Home
Class 12
PHYSICS
Two identical capacitors each of capacit...

Two identical capacitors each of capacitance `5 muF` are charged to potentials 2 kV and 1 kV respectively.The -ve ends are connected together. When the +ve ends are also connected together, the loss of energy of the system is

A

160 J

B

0 J

C

5 J

D

`1.25` J

Text Solution

Verified by Experts

The correct Answer is:
D
Loss of energy ` = (1)/(2) (C_(1)C_(2))/((C_(1) + C_(2)))(V_(1)-V_(2))^(2)`
` = (1/2)(5 xx 5 xx 10^(-12))/((5 + 5)xx 10^(-6))(2000-1000)^(2)`
` = (5xx5xx1)/(2xx10) = (5)/(4) = 1.25 `J.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KARNATAKA CET 2006

    KCET PREVIOUS YEAR PAPERS|Exercise PHYSICS|60 Videos

  • KARNATAKA CET 2008

    KCET PREVIOUS YEAR PAPERS|Exercise PHYSICS|60 Videos

Similar Questions

Explore conceptually related problems

Two capacitors 3muF and 5muF are charged to 18V and 25V respectively. These are then connected in series. Calculate the net voltage and net charge. Find the loss of energy after connecting them in series.

Two capacitors 1muF and 3muF are charged individually to voltages of 100V and 200V respectively. These are then connected in a parallel combination. Find the loss in the energy stored in the capacitors.

Knowledge Check

  • Two identical capacitors , have the same capacitance C . One of them is charged to potential V_(1) and the other to V_(2) .The negative needs of the capacitors are connected together . When the positive ends are also connected , the decrease in energy of the combined system is :

    A
    `1/4 C (V_(1)^(2)-V_(2)^(2))`
    B
    `1/4 C (V_(1)^(2)+V_(2)^(2))`
    C
    `1/4 C (V_(1)-V_(2))^(2)`
    D
    `1/4 C (V_(1)+V_(2))^(2)`

  • Two capacitors of capacitances 3 muF and 6muF are charged to a potential of 12V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

    A
    4V
    B
    6V
    C
    zero
    D
    3V

  • Two capacitors C_(1)and C_(2) are charged to 120 V and 200 V respectively . It is found that by connecting them together the potential on each one can be made zero . Then :

    A
    `3C_(1)=5C_(2)`
    B
    `3C_(1)+5C_(2)=0`
    C
    `9C_(1)=4C_(2)`
    D
    `5C_(1)=3C_(2)`

    • Similar Questions

    Explore conceptually related problems

    A capacitor of capacitance 5 muF is charged to potential of 500 V. Then it is disconnected from the battery and connected to uncharged capacitor of capacitance 3 muF . Calculate the common potential, charge on each capacitor and the loss of energy.

    A capacitor of capacity 2 muF is charged to a potential difference of 12V. It is then connected across an inductor of 0.6 mH. Current in the circuit at a time when P.D. across the capacitor be 6V is :

    A capacitor of capacitance C_(1) = 4 muC is charged to V_(1) = 80 V and another capacitor of capacitance C_(2) = 6 muC is charged to V_(2) = 30 V . When they are connected together, the energy lost by the 4 mu F capacitor is :

    Two parallel plate capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V by a battery. The battery is then disconnected and the space between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes :

    A system of 2 capacitors of capacitance 2muF and 4muF is connected in series across potential difference of 6 V. The electric charge and energy stored in the system are

    Home
    Profile