show that there is always loss of energy when two charged capacitors having a different capacities and potentialy are joined in parallel.
Answers
Answer:
Two capacitors of capacitances C1 and C1 have charge Q1 and Q2. How much energy, Δw, is dissipated when they are connected in parallel. Show explicitly that Δw is non-negative."
I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:
Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let Q′1 and Q′2 be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor
Ceq=C1+C2= Q′1+Q′2Vf
conservation of charge:
Q′1+Q′2=Q1+Q2,
Ceq= Q1+Q2Vf
Vf=Q1+Q2C1+C2
The initial energy of the capactiors is:
U0=Q212C1+Q222C2
Uf=12(C1+C2)V2f=(Q1+Q2)22(C1+C2)
ΔU=(Q1+Q2)22(C1+C2)-Q212C1−Q222C2
However, I don't see how this is necessarily non-negative.