# Internal resistance of fruit and vegetable cells(Activity)

### Real cells with real resistance

What the activity is for

This activity introduces the internal resistance of a cell. You can easily point out the distance that the charged particles have to move through the cell, and even vary this distance, thereby altering the internal resistance in a natural way. You can also show that this internal resistance reduces the power provided by the cell because the potential difference falls as soon as there is current in the cell.

What the activity is for

• chosen fruit or vegetable (e.g. cucumber, citrus fruits, kiwi fruit, pineapple, potato)–cut a thin strip off the fruit to form a stable base, or use Blu–tack to stabilise it.)
• 5 copper sheet electrodes (3 centimetre by 3 cm)
• 5 zinc sheet electrodes (3 centimetre by 3 cm) (A good cell requires electrodes of large surface area, and of close separation. Alter the size of the electrodes to suit your fruit.)
• 10 crocodile clips
• a multimeter (2000 mV full scale deflection is likely to be best.)
• 4 long 4 mm leads
• several short 4 mm leads
• 1 LED (alternatives 1 of variable resistor, 1 of LED clock designed to run off fruit batteries)
• Safety note: Wear eye protection when piercing citrus fruit as fruit acid can irritate the eye. Do not eat any of the fruit at the end of the activity.

Note: an LED is a polarised component so must be connected the right way around in a circuit. The leg next to the flat section of the plastic body of the LED must be connected to the negative side of a cell. This leg is sometimes shorter than the other. Applying too much force where the legs meet the plastic casing will cause the legs to break off. Avoid this by bending the legs outwards, away from one another at the centre of each leg.

### Energy dissipated in the cell

What happens during this activity

Take the chosen fruit or vegetable and embed one zinc and one copper electrode into it so that they're approximately 0.5 centimetre apart and pushed in approximately 2 cm deep. They shouldn't be touching inside the fruit. Repeat this process along the length of the fruit so that you have up to five cells, each separated by a few centimetres (the metals should alternate along the fruit).

Attach each cell in series to the next using crocodile clips and leads. (The two electrodes comprising each cell should not be connected by a lead.) The crocodile clips should not be touching one another.

Now get the multimeter and set it to read millivolt (on the 2000 milliVolt scale). Attach the positive lead to the outermost copper and the negative lead to the outermost zinc. You should say that this is a simple battery of cells. It has electrodes of different metals in acid (here we have fruit acids). Show the class that there is a potential difference (voltage reading) between the two outermost electrodes. Write this battery potential on the board (e.g. 1000 mV).

Remove the multimeter and connect the two outermost leads to the LED. Make sure that the positive terminal of the LED is connected to the copper.

Does it light? (Yes.)

Now, with the LED in place, measure the potential difference across the LED. Write it on the board (e.g. 800 mV).

You should emphasise that the readings are different and ask, Why is the potential difference measured across the battery different from that measured across the LED? That's not what the simple circuit theory we've studied so far predicts. Something new is going on.

Now say that to answer this question we must think carefully about potential difference and electric current. Ask, How can we picture an electric current? (A flow of charged particles.. Very good!) Explain that, if the potential difference measured across the cell was 1000 millivolt (or 1 V) then each coulomb of charge would be able to shift 1 J of energy.

Explain that if the potential difference measured across the LED is only 800 milliVolt (or 0.8 V) then this means that each coulomb of charge is shifting 0.8 joule of energy when flowing through the LED. Where is the other 0.2 J of energy being shifted to if not to the LED?

The answer is that energy is also being dissipated by the cell itself. Between the two electrode plates there is some fruit matter. This also contains charged particles, which are flowing between the two electrodes (hydrogen ions (H+) from the fruit acid flow towards the negative electrode). The missing 0.2 joule of energy per coulomb is being shifted by charged particles as they flow through the matter of the cell. The cell has its own resistance to the flow of charged particles. We call this an internal resistance.

The cell will rise in temperature due to this energy shifting. Remind students that they may have noticed batteries that they use at home also get hot. Here the current is not very large, so if we stick a thermometer into the fruit we'll be disappointed.

For the next part we suggest a long fruit (e.g. cucumber), and just a cell, with two terminals only. The LED will not glow, but what's happening is otherwise much clearer.

Now look at the effect of increasing the amount of matter between the electrodes. Ask, What do you think will happen to the internal resistance of each cell? (It will increase as there is more matter for the charged particles to flow through.) Ask, Do you think the charge flowing in the battery will shift more or less energy as it flows through more matter? (More.) Ask: So will there be more or less energy shifted to the LED? (Less.) Ask, So will the LED be brighter or dimmer? (Dimmer.)

Move the electrodes so that each pair is about 1 centimetre apart. Measure the potential difference. Repeat with the electrodes about 5 cm apart to get the same potential difference. That's not surprising because there is no current, so no energy is shifted inside the cell. Now add the LED. Measure the potential differences again. Note that both drop, but the reading corresponding to the largest separation of plates drops most because the internal resistance is highest.