Question

A 220V, 50 Hz a.c. generator is connected to an inductor and a $50\Omega$ resistance in series . The current in the circuit is 1.0A . The P.D. across inductor is

• A. 102.2V
• B. 186.4V
• C. 213.6V
• D. 302V

Answer 1

Answer: (C)

213.6V

Answer 2

$I=\frac{E}{Z},\therefore I=\frac{200}{Z},Z=220 \Omega$ $Z^2=R^2+X^2_L\,\therefore X_L=\sqrt{Z^2-R^2}$ $L=\frac{1}{\omega}\sqrt{Z^2-R^2}$ $\therefore L=\frac{1}{2\pi\,f}\sqrt{Z^2-R^2}=0.68H$ $\therefore V_L=\omega\,L I =2\pi \times 0.5 \times 0.68\times 1=213.6$