Energy After Parallel ConnectHardQuestion 31564
Question
After connecting the capacitors in the previous question, energy stored is:
- A16 mJCorrect
- B40 mJ
- C20 mJ
- D8 mJ
Correct answer
16 mJ
Explanation
U_f = ½×5μF×(80)² = ½×5×10⁻⁶×6400 = 16 mJ. Energy lost = 24 mJ (heat).