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. 2.0A
B. 4.0A
C. 8.0A
D. `(20)/(sqrt13) A`

Answer

B

Explanation

`because X_L = 2pi v L`
`therefore (X_L)/(X._L) = (v)/(v.)`
`X._L = (v.)/(v) xx X_L = (100/50) xx 20 Omega = 40 Omega`
Impedance , ` Z = sqrt(R^2 + (X._L)^2)`
` = sqrt((30)^2 + (40)^2) = 50 Omega`
Current in the coil , `I = V/Z = (200V)/(50 Omega)= 4A`