Question

A 14-µF capacitor and a capacitor of unknown capacitance are both charged to 3.25 kV. After charging, the two capacitors are disconnected from the voltage source. The capacitors are then connected to each other--positive plate to negative plate and negative plate to positive plate. The final voltage across the terminals of the 14-µF capacitor is 1.00 kV.

a)What is the capacitance of the second capacitor?

b)How much energy is dissipated when the connection was made?

Answer 1

Now from the above process we knew that the Value of other capacitor is 5 uF.

Voltage ttlid 3,25 kV 12 tt中 that othe- Catta An In anden to On 2 CI

(b) For energy dissipation. We just have to Subtract the final energy from initial energy. $\Delta E = \frac{1}{2}\left ( C_{1}-C_{2} \right )V^{2} - \frac{1}{2}\left ( C_{1}-C_{2} \right )V_{f}^{2}$ = 38.25 Joules