Is a 15 A Circuit Breaker Capable of Handling 10.6 A Maximum Current?

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A 15 A circuit breaker can handle a maximum current of 10.6 A when calculated incorrectly as I(max) = 15(rms) / square root of 2, but the correct formula is I(max) = 15(rms) * square root of 2. The average power supplied by the circuit is correctly calculated at 1.80 kW. The discussion highlights the importance of understanding the relationship between maximum and average current. Clarifications were made regarding the calculations, emphasizing the need for accuracy in electrical assessments.
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A circuit breaker is rated for a current of 15 A rms at a voltage of 120 V rms. Answer the following questions:
All answers must have three significant digits unless otherwise specified.

1) What is the largest value of I(max) that the breaker can carry? (Unit: A)

answer:

I(max) = 15(rms) / square root of 2
= 10.6A

2) What average power can be supplied by this circuit? (Unit: kW)

P(rms) = I (rms) * V (rms)
= 1.80 kW

pls correct me if I was wrong and pls direct me to the right path...
thanks
cseet
 
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Argh, stupid me.

I(max) = 15(rms) / square root of 2 is not correct, it should be:
I(max) = 15(rms) * square root of 2 if I'm not mistaken.
 
Last edited:
Hi chen,

what about the 2nd answer? is it correct?

thanks
cseet
 
Yes, 1.8kW. By the way, you could have noticed that your first answer is wrong because the maximum current cannot be lower than the average current. :smile:
 
Thanks Chen... I see that now...
you have a good day
cseet
 
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