Question

A coil of inductance 0.1H is connected to 50V, 100Hz generator and current is found to be 0.5A. The potential difference across resistance of the coil is

• A. 15V
• B. 20V
• C. 25V
• D. 39V

Answer 1

Answer: (D)

39V

Answer 2

$I=\frac{E}{Z};0.5=\frac{50}{Z}Z=100 \Omega$ $Z^2 =R^2+\omega^2L^2,then\,R=78 \Omega$ $Now\,V_R=\sqrt{V^2_{LR}-V^2_L}=23V$ $[V^2_R+V^2_{LR}]$