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Question
In the circuit shown in figure, initially K1 is closed and K2 is open. What are the charges on each capacitors.
Then K1 was opened and K2 was closed (order is important), What will be the charge on each capacitor now? [C = 1µF]
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Solution
Initially: `V oo 1/C` and `V_1 + V_2 = E`
⇒ `V_1 = 3V` and `V_2 = 6V`
∴ `Q_1 = C_1V_1 = 6C xx 3 = 18muC`
`Q_2 = 9muC` and `Q_3 = 0`
Later: `Q_2 = Q"'"_2 + Q_3`
With `C_2V + C_3V = Q2`
⇒ `V = Q_2/(C_2 + C_3) = (3/2)V`
`Q"'"_2 = (9/2)muC` and `Q"'"_3 = (9/2)muC`
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