A) U
B) \[\frac{U}{2}\]
C) 2U
D) \[\frac{3}{2}U\]
Correct Answer: B
Solution :
Key Idea: On removing the battery after charging, the charge stored in eke capacitor remains constant. When a capacitor is charged by connecting a battery across its plates, the initial energy stored, \[U=\frac{{{q}^{2}}}{2C}\] When the battery is disconnected, then the charge remains constant i.e., q = constant. Now another identical capacitor is connected across it i.e., the capacitors are connected in parallel, so the equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=C+C=2C\] Thus, final energy stored by the system of capacitors, \[U'=\frac{{{q}^{2}}}{2{{C}_{eq}}}\] \[=\frac{{{q}^{2}}}{2(2C)}=\frac{1}{2}U\] \[\therefore U'=\frac{U}{2}\]You need to login to perform this action.
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