Question

When an alternating voltage given by \(E = 100 \sin(10^2 t)\), where \(E\) is in volts and \(t\) is in seconds, is applied across a capacitor of capacitance \(20 \, \mu F\), the peak current flowing in the circuit is

• A. \(20 \, A\)
• B. \(2 \, A\)
• C. \(0.2 \, A\)
• D. \(0.02 \, A\)

Hint:

For a capacitor in an AC circuit, the peak current is given by: \[ I_0 = C \omega V_0 \] where \( \omega \) is the angular frequency. Ensure that capacitance is converted into Farads before substituting.

Answer 1

The Correct Option is C

Step 1: Using the Formula for Peak Current The peak current \(I_0\) in a capacitive circuit is given by: \[ I_0 = C \omega V_0 \] where: 
- \(C = 20 \, \mu F = 20 \times 10^{-6} \, F\) (capacitance), 
- \(\omega = 10^7 \, rad/s\) (angular frequency from the given equation), 
- \(V_0 = 100 \, V\) (peak voltage).

 Step 2: Substituting Values \[ I_0 = (20 \times 10^{-6}) \times (10^7) \times (100) \] \[ I_0 = 20 \times 10^{-6} \times 10^7 \times 10^2 \] \[ I_0 = 20 \times 10^{3} \] \[ I_0 = 0.2 \, A \] Thus, the correct answer is \( \mathbf{(3)} \ 0.2 \, A \).