Two condensers $\mathrm{C}_1 \& \mathrm{C}_2$ in a circuit are joined as shown in the figure. The potential…

Two condensers $\mathrm{C}_1 \& \mathrm{C}_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of point $B$ is $V_2$. The potential at point $D$ will be

1. $\frac{1}{2}\left(\mathrm{~V}_1+\mathrm{V}_2\right)$
2. $\frac{C_2 V_1+C_1 V_2}{C_1+C_2}$
3. $\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$
4. $\frac{C_2 V_2-C_1 V_2}{C_1+C_2}$

Solution

In series combination, $\mathrm{Q}_1=\mathrm{Q}_2$ $\Rightarrow C_1\left(V_1-V\right)=C_2\left(V-V_2\right)$ $\Rightarrow \mathrm{V}\left(\mathrm{C}_1+\mathrm{C}_2\right)=\mathrm{C}_1 \mathrm{~V}_1+\mathrm{C}_2 \mathrm{~V}_2 \Rightarrow \mathrm{~V}=\frac{\mathrm{C}_1 \mathrm{~V}_1+\mathrm{C}_2 \mathrm{~V}_2}{\mathrm{C}_1+\mathrm{C}_2}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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