow. Even if you were to remove one fuse you could still pull 60 amps from the other leg. The two 120 volt legs are the key to understanding this, not some extraneous number.

BTW, every time you post Diagram 1, you reinforce my position.

[Roland Miller]

Thanks for dropping the 120 amp stuff(you couldn't prove it anyway) and agreeing with me again. You are up to a D-, barely.

You don't seem to understand what this discussion has been all about--120/240 volt, single phase service with a 2-pole 60 amp breaker (close to diagram #2). Or do you even know what three elements are required to make up a circuit?

I know I should have stopped when my BS meter needle pegged and was bent, but when you accused me (and Vern) of costing some poor home owner money because we were spreading mis-information, I guess you raised my hackles. When in fact you and Robert are the ones doing this. What office are you guys running for? Are you Jerry's and HG's apprentices?

My suggestion for you two is to stick to topics you have some correct knowledge of and leave the rest to someone else.

Read my signature--and live it. You won't run into so many rocks:mad:

[Jim Port]

Since you seem to like oranges. Suppose you have an orange juice factory with one machine that can produce 10 gallons of OJ per hour. You have plenty of extra oranges and enough floor space to add a second identical machine. How many gallons of OJ can you produce using both machines for one hour? I bet your answer is 10 gallons.

[Roland Miller]

Since you seem to like oranges. Suppose you have an orange juice factory with one machine that can produce 10 gallons of OJ per hour. You have plenty of extra oranges and enough floor space to add a second identical machine. How many gallons of OJ can you produce using both machines for one hour? I bet your answer is 10 gallons.


“You are not entitled to your opinion. You are entitled to your informed opinion. No one is entitled to be ignorant.”
― Harlan Ellison (http://www.goodreads.com/author/show/7415.Harlan_Ellison)

You seem to take entitlement very seriously:flypig:

[Jim Port]

No, I feel that the truth and facts need to be told. I am kind of like the kid that told the emporer that he wasn't wearing any clothes. The truth doesn't change.

Thank you for this thread. I will use it as an example of what happens during so many home inpsection reports and why the "defects and issues" are so easy to refute. Thankfully all HI's are not so lacking.

BTW, the question posted in #122 remains unanswered. Quite a stumper wasn't it?

[Roland Miller]

No, I feel that the truth and facts need to be told. I am kind of like the kid that told the emporer that he wasn't wearing any clothes. The truth doesn't change.

Thank you for this thread. I will use it as an example of what happens during so many home inpsection reports and why the "defects and issues" are so easy to refute. Thankfully all HI's are not so lacking.

BTW, the question posted in #122 remains unanswered. Quite a stumper wasn't it?

You are more then welcome. Make sure they get the whole thread to read because anyone with average intelligence and a 5th grade education will figure out you are wrong. Although they will need a Phd in Bull Crap to figure out how you got it so wrong.

- - - Updated - - -


Misinformation, definitely Jim nailed it. Actually Jim and I are trying to keep a few of you from embarrassing yourselves any further with more nonsense. Either there's a high level of ignorance coming from a select few in this thread or there are just trolls lurking about. I would vote for the former since even a troll would have quit long ago. :rolleyes:

Both you guys are arguing with math and science. Hows it going? I am just pointing out the facts. One of which is --it is you that are doing a disservice by spreading misinformation.

I was once told by a very wise man--It is better to be thought of as a fool then to open one's mouth and remove all doubt like you two have...

[Jim Port]

And once again you ignored a another direct question asked of you. How much OJ can you get from the 2 machines? Can you get twice the water out of the two hoses attached to Hoover Dam as you would with just one?

You are arguing against the math facts as presented. You don't understand them so you think they are wrong despite numerous explanations and analogies to help you understand.

[Vern Heiler]

And once again you ignored a another direct question asked of you. How much OJ can you get from the 2 machines? Can you get twice the water out of the two hoses attached to Hoover Dam as you would with just one?

You are arguing against the math facts as presented. You don't understand them so you think they are wrong despite numerous explanations and analogies to help you understand.

Jim, take diagram or circuit #1 that you are so proud of and complete each side by connecting a load to each. You do know that no circuit will work unless it is a closed loop, right! Now counting the number of amps. With your method there are 240 amps (ya got to count each leg of each circuit with your method) Wow! now we have 240 amps. If we keep this up we can put Duke Power out of business in no time.:D Now because we can see this will not work, lets try circuit #2 . Put a load on between the neutral and one of the legs that draws 60 amps. Now all 60 amps obviously is gong through the neutral. That's ok, all is good, but do we have 60 amps or 120 amps? (Oh I do hope you said 60 amps) Now put a load across the other leg to neutral that draws 60 amps. Does any of the current flow through the neutral? No! So where does it go? It goes through the other leg. Should we measure it again? It didn't sound like a good idea for circuit #1 did it? No one has said you could get 240 amps so why would we now say we can get 120 amps?

[Jim Port]

Keep trying Vern, wrong on both counts. You still have 60 amps on each ungrounded conductor. Each ungrounded conductor in the 120 volt branch circuit is also carrying the 60 amp return current. You do not count this towards the panel capacity. In the 240 example there would be no current on the system neutral if both legs are balanced. Also the current on the neutral has nothing to do with the capacity of the fuse or breaker in this example. The OP asked how much could the 2 pole 60 amp fused pullout supply.

No one has said anything about having 240 amps because you do not. The fuse or breaker would have opened at 60 amps and you do not count system neutral current unless sizing a service or feeder. Your attempts to disprove myself and Robert and the others gets more ridiculous each time.

It is a shame that two independent 120 volt 2 wire circuit has you so confused. Maybe you can answer the question in Post #122? How much current is needed to trip a 20 amp branch circuit?

[Vern Heiler]

What the circuit will support is right on the breaker handle or fuse label. Why you and Vern will not believe this is beyond me.

Look they match.
I guess you have made that pretty clear:D.

[Jim Port]

Just another comment taken out of context. Let me spell it out for you. I will type slowly.
How two people can ignore how much current a fuse or breaker will allow before opening after numerous and different explanations defies logic. It is clearly labeled. I tried to give you the benefit of the doubt, but I see that was wasted.

Simple questions have been asked like how much current will a 20 amp breaker pass, how much will two pass and the answers, if given, are about the neutral or transformer winding or some other extraneous BS not germane to the original question.

Both your and Roland's attempts to prove the world to be flat again have failed.

[Roland Miller]

You have never answered this from post #9.

I guess you will have to show me where you could measure 120 amps on a 3-wire, 120/240, single phase transformer with the 120 volt phase/circuit each pulling 60 amps;)???

And don't get confused about the transformer thing. We all know it is way over your heads. This is the circuit under discussion, ie. apples..

In fact there are numerous questions Jim and Robert have failed to answer. They would all show their logic is screwy...

[Vern Heiler]

Do you really need someone to tell you that running equipment from a panel places a load on it? Amps are a measure of the load. A demand load calculation for a service gives you the expected load in amps.

The current got there by the equipment being used. If it were turned off there would be no load.

Go ahead and sit out. Robert and myself are sure of our answers and have given numerous examples to prove ourselves correct.

Kirchhoff's First Rule (Junction Rule) is based on the conservation of charge. It states that:

“At any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.”

This means that when current reaches the branches in a circuit, it will split up and take different routes. When the branches come back together, the currents will add back together too.


http://www.etap.org/demo/physics_files/lesson7/16.jpg

[Jim Port]

I guess you will have to show me where you could measure 120 amps on a 3-wire, 120/240, single phase transformer with the 120 volt phase/circuit each pulling 60 amps;)???



You just answered this yourself. What does each leg pulling 60 amps add up to? A very simple thing you keep ignoring. 60 + 60 = ?????

[Vern Heiler]

You just answered this yourself. What does each leg pulling 60 amps add up to? A very simple thing you keep ignoring. 60 + 60 = ?????
You've done the easy part, now make this work: When the branches come back together, the currents will add back together too.

[Jim Port]

Again you show your lack of understanding of the topic that has been debated for well over 100 posts. This has NOTHING to do with transformers, combining currents or anything else. Clear your mind of your preconceptions and read what is written. The statement was made by myself and others that the panel could SUPPLY 120 amps at 120 volts. You and your friend do not understand the basis of this statement and I do not think you ever will. You were given a chance to have your theory proven on a professional electrical forum but did not avail yourself of that opportunity, but as Robert said, you did avoid the embarrassment.

[Roland Miller]

Again you show your lack of understanding of the topic that has been debated for well over 100 posts. This has NOTHING to do with transformers, combining currents or anything else. Clear your mind of your preconceptions and read what is written. The statement was made by myself and others that the panel could SUPPLY 120 amps at 120 volts. You and your friend do not understand the basis of this statement and I do not think you ever will. You were given a chance to have your theory proven on a professional electrical forum but did not avail yourself of that opportunity, but as Robert said, you did avoid the embarrassment.


THis was never the original question. You and Robert introduced this series of unrelated questions (oranges) to prove your ignorant answer. I will keep bringing you back to the original question which you have never answered. 120/240 volt, single phase AC source(can't use transformer cause it scares them) with 2-pole 60 amp breaker with a 60 amps on each phase..

It's simple addition why can't you solve it? Use Kirchhoff's circuit laws. They are simple addition also.

[Jim Port]

I simply corrected your post where you said,

From Post #5.

At 120 volts you still only have 60 amps available.:)

I then corrected you by saying this in Post #8


Fixed this for you. At 120 volts you still only have 60 amps available per leg.

From Post #4;


Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder.

Since that point you have failed to understand. We have simply corrected your error.

[Vern Heiler]

Again you show your lack of understanding of the topic that has been debated for well over 100 posts. This has NOTHING to do with transformers, combining currents or anything else. Clear your mind of your preconceptions and read what is written. The statement was made by myself and others that the panel could SUPPLY 120 amps at 120 volts. You and your friend do not understand the basis of this statement and I do not think you ever will. You were given a chance to have your theory proven on a professional electrical forum but did not avail yourself of that opportunity, but as Robert said, you did avoid the embarrassment.
I have removed the transformer for you. I have also double-ended the arrows for you. Before you and your miss guided flock can analyse a circuit you will have to understand the basics of electricity.

[Jim Port]

You obviously do not want to understand this simple subject. Hopefully your eyes will open one day.

[Jim Port]

I have removed the transformer for you. I have also double-ended the arrows for you. Before you and your miss guided flock can analyse a circuit you will have to understand the basics of electricity.

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29208d1376137090-determining-amperage-mat-main-disconnect-fuse-block-max-smoke.jpg

This is not about the current on the neutral.

I will ask you a straight forward question. How much current does it take to open a 60 amp fuse or breaker? Does this matter whether the breaker is a single or double pole? If BOTH legs of the double pole breaker opened due to overload, how much current was flowing?

[Jim Port]

The source of all the contention was started in Post #9.


I guess you will have to show me where you could measure 120 amps on a 3-wire, 120/240, single phase transformer with the 120 volt phase/circuit each pulling 60 amps;)

This is also where the transformer sidetrack came from.

No one ever said you could measure 120 amps on ONE conductor. It was said that there was 120 amps of capacity available.


Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder.

[Vern Heiler]

Originally Posted by Jim Port http://www.inspectionnews.net/home_inspection/images/buttons/viewpost-right.png (http://www.inspectionnews.net/home_inspection/electrical-systems-home-inspection-commercial-inspection/36592-determining-amperage-mat-main-disconnect-fuse-block-post229894.html#post229894)
"Fixed this for you. At 120 volts you still only have 60 amps available per leg."

You do see 60 amps on each leg, but they are the same 60 amps as we have been saying since the beginning! NOT PER!

You asked for and authority to settle this, I thought Kirchhoff would be enough for even you!

[Jim Port]

Robert, I hope you don't expect an answer for your latest question. The question you posted days ago still has not been answered.

Vern, how do you explain the that half the panel will still function properly even with one fuse removed? How much capacity does that half have?

[Vern Heiler]

No laws of physics required, no diagrams required. I'll try one more example.

60 amp, 1Ø, 3 wire, 120/240 volt service with no loads:

How many 120 volt 60 amp heaters can I connect?

How many 240 volt 60 amp heaters can I connect?

The laws of physics does not apply to you, very interesting!

You are talking power again, you know I have already asked you to talk power OR current.

I am going to do some vector analysis. CE (center of effort) plotted against CLR (center of lateral resistance, to find DMG (distance made good).....I m going sailing:D.

[Jim Port]

The laws of physics does not apply to you, very interesting!

.

Again an answer is taken out of context. It does not require the application of any laws of physics. It does not mean they do not apply. All you need is written on the breaker handle or fuse label and a calculator. This has always been a simple math problem.

[Chris McIntyre]

I have followed this thread very closely and have found very educational....and confusing :).

My question is for Jim & Robert to help me, and probably others, understand the argument. Are you saying that a 120 amp balanced load would not trip the 60 amp OCPD?

I apologize in advance if this is a stupid question.:o

[Jim Port]

I have followed this thread very closely and have found very educational....and confusing :).

My question is for Jim & Robert to help me, and probably others, understand the argument. Are you saying that a 120 amp balanced load would not trip the 60 amp OCPD?

I apologize in advance if this is a stupid question.:o

Not a stupid question. It appears you understand this better than others.

There are 2 ungrounded legs in a typical home panel that feed alternating fingers on the bus bars. Using a balanced 120 amp 120 volt load across the 2 legs would have each leg carrying 1/2 the current or 60 amps. This could be as simple as three fully loaded 20 amp 120 volt circuits operating on each leg of the panel.

If the load were 240 volts the breaker or fuse will open when the current exceeds 60 amps.The breaker or fuse would open when either or both sides saw more than 60 amps. In an extremely unbalanced load where everything was on one leg and exceeded 60 amps the breaker would open even with the other leg carrying 0 amps.

The key factor that has been ignored numerous times is that the number of amps will change with the voltage. For example 600 watts at 120 volts = 5 amps, change the voltage to 240 volts and the amps drop to 2.5. Using a higher voltage allows for smaller wire sizes to be used.

[Vern Heiler]

Not a stupid question. It appears you understand this better than others.

There are 2 ungrounded legs in a typical home panel that feed alternating fingers on the bus bars. Using a balanced 120 amp 120 volt load across the 2 legs would have each leg carrying 1/2 the current or 60 amps. This could be as simple as three fully loaded 20 amp 120 volt circuits operating on each leg of the panel.

If the load were 240 volts the breaker or fuse will open when the current exceeds 60 amps.The breaker or fuse would open when either or both sides saw more than 60 amps. In an extremely unbalanced load where everything was on one leg and exceeded 60 amps the breaker would open even with the other leg carrying 0 amps.

The key factor that has been ignored numerous times is that the number of amps will change with the voltage. For example 600 watts at 120 volts = 5 amps, change the voltage to 240 volts and the amps drop to 2.5. Using a higher voltage allows for smaller wire sizes to be used.
Still mixing power and current huh!

In practical terms, the ampere is a measure of the amount of electric charge (https://en.wikipedia.org/wiki/Electric_charge) passing a point in an electric circuit per unit time with 6.241 × 1018electrons (https://en.wikipedia.org/wiki/Electron), or one coulomb (https://en.wikipedia.org/wiki/Coulomb)per second constituting one ampere.

59 amps will pass through a 60 amp breaker just fine. It does not matter if it is being pushed by 1 volt or 10,000 volts it will not trip. When it reaches 60 amps it will trip. Voltage is irrelevant. Tripping of a breaker or blowing of a fuse is due to power developed across the OCP device. All conductors have resistance and the power is the product of current squared times resistance. Resistance and current are the only two things a fuse knows. The voltage ratting on a fuse or circuit breaker is the voltage that the blown fuse will not arc across after it has blown.

This discussion has been about amps from the beginning, even though Jim & Robert keep asking questions regarding or insisting that power calculations be used.

There will never be more than 60 amps at one time on a 3-wire circuit protected by 60 amp breakers and I don't care if there are 100 breakers or fuses in the circuit.

[Jim Port]

Voltage is not irrelevant. The voltage something operates at determines the amp draw as I illustrated above showing the 600 watt load and the difference in amps being halved by the doubling of the voltage.

BTW, you appear to have little or no knowledge of breaker trip curves or the UL allowance for variation. Depending on the rate of rise a breaker will hold 125% for over an hour. It does not trip as soon as it sees 60.1 amps.

[Vern Heiler]

Voltage is not irrelevant. The voltage something operates at determines the amp draw as I illustrated above showing the 600 watt load and the difference in amps being halved by the doubling of the voltage.

BTW, you appear to have little or no knowledge of breaker trip curves or the UL allowance for variation. Depending on the rate of rise a breaker will hold 125% for over an hour. It does not trip as soon as it sees 60.1 amps.
Still can't see or admit that you are talking about power after I have posted the definition of current:confused:. And yes I do know about breaker ratings and the difference in a standard and slow blow fuse. Clouding the discussion with this or PF or current lag, would not help for anyone to understand the basics, and certainty didn't help you! Talk current or go home!

[Vern Heiler]

No such thing as a stupid question. Since you've been following this thread very closely and choose to ask Jim and myself for further information I'll assume that you value our knowledge. If you want distractions and hyperbole try to follow some of the other drawings, charts, copy and paste stuff from the Internet and other misinformation in this thread.

What we have said from the beginning is that if you have a 60 amp 1Ø, 120/240 volt system that you could connect either one 60 amp, 240 volt load or two 60 amp, 120 volt loads.

Post #4

"Those two guys are incorrect and you're right. The feeder is 60 amp @ 240 volts (or 208). At 120 volts you would have a capacity of 120 amps but that nothing to do with the rating of the feeder."

Post #10

"Think about it, you could try to pull 75 amps on one leg of the 2 pole breaker while the other leg is at 0. Depending on the trip curve, the breaker will trip. You could also have each of the 2 hots pulling 60 amps and the breaker does not trip. 60 + 60=120."

[Roland Miller]

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29208d1376137090-determining-amperage-mat-main-disconnect-fuse-block-max-smoke.jpg

This is not about the current on the neutral.

I will ask you a straight forward question. How much current does it take to open a 60 amp fuse or breaker? 60 Amps Does this matter whether the breaker is a single or double pole You don't know the difference? It would need to be a 2-pole....If BOTH legs of the double pole breaker opened due to overload, how much current was flowing? 60 Amps
It would be 60 amps, if it were more you could measure it.. Only in New England would it take more than 60 amps.............
:crazy:

- - - Updated - - -



:crazy:

[Jim Port]

Sorry Roland, that diagram you posted was by Vern, not me.

It would not have to be a double pole in order to trip. The OP asked about a fused pullout disconnect. If there was more than 60 amps on one leg the fuse would still open. More than 60 amps on a double pole breaker would also cause the breaker to open.

[Roland Miller]

Sorry Roland, that diagram you posted was by Vern, not me.

It would not have to be a double pole in order to trip. The OP asked about a fused pullout disconnect. If there was more than 60 amps on one leg the fuse would still open. More than 60 amps on a double pole breaker would also cause the breaker to open.


Thanks for agreeing with me. That's 4 times now.:D Are you convinced or still thrashing around with the laws of physics?

[Vern Heiler]

Sorry Roland, that diagram you posted was by Vern, not me.

It would not have to be a double pole in order to trip. The OP asked about a fused pullout disconnect. If there was more than 60 amps on one leg the fuse would still open. More than 60 amps on a double pole breaker would also cause the breaker to open.

Jim, don't try to spin this around! That was a simple diagram showing that YOUR interpretation of what is going on (120 amps) violates Kirchhoff's first law!

[Jim Port]

I understand this issue quite well. Some just can't get past the fact that no one said you could measure the 120 amps on one conductor and that two conductors carrying 60 amps each equal 120 amps of capacity.

[Vern Heiler]

I understand this issue quite well. Some just can't get past the fact that no one said you could measure the 120 amps on one conductor and that two conductors carrying 60 amps each equal 120 amps of capacity.
If you understand this so well, just show us where each of the 60 amps goes. You can use my diagram or the one I stole from Code Check, we can probably get a group rate from the lawyer:D.

[Roland Miller]

I understand this issue quite well. Some just can't get past the fact that no one said you could measure the 120 amps on one conductor and that two conductors carrying 60 amps each equal 120 amps of capacity.

Some of you may be wondering what this discussion is all about. At times I wonder myself.

Jim and Robert are using a mix of amps and volt-amps to try to prove that 120 amps of "loads or capacity" exists somewhere in the circuit under discussion(120/240 volt-single phase), but admit they cannot measure it. This should raise a big question mark for all of you...

Here are two definitions of "load" from two major publication textbooks.(one of which I am a technical editor of).

"Load-the power consumed by a piece of equipment or a circuit while performing its function."

"Load-The current demand on the output of a circuit."

And a quote from one--"There is a limit on the LOAD (current demand) that can be placed on any circuit, in many cases, this limit is indicated by the current rating of the circuit's fuse or circuit breaker."

The NEC uses a mix of amps and volt-amps when defining "load". Why is this? During this calculation one would use the system voltage available where the load is being used. But when everything is put in volt-amps you can take that total and use it with any of the NEC system voltages. It would be noted that the NEC typically only uses dual voltage systems, but 120 is available (a single phase). which already makes R&J's answer wrong. The total load in amps is always stated from the higher voltage's perspective. A 120/240 system would be stated in amps at 240 volts..Not at 120 volts. Why is this--because this load is what is used to size the service entrance conductors and stating the service size...So for example you could not have 5-20 amp 120 volt loads and claim 100 amps of total load on a 120/240 system(the one under discussion). In fact if the loads were balanced between the 2 phases, there would be 20 amps on the neutral with only 50 amps of total connected load (is it starting to look familiar?). The Volt-amps would be the same for 120 or 240 (so you could find total load even at say 480 volts)..And lastly, you can build any electrical circuit in a lab and prove out your calculations by actual measurements (it is a science after all)..So be very cautious when someone tells you it exists but you can't actually measure it. This is the principle that lead to the banking mortgage collapse....ie. the money exists but you can't actually hold it in your hands..

quotes from --Delmar's Standard Textbook of Electricity and Introduction to Electricity-Paynter & Boydell

It was pointed out to me that my example might be confusing. I kept it real-world so 5- 20 amp circuit breakers or fuses. Three on A phase and two on B phase results in the loading I described. Certainly a completely balanced circuit would result in 0 amps on the neutral..

[Roland Miller]

Of course, the other issue here is the one of "self appointed experts" and who should you believe.

Jim, Robert, Jerry Peck, and HG all answer questions and promote them selves as "experts". The difference comes when someone disagrees with them. They then spin the question, redirect the issue, try to send you off to check with "someone that knows", accuse you of costing some innocent group money, say you can't even answer simple questions and are spreading misinformation. And call you a TROLL. And often they resort to insults and directly questioning whether you know anything about the subject. ..and add to that--They have checked with their buddy's and they agree with them.

This sure sounds a lot like BULLYING to me!

[Jim Port]

You were right Robert, here we are over 200 posts and some still do not understand and are now blaming us for bullying them. If telling people the facts is bullying, so be it.

[Vern Heiler]

You know Jim I just asked a few simple questions in this thread and for the most part have yet to get a direct response. For example, here are two direct questions, look at the response:





I can make this one even easier, only a one word answer required:

TRUE or FALSE?

60 amp, 1Ø, 3 wire, 120/240 volt service with no loads:

This service have the capacity to supply one 240 volt, 60 amp heater or two 120 volt, 60 amp heaters.
E X I = P and the answer when asked of POWER is yes. All done with the same 60 amps of current.

Now answer the one I have been asking from the beginning. Can you show us where the two 60 amp currents you say are available go in a circuit?

[Roland Miller]

You were right Robert, here we are over 200 posts and some still do not understand and are now blaming us for bullying them. If telling people the facts is bullying, so be it.


Of course, the other issue here is the one of "self appointed experts" and who should you believe.

Jim, Robert, Jerry Peck, and HG all answer questions and promote them selves as "experts". The difference comes when someone disagrees with them. They then spin the question, redirect the issue, try to send you off to check with "someone that knows", accuse you of costing some innocent group money, say you can't even answer simple questions and are spreading misinformation. And call you a TROLL. And often they resort to insults and directly questioning whether you know anything about the subject. ..and add to that--They have checked with their buddy's and they agree with them.

This sure sounds a lot like BULLYING to me! yup:confused:

- - - Updated - - -


You know Jim I just asked a few simple questions in this thread and for the most part have yet to get a direct response. For example, here are two direct questions, look at the response:





I can make this one even easier, only a one word answer required:

TRUE or FALSE?

60 amp, 1Ø, 3 wire, 120/240 volt service with no loads:

This service have the capacity to supply one 240 volt, 60 amp heater or two 120 volt, 60 amp

heaters.


Some of you may be wondering what this discussion is all about. At times I wonder myself.

Jim and Robert are using a mix of amps and volt-amps to try to prove that 120 amps of "loads or capacity" exists somewhere in the circuit under discussion(120/240 volt-single phase), but admit they cannot measure it. This should raise a big question mark for all of you...

Here are two definitions of "load" from two major publication textbooks.(one of which I am a technical editor of).

"Load-the power consumed by a piece of equipment or a circuit while performing its function."

"Load-The current demand on the output of a circuit."

And a quote from one--"There is a limit on the LOAD (current demand) that can be placed on any circuit, in many cases, this limit is indicated by the current rating of the circuit's fuse or circuit breaker."

The NEC uses a mix of amps and volt-amps when defining "load". Why is this? During this calculation one would use the system voltage available where the load is being used. But when everything is put in volt-amps you can take that total and use it with any of the NEC system voltages. It would be noted that the NEC typically only uses dual voltage systems, but 120 is available (a single phase). which already makes R&J's answer wrong. The total load in amps is always stated from the higher voltage's perspective. A 120/240 system would be stated in amps at 240 volts..Not at 120 volts. Why is this--because this load is what is used to size the service entrance conductors and stating the service size...So for example you could not have 5-20 amp 120 volt loads and claim 100 amps of total load on a 120/240 system(the one under discussion). In fact if the loads were balanced between the 2 phases, there would be 20 amps on the neutral with only 50 amps of total connected load (is it starting to look familiar?). The Volt-amps would be the same for 120 or 240 (so you could find total load even at say 480 volts)..And lastly, you can build any electrical circuit in a lab and prove out your calculations by actual measurements (it is a science after all)..So be very cautious when someone tells you it exists but you can't actually measure it. This is the principle that lead to the banking mortgage collapse....ie. the money exists but you can't actually hold it in your hands..

quotes from --Delmar's Standard Textbook of Electricity and Introduction to Electricity-Paynter & Boydell

It was pointed out to me that my example might be confusing. I kept it real-world so 5- 20 amp circuit breakers or fuses. Three on A phase and two on B phase results in the loading I described. Certainly a completely balanced circuit would result in 0 amps on the neutral..


Good job Vern!!

[Vern Heiler]

I posted a simple question requiring a one word answer, TRUE or FALSE, and again neither of the dynamic duo would answer it. I'm done with the dog chasing his tail routine. You guys are free to beat up on Jim for a while. :boink:
Oh, I'm so sorry! Yes/True, No/False.... The answer was True to a loaded question given by a simpleton. Now answer mine!

[Roland Miller]

You know Jim I just asked a few simple questions in this thread and for the most part have yet to get a direct response. For example, here are two direct questions, look at the response:





I can make this one even easier, only a one word answer required:

TRUE or FALSE?

60 amp, 1Ø, 3 wire, 120/240 volt service with no loads:

This service have the capacity to supply one 240 volt, 60 amp heater or two 120 volt, 60 amp

heaters.


Here's the correct math AGAIN:smash:

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910t-determining-amperage-mat-main-disconnect-fuse-block-math.jpg (http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910-determining-amperage-mat-main-disconnect-fuse-block-math.jpg)

[Jim Port]

That math was posted 200 posts ago and you said it was wrong, now you say it was right. Robert and I both said that two 60 amp 120 circuits had the same capacity of one 240 volt. Amazing.

Someone asks a question analogous to how much water can two pipes supply to a house and some want to know how the water gets to the treatment plant.

[Roland Miller]

That math was posted 200 posts ago and you said it was wrong, now you say it was right. Robert and I both said that two 60 amp 120 circuits had the same capacity of one 240 volt. Amazing. I didn't say the correct math was wrong! And thanks for agreeing with me, that is 5 times now.
Someone asks a question analogous to how much water can two pipes supply to a house and some want to know how the water gets to the treatment plant.

Your analogy is screwy--to correct it you would need one water pipe going in (+) and one returning (-). Get it now:confused:

[Vern Heiler]

That math was posted 200 posts ago and you said it was wrong, now you say it was right. Robert and I both said that two 60 amp 120 circuits had the same capacity of one 240 volt. Amazing.

Someone asks a question analogous to how much water can two pipes supply to a house and some want to know how the water gets to the treatment plant.
You gotta have coaster breaks to back peddle like that dude!

[Jim Port]

Your analogy is screwy--to correct it you would need one water pipe going in (+) and one returning (-). Get it now:confused:


The question was about supply. It was not about the neutral, never was.

[Vern Heiler]

Fixed this for you. At 120 volts you still only have 60 amps available per leg.


The load on the neutral doesn't matter in this example. Leg A carries 60 amps, leg B carries 60 amps equaling 120 amps of power. The 120 power is not returning on the other leg. It is on the neutral.


The question was about supply. It was not about the neutral, never was.
Really Jim:confused:.

I see your basic knowledge of plumbing is as weak as your knowledge of electricity. For electricity to do work it must return to its source immediately. Water can do work, e.g. turn a water wheel, and dump into the sea, not to return to the headwaters of the river for thousands of years.

[Jim Port]

Still confused by a 120 circuit I see.

[Mike Borchardt]

I have followed this thread very closely and have found very educational....and confusing :).

My question is for Jim & Robert to help me, and probably others, understand the argument. Are you saying that a 120 amp balanced load would not trip the 60 amp OCPD?

I apologize in advance if this is a stupid question.:oI believe you have it. A total of a 120 amp load at 120 volts is achievable if you are drawing 60 amps per leg. Leg (A) has a 60 amp load and leg (B) has a 60 amp load. Vern continues to ask where we can actually measure 120 amps. The answer is you can't. There isn't anyplace in the panel you can place an amp meter to measure 120 amps. Even though 120 amps can't be measured, It can be calculated by adding the total ampacity of each leg together. 60 amps + 60 amps. If 60 amps total was all there was available as Vern keeps insisting, you would not have enough current to run a toaster, electric stove, coffee pot, microwave oven, nesco roaster, TV, lights, and furnace all at the same time like we do come thanks giving.

[Roland Miller]

I believe you have it. A total of a 120 amp load at 120 volts is achievable if you are drawing 60 amps per leg. Leg (A) has a 60 amp load and leg (B) has a 60 amp load. Vern continues to ask where we can actually measure 120 amps. The answer is you can't. There isn't anyplace in the panel you can place an amp meter to measure 120 amps. Even though 120 amps can't be measured, It can be calculated by adding the total ampacity of each leg together. 60 amps + 60 amps. If 60 amps total was all there was available as Vern keeps insisting, you would not have enough current to run a toaster, electric stove, coffee pot, microwave oven, nesco roaster, TV, lights, and furnace all at the same time like we do come thanks giving.


Its the same 60 amps. If 120 was available you would be able to measure it because electricity is a science. See my other posts for the correct math and theory...:) Or read Vern's post here..he's the only other one that understands the math and theory so far..

- - - Updated - - -


Still confused by a 120 circuit I see.


Of course, the other issue here is the one of "self appointed experts" and who should you believe.

Jim, Robert, Jerry Peck, and HG all answer questions and promote them selves as "experts". The difference comes when someone disagrees with them. They then spin the question, redirect the issue, try to send you off to check with "someone that knows", accuse you of costing some innocent group money, say you can't even answer simple questions and are spreading misinformation. And call you a TROLL. And often they resort to insults and directly questioning whether you know anything about the subject. ..and add to that--They have checked with their buddy's and they agree with them.

This sure sounds a lot like BULLYING to me!

Thanks for the good example of BULLYING!

[Roland Miller]

I have followed this thread very closely and have found very educational....and confusing :).

My question is for Jim & Robert to help me, and probably others, understand the argument. Are you saying that a 120 amp balanced load would not trip the 60 amp OCPD?

I apologize in advance if this is a stupid question.:o

YES, a 120 amp balanced load would trip a 60 amp OCPD. Anything that exceeds the 60 amps should trip it. And if 120 amps actually existed anywhere in this circuit you could calculate it and measure it in a lab. The dark theory Jim and Robert presented does not exist in this world...

[Jim Port]

Your math shows the same thing that others posted 200 posts ago, but you say yours is correct and theirs is wrong?

Lets try this; Step 1: Install three fully loaded 20 amp breakers all on Leg A. How much load is on the incoming leg? How much load is on Leg B? Do the circuits function properly?

Step 2: install three more fully loaded 20 amp breakers all on Leg B. How much load was just added to the panel? How much load is on Leg A? Has this changed since the addition of the three new circuits? How much load is now on Leg B? What is the TOTAL load that both Leg A and B are supplying? Do all six circuits function properly without opening the fuse or breaker? Turn off all the breakers on Leg A. Has the load on Leg B changed?

As far as the bullying comment, I don't remember anyone besides you and Vern using words like simpleton or advising a career change.

- - - Updated - - -


YES, a 120 amp balanced load would trip a 60 amp OCPD. Anything that exceeds the 60 amps should trip it. And if 120 amps actually existed anywhere in this circuit you could calculate it and measure it in a lab. The dark theory Jim and Robert presented does not exist in this world...

Actually it would not. There would be 60 amps on each leg which is not over the trip limit of the breaker or fuse. In fact, your math even show the 7200 watts on each leg. Care to reconsider your answer?

[Roland Miller]

Your math shows the same thing that others posted 200 posts ago, but you say yours is correct and theirs is wrong?

Lets try this; Step 1: Install three fully loaded 20 amp breakers all on Leg A. How much load is on the incoming leg? How much load is on Leg B? Do the circuits function properly?

Step 2: install three more fully loaded 20 amp breakers all on Leg B. How much load was just added to the panel? How much load is on Leg A? Has this changed since the addition of the three new circuits? How much load is now on Leg B? What is the TOTAL load that both Leg A and B are supplying? Do all six circuits function properly without opening the fuse or breaker? Turn off all the breakers on Leg A. Has the load on Leg B changed?

As far as the bullying comment, I don't remember anyone besides you and Vern using words like simpleton or advising a career change.

- - - Updated - - -



Actually it would not. There would be 60 amps on each leg which is not over the trip limit of the breaker or fuse. In fact, your math even show the 7200 watts on each leg. Care to reconsider your answer?

Here's the correct math for your example----------AGAIN:smash:

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910t-determining-amperage-mat-main-disconnect-fuse-block-math.jpg (http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910-determining-amperage-mat-main-disconnect-fuse-block-math.jpg)

- - - Updated - - -

Jim--your are still missing the fact that A phase and B phase are 180 degrees out of phase which makes one + and one - in respect on one another at any give time. And you can't take the total wattage at the 240 volts and divide it by 120 and have anything meaningful. The correct problem equivalent of 240 is the (2 X 120).. not a single 120. This leads to a mathematical error and an incorrect assumption..

The 2- 60 amp loads on A and B phases are added using vector addition and since they are 180 degrees out of phase it is as easy as sliding your finger up and down a number line ( much like in 5th grade).

[Roland Miller]

As far as the bullying comment, I don't remember anyone besides you and Vern using words like simpleton or advising a career change.

- - - Updated - - -

Do you just make things up as you go along? Show me where I have said this..:confused:

[Jim Port]

Oh, I'm so sorry! Yes/True, No/False.... The answer was True to a loaded question given by a simpleton. Now answer mine!

Here is one example.

- - - Updated - - -


Here's the correct math for your example----------AGAIN:smash:

http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910t-determining-amperage-mat-main-disconnect-fuse-block-math.jpg (http://www.inspectionnews.net/home_inspection/attachments/electrical-systems-home-inspection-commercial-inspection/29207d1376066910-determining-amperage-mat-main-disconnect-fuse-block-math.jpg)

- - - Updated - - -

Jim--your are still missing the fact that A phase and B phase are 180 degrees out of phase which makes one + and one - in respect on one another at any give time. And you can't take the total wattage at the 240 volts and divide it by 120 and have anything meaningful. The correct problem equivalent of 240 is the (2 X 120).. not a single 120. This leads to a mathematical error and an incorrect assumption..

The 2- 60 amp loads on A and B phases are added using vector addition and since they are 180 degrees out of phase it is as easy as sliding your finger up and down a number line ( much like in 5th grade).

And you continue to miss that someone could add a 60 amp load to one leg or both legs and the panel will still function. This is the same premise from over 200 posts ago. This can be measured.

[Roland Miller]

Here is one example.

There you go--using a half-truth to make yourself look good. Again you are using oranges to prove and apple theory.

- - - Updated - - -


Here is one example.

- - - Updated - - -



And you continue to miss that someone could add a 60 amp load to one leg or both legs and the panel will still function. This is the same premise from over 200 posts ago. This can be measured.

Thanks for agreeing with me again.. Lets see that's 6 times now.... I have not missed that point---again you spin it and half-truth it to seem like you are right. And you are not!

[Jim Port]

And you continue to miss that someone could add a 60 amp load to one leg or both legs and the panel will still function. This is the same premise from over 200 posts ago. This can be measured.




Thanks for agreeing with me again.. Lets see that's 6 times now.... I have not missed that point---again you spin it and half-truth it to seem like you are right. And you are not!

So 60 amps on two legs does not equal 120 amps of capacity? You can run two 60 amp 120 volt loads from one panel but the panel will not support 120 amps of load? Interesting.

I will bow out and leave you and Vern to your confusion. As Robert said many posts ago, it didn't matter how many times and different ways that this would work there is no light at the end of the tunnel.

Sorry that you never took the challenge to post this topic on an electrical forum.

[Roland Miller]

Sorry that you never took the challenge to post this topic on an electrical forum.


After 45 years of teaching and doing the math and science of electricity ( a degree in Communication Electronics and Science and a life long practicing master electrician) I would have to say I have probably taught some of them about the NEC or the science of electricity. Really Jim--it is simple enough that someone with a 5th grade education could grasp. I have said "I could teach anyone to be an electrician" but I am wrong because there are at least 3 of you wouldn't make it....

[Joe Palmer]

Hello, all, I am new here. I would like to respond to the discussion about electrical cabinet capacity. I have been in the multifamily industry for nearly 40 years, and have a lot of experience with electrical circuits. This topic is a difficult one to grasp at times. The ampacity of the cabinet in question (or any cabinet, for that matter) is the rated ampacity for the main disconnect. In this case the disconnect is rated at 60 amps. This means that this cabinet has an ampacity of 60 amps. The "legs" (more correctly, "poles") are not additive. If one would employ a clamp-style ammeter, clamp around both poles (service drop conductors), one would see a reading on the ammeter of zero amps if both poles were utilizing 60 amps of current simultaneously. If one pole is using more current (for example, a 120 volt lighting circuit is being utilized in addition to a 240 volt HVAC system), the ammeter would display the current difference ONLY. I wrestled with this concept until I performed the action I described by clamping my digital ammeter around the main conductors in a service panel just to see the outcome. VOILA! My Four Rules for Living: #1- Don't panic (thanks Mr. Adams); #2 - Pay attention; #3 - Learn something every day; #4 - When all else fails, read the directions.

[Roland Miller]

Hello, all, I am new here. I would like to respond to the discussion about electrical cabinet capacity. I have been in the multifamily industry for nearly 40 years, and have a lot of experience with electrical circuits. This topic is a difficult one to grasp at times. The ampacity of the cabinet in question (or any cabinet, for that matter) is the rated ampacity for the main disconnect. In this case the disconnect is rated at 60 amps. This means that this cabinet has an ampacity of 60 amps. The "legs" (more correctly, "poles") are not additive. If one would employ a clamp-style ammeter, clamp around both poles (service drop conductors), one would see a reading on the ammeter of zero amps if both poles were utilizing 60 amps of current simultaneously. If one pole is using more current (for example, a 120 volt lighting circuit is being utilized in addition to a 240 volt HVAC system), the ammeter would display the current difference ONLY. I wrestled with this concept until I performed the action I described by clamping my digital ammeter around the main conductors in a service panel just to see the outcome. VOILA! My Four Rules for Living: #1- Don't panic (thanks Mr. Adams); #2 - Pay attention; #3 - Learn something every day; #4 - When all else fails, read the directions.


Hi Joe, Welcome to this forum. You will find lots of useful information here. I look forward to seeing you post more often. I might add that all threads are not like this one. Another qualified electrical guy like yourself is needed. You just have to be willing to sort out the BS and Bullying from the real information. ;)

[Dwight Doane]

If you insert one double pole 60 amp breaker and put a 60 amp load on it, you've maxed out the available amount of current on both legs and any more breakers whether it be single or double pole will blow a fuse. The amperage available, determined by the amperage of the fuses, is based on 240 volts, not 120 volts. If you exceed 60 amps at 120 volts on either leg, you will blow a fuse, If you exceed 60 amps at 240 volts, you will blow fuses. this is why its a 60 amp panel.[/QUOTE]


I wanted to quote this and clear it up - The difference between a double (or three breaker) and fuse cartridges is that the breaker [more than one pole] is tied to the other so if one leg exceeds the rating all poles trip - this is done to keep the load somewhat balanced between the hot legs.

You can draw 60 amps (using this example) off of each leg - so to make this more of a mess let's talk about industrial loads "Three Phase Power" In this situation you have three hot legs and each leg is 120 degree out of phase with each other, Sometimes you have a neutral and sometimes you do not. However Now you have three fuses rated at 60 amps (remember amps are amps no matter what the voltage is they are still amps) the rating of the service would be 60 amps because that is the most you can draw through a hot leg.

The return get's a little confusing in 120 v circuits when we start looking at the load of the neutral because the neutral is often tied to the ground and in a home the neutral coming off the street is earth grounded at the transformer with a grounding rod , just like in a house (often tied to a water pipe) so if you could isolate the neutral and the ground and actually measure it where you are drawing your power - you would actually have two current reading (I know what the code says about grounds not being load carrying conductors) and these two readings (done at the same time with two meters) would add up to 120 amps - if you were to draw 60 amps from each leg.

On a side note I have found this thread to be most amusing and really , stop pulling my leg. :p

[Roland Miller]

I wanted to quote this and clear it up - The difference between a double (or three breaker) and fuse cartridges is that the breaker [more than one pole] is tied to the other so if one leg exceeds the rating all poles trip - this is done to keep the load somewhat balanced between the hot legs. I will just leave this at "NO".


The return get's a little confusing in 120 v circuits when we start looking at the load of the neutral because the neutral is often tied to the ground and in a home the neutral coming off the street is earth grounded at the transformer with a grounding rod , just like in a house (often tied to a water pipe) so if you could isolate the neutral and the ground and actually measure it where you are drawing your power - you would actually have two current reading (I know what the code says about grounds not being load carrying conductors) and these two readings (done at the same time with two meters) would add up to 120 amps - if you were to draw 60 amps from each leg I will leave this with "only your wildest dreams with bad math, back to the drawing board."

On a side note I have found this thread to be most amusing and really , stop pulling my leg. :p

Pulling your leg about what? I have found it amazing what people believe and what actually can happen with electricity.

Thanks for posting--I will let Vern pick up from here.

[Jim Port]

Hello, all, I am new here. I would like to respond to the discussion about electrical cabinet capacity. I have been in the multifamily industry for nearly 40 years, and have a lot of experience with electrical circuits. This topic is a difficult one to grasp at times. The ampacity of the cabinet in question (or any cabinet, for that matter) is the rated ampacity for the main disconnect. In this case the disconnect is rated at 60 amps. This means that this cabinet has an ampacity of 60 amps. The "legs" (more correctly, "poles") are not additive. If one would employ a clamp-style ammeter, clamp around both poles (service drop conductors), one would see a reading on the ammeter of zero amps if both poles were utilizing 60 amps of current simultaneously. If one pole is using more current (for example, a 120 volt lighting circuit is being utilized in addition to a 240 volt HVAC system), the ammeter would display the current difference ONLY. I wrestled with this concept until I performed the action I described by clamping my digital ammeter around the main conductors in a service panel just to see the outcome. VOILA! My Four Rules for Living: #1- Don't panic (thanks Mr. Adams); #2 - Pay attention; #3 - Learn something every day; #4 - When all else fails, read the directions.


You do not clamp around both conductors at the same time.

[Roland Miller]

Hey Dwight, Use this diagram to show us what you are thinking..


29214

[Roland Miller]

You do not clamp around both conductors at the same time.


Really? I think Joe is on the right track. Why?? what would happen? I would say he would get some good information (scientific and all)...:first:

[Jim Port]

Why don't you share some more of your knowledge? I would have thought with all that experience you would know how to use a clamp on ampmeter.

I know the reading will be useless, even on a single pole circuit.

[Roland Miller]

Why don't you share some more of your knowledge? I would have thought with all that experience you would know how to use a clamp on ampmeter.

I know the reading will be useless, even on a single pole circuit.


Jim, Robert, Jerry Peck, and HG all answer questions and promote them selves as "experts". The difference comes when someone disagrees with them. They then spin the question, redirect the issue, try to send you off to check with "someone that knows", accuse you of costing some innocent group money, say you can't even answer simple questions and are spreading misinformation. And call you a TROLL. And often they resort to insults and directly questioning whether you know anything about the subject. ..and add to that--They have checked with their buddy's and they agree with them.

This sure sounds a lot like BULLYING to me!


You really do need to give it up.

- - - Updated - - -


Why don't you share some more of your knowledge? I would have thought with all that experience you would know how to use a clamp on ampmeter.

I know the reading will be useless, even on a single pole circuit.


Putting a clamp-on ammeter around the hot and neutral of a single pole? circuit will tell you if there are neutral to neutral faults in the circuit. If it reads anything except 0 amps the neutral current is going somewhere else. This creates, rather than cancels electric fields..:p

[Roland Miller]

I will add that a small portion of the population is physically sensitive to stray electric fields. They make them physically ill. They solve it by having all the neutral to neutral faults corrected and installing a special isolation transformer (scared yet?) to supply their residences. Not to mention stray electric fields have a direct interference to sensitive electronic equipment..

[Jim Port]

That method would not tell you the load on the panel which was the subject under discussion. We were not looking for neutral to neutral faults.

[Roland Miller]

That method would not tell you the load on the panel which was the subject under discussion. We were not looking for neutral to neutral faults.I was just addressing the questions you raised, or did you forget?

[Joe Palmer]

Of course, Mr.Port, I am well-versed in the use of clamp ammeters, having owned both analogue and digital flavors (one of the first on the market, the A.W. Sperry Digisnap 1000, ca. 1982). I used them mostly to test the three-pole 300 amp contactors on the electric water heaters at the multifamily communities where I supervised the HVAC/plumbing crews. My present digital ammeter is bluetooth-enabled so that the meter can drop data into my laptop regarding voltage/current/temperature over time. I also use it to test start and run capacitors in HVAC systems. My intent was to illustrate that a simultaneous 60 amp load ON BOTH POLES OF THE SERVICE DROP does not constitute a 120 amp panel. The question has been answered many times here. The cabinet is a 60 amp panel. By the bye, the reading on a single-pole circuit would be 60 amps if there is a 60 amp load; your logic is erroneous, as one would be using the meter correcly, having clamped only one conductor, as is required for ampacity measurement.

[Jim Port]

I read it as you were testing the load by clamping both conductors, not one. Sorry for the confusion.

- - - Updated - - -

I read it as you were testing the load by clamping both conductors, not one. Sorry for the confusion.

[Vern Heiler]

Well here we are at over 250 posts and Judge Judy is about to reign down her wrath:(. Before I am sent off to the hall I would like to apologize for loosing my cool to all of my friends on IN, including and most importantly to Jim and Robert.

I truly was just trying to help people understand what was happening inside the 3-wire circuit. It does not really matter if you understand it fully to be able to size breakers and wiring, using ohms law formulas. Knowing the maximum current that can be on each leg, you will come up with the correct answers, and the maximum current will be the breaker size.

Where the most problems came from in this thread was from the terms used. Jim and Robert used the term "amp load" throughout the discussion. Amp load is typically reserved for electric motor and battery loads. In this discussion the term "amp load" was used synonymously with "watts or power". The original question was directed at "current" which is only one part of the formula to find power.

The 3-wire circuit, as in all circuits, is a continuous loop. A good analogy of the loop would be that of the lazy river that is found in many amusement parks. All of the water leaves the pump and sends inner tubers down the river, some of the tubers may take a short-cut back to the pump station, because there is a traffic jamb on the return leg (less load or more resistance), but they are using just some of the same current that is taking the rest of the tubers the long way around. When the traffic (load) is the same on both legs of the river, no tubes take the short-cut (neutral) but return to the pump via the main return leg. We can get the current to do more work "power" by adding more tubers (loads) but it requires more current. The total current is still in a loop so whether we split it at the short-cut or not is irrelevant. The total current leaving the pump must equal the current returning to the pump.

I hope this has helped someone, and will now go take my beating for prolonging this thread;).

[Roland Miller]

Before I am sent off to the hall I would like to apologize for loosing my cool to all of my friends on IN, including and most importantly to Jim and Robert.



I think you can quite holding your breath on this one:p.

[Dwight Doane]

I'm not about to choose a side, but you guys are to be commended.
Nearly 100 post on this and no name calling, slanders, or put downs.


You know Rick, I too noticed the same thing , no name call, no back stabbing , no cheap shots below the belt

There is something just Un American about the whole thing - it disgusts me :)

Actually , I am really proud to see people acting in a professional way trying to help someone understand a difficult concept.

[Bill Kriegh]

I may not have seen the answer I was looking for in the volume of posts written here, but the size of an electric service is determined by the lowest rating of the following three things: rating of the service conductors, manufacturer's rating of the equipment, and the rating of the over current device(s).

The size of the over current device is determined by the maximum size of load per hot leg. All other things equal, a 60 AMP fuse would limit a service size to a "60 AMP service". Two hot legs with 60 AMP fuses do not make a 120 AMP service any more than 3 of them make a 180 AMP service. Service size refers to the maximum size of an individual connected load. On a 60 AMP single phase service you can hook up a 240 volt 60 AMP load or two 120 volt 60 AMP loads. You can't hook up a single 120 volt 120 AMP load - ergo you have a 60 AMP service.

The maximum capacity of the service will be determined by the total wattage allowed by the voltage and overload size.

[Brad Richter]

As I know the correct answer for the previous thread, let me ask a question. How can a MWBC using # 12 wire for all of the conductors be allowed to be supplied by 20 amp breakers if the potential current on the grounded conductor is 40 amps? Is such an installation cited as non- compliant?

[Vern Heiler]

As I know the correct answer for the previous thread, let me ask a question. How can a MWBC using # 12 wire for all of the conductors be allowed to be supplied by 20 amp breakers if the potential current on the ungrounded conductor is 40 amps? Is such an installation cited as non- compliant?
Please read 254 posts:frusty::frusty:

[Brad Richter]

Please read 254 posts:frusty::frusty:
I did read the posts. Can you answer the question or not? You're hurting your head.

[Jim Port]

A properly wired MWBC will have the two hots on opposite legs of the panel. The neutral will carry the difference, not the sum of the hots.

[Brad Richter]

A properly wired MWBC will have the two hots on opposite legs of the panel. The neutral will carry the difference, not the sum of the hots.
Thanks , Jim, I know the answer. Just posted the question to get answers from the inspectors that seem confused.

[Roland Miller]

A properly wired MWBC will have the two hots on opposite legs of the panel. The neutral will carry the difference, not the sum of the hots.


And so according to Jim you would need a 40 amp panel to support this 20 amp load. If you were to read the other posts.

Oh and thanks for agreeing with Vern and I for the 7th time.

[Brad Richter]

And so according to Jim you would need a 40 amp panel to support this 20 amp load. If you were to read the other posts.

Oh and thanks for agreeing with Vern and I for the 7th time.
That in no way is what Jim has been posting.

[Roland Miller]

Please read 254 posts:frusty::frusty:
I can certainly understand you probably had some difficulty following all these posts. Or you didn't take Vern's advice. Jim will try to tell you this configuration has 40 amps of capacity. It doesn't. :brick:

At least in regards to the original question.

[Brad Richter]

I can certainly understand you probably had some difficulty following all these posts. Or you didn't take Vern's advice. Jim will try to tell you this configuration has 40 amps of capacity. It doesn't. :brick:

At least in regards to the original question.
Sorry, Jim is correct. 20 amps on each leg @ 120 volts.

[Roland Miller]

Ignore the transformer and focus on 2 60 amp single pole circuits. Each can flow 60 amps. You say you can understand the math, but cannot accept that 60 x 2 = 120?
Why does he say this over and over again in regards to the orginal post??



U.S. Illiteracy Statistics
Data


Percent of U.S. adults who can’t read
14 %


Number of U.S. adults who can’t read
32 Million


Percent of U.S. adults who read below a 5th grade level
21 %


Percent of prison inmates who can’t read
63 %


Percent of high school graduates who can’t read
19 %

[Brad Richter]

Why does he say this over and over again in regards to the orginal post??



U.S. Illiteracy Statistics
Data


Percent of U.S. adults who can’t read
14 %


Number of U.S. adults who can’t read
32 Million


Percent of U.S. adults who read below a 5th grade level
21 %


Percent of prison inmates who can’t read
63 %


Percent of high school graduates who can’t read
19 %



Makes else to me and to most other electricians as well.

[Roland Miller]

Makes else to me and to most other electricians as well.


:DI see I have made my point--:confused:

[Billy Stephens]

Nice Try.

(http://www.inspectionnews.net/home_inspection/members/john-kogel.html) John Kogel (http://www.inspectionnews.net/home_inspection/members/john-kogel.html)
http://www.inspectionnews.net/home_inspection/images/statusicon/user-online.png Member

Join DateFeb 2009LocationSouthern Vancouver IslandPosts3,603
http://www.inspectionnews.net/home_inspection/images/icons/icon1.png Re: Square D panel
It appears to be a split bus panel with the 50 amp breaker protecting the lower half. But this reply will bump your thread ahead of that other one which has bored most of us for 3 weeks, so hopefully you will get your answer today. :D

[Jim Port]

That in no way is what Jim has been posting.

Thanks Brad.

Roland, I keep saying the same thing because it is the truth. Check Bill M's recent post. He can understand the issue also and agrees with Robert, myself, Brad and other's. You can't change the truth. Sorry you still don't understand.

[Roland Miller]

Thanks Brad.

Roland, I keep saying the same thing because it is the truth. Check Bill M's recent post. He can understand the issue also and agrees with Robert, myself, Brad and other's. You can't change the truth. Sorry you still don't understand.
You keeping saying the same thing trying to make it true, when it isn't. You can't dispute the math and science not matter how many times you say it. Vern as well as myself can prove our position, you can only state yours. You have a growing following though---3 or 4 now..

[Jim Port]

Others have also explained numerous times the error of your understanding of the issue. It is such a shame that someone of your professed experience cannot grasp this simple concept.

[Vern Heiler]

I think that we should send a link to these generator manufacturers so that they can all fire their engineers for getting the generator output wrong:


Honda EB10000 Portable Industrial 10000 Watt Generator (http://powerequipment.honda.com/generators/models/eb10000)



Order Model
Voltage
Hz
Rated amps3
LPV fuel
Rated amps3
NG fuel
Circuit Breaker
Cold
weather
equipped
Certification


20GSBC-6727B
120/240
60
154/77
140/70



Cummins Onan (http://www.cumminsonan.com/residential/products/homestandby/compare?gensetId=108&detail=true)


DuroMax RV Grade 4400 Watt 7.0 Hp Gas Generator w/ Electric Start & Wheel Kit (http://www.generatorfactoryoutlet.com/duromax-rv-grade-4400w-gas-generator-w-electric-start-wheel-kit-xp4400e?CAWELAID=839346570&cagpspn=pla&gclid=CLSsoaO4trkCFcue4AodMlEAqw)
They would just call us idiots for thinking a generator was a 3 wire circuit!

[Roland Miller]

I inspected an apartment in a building with 200 apartment units, The main disconnect for the sub panel of the apartment is located at the electrical meter in the basement of the building. There are two 60 amp cartridge type fuses in a fuse block as the disconnect. I stated that it was a 60 amp service and the building supervisor stated no it is a 120 amp service because you combine the two legs of the service. I am confusing my self, I always thought you rated the disconnect by the weakest link, which would be the 60 amp fuse. Which is correct, is it a 60 amp service or 120 amp service ? Thanks in advance.:confused:
Just so it is clear. The answer is a 60 amp service. The two legs are not added to give the capacity of the service. And when installed according to the NEC there will not be more then 60 amps present at any point in the circuit.

I see Jim has started with his bullying tactics. That's a real shame that you cannot rise above your mistake and your ego.

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Others have also explained numerous times the error of your understanding of the issue. It is such a shame that someone of your professed experience cannot grasp this simple concept.


Jim, Robert, Jerry Peck, and HG all answer questions and promote them selves as "experts". The difference comes when someone disagrees with them. They then spin the question, redirect the issue, try to send you off to check with "someone that knows", accuse you of costing some innocent group money, say you can't even answer simple questions and are spreading misinformation. And call you a TROLL. And often they resort to insults and directly questioning whether you know anything about the subject. ..and add to that--They have checked with their buddy's and they agree with them.

This sure sounds a lot like BULLYING to me!

Jim, you have agreed with me 7 times now but insist your statement is true. What's up with that?

[Jim Port]

Roland, obviously I do not agree with you since I have said from somewhere on page 1 that the panel has a capacity of 120 amps at 120 or 60 amps at 240. I simply corrected your statement about the capacity per leg. You keep saying the limit is 60 amps regardless of voltage. It seems from the post from Robert concerning generator capacities that the engineers for those companies too agree with me. There never was an issue with the correct size of the service.

I again will invite your to post the same question as asked by the OP on an electrical forum and post a link to it. Since you have refused multiple times I will assume you do not want to submit yourself to ridicule and further embarassment.

Speaking the truth and refusing to waver is not a matter of ego. Simply because you do not like the answer does not make it wrong.

And once again you accuse me of being the bully when the true issue is your inability to understand a simple issue. I never posted literacy figures trying to infer anyone's lack of knowledge.