Question

Two condensers $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be :

• A. $\frac{1}{2}(V_1+V_2)$
• B. $\frac{C_1{V}_2+C_2{V}_1}{C_1+C_2}$
• C. $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
• D. $\frac{C_2{V}_1+C_1{V}_2}{C_1+C_2}$

Answer 1

Answer: (C)

$\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

$V_1-V_2$ will be divided between $C_1$ and $C_2$ and $V_D$ is potential at point D in series. Potential difference across $C_1$ is
$V_1-V_D=\frac{C_2(V_1-V_2)}{C_1+C_2}$
$V_D=V_1-\frac{C_2(V_1-V_2)}{C_1+C_2}$
or $V_D=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$