Thread: Determining the amperage mat a main disconnect fuse block

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.

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

Originally Posted by 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.

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.

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

Originally Posted by Robert Meier

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.

Originally Posted by Roland Miller

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.

Originally Posted by Jim Port

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

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.

Originally Posted by 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.

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

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

Originally Posted by Jim Port

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

Having trouble up loading scanned file

Originally Posted by 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.

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.

Originally Posted by Jim Port

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.

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.

I think I have done it Yay

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.

Originally Posted by 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.

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.

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

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.

Originally Posted by 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.

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....

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Originally Posted by 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.

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

Originally Posted by 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.

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.

Originally Posted by Vern Heiler

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..

http://www.mikeholt.com/instructor2/...809-sample.pdf

Maybe this will help, hopefully.

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

Originally Posted by Roland Miller

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.

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.

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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.

Originally Posted by 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.

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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.

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.

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Repeated again from post #4.

Originally Posted by Robert Meier

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.

Originally Posted by 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.

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Repeated again from post #4.

Take the diagram I posted for you and draw out where the 120a comes from and goes to.

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.

Originally Posted by 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.

Draw it out. And remember; " Kirchhoff's current law (KCL) The current entering any junction is equal to the current leaving that junction.

Originally Posted by 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.

Originally Posted by Jim Port

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.

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

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?

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.

Originally Posted by 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.

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

Originally Posted by Robert Meier

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,

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.

Originally Posted by 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.

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!)

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.

Originally Posted by Robert Meier

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.

Originally Posted by 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.

Jim, you are absolutely correct. And the neutral would be zero volt.

Originally Posted by Vern Heiler

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.

Originally Posted by Mike Borchardt

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.

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.

Originally Posted by 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.

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

Originally Posted by Vern Heiler

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.

Originally Posted by Jim Port

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.

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

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.

Originally Posted by 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. [img]http://www.ibiblio.org/kuphaldt/elec...s/AC/02168.png[/img]

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.

Originally Posted by Robert Meier

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.

Originally Posted by Jim Port

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. 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.

Originally Posted by Robert Meier

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.

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.

Originally Posted by 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.

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.

Originally Posted by Vern Heiler

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 that describes a smooth repetitive oscillation. It is named after the function sine, of which it is the graph. It occurs often in pure and applied mathematics, as well as physics, engineering, signal processing and many other fields. Its most basic form as a function of time (t) is:
where:

• A, the amplitude, is the peak deviation of the function from zero.
• f, the ordinary frequency, is the number of oscillations (cycles) that occur each second of time.
• ω = 2πf, the angular frequency, is the rate of change of the function argument in units of radians per second
• φ, the phase, 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.

Originally Posted by Mike Borchardt

The sine wave or sinusoid is a mathematical curve that describes a smooth repetitive oscillation. It is named after the function sine, of which it is the graph. It occurs often in pure and applied mathematics, as well as physics, engineering, signal processing and many other fields. Its most basic form as a function of time (t) is:
where:

• A, the amplitude, is the peak deviation of the function from zero.
• f, the ordinary frequency, is the number of oscillations (cycles) that occur each second of time.
• ω = 2πf, the angular frequency, is the rate of change of the function argument in units of radians per second
• φ, the phase, 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.

Originally Posted by Robert Meier

Uh how is that almost correct, haven't you've said the same thing.

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.

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Originally Posted by Robert Meier

Uh how is that almost correct, haven't you've said the same thing.

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.

Originally Posted by Robert Meier

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.

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

Originally Posted by Robert Meier

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.

Originally Posted by Vern Heiler

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.

Originally Posted by Mike Borchardt

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."

Originally Posted by Vern Heiler

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

Originally Posted by JOHN PAVAN

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!