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A capacitor of capacitance \( 10 \mu \mathrm{F} \) is connected to an AC source and an AC Ammeter. If the
source voltage varies as \( V=50 \sqrt{2} \sin 100 t \), the reading of the ammeter is
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source voltage varies as \( V=50 \sqrt{2} \sin 100 t \), the reading of the ammeter is
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Verified Answer
The correct answer is:
\( 50 \mathrm{~mA} \)
Given, capacitance $=10 \mu F=10 \times 10^{-6} F ;$ source voltage $=50 \sqrt{2} \sin 100 t$ and $\omega=100$
We know,
$I_{\text {rms }}=\frac{V_{\text {rms }}}{X_{C}}$ and $V_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}}$
Now, $V_{\max }=50 \sqrt{2} ; X_{C}=\frac{1}{\omega C}$
Therefore,
$I_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}} \times \omega C=\frac{50 \sqrt{2}}{\sqrt{2}} \times 100 \times 10 \times 10^{-6}$
$I_{\text {rms }}=5 \times 10^{4} \times 10^{-6}=50 \times 10^{-3}$
Therefore, average value of ac current over a cycle is $50 \mathrm{~mA}$
We know,
$I_{\text {rms }}=\frac{V_{\text {rms }}}{X_{C}}$ and $V_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}}$
Now, $V_{\max }=50 \sqrt{2} ; X_{C}=\frac{1}{\omega C}$
Therefore,
$I_{\text {rms }}=\frac{V_{\max }}{\sqrt{2}} \times \omega C=\frac{50 \sqrt{2}}{\sqrt{2}} \times 100 \times 10 \times 10^{-6}$
$I_{\text {rms }}=5 \times 10^{4} \times 10^{-6}=50 \times 10^{-3}$
Therefore, average value of ac current over a cycle is $50 \mathrm{~mA}$
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