Question

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potentials of points $A$ and $B$ are $V_1$ and $V_2$ respectively. Then the potential of point $D$ will be

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

Let the potential at point $D$ be $V$ and suppose that $V_2>V_1$ .
Then the potential drop across $C_1$ is $(V-V_1)$ and $C_2$ is $(V_2-V)$ .
Charge stored on the capacitor of capacitance $C_1$ is
$q_1=C_1(V-V_1)$
Charge stored on the capacitor of capacitance $C_2$ is
$q_2=C_2(V_2-V)$
As $q_1=q_2$ [capacitors are in series]
$\therefore C_1(V-V_1)=C_2(V_2-V)$
$\Rightarrow V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$