The potential difference E and current I flowing through the ac circuit is given by `E=5 cos (omega t -pi//6) V and I=10 sin (omega)t A`. Find the average power dissipated in the circuit.

Explanation

`E=5 cos [omega t =(pi//6)], I=10 cos [ omega t=(pi//2)]`
Phase difference between E and I is given by:
`phi=(pi)/(2) - (pi)/(6) = (2 pi)/(6) = (pi)/(3)`
Avaerage power dissipatted:
`P=(E_(0)I_(0))/(2) cos phi =(5xx10)/(2) xx 1/2=12.5 W`.