Recent content by Shiuanchiam

Ok take note of that, Initial energy =0.02Joule Final energy =0.01786Joule Total loss of energy=0.02-0.01786 = 2.14×10^-3 Joule Is that correct?

Total Capacitance = 3μ+4μ =7μ So the final energy =1/2 (200μ+300μ)^2/7μ Then do the initial energy separately = (1/2 (200μ)^2/4μ )+(1/2(300μ)^2/3μ)

Thanks.Yes it is quoted exactly.I assume that the statement is trying to show that the capacitors are arranged in parallel?

By using that equation what we get is the capacity in series, but when positive plate is connected together, isn't it suppose to be parallel? Capacity =C1+C2

Homework Statement A 4μF capacitor is charged to a charge of 200 μC,and a 3μF capacitor is charged to a charge of 300μC. Both the capacitors are connected so that the positive plates are connected together. What is the total loss of energy in the process? Homework EquationsThe Attempt at a...