Question

A capacitor circuit consists of two $6\mu F$ and $3\mu F$ capacitors is initially charged to $100V$ and $50V$ respectively now connected to a source of emf $200V$ through the switches $S_1,S_2,S_3$ as shown in figure below. Charge on capacitor $P$, in steady state, when $S_1$ is closed.

Answer 1

Since switches 2 and 3 are kept open, The bottom 2 capacitors are not taken into account when we do our calculations.

Thus, we can say that the right plate of the top 3uF and the left plate of the top 6 uF are part of an isolated system. If we assume that charge flows from left to right then we can say that charge on the left plate of the 3 uF capacitor us 150+q and on the right plate is -150-q. The charge is then 600+q on the left plate of the 6uF and -600-q on the right plate. Therefore, we can form the equation 2(150+Q)+(600+Q)=12.

This implies that Q=100 coulombs, which means that the charge on P is 600+Q=700 C.