# Battery power

1. Sep 1, 2007

### chanller

Hi everyone,
Can anyone please tell me if a battery delivers a constant power in any circuit!!!!!!!
Thanks

2. Sep 1, 2007

### Staff: Mentor

No. The power delivered essentially depends on the resistance of the circuit. The lower the resistance, the greater the power dissipation. If you short a battery, it'll dissipate all its available power in a few seconds.

3. Sep 1, 2007

### chanller

Well I am in fact doing research on electrolysis of water (Chemistry) at school and I found out that there are basic things which I need to know first!!!!!

4. Sep 1, 2007

### chanller

Hi russ watters,
well what i understood is that a battery gives a maximum power which lasts for a short while! Is that what you meant?

Last edited: Sep 1, 2007
5. Sep 1, 2007

### chanller

Let's say I have a circuit composed of a battery and a rheostat in series. I put two conducting inert plates in a beaker of water so that they are a distance apart and put this apparatus in series in the circuit. Can I vary the power into the apparatus by varying the resistance of the rheostat? Is the sum of the power into the apparatus and that in the rheostat equal to the maximum power of the battery?

6. Sep 1, 2007

### Staff: Mentor

Yes, you vary the power with the rheostat. Unfortunatly, the rheostat also dissipates power itself, so there is a tradeoff.

And there is no maxumum power of a battery (within some limits). There is a total energy content, but that energy can be expended at just about any rate you want, determined by the resistance of the circuit.

Draw yourself a circuit diagram, plug in some resistances and do some quick calculations. You know Ohm's law, right...?

7. Sep 1, 2007

### chanller

Is the voltage from a battery constant? Is it the current withdrawn which determines the power taken from it?

8. Sep 1, 2007

### bernhard.rothenstein

The power delivered by a battery depends on its emf E, internal r and external R resistences. If I remember well there is a given value of the external resistence for which the delivered (external) power is maximum (r=R?)
It is the value of r which ensured the maximum of the function
P(ext)=E^2xR/(R+r)^2

9. Sep 1, 2007

### alvaros

from bernhard.rothenstein:
Its true for ideal batteries ( constant voltage source in series whit a constant resistance ).
In real batteries, I dont know.

10. Sep 1, 2007

### chanller

Well, Voltage from real batteries is not constant: V=(E.M.F*R)/(R+r). Made some research!!!! Thanks everyone!

11. Sep 1, 2007

### leright

Hi Chanller,

A battery is a voltage source which means it ATTEMPTS to produce a constant voltage across its terminals no matter what circuit it is connected to. This does not mean the power is constant, since the power supplied by the battery depends on both the voltage across the battery and the current flowing through the battery.

Generally, a battery will have a particular maximum current it can supply at its rated voltage. The product of the battery's rated voltage and the maximum current is called the 'power rating' or the battery, and it is the maximum power the battery is capable of delivering to a circuit. Of course, the higher the power the battery is required to deliver, the lower its operating life since the battery's stored chemical energy is limited to some value.

A battery can be modeled with an ideal voltage source in series with a time dependent resistance called the internal resistance of the battery. The internal resistance increases as the battery gets older. Consequently, when a current is supplied to a load resistance, as the internal resistance gets larger more and more of the battery's emp is dropped across the internal resistance and less is dropped across the load. When the internal resistance is really high almost no voltage is dropped across the load. Note that if the battery is not connected to a resistive load but it is instead connected to a high quality voltmeter no current will flow and therefore no voltage will be dropped across the internal resistance, which means the voltmeter will show the battery's rated voltage! But when you try to hook the battery up to a regular load you will find that the battery is a dud, since current is flowing and voltage is being dropped across the internal resistance of the battery, and no voltage is left for the load. Try this some time and you might be surprised. I thought I'd point this fact out.

12. Sep 1, 2007

### Staff: Mentor

Unless you are drawing a lot of amperage and draining a battery in just a few minutes, you can consider the internal resistance insignificant and the voltage constant.

13. Sep 1, 2007

### leright

that's right. R is the load resistance and r is the internal resistance of the battery. The current supplied to the load is given by ohm's law, or I = (EMF)/(R+r). The voltage across the load is then of course V = (EMF)*R/(R + r). For a young and healthy battery, r, the internal resistance, is approximately zero. But as the battery gets older and more sluggish, r gets bigger and the voltage dropped across the load (V) gets smaller and smaller. Note however that if you connect a dead battery with a very high internal resistance (say 500 ohms) to a voltmeter where R is very large (say 2,000,000 ohms), R is about equal to R + r and the voltmeter reading is approximately equal to the EMF of the battery, so someone that might not understand batteries would be mistakenly led to believe that the battery is not actually dead.

14. Sep 1, 2007

### chanller

What value do I get if i connect a battery with a voltmeter? What do i get with an ammeter? Can I find the power rating of the battery using P=IV for the values obtained?

15. Sep 1, 2007

### leright

If you connect a battery (regardless of whether it is dead or alive) to a good voltmeter it will always show approximately the rated EMF of the battery.

To find the power rating of the battery you will need a brand new battery. Then take a look at the voltage rating specs and take the lower limit of the tolerance range. Then connect various loads to the battery with various resistances. Use smaller and smaller resistors. When the resistor value gets too small, eventually the voltage across the resistor will drop. When it drops to the lowest value given by the specs (lower imit of the tolerance range) then record the current. Multiply this current by the voltage to get the power rating.

16. Sep 1, 2007

### chanller

Is the value obtained when an ammeter is connected to a new battery the maximum current?

17. Sep 1, 2007

### Staff: Mentor

"Maximum current" is not a concept that applies here. I said that already! Once again, for your purposes, the voltage can be considered constant and the amperage is a function of resistance. Everything else people are talking about here is special cases of extremely high amperage, which don't apply. You don't want to be running your device in an amperage range where your battery is burning itself up.

What type of battery are you using? Many give amp-hour capacity ratings....

18. Sep 1, 2007

### chanller

It is a hydrogen-oxygen fuel cell. I need to find its power rating. In this case we do not need to matter about the time it takes to give a maximum current as the flow rate of hydrogen and oxygen are constant and hence per second it does give a maximum current! So, can i measure the maximum current by just connecting my fuel cell to an ammeter?

19. Sep 1, 2007

### chanller

Hey ihave tried this question in Chemistry dep. but no one seems to be online!

Talking about electrolysis, what happens to the excess power input when gas production rate remains constant even though power is increased? Droplets of water settled at the top of my gas storage cylinder, does this mean excess power is converted into heat which vaporises water in the electrolyte?

20. Sep 2, 2007