JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A \(5\, \mu F\) capacitor is charged fully by a \(220\,V\) supply. It is then disconnected from the supply and is connected in series to another uncharged \(2.5\;\mu F\) capacitor. If the energy change during the charge redistribution is \(\frac{ X }{100} \;J\) then value of \(X\) to the nearest integer is\(.....\)
- A \(4\)
- B \(10\)
- C \(9\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(u _{ i }=\frac{1}{2} \times 5 \times 10^{-6}(220)^{2}\) Final common potential \(v=\frac{220 \times 5+0 \times 2.5}{5+2.5}=220 \times \frac{2}{3}\) \(u _{ f }=\frac{1}{2}(5+2.5) \times 10^{-6}\left(220 \times \frac{2}{3}\right)^{2}\) \(\Delta u = u _{ f }- u _{ i }\)…
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