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 \(\dots\) \(\text{mA}\).
Correct Answer: 22
Approach Solution - 1
The given alternating emf is \( E = 110\sqrt{2} \sin 100t \, \text{volt} \). First, we identify the peak voltage: \( E_0 = 110\sqrt{2} \). The capacitor value is \( C = 2 \times 10^{-6} \, \text{F} \). We need to find the rms value of the current. The formula for the rms current \( I_{\text{rms}} \) in a capacitive circuit is \( I_{\text{rms}} = E_{\text{rms}} \times \omega \times C \), where \( E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \) and \( \omega \, (\text{angular frequency}) = 100 \, \text{rad/s} \).
Calculate \( E_{\text{rms}} \) as follows:
\( E_{\text{rms}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110 \, \text{V} \)
Using the formula for \( I_{\text{rms}} \), we have:
\( I_{\text{rms}} = 110 \times 100 \times 2 \times 10^{-6} = 22 \, \text{mA} \)
Thus, the rms current \( I_{\text{rms}} \) is 22 mA, which correctly falls within the specified range of 22 to 22 mA.
Approach Solution -2
Given:
\[C = 2 \, \mu\text{F}, \quad E = 110\sqrt{2} \sin(100t).\]
The capacitive reactance is:
\[X_C = \frac{1}{\omega C},\]
where $\omega = 100 \, \text{rad/s}$ and $C = 2 \times 10^{-6} \, \text{F}$.
Substitute:
\[X_C = \frac{1}{100 \cdot 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega.\]
The peak current is:
\[i_0 = \frac{E_0}{X_C},\]
where $E_0 = 110\sqrt{2} \, \text{V}$.
Substitute:
\[i_0 = \frac{110\sqrt{2}}{5000}.\]
The RMS value of current is:\[i_{\text{rms}} = \frac{i_0}{\sqrt{2}}.\]
Substitute:\[i_{\text{rms}} = \frac{110\sqrt{2}}{5000\sqrt{2}} = \frac{110}{5000}.\]
Simplify:\[i_{\text{rms}} = \frac{110}{5000} \, \text{A} = 22 \, \text{mA}.\]
Thus, the RMS value of current in the circuit is:\[i_{\text{rms}} = 22 \, \text{mA}.\]
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