# EMF and Internal Resistance

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 ^^ Related Introductory Physics Homework Help News on Phys.org
Office_Shredder
Staff Emeritus
Gold Member
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.

this is why Electricity is my least favourite part of physics :grumpy:

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...

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).

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 sorta unknown. Or I am blind and/or stupid :p

I really appreciated the help ;)

Last edited:
Hootenanny
Staff Emeritus
Gold Member
mininirime said:
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 sorta 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?

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
Staff Emeritus