A parallel plate capacitor of capacitance 20 \mu F 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 :

Capacitor question

2019 · Q127

Capacitor question

2019 · Q127

View written solutionFree

Correct answer: C

As, q = CV

Differentiating above equation w.r.t. t, we get

${{dq} \over {dt}} = C{{dV} \over {dt}}$

$\Rightarrow$ i_c = $C{{dV} \over {dt}}$ [As i = ${{dq} \over {dt}}$ ]

= 20 $\times$ 10^-6 $\times$ 3

= 60 $\times$ 10^-6 A = 60 $\mu$ A

Also, conduction current in wires is equal to displacement current between the plates of capacitor.

$\therefore$ i_d = i_c = 60 $\mu$ A

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Capacitor question

2019 · Q127

Capacitor question

2019 · Q127

View written solutionFree

Correct answer: C

As, q = CV

Differentiating above equation w.r.t. t, we get

${{dq} \over {dt}} = C{{dV} \over {dt}}$

$\Rightarrow$ i_c = $C{{dV} \over {dt}}$ [As i = ${{dq} \over {dt}}$ ]

= 20 $\times$ 10^-6 $\times$ 3

= 60 $\times$ 10^-6 A = 60 $\mu$ A

Also, conduction current in wires is equal to displacement current between the plates of capacitor.

$\therefore$ i_d = i_c = 60 $\mu$ A