In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in the…
In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in the circuit is $I=I_0 \sin \omega t$. The voltage drop $\left(V_L\right)$ across $L$ is
$\omega L l_0 \sin \omega t$
$\frac{I_0}{\omega L} \sin \omega t$
$\frac{I_0}{\omega L} \cos \omega t$
$\omega L I_0 \cos \omega t$
Solution
$V_L$ leads current $I$ by $\frac{\pi}{2}$
$V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right)$
$V_0=I_0 X_L$
$\Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t)$
