A coil has resistance `30 ohm` and inductive reactance `20 ohm` at `50 Hz` frequency. If an ac source of 200 volts. `100 Hz`, is connected across the coil, the current in the coil will be

Options

A. `(20)/(sqrt(13))A`
B. 2.0A
C. 4.0A
D. 8.0A

Answer

C

Explanation

`X_(L) = 2 pi f L rArr 20 = 2 pi xx 50 L rArr L = (20)/( 2 pi xx 50)`
`X_(L) = 2 pi xx f. L = 2 pi xx 100 xx (20)/( 2 pi xx 50) = 40 Omega `
`:. I = (V)/(Z) = (200)/( sqrt(R^(2) + X_(L)^(2)))= (200)/( sqrt(30^(2) + (40)^(2))) = 4 A `