A coil has a iductance of 0.7H and is joined in series with a resistance of `220 Omega`. When an alternating emf of 220V at 50 cps is applied to it, then the wattless component of the current in the circuit is `(take 0.7 = 2.2)`

Options

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

Answer

B

Explanation

Wattless component of ac
`-(I_v) sin phi =(E_v)/(Z) (X_L)/(Z) =(E_(v)X_(L))/(Z^2) =(220 xx omegaL)/(R^(2)+omega^(2)L^(2))`
As `omega L =0.7 xx 2 pi xx 50 =220 Omega`
Hence wattless component of ac
`(200 xx(220))/((220)^(2)_(220)^(2))=0.5 A`.