A parallel plate capacitor of capacitance 20μ 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 :

Question

A parallel plate capacitor of capacitance 20μ 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 : (A) 60μ A, zero (B) zero, zero (C) zero, 60μ A (D) 60μ A 60μ A

Tardigrade Admin · Accepted Answer

Capacitance of capacitor C =20μ F = 20 ×10-6 F Rate of change of potential ((dV/dt)) =3v/s q=CV (dq/dt) =C (dV/dt) ic = 20×10-6 ×3 = 60×10-6 A = 60μ A As we know that id =ic = 60 μ A

Q. 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 :

$60 μ A$ , zero

zero, zero

zero, $60 μ A$

$60 μ A$ $60 μ A$

Capacitance of capacitor $C = 20 μ F$
$= 20 \times 1 0^{- 6} F$
Rate of change of potential $(\frac{d V}{d t}) = 3 v / s$
$q = C V$
$\frac{d q}{d t} = C \frac{d V}{d t}$
$i_{c} = 20 \times 1 0^{- 6} \times 3$
$= 60 \times 1 0^{- 6} A = 60 μ A$
As we know that $i_{d} = i_{c} = 60 μ A$

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

60μA, zero

zero, zero

zero, 60μA

60μA 60μA

Capacitance of capacitor C=20μF =20×10−6F Rate of change of potential (dtdV​)=3v/s q=CV dtdq​=CdtdV​ ic​=20×10−6×3 =60×10−6A=60μA As we know that id​=ic​=60μA

Q. 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 :

$60 μ A$ , zero

zero, $60 μ A$

$60 μ A$ $60 μ A$

Capacitance of capacitor $C = 20 μ F$
$= 20 \times 1 0^{- 6} F$
Rate of change of potential $(\frac{d V}{d t}) = 3 v / s$
$q = C V$
$\frac{d q}{d t} = C \frac{d V}{d t}$
$i_{c} = 20 \times 1 0^{- 6} \times 3$
$= 60 \times 1 0^{- 6} A = 60 μ A$
As we know that $i_{d} = i_{c} = 60 μ A$