Three capacitors each of capacitance 9 pF are connected in series. What is the potential difference across each capacitor, if the combination is connected to a 120 V supply?

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40 V

60 V

80 V

50 V

answer is A.

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Detailed Solution

1Cs=1C1+1C2+1C3=19+19+191Cs=39⇒Cs=3pFCharge, q=CsV=3×120=360pCPotential difference acrossC1,V1=qC1=3609=40V=V2=V3

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