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A capacitor $50 \mu$ Fis connected to a.c. source $e=220 \sin (50 t)(e$ in volt ,$t$ in second). The value of peak current is
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Verified Answer
The correct answer is:
$\frac{0.55}{\sqrt{2}} A$
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
$\therefore$ 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
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