A capacitor of capacitance 900μF is charged by a 100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x×10−2 J. The value of x is ______.
Answer: 225
Solution
The charged capacitor carries Q=CV=900×10−6×100=0.09C and stores Ui=2CQ2=2×900×10−6(0.09)2=4.5J. Joining its plates to an identical uncharged capacitor with like plates together shares the charge 0.09C over a combined capacitance 2C, so the stored energy falls to Uf=2(2C)Q2=2.25J. The energy lost is Ui−Uf=4.5−2.25=2.25J=225×10−2J, so x=225.
JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer
