Question 36 of 190

$AC$ voltage $V(t)=20\sin \omega t$ of frequency $50Hz$ is applied to a parallel plate capacitor. The separation between the plates is $2mm$ and the area is $1{m}^{2}$. The amplitude of the oscillating displacement current for the applied $AC$ voltage is $\left[\text{Take}{\epsilon }_{0}=8.85\times {10}^{-12}F{m}^{-1}\right]$

A$21.14\text{μA}$B$83.37\text{μA}$C$27.79\text{μA}$D$55.58\text{μA}$

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Answer: (C) $27.79\text{μA}$

From the given information,
$C=\dfrac{{\in }_{0}A}{d}=\dfrac{{ϵ}_{0}\times 1}{2\times {10}^{-3}}F$
$∴{X}_{C}=\dfrac{1}{\omega C}=\dfrac{2\times {10}^{-3}}{2\times 50\pi \times {\in }_{0}}=\dfrac{2\times {10}^{-3}}{25\times 4\pi {\in }_{0}}\Omega$
$∴{X}_{C}=\dfrac{2\times {10}^{-3}}{25}\times 9\times {10}^{9}=\dfrac{18}{25}\times {10}^{6}\Omega$
$∴{i}_{0}=\dfrac{{V}_{0}}{{X}_{C}}=\dfrac{20\times 25}{18}\times {10}^{-6}A=27.47\text{μA}.$
The value of amplitude of displacement current will be same as value of amplitude of conventional current.
Hence option $3.$