Question

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 ($V_L$) across $L$ is:

• A. $\omega L I_0 \sin \omega t$
• B. $\frac{I_0}{\omega L} \sin \omega t$
• C. $\frac{I_0}{\omega L} \cos \omega t$
• D. $\omega L I_0 \cos \omega t$

Answer 1

Answer: (D)

$\omega L I_0 \cos \omega t$

Answer 2

The voltage across an inductor ($V_L$) is given by:

$V_L = L \frac{dI}{dt}$

Given $I = I_0 \sin \omega t$:

$\frac{dI}{dt} = \frac{d}{dt}(I_0 \sin \omega t) = I_0 \omega \cos \omega t$

Therefore:

$V_L = L(I_0 \omega \cos \omega t) = \omega L I_0 \cos \omega t$