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Solve it, - Electrostatics - JEE Main-6

 Condenser A has a capacity of 15 \muF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 \muF with air between the plates. Both are charged separately by a battery of 100V. after charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is 

  • Option 1)

    400V

  • Option 2)

    800V

  • Option 3)

    1200V

  • Option 4)

    1600V

 
Answers (2)
102 Views
D Diya

Thank you, sir. 

As we have learned

Parallel Grouping -

C_{eq}=C_{1}+C_{2}+\cdots

- wherein

 

 Charge on capacitor A is given by  Q_1= 15*10^{-6}*100= 15*10^{-4}C
Charge on capacitor B is given by  Q_2= 1*10^{-6}*100= 10^{-4}C
Capacity of capacitor A after removing dielectric  = \frac{15*10^{-6}}{15}= 1 \mu F
Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq  = 1+1 = 2\mu F
So common potential  = \frac{(15*10^{-4}+1*10^{-4})}{2*10^{-6}}= 800V

 


Option 1)

400V

Option 2)

800V

Option 3)

1200V

Option 4)

1600V

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