Thread: Size of dedicated branch circuit

This is an ICC practice exam question:

A 1512-VA, 120 v cord and plug connected room air-conidtion will require what size dedicated branc circuit in amperes?

A) 15
B) 20
C) 25
D) 30

The answer is B.

1512/120 = 12.6....Why is not the answer 15 amps?

thanks as always for your help!

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From the 2006 IRC. (underlining is mine)
- E3602.3 Fifteen- and 20-ampere branch circuits. A 15- or 20-ampere branch circuit shall be permitted to supply lighting units, or other utilization equipment, or a combination of both. The rating of any one cord-and-plug-connected utilization equipment not fastened in place shall not exceed 80 percent of the branch-circuit ampere rating. The total rating of utilization equipment fastened in place, other than lighting fixtures, shall not exceed 50 percent of the branch-circuit ampere rating where lighting units, cord-and-plug-connected utilization equipment not fastened in place, or both, are also supplied.

15 x 80% = 12.0 amps

You that room a/c draws 12.6 amps, which exceeds the 80% limitation of 12.0 amps.

Oh, yeah... I knew that .

Guess I won't be taking the ICC exam any time soon.

The key info is dedicated branch circuit. This kicks you into §RE3602.12.1. You need to take your 12.6 and divide by 0.80 giving 15.74 amps. Then round it up to 20.

§RE3602.12 Branch circuits serving room air conditioners. A room air conditioner shall be considered as a single motor unit in determining its branch-circuit requirements where all the following conditions are met:
1. It is cord-and attachment plug-connected.
2. The rating is not more than 40 amperes and 250 volts; single phase.
3. Total rated-load current is shown on the room air-conditioner nameplate rather than individual motor currents.
4. The rating of the branch-circuit short-circuit and ground-fault protective device does not exceed the ampacity of the branch-circuit conductors, or the rating of the branch-circuit conductors, or the rating of the receptacle, whichever is less.

§RE3602.12.1 Where no other loads are supplied. The total marked rating of a cord-and attachment plug-connected room air conditioner shall not exceed 80 percent of the rating of a branch circuit where no other appliances are also supplied.

§RE3602.12.2 Where lighting units or other appliances are also supplied. The total marked rating of a cord-and attachment plug-connected room air conditioner shall not exceed 50 percent of the rating of a branch circuit where lighting or other appliances are also supplied.

When I started typing there were no replies yet. Quick-fingers Jerry beat the one handed typist to the draw.... But my code cite is even better than his.

Fascinating question!

Why can't you use Table E3604.2(1) where it says for fixed motors, the full-load current of motors plus 25% of the full load current of the largest motor.

The A/C has a "fixed motor" and since it is the only motor in this question, it is also the largest motor.


Therefore:

((1512/120))/.75 = 16.66 Amps which gets you to the same answer, 20 amps, when you round up.

Or is my logic totally flawed?

Robert,

You could, all of the above code sections apply, some apply "specifically" and only to that question while other apply on a more broader scale, including to that question - all give the same answer.

Heck, one could even state that that room unit is going to run 3 hours or longer and thus be a "continuous load", in which case 125% of the load is used, also resulting in the need for 20 amp circuit conductors.

Most, if not all, of those sample questions are from the pool of test questions and are question which were weeded out for being flawed (several have been flawed) or for having multiple answers to the same question (meaning multiple code sections producing the same correct answer).

Thanks , everyone!

This may sound silly, but I've never run across a central vacuum system....is it cord-n-plug or fixed wiring? Why do I ask, because if it is cord-n-plug, then I understand how to get the answer to a similar ICC exam question. If not, I'm lost again....

A 1,500-VA, 102-volt central vacuum system is located in the garage. The minimum requirements for supplying branch circut to this load are:

A) 15amp circuit with single receptacle.
B) 20amp circuit with single receptacle.
C) 20amp circuit with single receptacle, GFCI protected.
D) 20amp circuit with duplex receptacle, GFCI protected.

The answer is B....if a cental vacumm system is cord -n- plug , then the required min amp is 1500/120=12.5. 12.5/.8 = 15.74 amps, round to 20amp. E3801.9 requires only a single receptacle for a garage. Thus, I can get to answer B.....but if the vacuum system is not a cord-n-plug, then I'm lost again.

All of the thousands I've seen came with (to my best recollection) a cord and plug for plugging into a receptacle outlet.

Originally Posted by Brandon Chew

The key info is dedicated branch circuit. This kicks you into §RE3602.12.1. You need to take your 12.6 and divide by 0.80 giving 15.74 amps. Then round it up to 20.

Brandon,

Why are you dividing by 0.80? Is it not a multiplication factor, percent of?

(1512VA/120v)*.8 = 10 Amps?

I guess I'm confused on the formula.

W

The total marked rating of a cord-and attachment plug-connected room air conditioner shall not exceed 80 percent of the rating of a branch circuit where no other appliances are also supplied.

There are two ways you could solve the problem. Jerry gave the "trial & error" way, and I used the direct way.

Jerry started with the ampacity of the smallest conductor (15 amps) and multiplied by 0.80 to get a number (12 amps) that the A/C could not exceed on that conductor. Then he compared this to the actual A/C of 12.6 amps, said 15 would be too small, and bumped the conductor up to 20 amps. To check that, you'd then take 20 * 0.80 and get 16 amps ... compare that to 12.6 amps actual, and say the 20 amp conductor is OK.

I approached this from the opposite direction, starting at the A/C and heading for the minimum allowable conductor size, which means I need to divide the A/C by 0.80 (instead of multiplying the conductor by 0.80). Given the A/C load of (1512VA/120v) = 12.6 amps, dividing that by 0.80 gets me directly to 15.75 amps as the minimum conductor ampacity needed for that A/C ... no further calculations or checks are needed.

Formula is: (A/C amps) = (0.80) x (conductor amps)

All I did was re-arrange the formula to move the 0.80 from the conductor side to the A/C side. When you do this the formula becomes:

(A/C amps)/(0.80) = (conductor amps)

Either method gets you the correct answer. The only difference is that the way I did it gets you there faster. Do it whichever way makes the most sense for you.