Answer:
Option D
Explanation:
Given, e =200 $\sqrt{2}$ $\sin $ (100 t) .......(i)
Capacitance of capacitor (C) = 1 $\mu$ F= 1 x 10-6 F
The standard equation of voltage of AC is given by
V= V0 $\sin \omega$t ..........(ii)
On comparing Eqs.(i) and (ii) , we get
$V_{0}= 200\sqrt{2}$
$\omega$ = 100
We know that,
$l_{rms}= \frac{V_{rms}}{X_{c}}$
But $V_{rms}= \frac{V_{0}}{\sqrt{2}}$ $l_{rms}=\frac{V_{0}}{\sqrt{2}X_{c}}$
$l_{rms}=\frac{V_{0}\omega C}{\sqrt{2}}$ $\left(X_{c}=\frac{1}{\omega C}\right)$
$l_{rms}=\frac{200\times\sqrt{2}\times100\times1 \times 10^{-6}}{\sqrt{2}}$
$l_{rms}=20\times 10^{-3}A=20 mA$
