Two condensers ${C_1}$and ${C_2}$ in a circuit are joined as shown in figure. The potential of point $A$ is ${V_1}$ and that of $B$ is ${V_2}$. The potential of point $D$ will be
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(c) Charge on ${C_1}$ = charge on ${C_2}$
$==>$ ${C_1}({V_A} - {V_D}) = {C_2}({V_D} - {V_B})$
$==>$ ${C_1}({V_1} - {V_D}) = {C_2}({V_D} - {V_2})$ $==>$ ${V_D} = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
$==>$ ${C_1}({V_A} - {V_D}) = {C_2}({V_D} - {V_B})$
$==>$ ${C_1}({V_1} - {V_D}) = {C_2}({V_D} - {V_2})$ $==>$ ${V_D} = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
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