An alternating voltage of amplitude 40 ~V and frequency 4 kHz is applied directly across the capacitor of 12 μ F . The maximum displacement current between the plates of the capacitor is nearly :

Correct answer: B. 12 ~A Displacement current is same as conduction current in capacitor. aligned & X_C= 1 ω C = 1 2 π fC & = 1 2 π × 4 × 10^3 × 12 × 10^ -6 =3.317 aligned I=(V)/(X_C)=(40)/(3.317)=12 ~A

JEE Main 20245 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2024 · 5 April · Morning Shift · Physics · Electromagnetic Waves

An alternating voltage of amplitude 40 ~V and frequency 4 kHz is...

An alternating voltage of amplitude

40 \mathrm{~V}

and frequency

4 \mathrm{kHz}

is applied directly across the capacitor of

12 \mu \mathrm{F}

. The maximum displacement current between the plates of the capacitor is nearly :

Options

A $10 \mathrm{~A}$
B $12 \mathrm{~A}$
C $8 \mathrm{~A}$
D $13 \mathrm{~A}$

Correct answer

B. $12 \mathrm{~A}$

Step-by-step solution

Displacement current is same as conduction current in capacitor.

\begin{aligned} & X_C=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}} \\ & =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega \end{aligned}

I=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}$

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