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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]

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Solution

The correct option is **C** 9 μC, 9 μC

Let 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

Let 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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