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

B

Explanation

`L = 0.7 H, R = 220 Omega, V_(rms)=220 "volt"`
`X_(L) = 2pifL = 2 xx (22)/(7) xx 50 xx 0.7 = 220 Omega`
`I_(rms) = (V_(rms))/(sqrt(X_(L)^(2) + R^(2))) = (220)/(sqrt(2) xx 220) = (1)/(sqrt(2))`
hence `I_(0) = 1 A`
wattless current `= I_(0)sinphi`
`I_(0)sqrt(I-cos^(2)phi)`
`= I_(0)sqrt(I-(R^(2))/(Z^(2))) = (I_(0))/(sqrt(2)) = 0.7 A`