How can you calculate the internal resistance from circuit data?

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SUMMARY

This discussion focuses on calculating the internal resistance of a circuit using data from a 6V battery, ammeter, switch, and variable resistor. The user aims to derive the internal resistance from current measurements at various resistance settings, utilizing key equations such as V=IRinternal and P=I^2R. Graphical analysis of Current vs. Resistance and p.d. vs. Power is suggested as a method to extract the internal resistance from the data. The importance of comparing experimental results with theoretical values is emphasized to identify anomalies.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with electrical power calculations (P=I^2R)
  • Basic knowledge of circuit components (battery, ammeter, variable resistor)
  • Ability to analyze graphical data (Current vs. Resistance, p.d. vs. Power)
NEXT STEPS
  • Learn how to derive internal resistance from experimental data using graphical methods.
  • Research the significance of the gradient in Current vs. Resistance graphs.
  • Explore the impact of temperature on resistance measurements and calculations.
  • Investigate the use of Excel for organizing and analyzing circuit data effectively.
USEFUL FOR

Students and educators in physics, electrical engineering enthusiasts, and anyone involved in circuit analysis and experimentation.

Joshybosh
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This is my first post to the site.

Homework Statement



there's a circuit with 6V battery, ammetre, switch, and variable resistor.
We are provided with primary data.
We use the equipment to find the current at to 2 different resistances
I have to decide what i want find out from my results. I decided to choose to find the total internal resistance.

Problem:
1| The Variable resistor had 3 dials on it and each went from 1-9.
Above the 1st 2nd and 3rd dial were the titles, "1A", "0.6A" and "0.3A"
Ive tried researching this but have come out clueless.

2| how do i find the anormalies and explain them and what can i find by comparing the sets of data.

Homework Equations


Equations i think i will find useful:

V=IRinternal
I=E x 1/(Rin+Rex)
V=E-IRinternal
P=I^2R
P=E^2 x 1/4R

value of R is +/- 0.2% at 18 oC
Here is a table of results, the first 4 rows were provided, and i had to find the rest.
R/W I1/A I2/A I3/A Mean I/A V1/V V2/V V3/V Mean V/V P1/W P2/W P3/W Mean P/W Range/V 1/2 Range Resolution Uncer % Uncert.
1.0 1.529 1.519 1.467 1.505 1.529 1.519 1.467 1.505 2.338 2.308 2.153 2.266 0.062 0.031 0.001 0.031 2.060
1.5 1.468 1.498 1.498 1.488 2.203 2.248 2.248 2.233 3.234 3.368 3.368 3.323 0.045 0.023 0.001 0.023 1.008
2.0 1.318 1.313 1.368 1.333 2.637 2.627 2.737 2.667 3.476 3.450 3.745 3.557 0.110 0.055 0.001 0.055 2.063
2.5 1.202 1.202 1.220 1.208 3.004 3.004 3.049 3.019 3.609 3.609 3.718 3.646 0.045 0.023 0.001 0.023 0.745
3.0 1.101 1.112 1.096 1.103 3.304 3.337 3.289 3.310 3.640 3.713 3.607 3.653 0.048 0.024 0.001 0.024 0.725
4.0 0.934 0.936 0.953 0.941 3.737 3.745 3.813 3.765 3.491 3.506 3.634 3.544 0.076 0.038 0.001 0.038 1.009
5.0 0.826 0.825 0.812 0.821 4.128 4.123 4.058 4.103 3.407 3.399 3.293 3.366 0.070 0.035 0.001 0.035 0.853
6.0 0.733 0.717 0.658 0.703 4.400 4.304 3.948 4.217 3.226 3.087 2.598 2.970 0.451 0.226 0.001 0.226 5.350
7.0 0.652 0.649 0.658 0.653 4.564 4.543 4.606 4.571 2.976 2.949 3.031 2.986 0.063 0.032 0.001 0.032 0.689
8.0 0.587 0.595 0.597 0.593 4.693 4.757 4.773 4.741 2.753 2.828 2.847 2.809 0.080 0.040 0.001 0.040 0.844
9.0 0.547 0.536 0.543 0.542 4.926 4.827 4.890 4.881 2.697 2.589 2.657 2.648 0.099 0.050 0.001 0.050 1.014
10.0 0.503 0.504 0.493 0.500 5.030 5.040 4.930 5.000 2.530 2.540 2.430 2.500 0.110 0.055 0.001 0.055 1.100
15.0 0.362 0.357 0.361 0.360 5.423 5.348 5.408 5.393 1.961 1.907 1.950 1.939 0.075 0.038 0.001 0.038 0.695

The Attempt at a Solution


1| I've been told that it might have something to do with limiting the current

2| I have made graphs from Current vs resistance, p.d vs power, p.d vs current. But am still unsure what i can find from these graphs. I think one of them can give you the internal resistance by finding the gradient.Any help would be great thanks, this has troubled so many people i know, so i thought i would ask people with a better knowledge of the subject.
 
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1) Yes, my guess would be that the dials represented the amount of power the variable resistor could handle, and simply divided by R and took the square root (P=I^2 R). Some resistances are made by coil windings, and maybe they didn't want you to put more than a certain amount of current into the windings. Maybe it doesn't mean anything, hard to tell without much context.

2) What is p.d.? Just compare your results from the experiment to the theoretical, ideal, results of the circuit.

Sorry about not being more help, but without a circuit diagram its hard to interpret. The table you made is kind of messy too, maybe put it into excel and post a screen shot of the excel table.
 

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