Capacitors Test...

The amount of work done in increasing the voltage across the plates of a capacitor from 5V to 10V is W. The work done in increasing it from 10V to 15V will be

Three capacitors each of capacitance 1µF are connected in parallel. To this combination, a fourth capacitor of capacitance 1µF is connected in series. The resultant capacitance of the system is

A capacitor 4µF charged to 50 V is connected to another capacitor of 2µF charged to 100 V with plates of like charges connected together. The total energy before and after connection in multiples of (10^-2 J) is

Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

A condenser of capacity C₁ is charged to a potential V₀. The electrostatic energy stored in it is U₀. It is connected to another uncharged condenser of capacity C₂ in parallel. The energy dissipated in the process is

Two condensers C₁ and C₂ in a circuit are joined as shown in figure. The potential of point A is V₁ and that of B is V₂ . The potential of point D will be

Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

Two capacitors of capacitance 2µF and 3µF are joined in series. Outer plate first capacitor is at 1000 volt and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will be

In the figure a potential of + 1200 V is given to point A and point B is earthed, what is the potential at the point P

Two identical capacitors are joined in parallel, charged to a potential V and then separated and then connected in series i.e . the positive plate of one is connected to negative of the other