Question:medium

An alternating voltage \(E = 100\sqrt{2} \sin(50t)\) is connected to a \(2\mu\text{F}\) capacitor through an a.c. ammeter. The ammeter reading will be

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Always convert peak voltage to RMS before using AC formulas.
Updated On: May 14, 2026
  • \(10\text{ mA}\)
  • \(5\text{ mA}\)
  • \(20\text{ mA}\)
  • \(30\text{ mA}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
An AC ammeter measures the root mean square (rms) value of the current flowing through the circuit.
In a purely capacitive circuit, the current depends on the rms voltage and the capacitive reactance.
Step 2: Key Formula or Approach:
Voltage equation: \(V = V_0 \sin(\omega t)\). Here \(V_0 = 100\sqrt{2}\text{ V}\), \(\omega = 50\text{ rad/s}\).
RMS voltage: \(V_{\text{rms}} = V_0 / \sqrt{2}\).
Capacitive reactance: \(X_C = \frac{1}{\omega C}\).
Ohm's Law for AC: \(I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \cdot \omega C\).
Step 3: Detailed Explanation:
Calculate RMS voltage: \[ V_{\text{rms}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\text{ V} \] Calculate current \(I_{\text{rms}}\): \[ I_{\text{rms}} = 100 \times 50 \times (2 \times 10^{-6}) \] \[ I_{\text{rms}} = 100 \times 100 \times 10^{-6} \] \[ I_{\text{rms}} = 10^4 \times 10^{-6} = 10^{-2}\text{ A} \] Convert Amperes to milliamperes: \[ I_{\text{rms}} = 0.01 \text{ A} = 10\text{ mA} \] Step 4: Final Answer:
The ammeter reading will be \(10\text{ mA}\).
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