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

To find the peak current in a purely capacitive AC circuit, we use:

\[ I_0 = \omega C E_0 \]

Given voltage:

\[ E = 100 \sin(10^2 t) \]

So,

• \( E_0 = 100 \, \text{V} \)
• \( \omega = 10^2 \, \text{rad/s} \)
• \( C = 20 \,\mu\text{F} = 20 \times 10^{-6} \, \text{F} \)

Substitute values:

\[ I_0 = (10^2)(20 \times 10^{-6})(100) \]

\[ I_0 = 10^2 \times 100 \times 20 \times 10^{-6} \]

\[ I_0 = 10^4 \times 20 \times 10^{-6} \]

\[ I_0 = 20 \times 10^{-2} = 0.2 \, \text{A} \]

Final Answer: \( 0.2 \, \text{A} \)