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GUJCET · Physics · Electrostatic Potential and Capacitance
A capacitor is charged with a battery and energy stored is U. After disconnecting the battery, another identical uncharged capacitor is connected in parallel with it. The total energy of system of capacitors is _________.
- A \(\frac{3 U}{2}\)
- B \(\frac{3 U}{4}\)
- C \(\frac{U}{4}\)
- D \(\frac{ U }{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{ U }{2}\)
Step-by-step Solution
Detailed explanation
(D)
Let the initial change on capacitor be Q .
So, energy of capacitor \(=U=\frac{Q^2}{2 C}\)
When the battery is disconnected and capacitor connected in parallel,charge on each capacitor is \(\frac{ Q }{2}\).
energy of each capacitor
\(\begin{array}{l}=\frac{1}{2} \frac{\left(\frac{ Q }{2}\right)^2}{ C } \\ =\frac{1}{4} \frac{ Q ^2}{2 C } \\ =\frac{ U }{4}\end{array}\)
energy of system
\(=\frac{U}{4}+\frac{U}{4}\)
\(=\frac{ U }{2}\)
Let the initial change on capacitor be Q .
So, energy of capacitor \(=U=\frac{Q^2}{2 C}\)
When the battery is disconnected and capacitor connected in parallel,charge on each capacitor is \(\frac{ Q }{2}\).
energy of each capacitor
\(\begin{array}{l}=\frac{1}{2} \frac{\left(\frac{ Q }{2}\right)^2}{ C } \\ =\frac{1}{4} \frac{ Q ^2}{2 C } \\ =\frac{ U }{4}\end{array}\)
energy of system
\(=\frac{U}{4}+\frac{U}{4}\)
\(=\frac{ U }{2}\)
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