A capacitor of capacity `20microF` is charged upto `500V` and and disconnected from cell. Now this charged capacitor is connected to another capacitor of capacitance `C` If final common potential is `20V` then find the capacitance `C`.

A capacitor of capacity 2muF is charged to 100 V . What is the heat generated when this capacitor is connected in parallel to an another capacitor of same capacity?

A capacitor of capacitnace C is charged to a potential difference V and then disconnected from the battery. Now it is connected to an inductor of inductance L at t=0. Then

A capacitor of capacitance C is charged to a potential differece V_(0) . The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance C//2 . a. Compute the original charge of the system. b. Find the final potential differce across each capacitor. c. Find the final energy of the system. d. Calculate the decrease in energy when the capacitors are connected. e. Where did the ''lost'' energy go?

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 capacitor of capacitance C_0 is charged to potential V_0 . Now it is connected to another uncharged capacitor of capacitance C_0/2 . Calculate the heat loss in this process.

A parallel plate capacitor of capacitance C has been charged, so that the potential difference between its plates is V. Now, the plates of this capacitor are connected to another uncharged capacitor of capacitance 2C. Find the common potential acquired by the system and loss of energy.

A capacitor of capacitance 5muF is charged to potential 20V and isolated. Now, an uncharged capacitor is connected in parallel to it. If the charge distributes equally on these capacitors, find total energy stored in capacitors.

A capacitor of capacitance C carrying charge Q is connected to a source of emf E . Finally, the charge on capacitor would be

A capacitor of capacitance C charged by battery at V volt and then disconnected. At t = 0, it is connected to an uncharged capacitor of capacitance 2C through a resistance R. The charge on the second capacitor as a function of time is given by q=(alpha CV)/(3) (1-e^(-(3t)/(beta RC))) then fing the value of (alpha )/(beta) .

A capacitor of 20muF charged upto 500 V is connected in parallel with another capacitor of 10muF , which is charged upto 200 V. the common potential is