Question

When capacitors are connected in series, energy is lost?

Answer 1

plz mark it a brainlist .

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.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.

Answer 2

Answer:

$\bigodot\mathtt \pink{\frac { 1 } { C } = \frac { 1 } { C _ { 1} } + \frac { 1 } { C _ { 2 } } + \frac { 1 } { C _ { 3 } } + \cdots + \frac { 1 } { C _ { n } }}$