Finding internal resistance of a cell/battery

[wingnut]

Hi all. I am a very frustrated science teacher hoping someone here can help me with my homework

Every year my Gr 12's have to find the internal resistance of a cell using Emf=IR + Ir where R is the external circuit resistance and r is internal resistance of the cell .
I know very easy ways to find r but from this year the powers that be want us to do it with a battery with a voltmeter in parallel across it, and with a switch, variable resistor and ammeter in series completing the circuit.

Learners have to change the potentiometer and for various different values of R they get I and V readings. These they combine into a graph.

Now where is what I don't understand. The slope of the graph is supposed to equal to r. But the slope is negative, whereas r is positive. Here is what I wrote to a fellow teacher...

"Why should V/I = r since we are NOT applying Ohm's law by keeping R constant and changing V and measuring I.
Also the slope of the graph is NEGATIVE and r is not negative.

I want to tell the powers that be that their mandatory prac is shait.

If you could look over the prac and memo - I feel this is the least useful way to find r and it gives no reason why slope = r.

I know that Emf=IR + Ir or Emf/I=R + r so Emf/I -R = r

and R=0 when I = max current. V = max voltage (Emf) when I = 0

So it happens that Emf/I = r but that is not the same as slope because these are specific values that only work when R=0 or external voltage (terminal potential difference) = 0."

I have attached the practical file and the teachers memorandum, which only states that slope = r.

Science teachers always have questions (or at least I always have had) and luckily this question is on electricity, so when I wondered where on earth I could find help, I remembered my old friends at electronics point.

 

[Alec_t]

In the first pdf's circuit, is R supposed to be the battery's internal resistance or an external resistor? If the former, then the circuit is incorrectly drawn because R would have to be between the two terminals of the voltmeter. If the latter, then there are three resistance values to take into account: R, r, and the rheostat.
If the voltmeter is assumed to have infinite resistance the problem should state this, otherwise that would be another resistance to account for.
Whoever set the problem should go back to school.

 

[WHONOES]

Lets carry out a bit of a thought experiment which may be repeating what you already know.

Think of the circuit as two resistors. One of which is the load and the other is the battery's internal impedance. We shall designate the internal impedance as "Z" and the load resistance as "R".

In the circuit. the two resistors are in series.

Imagine them in a vertical arrangement with "R" at the bottom "Z" at the top. with the battery emf in parallel with them.

Now, connect your voltmeter across "R" so that the positive terminal is at the junction of "Z" and "R".

Next, think or "R" as being open circuit. Your Volt meter now will be measuring the battery's open circuit voltage. Make a note of this. and call it V1.

Now when "R" is back in circuit, the voltage at the junction of "Z" and "R" will drop. Make a note of this new voltage and call it V2.

Now. subtract V2 from V1. This gives you emf across "Z".

Now assuming that you have measured the load current you will know the load current through "Z", ohm's law will tell what the value of "Z" is.

The value of "Z" will not be negative but will likely be less than 1R.

I will leave you to do the sums.

 

[wingnut]

In the first pdf's circuit, is R supposed to be the battery's internal resistance or an external resistor? If the former, then the circuit is incorrectly drawn because R would have to be between the two terminals of the voltmeter. If the latter, then there are three resistance values to take into account: R, r, and the rheostat.
If the voltmeter is assumed to have infinite resistance the problem should state this, otherwise that would be another resistance to account for.
Whoever set the problem should go back to school.

I agree with you that the circuit is incorrectly drawn. There is only one external resistance viz. the rheostat. Voltmeters are assumed to have infinite resistance.

My problem is that I cannot figure out from any formula why the slope of the graph should equal internal resistance. The slope of the graph is V/I. But this is not an Ohm's law situation where R is constant. R is varying all the time. So if slope of the graph equalled any resistance, why should it equal that particular resistance which is internal resistance. Also the slope of the graph reflects an inverse relationship (does not go through the origin) so is a negative slope and internal resistance is a positive value, in Ohms.

 

[wingnut]

Lets carry out a bit of a thought experiment which may be repeating what you already know.

Think of the circuit as two resistors. One of which is the load and the other is the battery's internal impedance. We shall designate the internal impedance as "Z" and the load resistance as "R".

In the circuit. the two resistors are in series.

Imagine them in a vertical arrangement with "R" at the bottom "Z" at the top. with the battery emf in parallel with them.

Now, connect your voltmeter across "R" so that the positive terminal is at the junction of "Z" and "R".

Next, think or "R" as being open circuit. Your Volt meter now will be measuring the battery's open circuit voltage. Make a note of this. and call it V1.

Now when "R" is back in circuit, the voltage at the junction of "Z" and "R" will drop. Make a note of this new voltage and call it V2.

Now. subtract V2 from V1. This gives you emf across "Z".

Now assuming that you have measured the load current you will know the load current through "Z", ohm's law will tell what the value of "Z" is.

The value of "Z" will not be negative but will likely be less than 1R.

I will leave you to do the sums.

I get what you say, but why would taking multiple different readings and plotting a graph of the V/I values give us Z?

 

[The Electrician]

I agree with you that the circuit is incorrectly drawn. There is only one external resistance viz. the rheostat. Voltmeters are assumed to have infinite resistance.

My problem is that I cannot figure out from any formula why the slope of the graph should equal internal resistance. The slope of the graph is V/I. But this is not an Ohm's law situation where R is constant. R is varying all the time. So if slope of the graph equalled any resistance, why should it equal that particular resistance which is internal resistance. Also the slope of the graph reflects an inverse relationship (does not go through the origin) so is a negative slope and internal resistance is a positive value, in Ohms.

It's fairly well explained on page 23 of the Teacher's guide.

 

[Harald Kapp]

Trying to fix this mathematically:
This is the circuit:

V0 = EMF of the battery
r = internal resistance of the battery
R = external load resistance
V = measured voltage across the terminals of the battery
I measured current through the load.

Clearly we have
V = V0 - I × r
The graph to be drawn and analyzed is V versus I. Therefore the slope of the graph is dV/dI.:
dV/dI = d(V0 - I × r)/dI = -r

Your observation is therefore 100 % correct: The slope of the graph does not indicate r but the negative value -r!

If you were to plot V0 - V over I instead (after measuring V0 initially without load), then the we had:
V0 - V = V0 - (V0 - I × r) = I × r
The slope of this graph is d(V0 - V)/dI = r which is the positive value of the internal resistance. But obviously this is not what was asked for in the task description

 

[wingnut]

Trying to fix this mathematically:
This is the circuit:
View attachment 47440
V0 = EMF of the battery
r = internal resistance of the battery
R = external load resistance
V = measured voltage across the terminals of the battery
I measured current through the load.

Clearly we have
V = V0 - I × r
The graph to be drawn and analyzed is V versus I. Therefore the slope of the graph is dV/dI.:
dV/dI = d(V0-I × r)/dI = -r

Your observation is therefore 100 % correct: The slope of the graph doe not indicate r but the negative value -t!

If you were to plot V0-V over I instead (after measuring V0 initially without load), then the we had:
V0 - V = V0 - (V0 - I × r) = I × r
The slope of this graph is d(V0 - V)/dI = r which is the positive value of the internal resistance. But obviously this is not what was asked for in the task description

Thanks so much for this. Now it makes more sense. Just reading the teachers explanation and memo answer they say "The internal resistance can be obtained from the gradient of a potential difference versus current graph." Maybe, on reflection they are not saying that internal resistance IS the slope of the graph, but can be obtained FROM the gradient.

My biggest problem with the powers that be is - why do they have to choose the most mathematical way of trying to find r? In our country we have the problem that science teaching is in danger of being captured by mathematics teaching.

My way of finding r (speaking as someone who loves science) would be the quickest and easiest way of doing it practically. What I do is first attach just a voltmeter across the battery and read emf (V0) usually 9V. Then I attach an ammeter across the battery to create a short circuit and read the maximum current, usually 4 amps. Using emf=IR +Ir where R=0 we have emf=Ir or 9=4xr. Therefore r=9/4 or 2.25 ohms. Done, dusted and captures all the essentials of science which is to solve problems as easily and practically as possible. Of course I would not do this with a car battery . With a car battery I would include one external resistor of say 20 ohms and again use emf=IR + Ir where 12=2x20 + 2xr if current were measured at 2 amps. Again just takes a few minutes.

Thanks again to you Harald and all for the help.

 

[Harald Kapp]

In our country we have the problem that science teaching is in danger of being captured by mathematics teaching.

Science and math are inextricably intertwined.
When I studied, I learned most of the math in the engineering classes long before the subjects were treated in the math class.

Using emf=IR +Ir where R=0 we have emf=Ir or 9=4xr.

Imho that's the usual way. It is not different from the way your students are taught, you simply use the endpoint (V=max, I=0 and V=0, I=max) to determine the slope and hence the resistance. The advantage of the variable load method is that you don't have to go all the way to a short circuit. This is
1. less burden for the battery
2. less danger for the people doing the test. Imagine doing this test on a car battery with hundreds of amps of short circuit current (I'll leave the results to your imagination). But I see you are well aware of this issue.

The proposed "slope method" has one definite advantage: The inner resistance of a battery is not a fixed value. It depends on the battery's chemistry and varies e.g. with current. By using multiple V/I pairs, one can determine the resistance for different operating points.
Besides, the method demonstrates scientific methodical working.

Maybe, on reflection they are not saying that internal resistance IS the slope of the graph, but can be obtained FROM the gradient.

That may be the case and is a good (or should I say bad) example of stating a task in an unclear way. Unfortunately this is not limited to South Africa. I'm struggling with this kind of imprecise task descriptions in my son's schoolbooks, too. Only our teachers don't seem to care about this the same way you do. Thumbs up!

 

[Kiwi]

"...by an investigating the relationship..." Perhaps they should get an English teacher to proof read their documents?