In figure, the charges on C1,C2 and C3 are Q1,Q2,Q3, respectively. Then
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a
Q2=8 μC
b
Q3=12 μC
c
Q1=20 μC
d
Q2=12 μC
answer is A.
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Detailed Solution
Effective capacitance is C=4×59=209μFCharge on C1 is Q1=CV=209×9=20 μCcharge on C2 is Q2(C2C2+C3)Q1=23×20=8 μCcharge on C3 is Q3=(C3C2+C3)Q1=33×20=12 μCSince, Q2C2=Q3C3and Q1=Q2+Q3=Q2+C3C2Q2=Q2(C2+C3C2)Therefore,Q2=(C2C2+C3)Q1Q3=(C3C2+C3)Q1Thus, options (1), (2), and (3) are correct