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  • A parallel plate capacitor of plate separation 2 mm  is connected in electric circuit having source voltage 400V. If the display of area is <img alt="60cm^{2}" src="https://entrancecorner

A parallel plate capacitor of plate separation 2 mm  is connected in electric circuit having source voltage 400V. If the display of area is 60cm^{2} then the value of displacement current for 10^{-6}sec will be

Option: 1

1.062 A


Option: 2

1.062\times 10^{-2}A


Option: 3

1.062\times 10^{-3}A


Option: 4

1.062\times 10^{-4}A


Answers (1)

best_answer

As learned

Maxwell's Displacement Current -

I_{D}=\varepsilon _{o}\frac{d\phi_{E}}{dt}

- wherein

I_{D} = Displacement Current

\varepsilon _{o} = Electrical permittivity for vacuum

\phi _{E} = Total electric flux

Let the plate separation=d

 I_{D}=\epsilon_o \frac{\Delta \phi }{\Delta t}=\epsilon _{o}\frac{\Delta(EA)}{t}=\frac{\epsilon _{o} \Delta V}{d} \times \frac{A}{\Delta t}

I_{D}=\frac{8.85\times 10^{-12}\times 400\times 60\times 10^{-4}}{2\times 10^{-3}\times 10^{-6}}=1.062\times 10^{-2}

 

Posted by

Divya Prakash Singh

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