and two capacitors A and B with capacitances and , respectively. Suppose that the switch is closed and the switch is open, and sufficient time passes until the quantity of the electric charge on the capacitor A becomes In this state, we open the switch and close the switch 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.
The above diagram is a circuit that consists of a power supply of voltage
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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
So ,Q-1/Q-2=1/2