Finding the Ɛ and r of a cell from a graph

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The discussion focuses on determining the internal resistance and electromotive force (Ɛ) of a dry cell using both a quick method and a graphical method. The user has successfully calculated Ɛ as 1.490 V and internal resistance (r) as 0.2194 Ω using the quick method, but faces challenges with the graphical method, resulting in a negative resistance value. Clarifications are provided regarding the relationship between the slope of the graph and the internal resistance, emphasizing that a negative resistance is not physically plausible. The user expresses confusion about calculating uncertainties and acknowledges that the emf and r could be derived directly from the graph. Overall, the thread highlights the importance of understanding the equations and methods used in the experiment.
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Homework Statement



It is an experiment. Finding the internal resistance of a Dry cell. It includes 2 methods, the quick method and the graphical method. Work out and record 8 pairs of V and I reading by varying the slider of the rheostat and then work out the Ɛ (e.m.f) and r (internal resistance) by creating a graph. Then compare these (Ɛ and r) values with the values obtained from the quick method.

Homework Equations


m=∆y/∆x (m=change in y/change in x=(y1-y)/(x1-x))
y=mx+c
V=-Ir+Ɛ

The Attempt at a Solution


I've done the quick method. The obtained values of Ɛ and r from the quick method are Ɛ=1.490 v and r= 0.2194 Ω. And now now I'm stuck here. I have recorded 8 pairs of reading in the result table. And tried to do the graph, but the result is showing some weird figures. I probably am doing it wrong. Also, I'm quite confused with the equation "y=mx+c", I figured this equation is connected with (somehow) V=-rI+Ɛ, but not really sure how to apply them!
Here is the graph I was trying to work out. hope that helps.

http://www.freeimagehosting.net/uploads/0d89095d56.jpg

And here is the calculetion:

So, m= ∆y/∆x
m= (y1-y)/(x1-x)
m= (1.197-1.359)/ (0.055-0.025)= (-0.162)/0.030
= -5.4
And the equation of a straight line is y=mx+c
So, 1.359= {(-5.4)*(0.025)} +c [since y=1.359, x=1.197 and m= -5.4]
=>1.359= (-0.135) +c
=>c= 1.359+ 0.135
=>c= 1.494
By rearranging the equation of the straight line,
y= mx+c
V= -Ir+ Ɛ
So, Ɛ= 1.494 v and the internal resistance of the dry cell r= negative 5.4 Ω
Comparing with the quick method, Ɛ=1.490 v and r= 0.2194Ω (Is that even close?, the values from graphical method! )

I've marked RED for those areas that I'm unsure!

Thank you.
 
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shafi25 said:
. . . here is the calculetion:

So, m= ∆y/∆x
m= (y1-y)/(x1-x)
m= (1.197-1.359)/ (0.055-0.025)= (-0.162)/0.030
= -5.4
Good enough, since the data points all lie very close to a straight line. And you wisely chose the 2 points farthest apart, which minimizes the error in the calculation.

And the equation of a straight line is y=mx+c
So, 1.359= {(-5.4)*(0.025)} +c [since y=1.359, x=1.197 and m= -5.4]
Yes (though you really meant to say x=0.025)
=>1.359= (-0.135) +c
=>c= 1.359+ 0.135
=>c= 1.494
By rearranging the equation of the straight line,
y= mx+c
V= -Ir+ Ɛ
So, Ɛ= 1.494 v and the internal resistance of the dry cell r= negative 5.4 Ω
Correct for Ɛ. However, getting a negative resistance does not make physical sense in this case.
Note the following:
m is the slope in "y=mx+c"
-r is the slope in "V = -Ir + Ɛ​
Equate -r with the value you got for m, and see what you get.
Comparing with the quick method, Ɛ=1.490 v and r= 0.2194Ω (Is that even close?, the values from graphical method! )
I don't know what the quick method is, so I can't really comment on it. What is your opinion about comparing the values for the two methods?
 
Thank you for your reply. yeah, I should have demonstrated the method 1 or as mentioned "quick method". the comparing was actually for the sake of calculating uncertainties, but that's ok. although I am a little confused how to work out uncertainties from a set of values, but I'm working on it. :) and yeah, I found out I could've work out the emf and the r from the graph without even calculating. Thank you.

Shafi
 
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