Can the Slope of a Graph Determine EMF and Internal Resistance?

[mininirime]

I'm doing a few practicals at home this summer, and the one that is bugging me is about EMF. I have measured voltage and current at different resistances across a 4,5 V battery, and made a graph. But to find the emf, am I supposed to take the slope of the graph, or the gradient? Or is that the same, aka the derivative? English isn't my first language, but I go to an English school, so...

My graph gives a function y=-1,26x + 5,772. Is it possible for the slope to be negative, and thus the emf? I have current at x-axis and voltage at y-axis.

Thanks for all the help, I hope this doesn't seem like I want you to do my homework for me ^^

 

[Office_Shredder]

The slope of voltage vs. current is resistance, as given by the equation: V=IR The gradient is the slope is the derivative for straight lines

I don't see how you could have gotten a negative slope. But then again, I don't see how your voltage is capable of varying on a 4.5 volt battery.

 

[mininirime]

this is why Electricity is my least favourite part of physics

No, the battery is 4.5 volts, which measn that the perfect emf is 4,5 volts, right? And then I vary the length of resistance, or the wire, from 100 cm to 5 cm. I measure the potential difference and current, and plot those two... so the voltage shrinks because the resistance gets bigger. does that make sense?
e=rI + V <- I'm supposed to be using this.
Maybe the slope is negative because I plotted the biggest resistance results first, and then the smaller resistance results. This would mean that the highest voltage values would come first...

 

[J Hann]

The internal resistance of the cell is the "negative" of the slope when
plotting V vs I. V = E - IR where V is the ordinate (y-axis and I is
the abcissa (x-axis).

 

[mininirime]

Thanks, I plotted V along y-axis and I along X-axis. But to find the internal resistance r, should I use the formula V=Ir, or does that only work for "external" resistances? Because I can only use the V=E-Ir with one unknown, and currently the r is sort of unknown. Or I am blind and/or stupid :p

I really appreciated the help ;)

 

[Hootenanny]

Thanks, I plotted V along y-axis and I along X-axis. But to find the internal resistance r, should I use the formula V=Ir, or does that only work for "external" resistances? Because I can only use the V=E-Ir with one unknown, and currently the r is sort of unknown. Or I am blind and/or stupid :p

I really appreciated the help ;)

Okay, you've plotted your graph correctly, now compare your equation (in a slightly re-arranged form) with the standard equation of a straight line;

V = -rI + E
y = mx + c

Note, that here voltage is on the y-axis and current is plotted on the x axis. Which letter is r equivalent to in the second equation and what does this letter represent?

 

[mininirime]

well, when you put it that way... It makes much more sense ;) I've managed to find the internal resistance and E now! I think it's about seing the connections... which I'm really not that good at :s but thanks for the help!

 

[Hootenanny]

I think it's about seing the connections... which I'm really not that good at :s but thanks for the help!

That will come, in time. My pleasure.