Question

Two condensers $C_1$ and $C_2$ in a circuit are joined as shown in 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. $\frac{C_2{V}_1+C_1{V}_2}{C_1+C_2}$
• 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

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

As $C_1$ and $C_2$ are in series, the charge on $C_1$ is equal to the charge on $C_2$,i.e.,$Q_1=Q_2$
Let, $V_D$ is the potential at D.

therefore, $C_1(V_1-V_D)=C_2(V_D-V_2)$
$V_D(C_1+C_2)=C_1{V}_1+C_2{V}_2$
$V_D=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$