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Q.

Two condensers C1  and  C2 in a circuit are joined as shown in figure. The potential of point A is V1 and that of B is  V2.The potential of point D will be

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a

12(V1+V2)

b

C1V2+C2V1C1+C2

c

C1V1+C2V2C1+C2

d

C2V1−C1V2C1+C2

answer is C.

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Detailed Solution

V1−V2 will be divided between C1  and  C2. C1  and  C2 are  in series. Potential difference across C1 isV1−VD=C2(V1−V2)C1+C2or VD=C1V1+C2V2C1+C2
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