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Question

# For the circuit shown in figure, initially key K1 is closed and key K2 is kept open for a long time. Then K1 was opened and K2 was closed, what will be the charge on capacitors C2 & C3 finally? [C=1μF]

A
18 μC, 0 μC
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B
0 μC, 18 μC
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C
9 μC, 9 μC
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D
4.5 μC, 4.5 μC
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Solution

## The correct option is C 9 μC, 9 μCLet Q1, Q2 and Q3 be the charge on the capacitors, when K1 is closed and K2 is open Initially: V∝1C and V1+V2=E ⇒V1=3 V and V2=6 V ∴Q1=C1V1=6×1×3=18 μC Q2=18 μC and Q3=0 When K1 is open and K2 is closed, let Q′2 and Q′3 be the final charge on C2 and C3 respectively. Q2=Q′2+Q′3 ⇒Q2=C2V+C3V⇒V=Q2C2+C3=(186×1) V=3 V Q′2=3×1×3=9 μC Q′3=3×1×3=9 μC

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