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what should \mathrm{b r} the rate of change of potential difference across capacitor to establish an instantaneous displacement current of 1 \mathrm{~A} in the space between the two parallel plates of a 2 \mu \mathrm{F} capacitor?
 

Option: 1

2 \times 10^5 \mathrm{~V} / \mathrm{s}

 


Option: 2

3 \times 10^5 \mathrm{~V} / \mathrm{s}
 


Option: 3

4 \times 10^5 \mathrm{~V} / \mathrm{s}
 


Option: 4

5 \times 10^5 \mathrm{~V} / \mathrm{s}


Answers (1)

best_answer

Here, \mathrm{I_D=1 \mathrm{Amp}, and \: C=2 \mu \mathrm{F}}

So \mathrm{\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{C}}=\frac{1}{2 \times 10^{-6}}=5 \times 10^5\: \mathrm{volt} / \mathrm{sec}.}

Thus, a displacement current of \mathrm{1 \mathrm{Amp}}. can be set up by changing the potential difference across the parallel plates of the capacitor at the rate of \mathrm{5 \times 10^5 \mathrm{volt} / \mathrm{sec}.}

Hence option 4 is correct.

Posted by

Ritika Harsh

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