A) \[\frac{{{C}_{2}}V}{{{C}_{1}}+{{C}_{2}}}\]
B) \[\frac{{{C}_{1}}V}{{{C}_{1}}+{{C}_{2}}}\]
C) \[\left( 1+\frac{{{C}_{2}}}{{{C}_{1}}} \right)\]
D) \[\left( 1-\frac{{{C}_{2}}}{{{C}_{1}}} \right)V\]
Correct Answer: B
Solution :
\[\operatorname{Charge} on {{1}^{st}}capacitor = {{q}_{1}}, =\,\,{{C}_{1}}V\] \[\operatorname{Charge}\,\,on\,\,{{2}^{nd}}\,\,capacitor\,\,=\,\,{{q}_{2}}\,\,=\,\,0\] When they are connected in parallel the total charge \[q={{q}_{1}}+{{q}_{2}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\operatorname{q}={{C}_{1}}V\] And capacitance, \[\operatorname{C}={{C}_{1}}\,+\,\,{{C}_{2}}\] Let V? be the common potential difference across each capacitors, then \[\operatorname{q} = CV'\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\operatorname{V}'=\frac{q}{C}\,\,=\,\,\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}V\]
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