Question:

An alternating emf E=1102sin100tvolt E = 110\sqrt{2} \sin 100t \, \text{volt} is applied to a capacitor of 2μF 2\mu\text{F} , the rms value of current in the circuit is \dots mA\text{mA}.

Updated On: Mar 22, 2025
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Correct Answer: 22

Solution and Explanation

Given:
C=2μF,E=1102sin(100t).C = 2 \, \mu\text{F}, \quad E = 110\sqrt{2} \sin(100t).
The capacitive reactance is:
XC=1ωC,X_C = \frac{1}{\omega C},
where ω=100rad/s\omega = 100 \, \text{rad/s} and C=2×106FC = 2 \times 10^{-6} \, \text{F}.
Substitute:
XC=11002×106=12×104=5000Ω.X_C = \frac{1}{100 \cdot 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega.
The peak current is:
i0=E0XC,i_0 = \frac{E_0}{X_C},
where E0=1102VE_0 = 110\sqrt{2} \, \text{V}.
Substitute:
i0=11025000.i_0 = \frac{110\sqrt{2}}{5000}.
The RMS value of current is:irms=i02.i_{\text{rms}} = \frac{i_0}{\sqrt{2}}.
Substitute:irms=110250002=1105000.i_{\text{rms}} = \frac{110\sqrt{2}}{5000\sqrt{2}} = \frac{110}{5000}.
Simplify:irms=1105000A=22mA.i_{\text{rms}} = \frac{110}{5000} \, \text{A} = 22 \, \text{mA}.
Thus, the RMS value of current in the circuit is:irms=22mA.i_{\text{rms}} = 22 \, \text{mA}.

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