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Q.

In the circuit shown here C1 = 6μF, C2 = 3μF and battery B = 20V. The switch S1, is first closed. It is then opened and afterwards S2 is closed. What is the charge finally on C2?

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a

40 μC

b

80 μC

c

20 μC

d

120 μC

answer is C.

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Detailed Solution

Common potential, V=6×20+3×0(6+3)=1209 V

So, charge on 3μF capacitor (by closing S2)

                       Q2=3×10-6×1209=40μC

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