Conservation of energy for charging a capacitor

[casanova2528]

one capacitor is charged with a battery. then the capacitor is disconnected away from the battery. Then, the capacitor is hooked up to another capacitor. the result involves losing of energy to thermal radiation and light as the charges from one capacitor moves to the other capacitor.

another situation involves having two parallel capacitors hooked up to a battery to be charged. then, the battery is disconnected. This situation includes the summation of potential energies of both capacitors to figure out the total potential energy of the system.
this makes sense.

However, I can't seem to pin point at where in the equations involved in situation 1 do we see the loss of this energy as the charges flow to the other capacitor.
In other words, you cannot add up the potential energies of both capacitors and equate it to the potential energy of the 1 capacitor hooked up to the battery.

here's a bit of the math.

u = .5(q squared) / C1 = potential energy of the one capacitor hooked up to the battery.

u = .5 (c)(v squared) = potential energy of the two capacitors hooked up to each other without battery.

u = .5(C1 + c2)(Vsquared) = total energy of the connected capacitors.

V of the connected capacitors = C1V1/(C1 + C2) where V1 = electric potential of the battery.

u = .5(C1V1)squared/(C1 + C2)

this energy is less than u=.5(Qsquared)/C1 = .5(C1)(V1 squared)


WHERE IS THE LOSS OF ENERGY?? WHERE IN THIS EQUATION CAN WE DEDUCE THAT THERMAL ENERGY IS BEING LOST?

 

[m.s.j]

the result involves losing of energy to thermal radiation and light as the charges from one capacitor moves to the other capacitor

I is very clear, you mentioned the missed energy in that sentence.



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Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[tiny-tim]

However, I can't seem to pin point at where in the equations involved in situation 1 do we see the loss of this energy as the charges flow to the other capacitor.

Hi casanova2528!

The equation you need is energy lost = power x time = ∫I(t)² R dt for the current, I(t), which flows from one capacitor to the other, through wires of resistance R.

From the PF Library:

Inverse exponential rate of charging:

A capacitor does not charge or discharge instantly.

When a steady voltage [itex]V_1[/itex] is first applied, through a circuit of resistance [itex]R[/itex], to a capacitor across which there is already a voltage [itex]V_0[/itex], both the charging current [itex]I[/itex] in the circuit and the voltage difference [itex]V_1\,-\,V[/itex] change exponentially, with a parameter [itex]-1/CR[/itex]:

[tex]I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}[/tex]

[tex]V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}[/tex]

So the current becomes effectively zero, and the voltage across the capacitor becomes effectively [itex]V_1[/itex], after a time proportional to [itex]CR[/itex].

Energy loss:

Energy lost (to heat in the resistor):

[tex]\int\,I^2(t)\,R\,dt\ =\ \frac{1}{2}\,C (V_1\,-\,V_0)^2[/itex]

Efficiency (energy lost per total energy):

[tex]\frac{V_1^2\,-\,V_0^2}{V_1^2\,-\,V_0^2\,+\,(V_1\,-\,V_0)^2}\ =\ \frac{1}{2}\,\left(1\,+\,\frac{V_0}{V_1}\right)[/tex]

Accordingly, charging a capacitor through a resistor is very inefficient unless the applied voltage stays close to the voltage across the capacitor.

But there is no energy loss on charging a capacitor through an inductor, basically because the applied voltage then appears across the inductor instead of across the capacitor.