Question

An alternating emf \( E = 110\sqrt{2} \sin 100t \, \text{volt} \) is applied to a capacitor of \( 2\mu\text{F} \), the rms value of current in the circuit is \(\dots\) \(\text{mA}\).

Answer 1

Correct Answer: 22

The provided alternating electromotive force (emf) is \( E = 110\sqrt{2} \sin 100t \, \text{volt} \). The peak voltage, \( E_0 \), is identified as \( 110\sqrt{2} \, \text{V} \). The capacitance, \( C \), is \( 2 \times 10^{-6} \, \text{F} \). The objective is to determine the root-mean-square (rms) value of the current. For a capacitive circuit, the rms current, \( I_{\text{rms}} \), is calculated using the formula \( I_{\text{rms}} = E_{\text{rms}} \times \omega \times C \). Here, \( E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \), and the angular frequency, \( \omega \), is \( 100 \, \text{rad/s} \).

The calculation for \( E_{\text{rms}} \) is as follows:

\( E_{\text{rms}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110 \, \text{V} \)

Applying the formula for \( I_{\text{rms}} \):

\( I_{\text{rms}} = 110 \times 100 \times 2 \times 10^{-6} = 22 \, \text{mA} \)

Consequently, the rms current \( I_{\text{rms}} \) is 22 mA, which is consistent with the specified range of 22 to 22 mA.