An alternating voltage of amplitude 40 V and frequency 4kHz is applied directly across the capacitor of 12μF. The maximum displacement current between the plates of the capacitor is nearly:
Select the correct option:
A
13 A
B
8 A
C
10 A
D
12 A
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Displacement current is same as conduction current in capacitor.
$$
\begin{aligned}
& X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C} \\
& =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega \\
& \mathrm{I}=\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{40}{3.317}=12 \mathrm{~A}
\end{aligned}
$$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
