A capacitor has capacitance `10muF` audit is charged to a potential `150V`. A second capacitor has a capacitance of `20 muF` and it is charged to a potential of `300V`. After charging, the two capacitors are connected in parallel with their same polarity plates together by using wires of negligible capacitance. Find how much energy is dissipated?

Find the energy stored in a capacitor of capacitance 100muF when it is charged to a potential difference of 20 V.

A capacitor of 20muF is charged to a potential of 10kV. Find th charge accumulated on each plate of the capacitor.

A capacitor of capacitance of 2muF is charged to a potential difference of 200V , after disconnecting from the battery, it is connected in parallel with another uncharged capacitor. The final common potential is 20V then the capacitance of second capacitor is :

Two capacitors of capacitance 10 muF and 20 muF are charged to potential 20 V and 10 V respectively. If they are connected to share charges, then loss of energy is

Two capacitors of capacitances 3 muF and 6 muF , are charged to potentials 2V and 5V respectively. These two charged capacitors are connected in series. Find the potential across each of the two capacitors now.

A capacitor of capacitance 5 muF is charged to a potential difference of 10 volts and another capacitor of capacitance 9 muF is charged to a potential difference of 8 volts. Now these two capacitors are connected in such a manner that the positive plate of one is connected to the positive plate of other. Calculate the common potential difference across the two capacitors.

A capacitor of capacitance 1/3muF is connected to a battery of 300volt and charged. Then the energy stored in capacitor is

A 2muF capacitor is charged to a potential = 10 V. Another 4muF capacitor is charged to a potential = 20V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?