Internal Resistance

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  • #1
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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?
 

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  • #2
Dick
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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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  • #3
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How could you extrapolate that value from your data?
Good question. I'm trying to figure that out.
 
  • #4
Dick
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Well, how about this. If you make a graph of external resistance vs current, how could you extrapolate to zero external resistance?
 
  • #5
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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?
 
  • #6
Dick
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Yessssss. And what intercept will be of interest?
 
  • #7
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x-intercept?
 
  • #8
Dick
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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.
 
  • #9
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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.
 
  • #10
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I took your advice and switched the the axes. Thanks.
 
  • #11
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Never mind. I think I got it. Thanks for all of your help.
 
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  • #12
Dick
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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!
 
  • #13
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I seem to be confused at what you want me to graph, now.
 
  • #14
Dick
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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.
 
  • #15
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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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  • #16
Dick
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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.
 
  • #17
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r = .542 Ω
r = .714 Ω
r = .891 Ω
r = 1.455 Ω
r = .889 Ω

Quite a big difference.
 
  • #18
Dick
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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.
 
  • #19
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I've finally got it, thanks to you. According to my graph, the internal resistance is .5094 Ω.
 

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