A capacitor $50 \mu$ Fis connected to a.c. source $e=220 \sin (50 t)(e$ in volt ,$t$ in second). The value…
A capacitor $50 \mu$ Fis connected to a.c. source $e=220 \sin (50 t)(e$ in volt ,$t$ in second). The value of rms current is
$\frac{0.55}{\sqrt{2}} A$
$\sqrt{2}$
$\frac{\sqrt{2}}{0.55} A$
$0.55 \mathrm{~A}$
Solution
Power source $V=220 \sin 50 t$
$\therefore$ Peak voltage $V_0=220$ volt
So, rms voltage $\quad V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{220}{\sqrt{2}}$ 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 rmscurrent $I_{r m s}=\frac{V_{r m s}}{X_C}=\frac{220}{\sqrt{2} \times 400}=\frac{0.55}{\sqrt{2}}$ ampere