# Power loss in a battery

Discussion in 'General Electronics Chat' started by johnny91, Jun 14, 2015.

1. ### johnny91 Thread Starter New Member

Jun 14, 2015
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If a 1.5V AA battery is connected both ends together by a copper wire of 7cm long and 1mm diameter, what will be the power loss? (Resistivity of Cu = 1.72x10^-8)

Jul 31, 2013
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what is the unit of "power loss"
can you provide a circuit diagram ?

3. ### johnny91 Thread Starter New Member

Jun 14, 2015
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Unit of power loss is Watt

The circuit is simple, no loads connected, just a battery with a copper wire

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Jul 31, 2013
196
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I think you might need to factor in 2 variables. Change of resistance of the copper wire with temperature.
Internal resistance of the battery - which in this circuit will also be a variable.
where/how will you measure "power loss" and will the measurement involve Time "t" ?
What is the capacity of the battery in Ah or Mah ?
and no, I'm not going to work it out.

5. ### johnny91 Thread Starter New Member

Jun 14, 2015
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The internal resistance and other variables are not provided, so we need to assume them as zero.

I tried solving it by finding the resistance value first [R=(resistivity*L)/A]=0.0015337ohm, then i added the R value into the power equation (P=VI), but i'm getting a huge value as current which means that i'm wrong somewhere, the power I'm getting is 1.47kW which again i think is wrong

6. ### strantor AAC Fanatic!

Oct 3, 2010
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Of course that is wrong, because you cannot assume zero internal resistance because a AA battery has relatively high internal resistance. Internal resistance of a AA is many many times higher than that of the wire.

This question is absolutely retarded. I sincerely hope it was not asked this in any coursework that you are paying for. To add to what cornishlad said, there are more factors like battery chemistry, quality, and age.

7. ### DickCappels Moderator

Aug 21, 2008
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Assuming an ideal battery you have the correct answer.

8. ### johnny91 Thread Starter New Member

Jun 14, 2015
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I don't this so, even for an ideal battery, the current cannot be 977A, that's what the value of current I'm getting when i'm substituting the value of R in the Ohm's law equation. (where P = 1.47kW)

It is indeed a retarded question, but someone asked me and I'm confused now.

9. ### johnny91 Thread Starter New Member

Jun 14, 2015
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I don't *think so..

10. ### DickCappels Moderator

Aug 21, 2008
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An ideal battery would have zero ohms internal resistance.

As strantor indicated, a real-life battery would have an internal resistance that would limit the amount of power one could obtain.

11. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Those are indeed the correct theoretical values. Which proves how nonsensical the question is. Of course, the battery would go flat in next to no time, if the wire didn't vapourise first .

12. ### DickCappels Moderator

Aug 21, 2008
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You're right...the copper wire needs to be ideal too.

13. ### ErnieM AAC Fanatic!

Apr 24, 2011
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TS asked for "power loss" without defining what power used. Does this mean how much power is removed from the battery (best answer: lots) or how much power goes to heating the wire?

Thus the question has no answer without defining terms.

14. ### wayneh Expert

Sep 9, 2010
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The answer is, "all of it". All of the power in the battery is lost due to heating the wire and the battery itself.

There are only two outcomes; the wire vaporizes very soon and becomes a fuse, or it does not. A nicad AA will melt the wire but a "normal" AA cannot.

In the latter case, the loss continues to completion. A battery rated at 1.5V and 2200mAh will provide 3300mWh of energy, all lost.

I believe the battery will give up a bit more than its ratings, since it is going to complete discharge.

15. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Since this ideal battery has zero resistance it will not dissipate power in heating the battery.