In the circuit shown in figure, each capacitor has a capacitance C. The emf of the cell is E and circuit already in steady state. If the switch S is closed.
A
some positive charge will flow out of the positive terminal of the cell
B
some positive charge will enter the positive terminal of the cell
C
the amount of charge flowing through the cell will be CE
D
the amount of charge flowing through the cell will be 
CE
CE
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(a, d)
equivalent capacitance before switch closed is C eq = 
,
Total charge flow through the cell is q = 
equivalent capacitance after switch S closed is C eq = 2C
Total charge flow through the cell is q = 2CE
Therefore some positive charge flow through the cell after closing the switch is = q f – q i = 2CE – 
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