When two capacitors one of 3 μF and the other of 6 μF arc connected in series and the combination is charged to a potential difference of 120 volt, the potential difference across 3 μF capacitor is
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a
40 V
b
60 V
c
80 V
d
180 V
answer is C.
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Detailed Solution
V1+V2=120 volt and V1V2=C2C1=2 ∴ V1=2V2Now 2V3+V2=120 or V2=40 volt and V1=2×40=80 volt.