Question

In order to establish an instantaneous displacement current of $1\, mA$ in the space between the plates of $2\,\mu F$ parallel plate capacitor, the rate of change of potential difference is

• A. $100\, V\, s^{-1}$
• B. $200\, V\, s^{-1}$
• C. $300\, V\, s^{-1}$
• D. $500\, V\, s^{-1}$

Answer 1

Answer: (D)

$I_{D}= 1 \,mA = 10^{-3}\, A$; $C = 2\,\mu F = 2 \times 10^{-6}\,F$
$I_{D} = I_{C} = \frac{d}{dt}\left(CV\right) = C \frac{dV}{dt}$
Therefore, $\frac{dV}{dt} = \frac{I_{D}}{C} = \frac{10^{-3}}{2 \times 10^{-6}} = 500\,V\,s^{-1}$
Therefore, applying a varying potential difference of $500\,V\,s^{-1}$ would produce a displacement current of desired value.