# Manipulation of V=IR to produce a directly proportional graph.

1. Nov 3, 2012

### Kier

1. The problem statement, all variables and given/known data
Hello, in physics the other day we sat a test which asked up to measure current and resistance of different resistors in a circuit. After collecting the data and plotting the graph for current vs resistance and coming up with a curve indicating an exponential relationship, we were then asked to manipulate the data to produce a graph with a straight line relationship between I and R.

2. Relevant equations

V=IR

3. The attempt at a solution
Now looking at the equation, it's obvious there is no option to square either side like you might for a distance/time graph. But that doesn't get me any closer to the end result. I thought about plotting R=V/I, but that doesn't do anything productive seen as though it was all measured at 6 volts, and, well, plotting a constant wont give me the proportional relationship I require, my teacher suggested I focus on the equation, but I really don't see anything else I can do?

Any help would be very much appreciated, thanks for your time.

2. Nov 3, 2012

### SammyS

Staff Emeritus
Hello Kier. Welcome to PF !
It was probably hyperbolic rather than exponential.
This is exactly what you do.
So rather than squaring the current, I, or taking the square root of I, or ...

What do you need to do with the I values to make $\displaystyle R=V(1/I)$ a direct proportion?

3. Nov 3, 2012

### Staff: Mentor

It looks like V is not part of what your teacher is asking you. Plot I vs R on a log-log plot and see if it comes out to a straight line. You can either use log-log graph paper (if that still exists), or have your graphics package plot the data using log scales. If the graphics package has a curve fitting resource, use that too.

4. Nov 4, 2012

### SammyS

Staff Emeritus
That will produce a linear graph, with a non-zero y-intercept, so it won't be the graph of a direct proportion.

5. Nov 5, 2012

### Staff: Mentor

The OP didn't say anything about direct proportion. It just said a linear relationship. Incidentally, there is no such thing as a y-intercept on a log-log plot.

6. Nov 5, 2012

### SammyS

Staff Emeritus
The Title does refer to a direct proportion. In the attempt at a solution, OP does mention trying to achieve a proportional relationship.

You're right that a traditionally labelled log-log graph does not have a y-intercept -- nor any intercept at all for that matter.

But in the spirit of what OP discusses, producing the log-log graph would amount to plotting log(R) versus log(I) on a Cartesian graph. That certainly does have a y-intercept where log(I)=0, i.e. where I=1 .

7. Nov 5, 2012

### Staff: Mentor

Only in one set of units. If you change the units on I, without changing the units on R, the intercept changes.

The problem statement here is very problematic, and not very well defined. I feel that we have beat this one to death.

8. Nov 5, 2012

### SammyS

Staff Emeritus
Yes. Especially since OP has not responded.

9. Nov 5, 2012

### CWatters

The standard equation for a straight line is

y = mx + c

or ignoring c..

y = mx

In the experiment V is constant so substitute m = v in the above..

y = vx

which is similar to

i = v/r where r = 1/x

so plot

i vs 1/r and the result should be a straight line with gradient v.

at least I think so. I've just finished a three hour drive :-(

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook