Question

An AC voltage of $ 10 \sin \omega t $ volt is applied to a pure inductor of inductance $ 10 \, \text{H} $. The current through the inductor in ampere is:

• A. \( \frac{1}{\omega} \sin \left( \omega t - \frac{\pi}{2} \right) \)
• B. \( \omega \sin \left( \omega t - \frac{\pi}{2} \right) \)
• C. \( \frac{1}{\omega^2} \sin \left( \omega t - \frac{\pi}{2} \right) \)
• D. \( \omega^2 \sin \left( \omega t - \frac{\pi}{2} \right) \)

Hint:

In a pure inductor, the current lags the voltage by \( \frac{\pi}{2} \), and the amplitude of the current is determined by the inductive reactance \( \omega L \).

Answer 1

The Correct Option is A

We need to find the current through a pure inductor with an AC voltage applied.
Step 1: Write the voltage equation. \[ V = 10 \sin \omega t \] Step 2: Relate voltage to current. For an inductor: \[ V = L \frac{di}{dt} \implies 10 \sin \omega t = 10 \frac{di}{dt} \implies \frac{di}{dt} = \sin \omega t \] Step 3: Integrate to find the current. \[ i = \int \sin \omega t \, dt = -\frac{1}{\omega} \cos \omega t = \frac{1}{\omega} \sin \left( \omega t - \frac{\pi}{2} \right) \] Final Answer: \[ \boxed{\dfrac{1}{\omega} \sin \left( \omega t - \dfrac{\pi}{2} \right)} \]