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Two condensers $C_1$ and $C_2$ in a circuit are joined as shown below.
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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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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Verified Answer
The correct answer is:
$\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}$
$V_2-V_1$ will be divided between $C_1$ and $C_2$. Let $V_D$ be potential at point D.
Let $\mathrm{Q}$ be the charge on each capacitor. Then the potential drop can be written as:
$\begin{aligned} & V_2-V_1=V_2-V_D+V_D-V_1 \\ & \Rightarrow V_2-V_1=\frac{Q}{C_2}+\frac{Q}{C_1} \\ & \Rightarrow Q=\frac{C_1 C_2\left(V_2-V_1\right)}{C_1+C_2}\end{aligned}$
Potential difference across $C_1$ is
$V_D=V_1+\frac{C_2\left(V_2-V_1\right)}{C_1+C_2}=\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}$
Let $\mathrm{Q}$ be the charge on each capacitor. Then the potential drop can be written as:
$\begin{aligned} & V_2-V_1=V_2-V_D+V_D-V_1 \\ & \Rightarrow V_2-V_1=\frac{Q}{C_2}+\frac{Q}{C_1} \\ & \Rightarrow Q=\frac{C_1 C_2\left(V_2-V_1\right)}{C_1+C_2}\end{aligned}$
Potential difference across $C_1$ is
$V_D=V_1+\frac{C_2\left(V_2-V_1\right)}{C_1+C_2}=\frac{C_2\left(V_1+V_2\right)}{C_1+C_2}$
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