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 [MP PMT 1997]

\[\frac{{{C}_{2}}{{V}_{1}}-{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]

\[\frac{1}{2}({{V}_{1}}+{{V}_{2}})\]

\[\frac{{{C}_{2}}{{V}_{1}}+{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]

\[\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]

Two condensers \[{{C}_{1}}\]and \[{{C}_{2}}\] in a circuit are joined

1.	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 [MP PMT 1997]
A.	\[\frac{1}{2}({{V}_{1}}+{{V}_{2}})\]
B.	\[\frac{{{C}_{2}}{{V}_{1}}+{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]
C.	\[\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]
D.	\[\frac{{{C}_{2}}{{V}_{1}}-{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]
Answer» D. \[\frac{{{C}_{2}}{{V}_{1}}-{{C}_{1}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]