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

1. Fundamental relation for an inductor 

The voltage across an inductor $V_L$ is proportional to the rate of change of current:

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

Where $L$ = inductance of the inductor

2. Given current

For an AC circuit, let:

$$ I = I_0 \sin(\omega t) $$

Where $I_0$ = peak current, $\omega$ = angular frequency

3. Differentiate current w.r.t. time

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

$$ \frac{dI}{dt} = I_0 \omega \cos \omega t $$

Therefore, voltage across inductor:

$$ V_L = L \cdot I_0 \omega \cos \omega t $$ $$ V_L = \omega L I_0 \cos \omega t $$

Note: $V_L$ leads $I$ by $90^\circ$ since $\cos \omega t = \sin(\omega t + \pi/2)$