In the circuit shown in the figure, the steady state charge on 6μF capacitor is qμC. The value of q is:
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 A4μC 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(V4−V3)=2×(8−6)=4μC As they are in series so charge on each capacitor is equal to Qeq