Question

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

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

Answer 1

The correct option is C $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

Let the potential at D is $V$.
Potential across $C_1=(V_1-V)$ and potential across $C_2=(V-V_2).$
Charge on $C_1$ is $q_1=C_1(V_1-V)$ and charge on $C_2$ is $q_2=C_2(V-V_2)$.
As capacitors are in series so charges on each capacitor are same.
thus, $q_1=q_2\Rightarrow C_1(V_1-V)=C_2(V-V_2)$
or $C_1{V}_1+C_2{V}_2=V(C_1+C_2)\Rightarrow V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$