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In the circuit shown in the figure, initially, K1 is closed and K2

Question:

In the circuit shown in the figure, initially, K1 is closed and K2 is open. What are the charges on each capacitor. Then K1 was opened and K2 was closed. What will be the charge on each capacitor now?

Solution:

When K1 is closed and K2 is open, the capacitors C1 and C2 are connected in series with the battery

Therefore, the charge in capacitors C1 and C2 are

Q1 = Q2 = q = (C1C2/C1+C2)E = 18μC

When the capacitors C2 and C3 are placed in parallel,

C2V’ + C3V’ = Q2

V’ = Q2/C2 + C3 = 3V

Therefore,

Q2’ = 3CV’ = 9 μC

Q3’ = 3CV’ = μC

Q1’ = 18 μC

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