When a DC voltage of 200 V is applied to a coil of self inductance (2/`sqrt 3`/`pi`)H a current of 1A flows through it . But by replacing DC source with AC source of 200 V , the current in the coil is reduced to 0.5A . Then the frequency of AC supply is

Options

A. 100Hz
B. 75Hz
C. 60Hz
D. 50Hz

Answer

D

Explanation

Resistance of coil , `R = (200V)/(1A) = 200 Omega`
With ac source , `I = (200)/(sqrt(R^2 + X_L^2) ) " or " 0.5 = (200)/(sqrt(R^2 + X_L^2))`
` R^2 + (2pi v L)^2 = (400)^2`
`(2pi v xx (2sqrt3)/(pi))^2 = (400)^2 - (200)^2 = 200 xx 600`
` 4 sqrt3 v = 2sqrt3 xx 100 " or " v = 50 Hz `