Verified Solution • 2025–26 Session moderate

QNEET-2024-Ph-142: NEET 2024 Physics - Class 12 NEET-2024-Ph-142 Solution: In the circuit shown below the inductance L is connected to an…

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

\omega L I_0\sin\omega t

\omega L I_0\cos\omega t

\frac{I_0}{\omega L}\sin\omega t

\frac{I_0}{\omega L}\cos\omega t

Multiple Choice Options

A $\omega L I_0 \sin \omega t$ _ L I 0 _ sin _ t
B $\frac{I_0}{\omega L} \sin \omega t$ _ L I 0 _ _ sin _ t
C $\frac{I_0}{\omega L} \cos \omega t$ _ L I 0 _ _ cos _ t
D $\omega L I_0 \cos \omega t$ _ L I 0 _ cos _ t

Step-by-Step Verified Solution

V L leads current I by � 2 _ V L = V 0 sin _ ( _ t + � 2 ) ( _ I = I 0 sin _ _ t ) V 0 = I 0 X L _ V L = I 0 X L cos _ ( _ t ) = I 0 _ L cos _ ( _ t )

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