Three uncharged capacitors of capacitance C_1 | vedclass

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(B) Let the potential at point $O$ be $V_0$. Since the capacitors are initially uncharged and connected to the junction $O$,the total charge at the ju

Vedclass · Accepted Answer

$\frac{V_A C_1 + V_B C_2 + V_D C_3}{C_1 + C_2 + C_3}$

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(B) Let the potential at point $O$ be $V_0$. Since the capacitors are initially uncharged and connected to the junction $O$,the total charge at the junction $O$ must be zero due to the conservation of charge.The charge on capacitor $C_1$ is $q_1 = (V_A - V_0) C_1$.The charge on capacitor $C_2$ is $q_2 = (V_B - V_0) C_2$.The charge on capacitor $C_3$ is $q_3 = (V_D - V_0) C_3$.Since the junction $O$ is isolated,the sum of charges on the plates connected to $O$ must be zero:$q_1 + q_2 + q_3 = 0$Substituting the expressions for the charges:$(V_A - V_0) C_1 + (V_B - V_0) C_2 + (V_D - V_0) C_3 = 0$Expanding the terms:$V_A C_1 - V_0 C_1 + V_B C_2 - V_0 C_2 + V_D C_3 - V_0 C_3 = 0$Grouping the $V_0$ terms:$V_A C_1 + V_B C_2 + V_D C_3 = V_0 (C_1 + C_2 + C_3)$Solving for $V_0$:$V_0 = \frac{V_A C_1 + V_B C_2 + V_D C_3}{C_1 + C_2 + C_3}$

Three uncharged capacitors of capacitance $C_1$,$C_2$ and $C_3$ are connected as shown in the figure to one another and to points $A$,$B$ and $D$ at potentials $V_A$,$V_B$ and $V_D$. Then the potential at point $O$ will be

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