Icebreak,
The equations I think you're looking for are:
R = V / I (R in ohms, V in volts, I in amps)
V = I * R
W = V * I = I^2 * R = V^2 / R
And if you're going to calculate the power (watts) consummed by a bulb under various configurations, you need to use true values for voltage and current in the equations. True voltage is not the nominal, advertised voltage of either the battery or the bulb. Wilkey ably discusses the sag in a battery's voltage as current is increased. This is a property of the cell. How it "sags" is different for each battery chemistry. At Wilkey's invitation, let me amplify on that.
I have discussed Li-Ion properties in the
Pila current & voltage question thread. Below, I have pasted and modified that text to show the properties of NiMHs. It describes my findings for my PowerEx 2200 mAh AA NiMHs. You can probably extend the results to NiMH cells with similar capacities
I ran a battery of a few AA NiMH cells into a load at about 2.5 Amps. I continued the test for 42 minutes, at which time the voltage per cell had dropped to 0.97 volts under a load of 2.34 Amps.
Open Circuit Voltage (V.oc). I stopped the current drain at the one-minute mark and every 10-minute mark to test the no-load voltage.
t = 00 . . . . 1.43
t = 01 . . . . 1.37
t = 10 . . . . 1.28
t = 20 . . . . 1.27
t = 30 . . . . 1.25
t = 40 . . . . 1.20
t = 42 . . . . test ended
Voltage Under Load (V.cell & V.bat). During those pauses in the test above, at each mark (including the t = 0 mark), I tested the battery under varying load-- measuring its voltage and current simultaneously. I measured voltage under loads of approximately 0.0, 0,5, 1.0, and 2.5 Amps. From these data I determined that the voltage sag from the open circuit voltage is:
<ul type="square">[*]directly proportional to the current
[*]independent (within the accuracy of my measurements) of cell freshness or depletion[/list]
For these reasons, the voltage under load could reasonably be described by a model of the battery as a voltage source equal to the open circuit voltage in series with a constant resistance of 0.070 ohms. The under-load voltage of the AA cell can be given by the following:
V.cell = V.oc - 0.070 * I, where I is the current.
To get the total battery voltage, just multiply by the number of cells:
V.bat = #(cells) * { V.oc - 0.070 * I }
Bulb Voltage (V.bulb). It is possible to also derive an equation for the WA 01185 bulb, as I did in the
500 Lumen Mag 3D thread. This reference also describes the characteristics of the 123 cell.
The equation for one of the 1185 bulbs I have is
V.bulb = 5.3 * I.bulb – 6.95 (for current between 2.5 amps & 3.5 amps)
If you're into solving simultaneous equations, or if you're into charting (which is more fun), you can determine exactly the current and voltage under which the WA 01185 will operate. Plot the equation for V.bat. The specific equation will depend on V.oc, as given by cell depletion in the chart above. Then plot V.bulb. Where they intersect is where the bulb will operate. If you plot a series of equations for V.bat (for a series of depletion levels), you'll get a series of intersections, which show how the voltage and current drop with time.
While these equations work for me, and probably will work for you, I have to admit that there is some error in using them. This is because during my the tests there were other resistances in the circuit. The cells were tested within the Elektrolumens 3-to-D holders. I had modified them so that the resistance of each holder itself was 0.048 ohms for 3 cells. That would account for 0.016 ohms per cell. In addition, there was contact resistance between the cell and holder, and between the holders and the clip leads attached to them. These were the only high-current connections. I am guessing (and the guess is a little on the wild side) that the stray resistance in a flashlight would be about the same. This is my way of saying that I will continue to use an internal resistance of 0.07 ohms in my calculating gyrations.
I mention all of this because Wilkey has stated that his NiMHs have an internal resistance of 0.006 ohms. I'm hoping that's a typo. I have read that AA NiMH resistances are on the order of 0.020 to 0.060 ohms. I trust Wilkey will comment.
Icebreak, I hope this has been of help and not just a confusing waste of your reading time. /ubbthreads/images/graemlins/smile.gif
Paul