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
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$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}=\frac{\mathrm{q}}{\mathrm{C}_{1}} \quad$ or $\mathrm{q}=\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}\right) \mathrm{C}_{1}$
$\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{0}=\frac{\mathrm{q}_{1}}{\mathrm{C}_{2}}$ or $\mathrm{q}_{1}=\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{0}\right) \mathrm{C}_{2}$
$\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{0}=\frac{\mathrm{q}_{2}}{\mathrm{C}_{3}} \quad$ or $\quad \mathrm{q}_{2}=\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{0}\right) \mathrm{C}_{3}$
$\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2}$
$\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{0}\right) \mathrm{C}_{1}=\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{o}}\right) \mathrm{C}_{2}+\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{o}}\right) \mathrm{C}_{3}$
$\therefore \mathrm{V}_{0}=\frac{\mathrm{V}_{\mathrm{A}} \mathrm{C}_{1}+\mathrm{V}_{\mathrm{B}} \mathrm{C}_{2}+\mathrm{V}_{\mathrm{D}} \mathrm{C}_{3}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}$
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