A) \[\frac{U}{2}\]
B) \[\frac{U}{4}\]
C) 2 U
D) 4 U
Correct Answer: B
Solution :
If q be the charges on each capacitors \[\therefore \,\,\,Energy\,\,stored\,\,U=\,\,\frac{1}{2}\frac{{{q}^{2}}}{C}=\frac{1}{2}\,\,C{{V}^{2}}\] Now, when battery is disconnected and another capacitor of same capacity is connected in parallel to first capacitor, then charge will be equally divided, if V is common potential then, \[V'=\frac{q}{2C}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[as\,\,q=CV]\] \[\therefore \] Energy stored in each capacitors \[=\,\,\,\frac{1}{2}C{{V}^{2}}=\frac{1}{2}C{{\left( \frac{q}{2C} \right)}^{2}}=\frac{1}{4}\cdot \frac{1}{2}\frac{{{q}^{2}}}{C}=\frac{U}{4}\]
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