
#1
May2006, 03:53 PM

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?




#2
May2006, 03:59 PM

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PF Gold
P: 9,789

This may be of interest to you; http://hyperphysics.phyastr.gsu.edu...c/capeng2.html . At the bottom of the page look for the derivation of the equation for work done;
[tex]U = \frac{Q^2}{2C}[/tex] ~H 



#4
May2106, 07:01 AM

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PF Gold
P: 9,789

Work done by the battery?~H 



#5
May2106, 08:28 AM

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PF Gold
P: 16,101

Presumably he can figure out the power dissipated by the resistor, and get the amount of work done through that component of the system.




#6
May2106, 08:34 AM

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PF Gold
P: 9,789

~H 



#7
May2106, 08:36 AM

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PF Gold
P: 16,101

Right. But you already covered that part, and I didn't have anything to add to it.




#8
May2106, 08:55 AM

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PF Gold
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~H 



#9
May2106, 11:30 AM

P: 363

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 [tex] U = \frac{Q^2}{C} [/tex] It's more complicated to consider it from the resistor side than from the capacitor side I suppose. 



#10
May2106, 12:58 PM

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PF Gold
P: 16,101

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! 



#11
May2106, 02:04 PM

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 [tex] i^2 R dt [/tex]. Ratio of two energies will be always the same regardless of electromotive force of the source [tex]E[/tex],resistance [tex]R[/tex] of the circuit, and amount of capacitance [tex]C[/tex] in the circuit! 



#12
May2106, 03:32 PM

P: 1,235

I would reason like this:
Assuming the battery voltage is constant: [tex]\text{Work by the battery} = \int V_{battery} I dt = V_{battery} Q[/tex]and since at the end of the charging [tex]Q = C V = C V_{battery}[/tex]we get [tex]\text{Work by the battery} = C V_{battery}^2[/tex](I assumed the capacitor has no charge at the beginning) 



#13
May2106, 06:32 PM

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? [tex]E=\frac{1}{2}qV = \frac{1}(2}CV^2 = \frac{q^2}{2C} [/tex] Am I right? 



#14
May2206, 06:23 AM

P: 2,057





#15
May2206, 07:24 AM

P: 363

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 : [tex]W_{battery}=\int_{0}^Q Vdq=QV=W_{dissipated}+U_{capacitor}=W_{dissipated}+ \frac{VQ}{2}[/tex] 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. 



#16
May2206, 01:51 PM

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? 



#17
May2206, 03:47 PM

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.




#18
May2206, 03:58 PM

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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