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.

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Answer 1

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

Answer 2

Capacitance of capacitor

C = 20 μ F

= 20 \times 1 0^{- 6} F

Rate of change of potential is,

(\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