Question

Four unchanged capacitors $C_1,C_2,C_3\text{and}{C}_4$ are connected as shown in the figure. The point $\text{A, B, C and D}$are kept at potential $V_1,V_2,V_3\text{and}{V}_4$ respectively. Then the potential $V_0$ at point $O$ is

• A. $V_1+V_2+V_3+V_4$
• B. $\frac{C_1{V}_1+C_2{V}_2+C_3{V}_3+C_3{V}_4}{C_1+C_2+C_3}$
• C. $\frac{1}{2}(V_1+V_2+V_3+V_4)$
• D. $\frac{1}{2}\frac{C_1{V}_1+C_2{V}_2+C_3{V}_3+C_3{V}_4}{C_1+C_2+C_3+C_4}$

Answer 1

The correct option is B $\frac{C_1{V}_1+C_2{V}_2+C_3{V}_3+C_3{V}_4}{C_1+C_2+C_3}$

The potential at $O$ be the common potential of each capacitors i.e., $V_0$
Thus,

$Q_1=C_1{V}_0,Q_2=C_2{V}_0,Q_3=C_3{V}_0\text{and}{Q}_4=C_4{V}_0$

If $V_1,V_2,V_3\text{and}{V}_4$ are given to capacitor then charges are

$Q_1^{{}^{\prime }}=C_1{V}_1,Q_2^{{}^{\prime }}=C_2{V}_2,Q_3^{{}^{\prime }}=C_3{V}_3\text{and}{Q}_4^{{}^{\prime }}=C_4{V}_4$

As total charge is conserved so,

$Q_1+Q_2+Q_3+Q_4=Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}+Q_3^{{}^{\prime }}+Q_4^{{}^{\prime }}{C}_1{V}_0+C_2{V}_0+C_3{V}_0+C_4{V}_0=C_1{V}_1+C_2{V}_2+C_3{V}_3+C_4{V}_4$

$V_0=\frac{C_1{V}_1+C_{2}V2+C_3{V}_3+C_4{V}_4}{C_1+C_2+C_3+C_4}$

Hence, option $B$ is corrcet.