Question

Three uncharged capacitors of capacitance $C_{1},C_{2}$ and $C_{3}$ are connected as shown in figure to one another and to points A, B and D at potentials $V_{A},V_{B}$ and $V_{D}$ . Then the potential at O will be :-

• A. $\frac{V_{A} C_{1} + V_{B} C_{2} + V_{D} C_{3}}{C_{1} + C_{2} + C_{3}}$
• B. $\frac{V_{A} V_{B} V_{D}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}$
• C. $\frac{V_{A} + V_{B} + V_{D}}{C_{1} + C_{2} + C_{3}}$
• D. $\frac{V_{A} V_{B} + V_{B} V_{D} + V_{D} V_{A}}{C_{1} + C_{2} + C_{3}}$

Answer 1

Answer: (A)

$V_{A}-V_{0}=\frac{q}{C_{1}}$ or $q=\left(V_{A} - V_{0}\right)C_{1}$
$V_{B}-V_{0}=\frac{q_{1}}{C_{2}}$ or $q_{1}=\left(V_{B} - V_{0}\right)C_{2}$
$V_{D}-V_{0}=\frac{q_{2}}{C_{3}}$ or $q_{2}=\left(V_{D} - V_{0}\right)C_{3}$
$q+q_{1}+q_{2}=0$
$\left( V_{A} - V_{0}\right)C_{1}+\left(V_{B} - V_{0}\right)C_{2}+\left(V_{D} - V_{0}\right)C_{3}=0$
$\therefore V_{0}=\frac{V_{A} C_{1} + V_{B} C_{2} + V_{D} C_{3}}{C_{1} + C_{2} + C_{3}}$