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Question

Two condensers $$C_1$$ and $$C_2$$ in a circuit are joined as shown in figure. The potential of point A is $$V_1$$ and that of B is $$V_2$$. The potential of point D will be-
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A
12(V1+V2)
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B
C2V1+C1V2C1+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 $$\dfrac {C_1V_1+C_2V_2}{C_1+C_2}$$
As $$C_1$$ and $$C_2$$ are in series, the charge on $$C_1$$ is equal to the charge on $$C_2$$,i.e.,$$Q_1=Q_2$$
Let, $$V_D$$ is the potential at D.
therefore, $$C_1(V_1-V_D)=C_2(V_D-V_2)$$
$$V_D(C_1+C_2)=C_1V_1+C_2V_2$$
$$V_D=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}$$

Physics
NCERT
Standard XII

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