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# Medical

A parallel palte capacitor of capacitance 20 \mu F is being charged by a voltage source whose potential is changing at the rate of 3 V/s . The conduction  current through the connecting wires , and the dispalcement  current through the plates of the capacitor , would be  resectively : 

 

  • Option 1)

    zero, 60 \mu A

  • Option 2)

    60 \mu A , 60 \mu A

  • Option 3)

    60 \mu A , zero

  • Option 4)

    zero , zero 

Answers (1)
P Plabita

I_ d = \varepsilon _0 \frac{d \theta _\varepsilon }{dt} = \frac{\varepsilon _ 0 d ( E \cdot A )}{dt } \: \: \:......( \because E\cdot d = V \\\\ C = \frac{\varepsilon _0 A }{d }) \\\\ = \frac{\varepsilon _ 0 d ( VA )}{ddt } \\\\ I_D = C \frac{dv }{dt} = 20 \times 20 \times 10 ^ {-6 } \times 3 \\\\ I_D = 60 \mu A \\\\

For \: \: charged \: \: capacitor \: \: I_D = I_C \\\\ I_C = 60 \mu A


Option 1)

zero, 60 \mu A

Option 2)

60 \mu A , 60 \mu A

Option 3)

60 \mu A , zero

Option 4)

zero , zero 

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