In an ac circuit , `L = (0.4)/(pi) H` and `R = 30 Omega`. If the circuit has an alternating emf of 220 V, 50 cps, the impedance and the current in the circuit will be :

Options

A. `40.4 Omega, 4.4 A`
B. `50 Omega, 4.4 A`
C. `3.07 Omega, 6.2 A`
D. `11.4 Omega, 17.5 Omega`

Answer

B

Explanation

Here, `X_(L) = omega L = 2 pi v L`
`= 2 pi xx 50 xx (0.4)/(pi) = 40 Omega`
`R = 30 Omega, V = 220 V`
Impedence of the circuit
`Z = sqrt(R^(2) + X_(L)^(2)) = sqrt((30)^(2) + (40)^(2)) = 50 Omega`
current, `I = (V)/(Z) = (220 V)/(50 Omega) = 4.4 A`