Power of lamp (P)=1000W AC voltage (E)=200sin(310t+60∘) Comparing with E=E0sin(ωt+φ), we get, E0= peak value of voltage φ= phase difference rms value of supply voltage, Erms=
E0
√2
Erms=
200
√2
=100√2 Average power is given by, P=VrmsIrmscosφ 1000=(100√2)Irms(