Science Project Battery Test

[tombat]

My third grade son has to do a science project so I suggested he test the capacity of the four standard size alkaline batteries - AAA, AA, C, and D. Seems simple enough for a 3rd grader to do and understand.

We bought a battery holder for each battery size and crimped a spade terminal on each wire. We did the same with a bulb socket. This allows us to swap out battery holders and use the same bulb socket. Simple circuit - works fine.

The problem is with the choice of bulbs. I knew it would take a while for the battery to die, but I didn't expect the bulb to burn out before the battery died. I'm using a 1.2V, .22 amp, .26 watt bulb - looks like 112/E10 bulb.

What I need is a bulb that will suck the life out of the batteries fairly quickly and have a relatively long life. I'd like to run the test twice for each battery size. Cost and availability are also considerations.

Any suggestions? My back-up plan is an electric motor with a load applied.

Thanks
Tom

 

[Norm]

How about using your existing lamp and wiring a suitable resistor in parallel to increase the load.
Norm

 

[tombat]

How about using your existing lamp and wiring a suitable resistor in parallel to increase the load.
Norm

I guess I'm not smart enough to figure out a suitable size resistor. I didn't do real well in that EE class I had to take. Any suggestions on how to calculate a suitable size resistor?

 

[AnAppleSnail]

Did the 1.2v bulb burn out on one Alkaline cell? If so, I believe that two bulbs in parallel would work well, splitting the current available and draining the batteries faster.

A light bulb is a resistor. If you use a multimeter to measure its resistance (while warm) then you know its resistance. The light bulb and an equal resistor wired side-by-side (parallel) will see equal current. A smaller resistor would get more current than the larger one. The exact equation for the current each gets is a two-step process:

V = voltage
I = current (amps)
R = Resistor
R1 = resistor #1

1 / [(1/R1) + (1/R2)] = "Equivalent Resistance." This is the resistance of resistors in parallel.

Then, V=IR. V of two resistors in parallel is the same, so the I will be adjusted to match R. So if my light bulb is 1 ohm, and I put ten 1-ohm resistors in parallel with it, the battery will be drained about ten times as fast. If I put a lower-resistance resistor in, the bulb will be dimmer (less current, so less heat).

Where did you source bulbs? Mag Lite's 1-cell "Mag Solitaire" bulb costs a few bucks, is tiny and fairly bright, and should survive a D cell alkaline.

 

[tombat]

Yes, the 1.2V bulb burned out with a single C battery in less than 4 hours. I suspect infant mortality on this bulb because I have two other bulbs that each have about 10 hours them. I read somewhere that the bulbs should be good for about 10 hours. I bought the bulbs at Radio Shack.

The more I look into this, the more I think this was a poorly thought out plan. Wikipedia lists the typical capacity of alkaline batteries as:
AAA - 1200 mAh
AA - 2700 mAh
C - 8000 mAh
D - 12000 mAh

For some reason I thought they maxed out about 3000mAh.

The bulb I'm using is rated at .22 amps or 220 mA. If I run the numbers for runtime, I get:
AAA = 1200/220 = 5.45 hours (first test with a AAA gave 4.2 hours so this is consistent)
AA = 2700/220 = 12.3 hours (running right now)
C = 8000/220 = 36.4 hours ( had one running for about 6 hours)
D = 12000/220 = 54.5 hours

Do these numbers make sense? Am I calculating correctly? If so, I really need to speed up this process. Based on the suggestion by Norm and the explanation by AnAppleSnail, looks like a resistor may be the way to go.

I measured the resistance of the bulb and came up with about 2.8 ohms. So doubling the resistance in the circuit should half the runtime. Still not fast enough. While I was at Radio Shack today buying more bulbs, I picked up some 1/4 watt resistors. I picked 1/4 watt because the bulb is rated at .26 watts so it seemed to make sense to me. I bought some 10, 100, 1K, and 10K to try to cover the waterfront.

So using the equivalent resistance formula using a 10 ohm resistor:
1 / [(1/2.8) + (1/10)] = 2.2
so then I = 1.5/2.2 = .68
in the original circuit I = 1.5/2.8 = 0.54
So I have a little faster battery drain

If I use all 6 of the 10 ohm resistors:
1 / [(1/2.8) + (1/10) + (1/10) + (1/10) + (1/10) + (1/10) + (1/10)] = 1
so then I = 1.5/1 = 1.5
Or about 3 times faster battery drain.

If I had a 3 ohm resistor:
1 / [(1/2.8) + (1/3)] = 1.5
so then I = 1.5/1.5 = 1
Or about 2 times faster battery drain.

If I had two 3 ohm resistors:
1 / [(1/2.8) + (1/3) + (1/3)] = 1
so then I = 1.5/1 = 1.5
Or about 3 times faster battery drain.

Am I doing this right? Based on this formula, it seems the key to faster battery drain is to have equally sized resistors. So if I really wanted to speed this process up by a factor of 10, I would need to put ten 3 ohm resistors in parallel, as suggested above, right? This really is a more complex circuit than I wanted for a third grader.

Couple questions:
If I decide to go this route, I assume I can twist the ends of the resistors together and blob some solder on to connect them all. I would still have equal current flow through each resistor - correct?
Is there a reason I wouldn't want to connect a single resistor in series?

Thanks for all the input.
Tom

 

[Gregozedobe]

Another possibility is a higher wattage halogen globe (if you have a suitable low voltage one) - this will use more current and last longer than your typical (cheapo) incandescent globe.

 

[Norm]

More ideas here.

I'd wire a battery box with a push button switch, a change over switch (SPDT) with a buzzer on one side of the switch and a lamp on the other. Nothing too hard for the third graders to understand.

Norm