Tricky V I Graph Need help please

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The discussion centers on determining the slope of a voltage-current (V-I) graph related to a battery's internal resistance. Participants clarify that as the load resistor value decreases, the loop current increases, leading to a decrease in measured source voltage due to the internal resistance of the battery. The equation used is emf = V + Ir, which is rearranged to V = emf - Ir, indicating a negative slope in the graph. The participants emphasize the importance of accurately measuring voltage across the battery, including its internal resistance, to construct the graph correctly.

PREREQUISITES
  • Understanding of Ohm's Law and its application in circuit analysis.
  • Familiarity with the concept of internal resistance in batteries.
  • Knowledge of how to interpret V-I graphs in electrical circuits.
  • Basic skills in graphing and analyzing linear relationships.
NEXT STEPS
  • Study the relationship between load resistance and current in circuits.
  • Learn how to calculate internal resistance using V-I graphs.
  • Explore the concept of electromotive force (emf) in electrical circuits.
  • Investigate different methods for measuring voltage in circuit configurations.
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Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit design or analysis who seeks to understand the implications of internal resistance on battery performance.

Steve12345
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What do you mean by 'gradient'?

You might consider taking more than 1 measurement of I and V ...
 
Last edited:
Steve12345 said:
need help with this question, is the gradient negative or positive since the voltmeter is over the battery?
Here is the image (sorry for making you tilt your heads)https://www.dropbox.com/sh/v6ydqbfrg20whtb/ZWR-r9ZOSH?lst#f:q17.JPG

As the load resistor value gets smaller, the loop current increases, and the measured source voltage decreases because of the increasing voltage drop across the voltage source's internal resistance.

Does that make sense? Can you now show us your work on the graph?
 
berkeman said:
As the load resistor value gets smaller, the loop current increases, and the measured source voltage decreases because of the increasing voltage drop across the voltage source's internal resistance.

Does that make sense? Can you now show us your work on the graph?

I drew a line that made a triangle from the V axis to the I axis, so when the current was at its highest V was 0 and when V was at its highest I was 0, so it was a negative slope.

I used emf= V+Ir and rearranged it go get emf - Ir = V. I am confused at wether the graph should be the a positive slope since the voltmeter is measuring the battery and not the load resistor if you get where I am coming from
 
berkeman said:
As the load resistor value gets smaller, the loop current increases, and the measured source voltage decreases because of the increasing voltage drop across the voltage source's internal resistance.

Does that make sense? Can you now show us your work on the graph?

So I am saying I drew it like this http://www.s-cool.co.uk/a-level/assets/learn_its/alevel/physics/Resistance/internal-resistance-emf-and-potential-difference/Finding%20the%20internal%20resistance.gif

Does it make a difference to the graph that were measuring V at a different point
 
Last edited by a moderator:
Steve12345 said:
So I am saying I drew it like this http://www.s-cool.co.uk/a-level/assets/learn_its/alevel/physics/Resistance/internal-resistance-emf-and-potential-difference/Finding%20the%20internal%20resistance.gif

Does it make a difference to the graph that were measuring V at a different point

Your graph is fine as far as it goes.

I don't understand your question abot V being measured "at a different point". The problem shows exactly where V is measured - across the battery including its internal resistance. Your equation V = emf - I*r fully corresponds to the diagram of the problem.

Now, how do you propose to construct your graph? You don't know emf or r. You're supposed to use the graph to determine emf and r.
 
Last edited by a moderator:

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