Question

AC voltage, $V\left(t\right)=20\sin \left(\omega t\right)$, of frequency $50\text{Hz}$, is applied to a parallel plate capacitor. The separation between the plates is $2\text{mm}$ and the area of the plates is $1{\text{m}}^2$. The amplitude of the oscillating displacement current, for the applied AC voltage is -

Take ${ϵ}_0=8.85\times {10}^{-12}\text{F/m}$

• A. $83.37\mu \text{A}$
• B. $55.58\mu \text{A}$
• C. $21.14\mu \text{A}$
• D. $27.79\mu \text{A}$

Answer 1

The correct option is D $27.79\mu \text{A}$


From the given information,

$C=\frac{{ϵ}_{0}A}{d}=\frac{{ϵ}_0}{2\times {10}^{-3}}\text{F}$

Further,

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi \times 50\times \frac{{ϵ}_0}{2\times {10}^{-3}}}$

$=\frac{2\times {10}^{-3}}{25\times 4\pi {ϵ}_0}=\frac{2\times {10}^{-3}}{25}\times 9\times {10}^9$

$\therefore X_C=72\times {10}^4\text{F}$

Now,

$i_0=\frac{V_0}{X_C}=\frac{20}{72\times {10}^4}\approx 2.779\times {10}^{-5}\text{A}=27.79\mu \text{A}$

Hence, option $\left(C\right)$ is the correct answer.