Question:

An alternating voltage \(V=200\sqrt{2}\sin (100t)\) volt is connected to a \(1\) \(\mu\)F capacitor through an a.c. ammeter. The reading of the ammeter will be

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An a.c. ammeter reads the r.m.s. current, which is the r.m.s. voltage divided by the capacitive reactance.
Updated On: Nov 12, 2026
  • \(10\) mA
  • \(20\) mA
  • \(40\) mA
  • \(80\) mA
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept
A capacitor offers a reactance \(X_C=\dfrac{1}{\omega C}\) to alternating current.
An a.c. ammeter reads the r.m.s. value of the current.

Step 2: Find the r.m.s. voltage
The peak voltage is \(V_0=200\sqrt{2}\) V, so \(V_{rms}=\dfrac{V_0}{\sqrt{2}}=200\) V.
The angular frequency is \(\omega=100\) rad/s.

Step 3: Calculation
\(X_C=\dfrac{1}{100\times 10^{-6}}=10^{4}\,\Omega\)
\(I_{rms}=\dfrac{V_{rms}}{X_C}=\dfrac{200}{10^{4}}=0.02\) A \(=20\) mA

Step 4: Compare the options
The value 20 mA matches the second option.

Final Answer:
The ammeter reads 20 mA. \[ \boxed{\text{(B) }20\text{ mA}} \]
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