# Why does internal resistance increase as current decreases?

Our class conducted and experiment to investigate internal resistance and terminal pd, using a cell, fixed resistor and variable resistor. The results were as follows:

As the resistance of the variable resistor was increased, the current through the circuit decreased. (As using E=I(R+r), for emf E to remain constant, if R increases, current I must decrease). Note: E=emf, I=current, R=external resistance(I.e the fixed and variable resistor) and r=internal resistance.

As current decreasesd, voltage across the internal resistance of the cell, Ir, decreased (using V=IR), because internal resistance should remain constant. Therefore voltage across the terminals increased (as less is wasted across the internal resistor).

When plotting a graph of terminal pd(y) against current(x), a linear relationship was shown: as current increased terminal pd decreased.

The gradient gives the (-)internal resistance, which was constant.

However here's the problem, when calculating internal resistance using the values ( with r=(E-V)/I ), the internal resistance increased as current decreased (as external resistance increased). It did not remain constant. The individual internal resistances did not match the internal resistance calculated through the gradient, and I do not know why.

My physics teacher also has no idea why this is happening, so any help would be greatly appreciated. ![Graph ]1 • What kind of battery were you using? – garyp Apr 30 '16 at 13:40
• @garyp it was a battery pack with 4 batteries – John Apr 30 '16 at 13:53
• Did you find the emf of the battery from the intercept on the voltage axis of your graph? – Farcher Apr 30 '16 at 15:39
• @Farcher yes I did, the data is now uploaded. – John Apr 30 '16 at 17:48

## 2 Answers

UPDATE :

John : Thanks for data. Graph is ok. I note your intercept is E=3.94V but your calculations use E=4.5V. This explains the discrepancy in your results. If you use 3.94V you get r ranging from 1.59 to 1.76, close to slope value of 1.68 Ohms.

ORIGINAL ANSWER :

Your line of best fit gives an average internal resistance r based on all measurements. If data points do not lie exactly on this line then the value of r calculated for individual data points (measured pairs of V and I) will not be exactly the same as the slope of the line of best fit.

If you have drawn the line correctly some points will be above the line and some below, with about as many each side, and with the above and below points distributed randomly.

However, it sounds as though there is a consistent trend in your data points : eg all 'below' points at low current and all 'above' points at high current. This suggests that internal resistance was not in fact constant, within the limitations of experimental error. You do not say how big an effect this is : if small, you may be able to ignore it.

EMF and r should be measured when the current drawn is very small, ideally 0. Possibly you have taken readings at a high current, or you have taken a long time to take them. This can have two effects : (i) depleting the battery, reducing EMF, and (ii) increasing r because the battery is warming up and this increases internal resistance.

Your observation that internal resistance increased as current decreased suggests to me that you may have started readings with a high current then worked down to low current.

You will need to decide for yourself what went wrong, perhaps after consulting your teacher again and explaining how you took the readings.

• Our original readings did have a higher current than later readings, as this was an effect of increasing the resistance. I understand that the lobf gradient gives and average of all internal resistances, however the gradient had a value of (-)1.68ohms, whereas the individual internal resistances had values from 2.9 to 4.5 ohms with an average of approximately 3.8 ohms. Therefore the gradient is not close to the average internal resistance. Also in regards to the heating effect, if the current was decreasing throughout the experiment, then surely there would be little increase in heat? – John Apr 30 '16 at 15:45
• Thanks, John. The large and systematic discrepancy between slope and individual values suggests an error somewhere in the calculations or in plotting the graph. Even 1.68 Ohms is a bit big for 4 x AA cells (0.1-0.2 ohm each is more usual). It is difficult to spot the error without seeing your graph and the data. It is prudent to presume such discrepancies are the result of human error, rather than a real physical effect. – sammy gerbil Apr 30 '16 at 16:16
• Thanks for the quick response. I think the graph is fairly accurate (with the data given) as I also plotted it using a graphics calculator which gave the same gradient. I also forgot to mention, they were not 4 AA cells but 4 D cells. Would you like to see the data and graph, as the data shows a steady increase in internal resistance? – John Apr 30 '16 at 17:29
• Since Viakas also asks, yes it would help to see the data (V and I values), the value of E from your graph and an example of your calculation of r (2.9-4.5 ohms). Also the graph if easy to post, but not essential. – sammy gerbil Apr 30 '16 at 17:48
• I've now uploaded the data. – John Apr 30 '16 at 18:10

It is not generally realised that "internal resistance" is a myth which cam be evaluated only by practical experiment and calculation using Thevenin's theorem. But that does not mean that it is a useless concept. Indeed it is essential for design in ALL branches of engineering.

For example a plumber measures the efficincy of a water-supply by measuring the water-pessure with the delivery pipe blocked (tap turned off - the static- pressure) and then with the water freely flowing (tap fully open - the dynamic pressure). These two measurements yield (by calculation) the losses from friction between the water and the inside surface of that delivery system (i.e. the internal-resistance of that water system).

To understand the problem set out here it is essential to get to grips with the dual concepts of Thevenin's theorem (measure pressure - voltage ) and Norton's theorem (measure current - amperes). In the eletrical world these correspond to high-impedance and to low-impedance circuits. Ken Green