Question

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

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

Let the potential of point $\text{D}$ be $V$.


From the given figure, we can see that the capacitors $C_1$ and $C_2$ are in series.

Potential difference across both the capacitors:

$\Delta V_1=V_1-V;\Delta V_2=V-V_2$

In the series arrangement, the charge on capacitors will be equal.

Therefore, $q_1=q_2$

$\Rightarrow C_1(V_1-V)=C_2(V-V_2)$

$\Rightarrow C_1{V}_1-C_{1}V=C_{2}V-C_2{V}_2$

$\Rightarrow (C_1+C_2)V=C_1{V}_1+C_2{V}_2$

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

Hence, option $\left(c\right)$ is correct.

Why this question ? It must not be confused that why we have taken $V_1>V_2$ in calculations. The same result will be obtained if you consider $V_2>V_1$, just make sure to use it uniformly or throughout the solution.