General help understanding emf and p.d

In summary, a graph is used to show the variation of potential difference (V) across a battery with current (I) as the resistance (R) is changed. A line of best fit is drawn and its gradient is used to determine the emf of the battery and its internal resistance (r). The equation E=V+Ir is used to find the emf, and by rearranging it in the form of a straight line, the y-intercept represents the emf and the gradient represents the internal resistance. This concept is similar to the graph of an ohmic conductor, where V and I are directly proportional.
  • #1

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


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 [itex] \approx \frac{\Delta V}{\Delta I} = 1.67 \Omega [/itex]

Just don't know what to do really. Need an example and explanation of the concept.
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  • #2
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


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.
  • #3
[itex] V = f(I) [/itex]

[itex] V = -Ir + E [/itex]

y-intercept is emf, gradient is -r. Thanks.
  • #4
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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