Question:
An alternating voltage \( E = 200\sqrt{2} \sin(100t) \, \mathrm{V} \) is connected to a \( 1\,\mu\mathrm{F} \) capacitor through an AC ammeter. The reading of ammeter is:
An alternating voltage \( E = 200\sqrt{2} \sin(100t) \, \mathrm{V} \) is connected to a \( 1\,\mu\mathrm{F} \) capacitor through an AC ammeter. The reading of ammeter is:
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The current through a capacitor in an AC circuit is given by \( I = C \frac{dV}{dt} \).
Updated On: Apr 22, 2026
- 10 mA
- 20 mA
- 40 mA
- 80 mA
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The Correct Option is B
Solution and Explanation
Step 1: Understand the relationship between current, voltage, and capacitance in an AC circuit.
The current \( I \) in a capacitor connected to an AC source is given by: \[ I = C \frac{dV}{dt} \] where \( C = 1 \, \mu F \) is the capacitance and \( V = 200 \sqrt{2} \sin(100t) \) is the voltage.
Step 2: Differentiate the voltage.
The derivative of the voltage is: \[ \frac{dV}{dt} = 200 \sqrt{2} \times 100 \cos(100t) \] Thus, the current is: \[ I = 1 \times 10^{-6} \times 200 \sqrt{2} \times 100 \cos(100t) \] \[ I = 20 \times 10^{-3} \cos(100t) = 20 \, \text{mA} \]
Step 3: Conclusion.
The ammeter reads 20 mA, which is option (2).
The current \( I \) in a capacitor connected to an AC source is given by: \[ I = C \frac{dV}{dt} \] where \( C = 1 \, \mu F \) is the capacitance and \( V = 200 \sqrt{2} \sin(100t) \) is the voltage.
Step 2: Differentiate the voltage.
The derivative of the voltage is: \[ \frac{dV}{dt} = 200 \sqrt{2} \times 100 \cos(100t) \] Thus, the current is: \[ I = 1 \times 10^{-6} \times 200 \sqrt{2} \times 100 \cos(100t) \] \[ I = 20 \times 10^{-3} \cos(100t) = 20 \, \text{mA} \]
Step 3: Conclusion.
The ammeter reads 20 mA, which is option (2).
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