Alternating Current

JEE Main level, single correct, medium

95 of 190

Question 95 of 190

The RMS value of conduction current in a parallel plate capacitor is 6.9μA6.9\mu A. The capacity of this capacitor, if it is connected to 230V230V AC supply with an angular frequency of 600rads−1600rad{s}^{-1}, will be

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Answer: (B) 50pF50pF

Irms=6.9μA;Vrms=230V;ω=600rads−1{I}_{rms}=6.9\mu A;{V}_{rms}=230V;\omega =600rad{s}^{-1}
Current in capacitor Irms=VrmsXC{I}_{rms}=\dfrac{{V}_{rms}}{{X}_{C}}, where, XC{X}_{C} is capacitive reactance given by XC=1ωC{X}_{C}=\dfrac{1}{\omega C}
Now, current, Irms=(Vrms)×(ωC){I}_{rms}=\left({V}_{rms}\right)\times \left(\omega C\right)
So, capacitance is C=IrmsVrmsω=6.9×10−6230×600=50pFC=\dfrac{{I}_{rms}}{{V}_{rms}\omega }=\dfrac{6.9\times {10}^{-6}}{230\times 600}=50pF

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