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## Homework Statement

Students are given some resistors, a battery with internal resistance, and an ammeter. They are asked to determine the emf e and internal resistance of the battery using just this equipment. Working with the circuit shown at right, they insert each resistor into the circuit and measure the current I in the circuit each time they insert a resistor. From their data, students generate a graph of the inverse of their current (1/I) as a function of the resistance of each resistor. Find the internal resistance of the battery.

## Homework Equations

Kirchoff’s 2nd rule or loop rule is based on conservation of energy. The loop rule is that the sum of changes in potential around any closed loop of a circuit must be zero. For this circuit we have three changes in potential.

We have the battery, internal resistor, and the added resistor that all results in potential changes.

We will say the battery is positive and the resistors cause a negative (drop) in potential, so the sum is:[/B]

E-IR-Ir=0

where E=electrostatic motion, I=current, R=resistance, and r=internal resistance

## The Attempt at a Solution

E=IR+Ir

E=I(R+r)

E/I=R+r

r=(E/I)-R

Looking at the graph, some people say that the internal resistance would be the x-intercept. Here 1/I=0, so current would be infinite. This is the point where there is zero resistance in the system

But if I solve this equation for inverse of current I get

Inverse of current=R/E+r/E

The resistance would be our x-values for this graph. I was thinking that the 1/E would be like the slope while r/E is the y-intercept.

I am not sure what to do. Could someone help clarify?