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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 :
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The correct answer is:
$12 \mathrm{~A}$
Displacement current is same as conduction current in capacitor.
$\begin{aligned}
& X_C=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}} \\
& =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega
\end{aligned}$
$I=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}$
$\begin{aligned}
& X_C=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}} \\
& =\frac{1}{2 \pi \times 4 \times 10^3 \times 12 \times 10^{-6}}=3.317 \Omega
\end{aligned}$
$I=\frac{V}{X_C}=\frac{40}{3.317}=12 \mathrm{~A}$
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