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Question

Each capacitor shown in the figure has a capacitance of 10 μF. The emf of the battery is 1 V. How much charge will flow through AB , if the switch S is closed?



A
203 μC
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B
403 μC
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C
103 μC
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D
20 μC
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Solution

The correct option is B 403 μC
When S is open capacitors 1 and 2 are in parallel and they can be replaced by a capacitor whose capacitance is,

C=(10+10) μF=20 μF


Now C and C3 are connected in series,
Ceq =CC3C+C3= 10×2030 = 203μF
Therefore the charge supplied by the battery is,
Q=CeqV = 203 μC

When the switch S is closed, the capacitors 1 and 2 are in parallel but capacitor 3 becomes ineffective due to short circuit.
Ceq=(10+10) μF=20 μF
The charge on the equivalent capacitor will be,
Q=20 μC


Therefore the charge flown through AB will be,
ΔQ=QQ
ΔQ=(20203) μC = 403 μC

Hence, option (b) is the correct answer.

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