Lemon Battery Voltage: Why More Resistance?

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SUMMARY

The discussion centers on the performance of a lemon battery, specifically its voltage and current output when powering a 5mm red LED. Users noted that while the lemon battery produced approximately 3.7V, the LED was dim compared to when powered by a standard power supply at the same voltage. This discrepancy is attributed to the higher internal resistance of the lemon battery, which limits its current output. Key factors affecting internal resistance include the size and design of the electrodes and the electrolyte's efficiency.

PREREQUISITES
  • Understanding of Ohm's Law (R = V/I)
  • Knowledge of internal resistance in batteries
  • Familiarity with voltage drop measurements in circuits
  • Basic concepts of electrochemistry and battery design
NEXT STEPS
  • Research the effects of internal resistance on battery performance
  • Learn about different types of electrolytes and their impact on energy loss
  • Explore the design principles of lithium-ion batteries and their advantages
  • Study methods for measuring voltage drop across loads in electrical circuits
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Electronics enthusiasts, students studying electrochemistry, hobbyists experimenting with homemade batteries, and anyone interested in understanding battery performance and circuit design.

Alex Hughes
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So I was messing around and made a lemon battery using a zinc nail and a copper penny. I wired 4 of these lemons together and connected it to a 5mm red LED and got around 3.7V. It came on but was dim. However, when I wire my power supply and also set it to 3.7V, it is much brighter. Now i assume this has to be because the lemon battery has much less current. I'm just curious as to why this is. If it has less current at the same voltage, this would have to mean by lemon battery circuit had more resistance correct? Where does this extra resistance come from?
 
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Alex Hughes said:
So I was messing around and made a lemon battery using a zinc nail and a copper penny. I wired 4 of these lemons together and connected it to a 5mm red LED and got around 3.7V. It came on but was dim. However, when I wire my power supply and also set it to 3.7V, it is much brighter. Now i assume this has to be because the lemon battery has much less current. I'm just curious as to why this is. If it has less current at the same voltage, this would have to mean by lemon battery circuit had more resistance correct? Where does this extra resistance come from?
When you have the lemon batteries hooked to the LED, do you read 3.7v across the LED?
 
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phinds said:
When you have the lemon batteries hooked to the LED, do you read 3.7v across the LED?
Uh actually I'm not sure, I only measured the terminals and I've disabled the lemon battery since then.
 
Alex Hughes said:
If it has less current at the same voltage, this would have to mean by lemon battery circuit had more resistance correct? Where does this extra resistance come from?

the internal resistance of all the lemons (4) and all the extra electrodes

you also need to look at from the point of view that the lemon battery is just incapable of supplying the current required

phinds asked you this Q for a good reason...

phinds said:
When you have the lemon batteries hooked to the LED, do you read 3.7v across the LED?
He wanted to see if you noted a voltage difference when there was no load on the lemon battery compared to when there was a load ( the LED)
When a power supply ( regardless of what type of supply) cannot supply enough current, there will ALWAYS be a drop in supply voltageDave
 
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It is not just about an open circuit voltage, but also about the ohmic drop in the battery. The larger the internal resistance of the battery, the larger the drop and the lower the electromotive force (which you can think of as a "practical" voltage delivered by the battery under load).
 
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Borek said:
It is not just about an open circuit voltage, but also about the ohmic drop in the battery. The larger the internal resistance of the battery,

agree totally
yup, hence why I commented on that :smile:
 
davenn said:
agree totally
yup, hence why I commented on that :smile:

Yes, to some extent I wrote the same thing.
 
That makes sense. So you're both saying I should measure the voltage drop on my LED and not just the open circuit voltage of my lemons. Just curious, what causes a larger internal resistance in a battery?
 
Alex Hughes said:
That makes sense. So you're both saying I should measure the voltage drop on my LED and not just the open circuit voltage of my lemons. Just curious, what causes a larger internal resistance in a battery?
Alex,
Alex Hughes said:
So if making a battery bigger only results in more energy being stored and supplying a certain current for longer, how would you output more current?

Guineafowl said:
The maximum current a cell can supply is limited by its internal resistance, which a function of size and design, surface area and thickness of plates, etc.

On a car battery, this is indicated by the CCA or CA rating ((cold) cranking amps). Generally, a bigger or heavier-duty battery can supply a bigger current, as its internal resistance is lower by design and heft.

I think you need to slow down a little on the questions and take time to digest the answers...
 
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Alex Hughes said:
That makes sense. So you're both saying I should measure the voltage drop on my LED and not just the open circuit voltage of my lemons. Just curious, what causes a larger internal resistance in a battery?
You might find it easier if you realize that 'Resistance' is not a quantity in itself; it just represents the amount of Energy used in moving Charge through an object. This is where R = V/I comes from; the V and the I are actual quantities and R is just the ratio. It's strictly only in metals that R is the same for all values of current flowing. (Ohm's Law)
There is alway a loss of energy where the ions are formed, transported and deposited in an electrolyte. A 'good' electrolyte / electrode combination will have less energy loss. The area of the electrodes will also be very relevant. If you use a large area of plates, the internal resistance will be less.
One good feature of Lithium Ion batteries is that the Lithium ions are small and very mobile.
 
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  • #11
sophiecentaur said:
You might find it easier if you realize that 'Resistance' is not a quantity in itself; it just represents the amount of Energy used in moving Charge through an object. This is where R = V/I comes from; the V and the I are actual quantities and R is just the ratio. It's strictly only in metals that R is the same for all values of current flowing. (Ohm's Law)
There is alway a loss of energy where the ions are formed, transported and deposited in an electrolyte. A 'good' electrolyte / electrode combination will have less energy loss. The area of the electrodes will also be very relevant. If you use a large area of plates, the internal resistance will be less.
One good feature of Lithium Ion batteries is that the Lithium ions are small and very mobile.
Thank you that helped a lot.
 
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Guineafowl said:
Alex,
I think you need to slow down a little on the questions and take time to digest the answers...
Thanks. Sorry I'll keep that in mind. Forgot you answered that in one of my questions earlier.
 
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  • #13
Alex Hughes said:
Thanks. Sorry I'll keep that in mind. Forgot you answered that in one of my questions earlier.
Your curiosity far will get you...

These are difficult concepts, and the more you delve, the more complex a seemingly simple system becomes. As @sophiecentaur points out, don’t worry too much about the specifics of what the electrons are doing. You can build and fix complex circuits knowingly only the concepts of volts, amps, ohms, etc. But you need to know them well, and from every angle.
 
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  • #14
Guineafowl said:
But you need to know them well, and from every angle.
. . . . and have e confidence in the results of your calculations, using all the standard equations etc.. That is where true 'understanding' comes from.
 

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