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  • In order to establish an instantaneous displacement current of in the space between the p

In order to establish an instantaneous displacement current of 1 \mathrm{~mA} in the space between the plates of \mathrm{2 \mu \mathrm{F}} parallel plate capacitor, the potential difference need to apply is:
 

Option: 1

100 \mathrm{Vs}^{-1}
 


Option: 2

200 \mathrm{Vs}^{-1}
 


Option: 3

300 \mathrm{Vs}^{-1}

 


Option: 4

500 \mathrm{Vs}^{-1}


Answers (1)

best_answer

\mathrm{ I_D=1 \mathrm{~mA}=10^{-3} \mathrm{~A}, \mathrm{C}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F} }

\mathrm{ \mathrm{I}_{\mathrm{D}}=\mathrm{I}_{\mathrm{C}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{CV})=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}}

Therefore, \mathrm{ \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{C}}=\frac{10^{-3}}{2 \times 10^{-6}}=500 \mathrm{Vs}^{-1}}

Therefore, applying a varying potential difference of \mathrm{500 \mathrm{~V} \mathrm{~s}^{-1}}would produce a displacement current of desired value.

Hence option 4 is correct.

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mansi

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