MCQ
Instantaneous displacement current 1A in the space between the parallel plates of $1\mu\text{F}$ capacitor can be established by changing the potential difference at the rate of:
  • A
    $\frac{0.1\text{V}}{\text{s}}$
  • B
    $\frac{1\text{V}}{\text{s}}$
  • C
    $\frac{10^6\text{V}}{\text{s}}$
  • D
    $\frac{10^{-6}\text{V}}{\text{s}}$

Answer

  1. $\frac{10^6\text{V}}{\text{s}}$

Explanation:

In a capacitor of capacitance C,

$\text{V}=\frac{\text{q}}{\text{C}}$

$\Rightarrow\frac{\text{dV}}{\text{dt}}=\frac{\text{i}}{\text{C}}=\frac{1\text{A}}{1\mu\text{F}}=\frac{10^6\text{V}}{\text{s}}$

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