Emf applied and current in an A.C circuit are E = 10sqrt(2)sin(omegat) volt and I = 5sqrt(2)cos(omegat) ampere respectively. Average power loss in the circuit is??

1 Answer
Aug 9, 2017

The average power loss is zero.

Explanation:

The instantaneous power loss is, of course

P = EI = 100 sin(omega t) cos (omega t)

To find the average power loss, you must integrate this over one complete period

P_{av} = 1/T int_0^T 100 sin(omega t) cos (omega t) dt

where T={2 pi}/omega is one time period. It is easy to see that this integral vanishes

int_0^T 100 sin(omega t) cos (omega t) dt
= 50 int_0^T sin(2omega t) dt
=-50/{2 omega} (cos (2omega T)-cos 0)
=-50/{2omega} (cos(4 pi) - cos 0) = 0

Note that this is a case where the emf lags 90 degrees behind the current, as happens for a capacitor. This is an example of so called "wattles current".