The circuit shown in Figure is in steady state. Find the charges on the capacitors C1 and C2 respectively :
A
3μC,12μC
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B
4μC,9μC
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C
2μC,12μC
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D
3μC,0
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Solution
The correct option is A3μC,12μC In steady state, current will flow as shown in Figure, I=6/(10+10)=0.30A Potential difference across C1 is same as that across 10Ω on left side. So V1=10I=3V Charge on C1 is q1=C1V1=1×3=3μC We can see that potential difference across C2 is 6 V. So charge on C2 is q2=C2V2=2×6=12μC.