Question

Three uncharged capacitors of capacitane $C_1=1\mu \text{F},C_2=2\mu \text{F}$ and $C_3=3\mu \text{F}$ are connected as shown in figure to one another and to points A, B and D potential $V_A=10\text{V},V_B=25\text{V}$ and $V_D=20\text{V}$, Determine the potential $\left(V_0\right)$ at point O (in Volt).

Answer 1

By junction law,

$\Rightarrow q_1+q_2+q_3=0$

$\Rightarrow C_1(V_{\text{O}}-V_{\text{A}})+C_2(V_{\text{O}}-V_{\text{B}})+C_3(V_{\text{O}}-V_{\text{D}})=0$

$(C_1+C_2+C_3)V_{\text{O}}=C_1{V}_{\text{A}}+C_2{V}_{\text{B}}+C_3{V}_{\text{D}}$

$V_{\text{O}}=\frac{C_1{V}_A+C_2{V}_B+C_3{V}_D}{C_1+C_2+C_3}$

$=\frac{C_1{V}_{\text{A}}+C_2{V}_{\text{B}}+C_3{V}_{\text{D}}}{C_1+C_2+C_3}$

$=\frac{(1\times 10)+(2\times 25)+(3\times 20)}{1+2+3}$

$=\frac{10+50+60}{6}=20\text{Volt}$