A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changing at the rate of 3 V/s . The conduction - Physics MCQ NEET 2019 2026 | ExamTest.live

mediumMCQNEET 20192026Physics

1 attempts0% success rate4 marks (−1)

A parallel plate capacitor of capacitance 20 μ F is being charged by a voltage source whose potential is changing at the rate of 3 V/s . The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:

Solution & Step-by-step Explanation

Displacement current i_{d} is given by: i_{d} = C \frac{d V}{d t} = 20 \times 1 0^{- 6} \times 3 = 60 \times 1 0^{- 6} A = 60 μ A According to Ampere-Maxwell law, the conduction current (i_{c}) in the wires is equal to the displacement current (i_{d}) in the gap between the plates during the charging process. i_{c} = i_{d} = 60 μ A

Try it yourself before checking the explanation above.

A parallel plate capacitor of capacitance 20 μ F is being charged by a voltage source whose potential is changing at the rate of 3 V/s . The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:

Zero, 60 μ A

60 μ A, 60 μ A

60 μ A, zero

Zero, zero

Practice with timed mock tests and track your performance across Physics.