Question:
An alternating voltage \(\mathrm{V}=100 \sqrt{2} \sin (50 \mathrm{t})\) is connected to \(2 \mu \mathrm{~F}\) capacitor through a.c. ammeter. The reading of the ammeter is
An alternating voltage \(\mathrm{V}=100 \sqrt{2} \sin (50 \mathrm{t})\) is connected to \(2 \mu \mathrm{~F}\) capacitor through a.c. ammeter. The reading of the ammeter is
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The ammeter reads the rms current, found from the rms voltage and capacitive reactance.
Updated On: Nov 13, 2026
20 mA
2 mA
10 mA
0.2 mA
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
An a.c. ammeter shows the rms current \(I_{rms}=\dfrac{V_{rms}}{X_{C}}\).
Step 2: Read the data
\(V_{0}=100\sqrt{2}\ \text{V}\), so \(V_{rms}=100\ \text{V}\). \(\omega=50\ \text{rad/s}\).
Step 3: Reactance
\(X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{50\times2\times10^{-6}}=10^{4}\ \Omega\).
Step 4: Current
\(I_{rms}=\dfrac{100}{10^{4}}=10^{-2}\ \text{A}=10\ \text{mA}\).
Final Answer:
The ammeter reads 10 mA. \[ \boxed{\text{(C) }\text{10 mA}} \]
An a.c. ammeter shows the rms current \(I_{rms}=\dfrac{V_{rms}}{X_{C}}\).
Step 2: Read the data
\(V_{0}=100\sqrt{2}\ \text{V}\), so \(V_{rms}=100\ \text{V}\). \(\omega=50\ \text{rad/s}\).
Step 3: Reactance
\(X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{50\times2\times10^{-6}}=10^{4}\ \Omega\).
Step 4: Current
\(I_{rms}=\dfrac{100}{10^{4}}=10^{-2}\ \text{A}=10\ \text{mA}\).
Final Answer:
The ammeter reads 10 mA. \[ \boxed{\text{(C) }\text{10 mA}} \]
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