Question

An alternating voltage $e = 200 \sqrt{2} \, \sin(100 \, t)$ volt is connected to $1\, \mu F$ capacitor through AC ammeter. The reading of ammeter is

• A. $5\, mA$
• B. $10\, mA$
• C. $15 \,mA$
• D. $20\, mA$

Answer 1

Answer: (D)

$20\, mA$

Answer 2

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