# Find Inner Resistance of Power Source: 5 Trials

• Plasma
In summary, to find the internal resistance of a power source, one can calculate it from the current that would flow if there were no external resistance. This can be done by making a graph of external resistance vs. current and extrapolating to zero external resistance using a best fit line and the x-intercept. Alternatively, one can use the equation R_int+R_ext=E/I, where E is the open circuit voltage and I is the current. By taking multiple data points and using the slope of the graph or individual calculations, the internal resistance can be estimated. However, it is important to note that this method may be more complicated if the circuit involves a battery and concentration overvoltage.
Plasma
If I do 5 trials of measuring voltage and current of a circuit (with different resistors for each trial), how would I find the inner resistance of the power source?

Internal resistance can be calculated from the current that would flow if there were no external resistance. How could you extrapolate that value from your data?

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Dick said:
How could you extrapolate that value from your data?

Good question. I'm trying to figure that out.

Well, how about this. If you make a graph of external resistance vs current, how could you extrapolate to zero external resistance?

Dick said:
Well, how about this. If you make a graph of external resistance vs current, how could you extrapolate to zero external resistance?

A Best fit line?

Yessssss. And what intercept will be of interest?

x-intercept?

If the line corresponding to zero external resistance is the x-axis, yes. I would maybe graph it the other way around, but it's your choice.

We have the equation : Ri+Re=E/I
where Ri is internal resistance, Ro - external resistance, E emf and I : current.
E can be measured as open circuit voltage or when the current is zero. In fact if you measure the voltage, there's some current going through, but it is very small and can be negligible. So theoretically, with only one value of Re, you can calculate Ri.
If the circuit involves a battery, it may be more complicated for the concentration overvoltage.

Never mind. I think I got it. Thanks for all of your help.

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Hmm. I apologize - I've been having you graph and extrapolate a function which is not linear. Not the best way to reduce the data. I somehow had it upside down in my head. I think it's better to take Haiha's suggestion, (R_ext+R_int)=E/I. Your graph will also give you an approximate answer if you take the unloaded voltage of the battery and divide by your current. But it's probably better to take each data point independently. Sorry again!

I seem to be confused at what you want me to graph, now.

I don't know that I want you to graph anything. Do you have a measurement of the voltage without any resistors? If you want to graph something try measured currents vs measured voltages. That should be linear.

Dick said:
I don't know that I want you to graph anything. Do you have a measurement of the voltage without any resistors? If you want to graph something try measured currents vs measured voltages. That should be linear.

The Voltage without external resistors is 5 V

I also have the equation of the Best-fit line for Voltage vs. Current

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Good. Then let's take Haiha's suggestion. R_ext+R_int=E/I where the E is 5V. You know I for each value of R_ext. This will give you 5 estimates for R_int.

r = .542 Ω
r = .714 Ω
r = .891 Ω
r = 1.455 Ω
r = .889 Ω

Quite a big difference.

I'm really confusing you here, aren't I? Sorry. Start with Haiha's formula. R_ext+R_int=E0/I, where E0 is your 5V and I is your measured current. I*R_ext will be the voltage you measure across the resistor (call it E_ext). So we can rewrite this as E_ext=E0-I*R_int. So if you graph measured voltage vs measured current the slope will be -R_int. The intercept at I=0 will be (approximately), your 5V. Or you could just follow Haiha's suggestion and get an estimate of R_int from each of your 5 data points independently. Sorry again to make this so confusing.

I've finally got it, thanks to you. According to my graph, the internal resistance is .5094 Ω.

## 1. What is the purpose of finding the inner resistance of a power source?

The inner resistance of a power source is an important factor in understanding its overall performance and efficiency. By measuring the inner resistance, scientists can determine how much power is being lost within the source and make improvements to increase its efficiency.

## 2. How is the inner resistance of a power source measured?

The inner resistance of a power source can be measured by conducting five trials with different load resistances and recording the voltage and current readings for each trial. By plotting a graph of voltage versus current and finding the slope of the line, the inner resistance can be calculated using Ohm's Law (R = V/I).

## 3. Why is it necessary to conduct multiple trials to find the inner resistance?

Conducting multiple trials allows for more accurate and reliable results by reducing the impact of any external factors that may affect the measurements. It also helps to identify any inconsistencies or anomalies in the data, leading to a more precise calculation of the inner resistance.

## 4. What are some potential sources of error when measuring the inner resistance of a power source?

Some potential sources of error include fluctuations in the power supply, variations in the resistance of the load, and imprecise measuring equipment. It is important to control these factors as much as possible to obtain accurate results.

## 5. How can the knowledge of the inner resistance of a power source be applied in real-world situations?

Knowing the inner resistance of a power source can help in designing more efficient circuits and devices, as well as troubleshooting any performance issues. It can also aid in selecting the most suitable power source for a specific application, taking into account its inner resistance and overall performance.

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