How Does Graphing V vs. I Help Determine E.m.f and Internal Resistance?

[Andresx90]

Homework Statement

This problem relates to an experiment to determine the e.m.f and internal resistance of a battery using a variable load. Taking values for V & I a graphy of V vs. I was plotted.

Homework Equations

E=IR+Ir
V=IR

The Attempt at a Solution

After plotting the graph and calculating the gradient. I am not sure where the gradient and the equation y=mx+c fit in the equation E=IR+Ir, although I believe it could be rearraging the equation to look like IR=E-Ir. Hence y=IR, E=mx and -Ir=c

 

[Kurdt]

From the equations yu've given you could have: $E=V+Ir$. What would the gradient and intercept be then?

 

[andrevdh]

If you wqrite your original equation in this form

E = Ir + V

and rearrange you get

V = E - rI

this now corresponds with the linear equation for your graph:

V = c + mI

where c is the y-intercept and m is the gradient of your graph.