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Question

Two capacitors A and B of capacitance 6μF and 10μF respectively are connected in parallel and this combination is connected in series with a third capacitors C of 4μF. A potential difference of 100 volt is applied across the entire combination. Find the charge and potential difference across 6 μF capacitor.

A
120μC;20V.
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B
200μC;20V.
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C
320μC;80V.
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D
320μC;60V.
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Solution

The correct option is A 120μC;20V.
Since A and B are connected and Parallel,
Hence,
Equivalence capacitance of capacitorA and B=6μF+10μF=16μF

Now 16μF and capacitor C of 4μF are connected in series
Hence,
the equivalence capacitance (EC) will be given by
1EC=116+14
we get,
Equivalence capacitance (EC)=165μF
Charge=EC×P.D.
Charge=165×100
Charge(Q)=320 μF

Now,
QA=320×616μF
QA=120 μF
QB=320×1016μF
QB=200 μF

Now proceeding for P.D. across each capacitor,
VA=QACA

VA=1206=20V ,

VB=QBCB

VB=20010=20V ,

VC=QCCC

VC=3204=80V
this is the required solution.

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