Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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 :
PhysicsElectromagnetic WavesJEE MainJEE Main 2024 (05 Apr Shift 1)
Options:
  • A $10 \mathrm{~A}$
  • B $12 \mathrm{~A}$
  • C $8 \mathrm{~A}$
  • D $13 \mathrm{~A}$
Solution:
2823 Upvotes Verified Answer
The correct answer is: $12 \mathrm{~A}$
Displacement current is same as conduction current in capacitor.
$\begin{aligned}
& X_C=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}} \\
& =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega
\end{aligned}$
$I=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.