Question

I am using V+E-Ir. But I can't get the answer yet as 05 or 0.6 ohm? How?

Answer 1

`r~~0.59Omega`

Explanation:

The graph plotted follows the equation `V=epsilon-Ir`, which is equivalent to `y=mx+c`

`[(V,=,epsilon,-I,r),(y,=,c,+m,x)]`

So, therefore the gradient is `-r=-(DeltaV)/(DeltaI)~~-(0.30-1.30)/(2.00-0.30)=-1/1.7=-10/17`

`r=-(-10/17)=10/17~~0.59Omega`