Q136 / 200 · NEET 2023 (Manipur) · Physics

To produce an instantaneous displacement current of $2 \mathrm{~mA}$ in the space between the parallel plates of a capacitor of capacitance $4 \mu \mathrm{F}$, the rate of change of applied variable potential difference $\left(\frac{d V}{d t}\right)$ must be

  • A
    $200 \mathrm{~V} / \mathrm{s}$
  • B
    $400 \mathrm{~V} / \mathrm{s}$
  • C
    $800 \mathrm{~V} / \mathrm{s}$
  • Correct answer
    D
    $500 \mathrm{~V} / \mathrm{s}$

Solution

$\begin{aligned} & \text { Displacement current } i_d=c \frac{d V}{d t} \\ & \Rightarrow 2 \times 10^{-3}=4 \times 10^{-6} \times \frac{d V}{d t} \\ & \Rightarrow \frac{2 \times 10^{-3}}{4 \times 10^{-6}}=\frac{d V}{d t} \\ & \Rightarrow \frac{d V}{d t}=500 \mathrm{~V} / \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'D'.

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