Question

Two capacitors of capacitances $C_1$ and $C_2$ , in a circuit are joined as shown in the figure. The potential of point $\text{A}$ is $V_1$ and potential of point $\text{B}$ is $V_2$. The potential at point $\text{D}$ will be

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

Answer 1

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


Let's assume that , $V_1>V_2$


As the two capacitor are in series combination across $\text{A}$ and $\text{B}$, their equivalent capacitance will be
$$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$$
Potential difference across the equivalent capacitance, $\Delta V=V_1-V_2$

So, the charge flowing through it,

$Q=\left(\frac{C_1{C}_2}{C_1+C_2}\right)(V_1-V_2)$

On moving from $\text{A}$ to $\text{D},$ there will be potential drop of $\frac{Q}{C_1}$

Thus, $V_A-V_D=\frac{Q}{C_1}$

As, $V_A=V_1$

$\Rightarrow V_1-V_D=\frac{Q}{C_1}$

$\Rightarrow V_D=V_1-\frac{Q}{C_1}$

Substituting the value of $Q$,

$\Rightarrow V_D=V_1-\frac{C_2}{C_1+C_2}(V_1-V_2)$

$\Rightarrow V_D=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

Hence, option (c) is the correct answer.

Key Concept: Voltage across a capacitor, $V=\frac{Q}{C}$