An inductor of inductance 2H and a resistance of `10 Omega` are connected in series to an ac source of 1109 V, 60 Hz. The current in the circuit will be

Options

A. 0.32A
B. 0.15A
C. 0.48A
D. 0.80A

Answer

B

Explanation

`X_L = omega L = 2pi v L = 2pi xx 60 xx 2 = 240 pi Omega , R = 10 Omega `
` Z = sqrt(R^2 + X_L^2) = sqrt(10^2 + (240pi)^2) = 240 pi `
`I_(rms) = (V_(rms))/(Z) = (110)/(240 pi) = 0.15 A`