Question

An inductor of inductance $L=5\text{H}$ is connected to an $\text{AC}$ source having voltage $E=10\sin (10t+\frac{\pi }{6})$. Find the maximum value of current in the circuit.

Answer 1

Given:

$L=5\text{H};\omega =10\text{rad/s}$

$E=10\sin (10t+\frac{\pi }{6})$

Comparing the above equation with, $E=E_0\sin (\omega t+\varphi )$, we have $E_0=10\text{V}$

The inductive reactance of the circuit is,

$X_L=\omega L=10\times 5=50\Omega$

The maximum value of current in the circuit is

$I_{max}=\frac{E_0}{X_L}$

$=\frac{10}{50}=0.2\text{A}$

Hence, option $\left(A\right)$ is correct.