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Question

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After the long time, the charge on the capacitor is q2 μC.

The magnitude of q1 (in μC) is

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Solution

At steady state

at steady state current passing through capacitor is 0 then current in circuit I=213
I=13 A
Potential at point A=VA
Potential at point B=VB
Applying KVL in loop 1, we have
VA1(13)1=VB
VAVB=1+13=43V
Charge on capacitor, q1=C(VAVB)
q1=X=1×43μC
q1=1.33 μC

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