2024 Re-NEET Alternating CurrentAC circuit with a pure inductor Medium Question 7 of 60 in Alternating Current

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 $\dfrac{I_0}{\omega L}\sin\omega t$
C $\dfrac{I_0}{\omega L}\cos\omega t$
D $\omega L I_0\cos\omega t$

The answer is D.

Solution

In a pure inductor, voltage leads current by $\dfrac{\pi}{2}$. So $V_L=V_0\sin\left(\omega t+\dfrac{\pi}{2}\right)=V_0\cos\omega t$, where $V_0=I_0X_L=I_0\omega L$. Hence $V_L=\omega L I_0\cos\omega t$.

Concept AC circuit with a pure inductor

Alternating Current in past papers

60 questions from this chapter have appeared across 24 exam years.