Solution:
When Switch is open Ceq= C/4 Charge given by the Battery is CE/4.
When Switch is open Ceq= C/3 Charge given by the Battery is CE/3.
Extra Charge flowing through the circuit it = \( \frac{CE}{3}-\frac{CE}{4}= \frac{CE}{12}\)
In the given figure, find the charge flowing through section AB when switch S is closed:
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