Question

A 220V, 50Hz 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 the inductor is:

• A. 102.2 V
• B. 186.4 V
• C. 214.24 V
• D. 302 V

Answer 1

Answer: (C)

214.24 V

Current = $\frac{V}{Z}$

$1=\frac{220}{Z}$

$\Rightarrow Z=220\Omega$

$Z=\sqrt{X_L^2+R^2}$

$(220{)}^2=X_L^2+(50{)}^2$

$X_L^2=45,900$

$X_L=214.24$

$V_L=i{X}_L$

$=\left(1\right)214.24$

$=214.24V$