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Question

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


A
12(V1+V2)
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B
C1V2+C2V1C1+C2
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C
C1V1+C2V2C1+C2
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D
C2V1C1V2C1+C2
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Solution

The correct option is C C1V1+C2V2C1+C2
Let the potential of point D be V.


From the given figure, we can see that the capacitors C1 and C2 are in series.

Potential difference across both the capacitors:

ΔV1=V1V; ΔV2=VV2

In the series arrangement, the charge on capacitors will be equal.

Therefore, q1=q2

C1(V1V)=C2(VV2)

C1V1C1V=C2VC2V2

(C1+C2)V=C1V1+C2V2

V=C1V1+C2V2C1+C2

Hence, option (c) is correct.
Why this question ?
It must not be confused that why we have taken V1>V2 in calculations. The same result will be obtained if you consider V2>V1, just make sure to use it uniformly or throughout the solution.

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