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 current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :
(1) zero, 60 μA
(2) 60 μA, 60 μA
(3) 60 μA, zero
(4) zero, zero
Answers
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
- Capacitance = 20×10^-6 F
- Rate of change of potential,
dV/dt = 3 V/s
3. q = C×V
dq/dt = C×dV/dt
4. Ic = dq/dt = C×dV/dt
= 20×10^-6×3
= 60×10^-6 A
5. We know that, Ic = Id
Id = 60×10^-6
Hence, Conduction current and displacement current both are equal to 60 microfarad
The displacement current through the plates of the capacitor, would be 60 μA, 60 μA respectively.
Option (2) is correct.
Explanation:
Now capacitance is 20 x 10^-6 f
Rate of change of potential is dv/dt = 3 v/s
q = cv
dq/dt = c dv/dt
dq/dt = 20 x 10^-6 x 3
dq/dt = 60 x 10^-6
I c = 60 μ A
As we know that
id = ic = 60 μ A
Thus the displacement current through the plates of the capacitor, would be 60 μA, 60 μA respectively.
Also learn more
A 0.3 H inductor 60 micro farad capacitor & a 50 ohm resistor are connected in series with a 120 V , 60 Hz supply . Calculate (1) Impedance of the circuit & current flowing in the circuit. ?
https://brainly.in/question/1134892