Electromagnetic Waves

JEE Main level, single correct, medium

33 of 165

Question 33 of 165

An alternating voltage of amplitude 40 V40 \mathrm{~V} and frequency 4kHz4 \mathrm{kHz} is applied directly across the capacitor of 12μF12 \mu \mathrm{F}. The maximum displacement current between the plates of the capacitor is nearly :

Show answer and solution

Answer: (B) 12 A12 \mathrm{~A}

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=VXC=403.317=12 AI=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Waves and Maxwell's equations), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.