Question

220 V, 50 Hz, AC source is connected to an inductance of 0.2 H and a resistance of 20 $\Omega$ in series. What is the current in the circuit?

• A. 3.33 A
• B. 33.3 A
• C. 5 A
• D. 10 A

Answer 1

The correct option is A 3.33 A

The frequency of AC source $=f=50Hz$

Thus the impedance of the inductor, $X_L=\omega L=\left(2\pi f\right)L=20\pi \Omega$

Resistance of the resistor $=20\Omega$

Thus, the net impedance of the circuit $=\sqrt{X_L^2+R^2}=65.94\Omega$

Thus, the current in the circuit $=\frac{V}{X}=\frac{220V}{65.94\Omega }=3.33A$