[Lab] Odd Question
Michael
krazatchu at hotmail.com
Fri Jan 20 15:00:29 EST 2012
This is a interesting challenge..
The supply droop with resistive load will be a factor of the batteries
internal resistance.
Which is an effect of temperature, chemistry and state of charge.
Internal resistance can be measured by subjecting the battery to a known
load ...
http://en.wikipedia.org/wiki/Internal_resistance
Then applying (Nominal V - Loaded V) * (Load R) / (Nominal V)
Or you can just figure it's about 0.2 Ohms per Alkaline cell.
A 9V is sometimes made of 6 cells, so about 1.2 Ohms.
Once you have the internal resistance figured...
The circuit is a 9v voltage source with two series resistances, internal
+ load.
The voltage drop over the internal resistance will be 9 - 6 = 3v and
using i = V/R we get the current over the internal resitance...
i = 3/1.2 = 2.5 Amps (there's going to be some serious self heating
which will skew the results)
As the circuit is series, the current thru the internal resistance is
the same as the current thru the load...
And the final voltage drop over the unknown load is the remainder, 6
volts. Using R = V/i...
R = 6 / 2.5 = 2.4 Ohms ...
Finally heat is a factor of power dissipation...
Power is V*i = 6 * 2.5 = 15 Watts...
I just tested this with a Rayovac 9v and a 2.7 ohm load...
The results were a bit off, the internal resitance of the rayovac must
be higher than 1.2 ohms
I got a supply drop to 3.4 volts and a 1.15 amps current.
Using the above calculation, the rayovac has an internal resistance of
1.73 ohms..
So we could probably guess between 10 to 12 watts of power lost into load...
Temperature rise is a factor of thermal conductivity and geometry...
I'd guess that the sugar cell has a pretty low thermal conductivity and
that 10 watts could heat a sugar cell to over 100c given enough time..
Combustion is going to require an oxidizer so YMMV....
Michael
http://n0m1.com/
http://krazatchu.ca/
On 1/20/2012 1:49 PM, Darcy Whyte wrote:
>
> I suppose if you can try it on a couple and see what the current is,
> then you can know how many Watts. That would be a start. Then you just
> have to figure out how much of the heat is in the load and how much is
> in the battery. You can probably do that by knowing the resistance of
> the load.
>
>
>
>
> On Fri, Jan 20, 2012 at 1:33 PM, Emily Daniels
> <emily.daniels at gmail.com <mailto:emily.daniels at gmail.com>> wrote:
>
> Hi All,
>
> Does anyone know how much heat is produced by a 9 volt battery
> when the resistance of the material it flows through reduces it to
> 6 volts? What would the temperature be inside the material? Thanks!
>
> Emily
>
> --
> Emily Daniels | emilydaniels.com <http://emilydaniels.com> |
> @emdaniels | awesomefoundation.org <http://awesomefoundation.org>
>
>
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