The network shown in the figure is assembled with uncharged capacitors X, Y, and Z, with
, and
And open switches, S₁ and S₂. A potential difference
V_ab= +120 V is applied between points a and b. After the network is assembled, switch S₁ is closed for a long time, but switch S₂ is kept open. Then switch S₁ is opened and switch S₂ is closed. What is the final voltage across capacitor X?

Question

The network shown in the figure is assembled with uncharged capacitors X, Y, and Z, with

, and

And open switches, S₁ and S₂. A potential difference
V_ab= +120 V is applied between points a and b. After the network is assembled, switch S₁ is closed for a long time, but switch S₂ is kept open. Then switch S₁ is opened and switch S₂ is closed. What is the final voltage across capacitor X?

Quizplus · Accepted Answer

Initially, when S₁ is closed and S₂ is open, the capacitors X and Y will be in series and their total capacitance will be CXY = CXC Y / (CXC + CY) = 4 μF. The voltage across CXY will be equal to the potential difference applied, i.e., VXY = 120 V. When S₁ is opened and S₂ is closed, capacitor Z will be charged to the potential difference VXY through the closed switch. The charge transferred from CXY to CZ will be Q = CXY VXY = 480 μC. After the switches change positions, the capacitors X and Z are in series and their total capacitance is CXZ = CXCZ / (CXC + CZ) = 3 μF. The total charge on CXZ will be Q = CZ(VXY - VXZ) = 80 μC. Solving the above equations, we get VXZ = 26 V. Therefore, the final voltage across capacitor X will be 120 V - 26 V = 94 V. Thus, the answer is choice A.

The Network Shown in the Figure Is Assembled with Uncharged