General help understanding emf and p.d

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SUMMARY

This discussion focuses on understanding the relationship between electromotive force (emf), potential difference (p.d.), and internal resistance in a battery circuit. The key equation derived from Kirchhoff's second law is E = V + Ir, where E represents the emf, V is the p.d. across the load, and r is the internal resistance. The gradient of the plotted graph of p.d. (V) against current (I) is approximately 1.67 Ω, indicating the internal resistance of the battery. The y-intercept of the graph represents the emf of the battery.

PREREQUISITES
  • Understanding of Kirchhoff's laws, particularly the second law.
  • Familiarity with the concepts of electromotive force (emf) and internal resistance.
  • Basic knowledge of graphing linear equations and interpreting gradients.
  • Experience with electrical circuits and components, specifically batteries.
NEXT STEPS
  • Study the derivation of the equation E = V + Ir in detail.
  • Learn how to plot and interpret graphs of p.d. versus current for different circuit configurations.
  • Explore the concept of internal resistance in various types of batteries.
  • Investigate the differences between ohmic and non-ohmic conductors in terms of their V-I characteristics.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone interested in understanding battery performance and circuit analysis.

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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