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Three capacitor each of capacitance 9 pF...

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

A

40v

B

120v

C

80v

D

60v

Text Solution

Verified by Experts

The correct Answer is:
40v
Here `C_(1)=C_(2)=C_(3)=9 pF`
Voltage `V_(2)=120V`
Total charge `q=Vxx C_(s)`
where `V=120V`
`q=(120xx3xx10^(-9))`
`=360pC`
`:.` Potential difference across `C_(1),V_(1)=(q)/(C_(1))`
`=(360xx10^(-12))/(9xx10^(-12))=40V`
`V_(2)=(q)/(C_(2))=(360)/(C_(2))=(360)/(9)=40V`
`V_(3)=(q)/(C_(3))=(360)/(9)=40V`
`:.` Potential difference across each capacitor is 40 V
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