Question

A capacitor $50$ $\mu F$ is connected to a power source $V=220\sin 50t$ ($V$ in $volt$, $t$ in $second$). The value of $rms$ current is (in $ampere$):

• A. $\frac{\sqrt{2}}{0.55}$
• B. $0.55$
• C. $\sqrt{0.55}$
• D. $\frac{\left(0.55\right)}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{\left(0.55\right)}{\sqrt{2}}$

Power source $V=220$ $sin50t$

$\therefore$ Peak voltage $V_o=220volt$

So, rms voltage $V_{rms}=\frac{V_o}{\sqrt{2}}=\frac{220}{\sqrt{2}}volt$

Also $\omega =50$

$\therefore$ capacitive reactance $X_C=\frac{1}{\omega C}=\frac{1}{50\times 50\times {10}^{-6}}=400\Omega$

Value of rms current $I_{rms}=\frac{V_{rms}}{X_C}=\frac{220}{\sqrt{2}\times 400}=\frac{0.55}{\sqrt{2}}ampere$