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General help understanding emf and p.d

  • Thread starter Gregg
  • Start date
  • #1
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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 [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.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
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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.
 
  • #3
459
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[itex] V = f(I) [/itex]

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


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