Question

uploaded imageThe network shown is assembled with uncharged capacitors X, Y, and Z, with CX = 4 ?F, CY = 3 ?F, and CZ = 3 ?F, and open switches, S1 and S2. A potential difference Vab = +120 V is applied between points a and b. After the network is assembled, switch S1 is closed, but switch S2 is kept open. Switch S1 is opened, and then switch S2 is closed. What is the final potential difference across capacitor X?

Answer 1

When S1 is closed and S2 is open, the voltage across X is 120V, and 60V across Y and Z. In this case, the charge on X is 120V * 4*10^-6F = 4.8*10^-4 C, while the charge on Z is 60V * 3*10^-6F = 1.8*10^-4 C.

When S1 is open and S2 is closed, X and Z are connected in parallel. The equivalent capacitance is 4uF + 3uF = 7uF. And the total charge on the equivalent capacitor is 4.8*10^-4C + 1.8*10^-4C = 6.6*10^-4C. So the voltage across is 6.6*10^-4C/7*10^-6F = 94V.

Hope it's clear.