When a coil is connected to AC supply of frequency 50 Hz , current of 4 A flows in it and it consumes 240 w power . If the potential difference across the coil is 100v . Then the inductance value of the coil is

Options

A. `L = (5 pi) H`
B. ` L = (pi)/(5) H`
C. `L = (1)/(5pi) H`
D. `L = (1)/(25 pi) H`

Answer

C

Explanation

`I_(2) R = 240 W, I = 4 A `
`:. R = (240)/(16) = 15 Omega`
Now , for the coil (L-R) circuit,
` V = IZ = I sqrt(X_(L)^(2) + R^(2))`
`rArr (V^(2))/(I_(2)) = X_(L)^(2) + R^(2)`
or `X_(L)^(2) = (100 xx 100)/(4 xx 4) - 225`
= 625 - 225 = 400
`:. X_(L) = 20 = Lomega`
`:. L = (20)/(2 pi f) = (20)/(2 pi xx 50)`
`:. = (1)/(5 pi) H` .