Tricky V I Graph Need help please

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The discussion revolves around determining the gradient of a graph related to voltage (V) and current (I) measurements across a battery. Participants clarify that as the load resistor value decreases, the loop current increases, leading to a decrease in measured source voltage due to internal resistance. The confusion arises regarding whether the slope of the graph should be positive or negative since the voltmeter measures across the battery. It is emphasized that the graph should reflect the relationship defined by the equation V = emf - I*r, which corresponds to the battery's internal resistance. Ultimately, the participants encourage the poster to clarify their graph construction based on these principles.
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What do you mean by 'gradient'?

You might consider taking more than 1 measurement of I and V ...
 
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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
 
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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.
 
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