# Work done by the battery?

by willydavidjr
Tags: battery, work
 P: 67 Can someone give me an idea how to compute or how to get the work done by the battery if there is a circuit given with resistor and a capacitor?
 Emeritus Sci Advisor PF Gold P: 9,772 This may be of interest to you; http://hyperphysics.phy-astr.gsu.edu...c/capeng2.html . At the bottom of the page look for the derivation of the equation for work done; $$U = \frac{Q^2}{2C}$$ ~H
 Emeritus Sci Advisor PF Gold P: 16,091 Power relates to work, right? ...
Emeritus
PF Gold
P: 9,772
Work done by the battery?

 Quote by Hurkyl Power relates to work, right? ...
What are you getting at?

~H
 Emeritus Sci Advisor PF Gold P: 16,091 Presumably he can figure out the power dissipated by the resistor, and get the amount of work done through that component of the system.
Emeritus
PF Gold
P: 9,772
 Quote by Hurkyl Presumably he can figure out the power dissipated by the resistor, and get the amount of work done through that component of the system.
Yeah, he could. But it also takes work to place charge on the capacitor plates.

~H
 Emeritus Sci Advisor PF Gold P: 16,091 Right. But you already covered that part, and I didn't have anything to add to it.
Emeritus
PF Gold
P: 9,772
 Quote by Hurkyl Right. But you already covered that part, and I didn't have anything to add to it.
Oh right, I thought I was missing something, like you could just work out the problem by considering power dissapated by the resistors. Thanks

~H
P: 363
 Quote by Hootenanny Oh right, I thought I was missing something, like you could just work out the problem by considering power dissapated by the resistors. Thanks ~H
Matter of fact you could!
The energy dissipated during the charging of a capacitor equals the energy stored in the capacitor at the end of the process.
So the total energy that battery gave away is

$$U = \frac{Q^2}{C}$$

It's more complicated to consider it from the resistor side than from the capacitor side I suppose.
Emeritus
PF Gold
P: 16,091
 Quote by tehno Matter of fact you could! The energy dissipated during the charging of a capacitor equals the energy stored in the capacitor at the end of the process.
It's clear that when the battery is disconnected, the power subsequently dissipated by the resistor will be exactly the energy stored in the capacitor.

But I don't see why there should be such a trivial relationship between the power dissipated by the resistor with the power stored in the capacitor during the time the capacitor is charging!
 P: 363 It's a just a variant of Kelvin's rule applied to electrical circuits. If you don't trust me, write down differential equation ,solve it in current , and take the integral of $$i^2 R dt$$. Ratio of two energies will be always the same regardless of electromotive force of the source $$E$$,resistance $$R$$ of the circuit, and amount of capacitance $$C$$ in the circuit!
 P: 1,235 I would reason like this: Assuming the battery voltage is constant: $$\text{Work by the battery} = \int V_{battery} I dt = V_{battery} Q$$and since at the end of the charging $$Q = C V = C V_{battery}$$we get $$\text{Work by the battery} = C V_{battery}^2$$(I assumed the capacitor has no charge at the beginning)
 P: 67 Meaning the energy that occurs in the capacitor is also the same as the energy use by the battery? And it goes like this? $$E=\frac{1}{2}qV = \frac{1}(2}CV^2 = \frac{q^2}{2C}$$ Am I right?
P: 2,056
 Quote by willydavidjr Can someone give me an idea how to compute or how to get the work done by the battery if there is a circuit given with resistor and a capacitor?
I don't CARE what work the battery is doing, the battery needs to get its smelly arse back to Mexico!!!!
P: 363
 Quote by willydavidjr Meaning the energy that occurs in the capacitor is also the same as the energy use by the battery?
No!You are missing the point.
The energy that is delivered by the battery is twice as high as the energy of the capacitor.
Other half is dissipated by the resistor during the charging as the heat.
You have a simple energy bilance :
$$W_{battery}=\int_{0}^Q Vdq=QV=W_{dissipated}+U_{capacitor}=W_{dissipated}+ \frac{VQ}{2}$$
Maybe I should emphasize that prior to the stationary state established,more energy is spent as a Joule heat than being stored in the capacitor. Also easy to show that.
 P: 1,235 Hi, I think something is still missing in this thread: what is the energy stored in the capacitor, and how this can be proved/calculated. My suggestion on how to do that: once the capacitor is charged, calculate how much energy it will dissipate in the resistor if the battery is removed and replaced by a shortcut. The calculation is similar to the previous work done by the battery except that the voltage on the capacity will decrease during this process. The outcome is then the 1/2 factor.But I think there is a better way to proof the formula for the capacitor energy storage. I don't remember, help me. Michel PS: The fact that the work done goes in equal amounts to dissipation and to storage in the capacity is strinking. Is it related to a more general rule?
 P: 67 energy=1/2q^2/c= 1/2C^2V.... Can I ask, do you know what is the formula for the work done by the battery? Thanks.
 P: 67 Techno you mean, if the energy in capacitor is 20 joules, the energy in the battery is 40 joules? Am I right?

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