Question

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??

Answer 1

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".