Electric circuit

Electricity and Magnetism Level 2

The above diagram is a circuit that consists of a power supply of voltage $V$ and two capacitors A and B with capacitances $C$ and $2C$ , respectively. Suppose that the switch $S_1$ is closed and the switch $S_2$ is open, and sufficient time passes until the quantity of the electric charge on the capacitor A becomes $Q.$ In this state, we open the switch $S_1$ and close the switch $S_2$ and let sufficient time pass. Then what will be the ratio of the quantities of the electric charges on the capacitors A and B , i.e. $Q_A:Q_B?$

2:1 1:1 1:4 1:2

2 solutions

Arghyanil Dey
Apr 16, 2014

At first the charge in the capacitor A is CV,Let CV=Q

On closing 2nd switch where the first switch is open and give the circuit proper time to rearrange the charge .

Applying Kirsoff's voltage law we get

                   Q-q/C=q/2C[let, on closing 2nd switch q amount of charge come to second 
                                                    capacitor from 1st one]
                   q=2Q/3

So ,Q-1/Q-2=1/2

Mudit Jha
Aug 3, 2014

When S1 is opened and S2 is closed, the two capacitors come to the same potential. Since C = Q/V, the charge on them distributes in the ratio of capacitance

Electric circuit

2 solutions

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