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 th...

Answer Details

When the key is open, the two capacitors are not connected and the charge on each of them is zero. The energy stored in capacitor C1 is given by the formula: W = (1/2) * C1 * V^2, where V is the voltage across capacitor C1. When the key is closed, the two capacitors are connected in parallel, and a charge Q will flow from one plate of C1 to the other plate of C2 until the voltage across both capacitors is the same. Let the voltage across both capacitors be V. Then, the charge on C1 will be Q/2, and the charge on C2 will be Q. The total energy stored in the system can be calculated as follows: W(total) = (1/2) * C1 * V^2 + (1/2) * C2 * V^2 Substituting C2 = 2C1 and Q = CV, we get: W(total) = (1/2) * C1 * V^2 + (1/2) * 2C1 * V^2 = (1/2) * 3C1 * V^2 = (3/2) * W Therefore, the total energy stored in the system when the key is closed and the system is allowed to attain electrical equilibrium is (3/2) times the energy stored in C1 when the key is open. Hence, the correct option is 3W.