# E.m.f and internal resistance

1. Oct 22, 2007

### Andresx90

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
E=IR+Ir
V=IR

3. 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

2. Oct 22, 2007

### Kurdt

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

3. Oct 23, 2007

### 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.