KCET · Physics · Alternating Current
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
- A \( 50 \mathrm{~mA} \)
- B \( 70.7 \mathrm{~mA} \)
- C \( 5.0 \mathrm{~mA} \)
- D \( 7.07 \mathrm{~mA} \)
Answer & Solution
Correct Answer
(A) \( 50 \mathrm{~mA} \)
Step-by-step Solution
Detailed explanation
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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