I'm a tutor helping a student with this project. We're having some difficulties with this project.
First, we had various problems with the meter - the first one didn't have the capability to measure current and I finally realized I had to buy another meter. It might be better if you specified what capabilities the meter should have.
However, a real problem is measuring internal resistance in the battery, and this is my first QUESTION. We're getting between .69 and .95 V for various fruits and vegetables. The directions state to use a 1Kohm resistor in series and see how much the voltage goes down. Then add another resistor. However, when the first resistor was added, the voltage went to zero. When I bought the "various resistors" listed in the materials section, I hadn't seen the 1Kohm figure, so I had bought various clay resistors from .25 ohm to 5 and 8 ohm at an old-fashioned electronics store. I then went to Radio Shack and bought 1Kohm resistors. When the first 1Kohm took the voltage to zero, we tried it with the others, including the .25 ohm, and with all of them the voltage went to zero. THE QUESTION: WHAT DOES THIS MEAN? AND HOW DO YOU COMPUTE THE INTERNAL RESISTANCE OF THE BATTERY?
Another anomaly we found was that, as we measured the voltage and tried to measure current with two and three potatoes in series and parallel, the meter suddenly started acting very strangely. Instead of showing the voltage (we couldn't get current because the meter didn't measure current), it started just going up and up and up and up. (We reversed the wires and it went down and down and down and down, so now I'm not really sure which it was doing first.) That day I gave up on it until I could talk to someone who knows more about it, and a knowledgeable friend said the potato was acting as a battery and storing electricity! If that is so, the possibility of this happening should also be included in the project description. CAN YOU GIVE US A REFERENCE WHERE WE MIGHT FIND MORE TO UNDERSTAND THIS PHENOMENON?
I am not very familiar with electricity and electronics, but after 35 years' tutoring experience, I was fairly sure I could figure it out. I assumed that a project written for 6-8th grade should be pretty fool-proof. It seems that there should be some kind of warning or disclaimer or way to look up the weird phenomena we encountered. In my opinion, although it appeared pretty straightforward, a child working independently or with the help of a novice parent could not do this experiment.
Sincerely, Sara Smith
Fruit and Veggie Power
Moderators: AmyCowen, kgudger, bfinio, MadelineB, Moderators
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sjs4BetterEd
- Posts: 1
- Joined: Sun Mar 09, 2008 5:58 am
- Occupation: Tutor
- Project Question: fruit and vegetable electricity
- Project Due Date: 3/13/08
- Project Status: I am conducting my experiment
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deleted-71827
- Former Expert
- Posts: 404
- Joined: Tue Sep 18, 2007 3:27 pm
- Occupation: Research Assistant
- Project Question: Neuroregeneration
- Project Due Date: N/A
- Project Status: Not applicable
Re: Fruit and Veggie Power
Hi Sara!
I'm not an expert on this particular topic but here are a couple of links that might help-
http://www.madsci.org/posts/archives/20 ... .Ph.r.html
http://www.sasked.gov.sk.ca/forum/scien ... /5115.html
The potato is not really storing the electricity, but rather serving as the electrolyte and helping the ions (zinc and cooper) get transported. Hope this helps!
I'm not an expert on this particular topic but here are a couple of links that might help-
http://www.madsci.org/posts/archives/20 ... .Ph.r.html
http://www.sasked.gov.sk.ca/forum/scien ... /5115.html
The potato is not really storing the electricity, but rather serving as the electrolyte and helping the ions (zinc and cooper) get transported. Hope this helps!
"There is a single light of science, and to brighten it anywhere is to brighten it everywhere." -Isaac Asimov
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deleted-71588
- Former Expert
- Posts: 1297
- Joined: Mon Oct 16, 2006 11:47 am
Re: Fruit and Veggie Power
I'm reading and trying to understanding your post correctly, you placed a 1 Kohm resistor between the two electrodes of your battery and your meter measured 0v across the resistor. Without the "1 Kohm load resistor", you got between .69 v and .95 v depending on the fruit.
Based on your meter observations, I'm going to assume that your meter is capable of measuring 0.01 volts.
If these facts and assumptions are correct, then the current flowing through the 1 Kohm resistor is less than 0.01 mA (< 0.01 V / 1 Kohm < 0.01 mA from ohms law).
Lets assume that 0.01 mA is flowing with 0.01 V across the 1 Kohm resistor attached to the 0.95 V open circuit battery, then 0.94 V is dropped across the internal resistance of that battery. By ohms law, the internal resistance would be 94 Kohms. Since we know that less than 0.01 mA is flowing because you didn't see 0.01 V across the 1 Kohm resistor, the internal resistance must be > 94 Kohms.
To be able to measure the internal resistance of this fruit battery, you are going to need a much higher known resistor value. Say 47 Kohms or 100 Kohms.
Now there is a possibility that your meter is an inexpensive 1 Kohms / volt meter that won't work for your setup. If the internal resistance of the meter is 1 to 5 Kohms, then it isn't going to work for your existing batteries even if you acquire a 47 K to 100 K resistor because the internal meter resistance will be much smaller and predominate.
Search the archives for some earlier posts of mine helping others with this project. Instead of improving your measuring equipment, you can work on improving your battery. By using larger area electrodes spaced closer together and hydrating (soaking the fruit / vegatable in water) you can lower the internal resistance of your battery so that it will work with your setup.
Based on your meter observations, I'm going to assume that your meter is capable of measuring 0.01 volts.
If these facts and assumptions are correct, then the current flowing through the 1 Kohm resistor is less than 0.01 mA (< 0.01 V / 1 Kohm < 0.01 mA from ohms law).
Lets assume that 0.01 mA is flowing with 0.01 V across the 1 Kohm resistor attached to the 0.95 V open circuit battery, then 0.94 V is dropped across the internal resistance of that battery. By ohms law, the internal resistance would be 94 Kohms. Since we know that less than 0.01 mA is flowing because you didn't see 0.01 V across the 1 Kohm resistor, the internal resistance must be > 94 Kohms.
To be able to measure the internal resistance of this fruit battery, you are going to need a much higher known resistor value. Say 47 Kohms or 100 Kohms.
Now there is a possibility that your meter is an inexpensive 1 Kohms / volt meter that won't work for your setup. If the internal resistance of the meter is 1 to 5 Kohms, then it isn't going to work for your existing batteries even if you acquire a 47 K to 100 K resistor because the internal meter resistance will be much smaller and predominate.
Search the archives for some earlier posts of mine helping others with this project. Instead of improving your measuring equipment, you can work on improving your battery. By using larger area electrodes spaced closer together and hydrating (soaking the fruit / vegatable in water) you can lower the internal resistance of your battery so that it will work with your setup.
-Craig

