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Alternating Current question

2024 · Q160
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Alternating Current question

2024 · Q160

NEETPhysicsAlternating CurrentMCQ+4 / −1

In the circuit shown below, the inductance LLL is connected to an ac source. The current flowing in the circuit is I=I0sin⁡ωtI=I_0 \sin \omega tI=I0​sinωt. The voltage drop (VL)\left(V_L\right)(VL​) across LLL is

NEET 2024 (Re-Examination) Physics - Alternating Current Question 2 English

  1. A
    ωLI0sin⁡ωt\omega L I_0 \sin \omega tωLI0​sinωt
  2. B
    I0ωLsin⁡ωt\frac{I_0}{\omega L} \sin \omega tωLI0​​sinωt
  3. C
    I0ωLcos⁡ωt\frac{I_0}{\omega L} \cos \omega tωLI0​​cosωt
  4. D
    ωLI0cos⁡ωt\omega L I_0 \cos \omega tωLI0​cosωt
View written solutionFree

Correct answer: D

$V_L$ leads current $I$ by $\frac{\pi}{2}$

$$\begin{aligned} & \therefore V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right) \\ & V_0=I_0 X_L \\ & \Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t) \end{aligned}$$

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