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 seriesHence,the equivalence capacitance (EC) will be given by1EC=116+14we get,Equivalence capacitance (EC)=165μFCharge=EC×P.D.Charge=165×100Charge(Q)=320 μFNow,QA=320×616μFQA=120 μFQB=320×1016μFQB=200 μFNow proceeding for P.D. across each capacitor,VA=QACAVA=1206=20V ,VB=QBCBVB=20010=20V ,VC=QCCCVC=3204=80Vthis is the required solution.

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