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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+C2Let 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=V1−V; ΔV2=V−V2 In the series arrangement, the charge on capacitors will be equal. Therefore, q1=q2 ⇒C1(V1−V)=C2(V−V2) ⇒C1V1−C1V=C2V−C2V2 ⇒(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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