Home
Class 12
PHYSICS
A capacitor of capacitance C is charged ...

A capacitor of capacitance C is charged to a potential difference `V_(0)`. The charged battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance `C_(x)`. The potential difference across the combination is V. The value of `C_(x)` should be

A

`(C(V_(0) - V))/(V)`

B

`(C(V - V_(0)))/(V)`

C

`(CV)/(V_(0))`

D

`(CV_(0))/(V)`

Text Solution

Verified by Experts

The correct Answer is:
A
The initial charge on first capacitor is `Q = CV_(0)`
After connection, voltages of both capacitors become same, i.e. V
`:. Q = q_(1) + q_(2) or CV_(0) = CV + C_(x)V`
`V = (CV_(0))/(C + C_(x)) :. C_(x) = (C(V_(0) - V))/(V)`
Promotional Banner

Topper's Solved these Questions

  • ELECTRIC CAPACITOR

    BITSAT GUIDE|Exercise BITSAT archives|6 Videos

  • CURRENT ELECTRICITY

    BITSAT GUIDE|Exercise Bitsat Archives|26 Videos

  • ELECTRIC CHARGE

    BITSAT GUIDE|Exercise All Questions|28 Videos

Similar Questions

Explore conceptually related problems

A capacitor C is charged to a potential difference V and battery is disconnected. Now if the capacitor plates are brough close slowely by some distance:

An ideal capacitor of capacitance 0.2muF is charged to a potential difference of 10V . The charging battery is than disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH . The current at a time when the potential difference across the capacitor is 5V is :

A capacitor of capacitance 10 mu F is charged to a potential 50 V with a battery. The battery is now disconnected and an additional charge 200 mu C is given to the positive plate of the capacitor. The potential difference across the capacitor will be.

A 100pF capacitor is charged to a potential difference of 80.0V and the charging battery is disconnected. The capacitor is then connected in parallel with a second capacitor. If the potential difference across the first capacitor drops to 35.0 V what is the capacitance of this second capacitor?

A capacitor of capacitance C is charged to a potential V . The flux of the electric field through a closed surface enclosing the capacitor is

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now.

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is