A coil has an inductance of `0.7 H` and is joined in series with a resistance of `220 Omega`. When an alternating e.m.f of `220 V` at 50 c.p.s. is applied to it, then the wattless component of the current in the circuit is

Options

A. 5 ampere
B. 0.5 ampere
C. 0.7 ampere
D. 7 ampere

Answer

2

Explanation

Watt less component of
`A.C = I_(V) sin theta = (E_(v))/(Z) sin theta`
`=(200)/(sqrt(R^(2) + L^(2) omega^(2))) xx (L omega)/(sqrt(R^(2) + L^(2) omega^(2)))` `:. L 0.7 xx 2 pi xx 50`
`=(220 xx L omega)/((R^(2) + L^(2) omega^(2))) = 0.7 xx 2 xx (22)/(7) xx 50`
`= (220 xx (0.7 xx 2pi xx 50))/((220^(2) + 220^(2))) = 220 Omega`
`= (220 xx 220)/(220^(2) (2)) = (1)/(2) = 0.5`