## Veggie Power (Fruit Battery)

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### Veggie Power (Fruit Battery)

Hello,

I am in 5th grade and doing a Veggie Power project for the science fair. My question is
"What produce requires the least amount (in mass) to make a battery that lasts the longest?".
His hypothesis is that the citrus produce will make the best battery. I am using 2.6V LED to light
up using this battery.

When I set up the trials using Lemons, Oranges, Potatoes and apples, I measured both open and closed circuit
voltages for 1 cell, 2 cells etc using digital multimeter.. I am not using any external resistor and also the LED is not having any resistance. The open and close circuit voltages were same when only 1 cell is there in the circuit. But when a 2nd cell is added to the circuit the open and close circuit voltages are coming out different. The LED was able to light up with 2 cells battery. Also the current was starting out with one number and continuously going down. It was not stabilizing even after waiting for few minutes.. I would like to use closed circuit voltage and current to calculate power.

Could you please help explain why the open and close circuit voltages are different with 2 cells battery and current is being varied all the time?

thanks
Rahil
arpitajmodi

Posts: 4
Joined: Fri Jan 21, 2011 9:19 am
Occupation: Software Engineer
Project Question: Why does refrigerator cooling(27 deg F) of saturated solution produces crystals faster than the room temperature (68 deg F) cooling or cooling on a heating pad (100 deg F)? Also thought that cooling on heating pad would produce crystals slower than room temperature cooling but it created crystals faster than the room temperature cooling. Why is that so?
Project Due Date: Feb 1st 2011
Project Status: Not applicable

### Re: Veggie Power (Fruit Battery)

First, I think an LED probably draws too much current to serve as a test device in this experiment. If you re-read the "Veggie Power" write-up, you will see that it is recommended that you "look for things that draw less than 0.5 mA" for your loads. A typical LED draws about 20 mA when properly connected, where by properly connected I mean with a series resistor that limits the current to that value. If you increase the current through the LED it will shine more brightly, then at some maximum allowed current, burn out. Also, LEDs vary in their current needs: a low-power LED might need only 2 mA, while a high-brightness LED might consume 20 mA or more.

Second, you need to become familiar with the concept of "internal resistance" of the battery. An "ideal" battery would produce a constant voltage no matter what the external resistance was. You can see that such an ideal battery cannot really be made, since if you put a very low resistance, such as a "short", across its terminals, a very large current would flow. Real batteries can be approximated as an ideal battery with an internal series resistor. The size of this resistance will vary according to the detailed construction of the battery; it will be low for batteries designed to provide high currents, for example a car battery, and higher for devices intended to drive low-power loads, for example the kind of 9 V battery used by most smoke detectors. I would GUESS the internal resistance of your veggie battery cell would be somewhere between 100 and 1000 ohms, but I could well be wrong. There are some references to internal resistance in the project write-up.

Because of the internal resistance of your veggie battery, as you draw more current out of the battery the voltage drop across the internal resistance will increase, with the result that the voltage available at the terminals of the cell wil decrease. If, as an experiment, you measure the battery voltage at two different current draws, you can deduce the magnitude of the internal resistance. (Note: you cannot simply measure this resistance using the ohm meter function of your multimeter.)

Third, another feature of real batteries versus ideal voltage sources is that real batteries will discharge as current is withdrawn from them. If you look at the specification for a commercial battery (do a Google search on the name and type of the battery to find its specifications) you will see it rated in "amp hours", that is how many amps it can deliver times the number of hours it can deliver it. Batteries provide energy when they supply current; the rate of energy supplied is the power, given by current x voltage. Energy is conserved, so the energy supplied must come from somewhere. For a battery it is the energy stored in latent form by chemical potential energy within the battery. As the battery is used, chemical reactions occur in the internal chemicals that convert a mixture of chemicals from one with higher chemical potential energy to one with lower potential energy. As the "active" chemicals are depleted by the reactions the internal available energy of the battery decreases; eventually the active chemicals no longer exist in enough concentration to provide useful power. The external sign of this is a decrease in the open circuit voltage after the battery has been used for a while, often accompanied by an increase in the internal resistance of the battery. (Note: these chemical reactions tend to occur slowly even if the battery is not being used; hence batteries have a limited "shelf life".)

The voltage and internal resistance of a battery cell also depends on the temperature of the cell, since chemical reactions are temperature sensitive. And if you draw a lot of current through the battery, the internal resistance will develop power (P=I^2 R) which will tend to heat up the battery. In order to minimize this heating effect it may be wise to make measurements of things like voltage as a function of lead resistance as fast as possible. A water-ice bath could also help, being careful not to short the system with water, which is a conductor (distilled water, available at supermarkets, conducts electricity much, much more poorly than tap water).

John Dreher
Expert

Posts: 294
Joined: Sun Dec 25, 2011 8:33 am
Occupation: Astronomer, Professor of Physics, SETI Researcher (retired)
Project Question: n/a
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Project Status: Not applicable

### Re: Veggie Power (Fruit Battery)

John,
Thanks for the detailed information. Really helped me understand things better now.

As I mentioned earlier, the current was kept on changing with 1 cell or 2 cells veggie battery. I measured the closed circuit current (in micro amps) with 1 cell and 2 cells. It would always start with one number like 282 and then keep going down constantly and wasn't getting stabled at one number. So should I use that starting number of current reading to calculate power using the closed circuit voltage?

Also my battery cells are connected in series so I was thinking that the current would stay same regardless the number of cells. Can you help understand this?

thanks
Rahil
arpitajmodi

Posts: 4
Joined: Fri Jan 21, 2011 9:19 am
Occupation: Software Engineer
Project Question: Why does refrigerator cooling(27 deg F) of saturated solution produces crystals faster than the room temperature (68 deg F) cooling or cooling on a heating pad (100 deg F)? Also thought that cooling on heating pad would produce crystals slower than room temperature cooling but it created crystals faster than the room temperature cooling. Why is that so?
Project Due Date: Feb 1st 2011
Project Status: Not applicable

### Re: Veggie Power (Fruit Battery)

My deepest apologies for not answering your last post. I am new to Science Buddies and did not understand the function that tracks my own threads. I've fixed that now. I see that your due date has passed -- I hope your project went well despite my lapse. Again, I'm so sorry that it's too late to help. Just to satisfy your curiosity, the reason the voltage kept decreasing is that the excessive current draw was discharging the batteries. For a commercial battery, once a cell has discharged it won't recover, but it's possible that a veggie battery might slowly regain part of it's capacity after a rest, but I suspect eventually it too will be dead. Your other question was regarding series versus parallel configuration of the batteries. In series both the voltages and the internal resistance will add, while in parallel the voltage will stay the same but the effective internal resistance will be reduced (in the case of two cells in parallel the net resistance will be halved). For an application needing the best current the parallel configuration is better than the series.
John Dreher
Expert

Posts: 294
Joined: Sun Dec 25, 2011 8:33 am
Occupation: Astronomer, Professor of Physics, SETI Researcher (retired)
Project Question: n/a
Project Due Date: n/a
Project Status: Not applicable