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JEE Main 20241 FebruaryMorning Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2024 · 1 February · Morning Shift · Physics · Alternating Current

A parallel plate capacitor has a capacitance C = 200pF . It is...

A parallel plate capacitor has a capacitance C=200pFC = 200pF. It is connected to 230V230V ac supply with an angular frequency 300rads−1300rad{s}^{- 1}. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

Options

  1. A1.38μA1 .38 \mu A and 1.38μA1 .38 \mu A
  2. B14.3μA14 .3 \mu A and 143μA143\mu A
  3. C13.8μA13 .8 \mu A and 138μA138\mu A
  4. D13.8μA13 .8 \mu A and 13.8μA13 .8 \mu A

Correct answer

D. 13.8μA13 .8 \mu A and 13.8μA13 .8 \mu A

Step-by-step solution

By Maxwell's equation for a time varying voltage (AC voltage), both conduction and displacement currents are the same. Therefore,

Idisplacement=Iconduction=VrmsXC=Vrms1\omegaC=230×300×200×10−12=13.8μA{I}_{displacement} = {I}_{conduction} = \dfrac{{V}_{rms}}{{X}_{C}} = \dfrac{{V}_{rms}}{\dfrac{1}{\omegaC}} = 230 \times 300 \times 200 \times {10}^{- 12} = 13 .8 \mu A

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