Q 12-02-030NEETJEE MainMedium
A $4\ \mu$F capacitor charged to $50$ V is connected to an uncharged $6\ \mu$F capacitor. The common potential and the energy lost are
Answer: (C) $20$ V and $3$ mJ
Charge is conserved: $V = \dfrac{4 \times 50}{4 + 6} = 20$ V.
Energy lost $= \dfrac{1}{2}\dfrac{C_1C_2}{C_1 + C_2}(V_1 - V_2)^2 = \dfrac{1}{2} \times 2.4 \times 10^{-6} \times 2500 = 3$ mJ.
Solution by Sreeraj P, M.Sc Physics