Two capacitors C1 and C2 are connected as shown in the diagram. The capacitance C2 is twice C1 when the key is opened the energy stored up in C1 is W. If the key is later closed and the system is allowed to attain electrical equilibrium, the total energy stored in the system will be
- \(\frac{1}{2}\)W
- \(\frac{2}{3}\)W
- W
- 2w
-
3W
The correct answer is: E
Explanation
E = \(\frac{1}{2} Cr^2 = \frac{1}{2}(C-1 + C_2)V^2 = (3C_1)V^2\)= 3(\(\frac{1}{2}C_1v^2\)) = 3W