In the circuit shown below, the inductance L is connected to an ac source. The current flowing in the circuit is I = I₀ sin ω t. The voltage drop (VL) across L is (a) ω L I₀ sin ω t (b) (I₀)/(ω L) sin ω t (c) (I₀)/(ω L) cos ω t (d) ω L I₀ cos ω t

Answer: (d). VL = LdI/dt = Ld/dt(I₀ sin ω t) VL = ω L I₀ cos ω t (voltage leads current by 90°) So the answer is (d) ω L I₀ cos ω t.

In the circuit shown below, the inductance L is connected to an ac… | NEET 2024

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Pure R, L and C Circuits

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

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Answer: (d)

$V_L = L\frac{dI}{dt} = L\frac{d}{dt}(I_0 \sin \omega t)$
$V_L = \omega L I_0 \cos \omega t$ (voltage leads current by $90^\circ$ )
So the answer is (d) $\omega L I_0 \cos \omega t$ .

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Pure R, L and C Circuits question from NEET 2024

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