Q.
A capacitor 4μF charged to 50,V is connected to another capacitor of 2μF charged to 100V with plates of like charges connected together. The total energy before and after connection in multiples of (10−2J) is
2979
179
Electrostatic Potential and Capacitance
Report Error
Solution:
The total energy before connection =21×4×10−6×(50)2+21×2×10−6×(100)2 =1.5×10−2J
When connected in parallel, 4×50+2×100=6×V ⇒V=3200
Total energy after connection =21×6×10−6×(3200)2 =1.33×10−2J