JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2020
- Medium
- Numerical
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\(.....\)
Pick an option to check your answer.
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
\(v=\frac{220 \times 5+0 \times 2.5}{5+2.5}=220 \times \frac{2}{3}\)
\(u _{ f }=\frac{1}{2}(5+2.5) \times 10^{-6}\left(220 \times \frac{2}{3}\right)^{2}\)
\(\Delta u = u _{ f }- u _{ i }\)
\(\Delta u =-403.33 \times 10^{-4}\)
\(\Rightarrow-403.33 \times 10^{-4}=\frac{ X }{100}\)
\(X=-4.03\)
or magnitude or value of \(X\) is approximate \(4\)
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