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

The Correct Option is D

The voltage across an inductor, denoted as $V_L$, is defined by the equation:

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

If the current $I$ is represented by $I = I_0 \sin \omega t$:

The rate of change of current, $\frac{dI}{dt}$, is calculated as: $\frac{dI}{dt} = \frac{d}{dt}(I_0 \sin \omega t) = I_0 \omega \cos \omega t$

Consequently:

The inductor voltage becomes: $V_L = L(I_0 \omega \cos \omega t)$, which simplifies to $V_L = \omega L I_0 \cos \omega t$