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. 30Hz
B. 60Hz
C. 75 Hz
D. 50Hz

Answer

D

Explanation

Coil resistance ` R = (200)/(1) = 200 Omega `
For AC source
Current (I) = `(200)/( sqrt(R^(2) + X_(L)^(2))) rArr 0.5 = (200)/( sqrt(R^(2) + X_(L)^(2)))`
`rArr sqrt(R^(2) + X_(L)^(2)) = 400 `
`rArr (200) ^(2) + (2 pi f L)^(2) = (400)^(2) " " as R = 200 Omega `
`rArr ( 2 pi f xx ( 2 sqrt(3))/( pi))^(2) = 12 xx 10^(4)`
`rArr 4 f xx sqrt(3) = 346 rArr f = 50 Hz`