Potato Battery Challenge

The filament of the bulb serves as a short circuit and the open circuit potential is too low to over come the heavy drain due the ESR of the "battery". Drawing the schematic with ESR in the leads from the battery symbol while depict this.

That one wasn't quite hard enough Jorrie. How about this one: You need to measure a battery's internal resistance. 12V battery 12AH lead-acid. You have a copper bar and a voltmeter. How do you do it?

Steve

Hi Steve, but Ron above did not answer the full question. First that and then your one!

OK, The emf potential comes from a chemical reaction between the dissimilar plates and the acid in the potato. This "battery" is only able to supply or source a few mA, so when you put an extremely low resistance as the load, the potato cannot supply the required current, so the bulb does not light and the potential reads almost zero volts. Close enough for 7:00 AM? (need more cofffeee...)

Wondefully complete explaination. Great job!!

For us EE types, it the high internal resistance of the source (the potato battery) that's the culpit. That limits the maximum current that can be drawn. So when the lightbulb is connected, the real circuit is the bulb in series with the internal resistance connected to the source voltage. So you have a voltage divider.

Just another way to say the same thing that ya'll already said.

All that plus the potato has a cell structure and an almost neutral Ph. Ion exchange and osmotic migration between the electrodes is impaired by the cell structure. The electrolyte in contact with the electrodes is quickly depleted, disconnect it and allow equilibrium to be achieved and open circuit "nine lives" occurs. It all adds up to the same thing.

OK since nobody else has bothered to draw the circuit here it is.

Any real power supply can be represented circuit wise by a perfect voltage source (the circle with the V in it) and a resistor in series with it that represents the internal resistance of the battery/poser supply.

So to answer both questions so far if we know the open circuit voltage V_O and load voltage V_L taken across T1 and T2 and the load resistance R_L then we can calculate the internal resistance using the following formula

R_I = R_L(V_O/V_L - 1)

So that answers Steve_o and the answer to Jorrie's question is that the internal resistance is where all the voltage went.

For those that are interested here is the derivation of the formula.

I = V_L/R_L …………………….….Eq 1

V_RI = V_O – V_L …………………..Eq 2

V_RI = I x R_I ……. R_I = V_IR/I ……Eq 3 Now substituting 2 gives

R_I = (V_O – V_L)/I ………..……….Eq 4 Now substituting 1 gives

R_I = (V_O – V_L) x R_L/V_L = V_OR_L/V_L – V_LR_L/V_L = R_L(V_O/V_L -1)

Nicely done Masu!

Now, with your equations in hand, can we we solve steve-o's question of post 2 above? He said: "... You need to measure a battery's internal resistance. 12V battery 12AH lead-acid. You have a copper bar and a voltmeter. How do you do it?"

Would have been a lot easier (and safer) of we had an multimeter and a more decent load than the copper bar! I suppose one can calculate the copper bar's resistance per mm from its dimensions, but still, the battery might explode in your face!

I need to think about Stev-o's problem. I have a feeling that the answer is to do with the fact that a volt meter is actually a galvanometer with a high value resistor in series with it. I havn't played with the theory behind galvanometers since 1976 so its definitely going to be a stretch.

A rough cut at the second problem, posed by steve-o, would be to measure the unloaded output voltage of the battery, then short the battery with the copper bar while measuring the battery voltage between the terminals. DON'T DO THIS WITH A CAR BATTERY!!!

This would normally be a no-no, as lead-acid battery have very low internal resistance and can produce extremely high current, a necessary characteristic for cranking a car's starter motor. It would cause dangerously high temperatures in the battery and the copper bar, high enough to rupture the battery and explose. I'm thinking the 12 amp-hour rating suggest a very small lead-acid battery, with which one might could get away with this abuse. And since this is a paper exercise, no harm done.

Anyhoo, you would then measure the cross-sectional are of the copper and distance between battery terminals so you could calculate the circular-mil-foot of the copper bar between the terminals and using the standard cir-mil-ft = 10.37 ohm, find the resistance of the copper. Remember that the cir-mils are in parallel with one another.

Now you have external resistor, external voltage, and internal voltage for the equations and you find the internal resistance.

Again, not a safe way to determine the internal resistance, but one way.

The internal resistance of the potato battery is much higher than you load (bulb).

Let see it this way!

If we make this:

1V = 100m high (like waterfall)

1 mA = 100 ltr

1 ohm = Mi hand

What i mean, You can have the same potential like 1ltr of water and 1000ltrs of water at 100 mtrs high. Correct?. But when we want to move things, like a turbine generator or simple my body in a surf table I can not doit with 1 liter fulling from 100mtrs high as 1000 ltrs at 100 mtrs high.

I have the same potential in both samples, water at 100 mtrs. But I have not the same amount of water, that makes the difference.

Going to potato batery compared to a transformer 110V:1V. Both had the same potential (1V) to supply the bulb. But the first one with not have the enough current to turn the bulb on because it has not enough current.

About why the potential falls after you connect the bulb... if falls one liter of watter from 100mtrs it will be present just in a few seconds. All the potential is gone because the 1 ltr of water change from 100mtrs high to 0mtrs high. So the potential is gone. But in case you release 1ltrs of water of the 1000 ltrs then you have 999ltrs at 100mtrs bigh to continue with the process, thus the potential is still 100mtrs.

I hope I was enough explicit. If not let me know :).

Delmar

so how the hell do you do it so it lights up

You either cheat or need to modify the battery to produce more current.

You need to increase the cross sectional area of the electrodes that are inserted into the potatoes. The voltage a battery produces is controlled by the chemistry of the electrolyte and electrodes but the current and internal resistance are controlled by the surface area of the electrodes. The larger the surface area the lower the internal resistance and the greater the current the battery can supply.

graccia tando

plz can u explan bit more about emf of a potato cell besue my sir doesnot help. ur my hop guru.