Question

An inductor of inductance $L=5H$ is connected to an AC source having voltage $V=10\sin \left(10t\right)$. What is the value of peak current?

• A. $\frac{1}{2}A$
• B. $\frac{1}{3}A$
• C. $\frac{1}{4}A$
• D. $\frac{1}{5}A$

Answer 1

The correct option is D $\frac{1}{5}A$

$V=V_0\sin \left(\omega t\right)=10\sin \left(10t\right)$
For an inductor in an a.c circuit, voltage leads current.

Drawing the phasor diagram,



Current through the inductor $i=i_0\sin (\omega t-\frac{\pi }{2})$

where $i_0=\frac{V_0}{X_L}$
$X_L$ (inductive reactance) $=\omega L=10\times 5$
$=50\Omega$

$\therefore i_0=\frac{10}{50}=\frac{1}{5}A$