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Question

In the circuit shown in the figure, the steady state charge on 6μF capacitor is qμC. The value of q is:
333610_58a514499b92487fb19b67916db2b0ec.png

A
4μC
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B
5μC
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C
6μC
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D
7μC
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E
None of these
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Solution

The correct option is A 4μC
In steady state, no current flow in the capacitor branches.
The current flow in the top loop is It=103+2=2A
The current flow in the bottom loop is Ib=204+6=2A
Potential across 3Ω is V3=3It=6V
Potential across 4Ω is V4=3Ib=8V
Now we can redraw the circuit as shown in figure.
here, the capacitors are in series so Ceq=3×63+6=2μF
and Qeq=Ceq(V4V3)=2×(86)=4μC
As they are in series so charge on each capacitor is equal to Qeq
369527_333610_ans_a1347fdd07194db2b7bae6317c51da54.png

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