A coil of inductance 0.12 H and resistance 4Ω, is connected across a 240V, 50Hz supply. Calculate the current through it. [π = 3.142]

A. 6.3A
B. 33.3A
C. 37.2A
D. 40.0A

Correct Answer: Option A

A. 6.3A

Explanation

XL = 2πfL

= 2 x 3.142 x 50 x 0.12

= 37.68Ω

Z = √R²+X²L = √4²+37.68²

I = Vz=240/37.89

= 6.3A