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Q.

Current through the battery in the circuit shown in the figure

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a

immediately after the switch S is closed is $ε / R$

b

immediately after the switch S is closed is $ε / 2 R$

c

after long time is $ε / 2 R$

d

after long time is $3 ε / 4 R$

answer is A, C.

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Detailed Solution

Immediately after closing the switch, the capacitor will behave like conducting wires, so all the resistances between B and G will be short-circuited and current will flow through ABFEG. Hence current $ε / R .$ After long time when the capacitors are fully charged, current will go through ABDG. Hence current $ε / 2 R .$

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