General help understanding emf and p.d

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Homework Help Overview

The discussion revolves around understanding the concepts of electromotive force (emf) and potential difference (p.d.) in the context of a battery's internal resistance. Participants are examining a graph that illustrates the relationship between p.d. across a battery and current as resistance is varied.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between emf, p.d., and internal resistance, referencing Kirchhoff's laws. There are attempts to derive equations from the graph and interpret the gradient and intercept. One participant raises a question about the proportionality of V and I in different contexts.

Discussion Status

Some guidance has been provided regarding the interpretation of the graph and the mathematical relationships involved. Multiple interpretations of the relationship between voltage and current are being explored, particularly in relation to different types of conductors.

Contextual Notes

There is an indication of confusion regarding the concepts of internal resistance and the behavior of V and I in different scenarios, as well as a request for examples and explanations to clarify these concepts.

Gregg
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Circuit1.jpg


Graph to show the variation of the p.d. V across the battery with current I as R is varied.


Table1.jpg


i) Draw line of best fit.

ii) Use it to determine the emf of the battery;

and the internal resistance r of the battery.


The gradient \approx \frac{\Delta V}{\Delta I} = 1.67 \Omega

Just don't know what to do really. Need an example and explanation of the concept.
 
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Let the internal resistance of the battery be r.
The same current,I, passes through both r and R right?

So by Kirchoff's 2nd law

the emf of the battery=sum of the pd's around the loop
If V is the pd across the load then

E=V+Ir.

Since they plotted V against I, rearrange the equation it in the form a straight line y=mx+c

you calculated the gradient correctly to give a resistance. So rearrange the equation and you will see what that gradient represents and what the intercept represents as well.
 
V = f(I)

V = -Ir + E


y-intercept is emf, gradient is -r. Thanks.
 
Hi I had a similar question, which I posted at the the thread 'internal resistance graph'. It was basically why are V are I inversely proportional in this graph, but proportional in the graph of an ohmic conductor?
 

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