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 $O$ will be

• A. $C_1{V}_1+C_2{V}_2$
• B. $\frac{C_2{V}_1+C_1{V}_1}{C_1+2C_2}$
• C. $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
• D. $\frac{V_1+V_2}{2}$

Answer 1

Answer: (C)

Both capacitors are joined in series combination in which charge remain same in all capacitors
So, charge on $C_1$ = charge on $C_2$
$\therefore C_1\Delta V=C_2\Delta V$
$\therefore C_1\left(V_A-V_{\Delta }\right)=C_2\left(V_{\Delta }-V_B\right)$
$C_1\left(V_1-V_{\Delta }\right)=C_2\left(C_{\Delta }-V_2\right)$
$\therefore C_1{V}_1-C_1{V}_{\Delta }=C_2{V}_{\Delta }-C_2{V}_2$
$\therefore C_1{V}_1+C_2{V}_2=C_2{V}_{\Delta }+C_1{V}_{\Delta }$
$V_{\Delta }=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
It means there is a common potential at point $O$