Homework Answers
Part A.
Average power in ac circuit is given by:
P_avg = I_rms^2*R
Given that: I_rms = 6.00 A
R = 16.0 ohm
So,
P_avg = 6.00^2*16.0
P_avg = 576 W
Part B.
Since we know that:
P_avg = I_rms^2*R
And, I_rms = I_peak/sqrt 2, So
P_peak = (I_peak)^2*R
P_peak = (I_rms*sqrt 2)^2*R
P_peak = 2*P_avg
P_peak = 2*576
P_Peak = 1152 W
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