Error Propogation in Measuring Resistance of I-V Graph

[Gregg]

I plan to set up a circuit to measure the resistance and basically verify Ohm's Law, I am to measure the current and potential difference across a resistor and plot an I-V graph with error bars included. The resistance on this graph is going to be $\frac{\Delta V}{\Delta I} = \frac{1}{\text{gradient}}$ The question is, is how am I going to estimate an error on the evaluation of the resistance of this graph?

The error on $$\alpha_{\Delta X} = \sqrt{ (\alpha_{x_1})^2+(\alpha_{x_2})^2}$$

Where in general $\alpha_x$ is the error of the variable $x$

And the function is of the form

$$R=\frac{\Delta V}{\Delta I} \Rightarrow \alpha_R=R\sqrt{ (\frac{\alpha_{\Delta I}}{\Delta I})^2+(\frac{\alpha_{\Delta V}}{\Delta V})^2 }$$

The problem I have is this:

For say $\Delta V = V_2-V_1$ there is an error $$\sqrt{2(\alpha_V)^2}$$ from the equipment used to take the reading. But with the graph, depending on the separation. One division in the graph paper being x units gives further error $$\pm x$$ for $$\Delta V$$. How am I supposed to combine them?

 

[Valerie Park]

The best approach would be to combine the two errors by taking the square root of the sum of the squares of the individual errors. In this case, you would calculate the error on the resistance as:\alpha_R=R\sqrt{ (\frac{\alpha_{\Delta I}}{\Delta I})^2+(\frac{\alpha_{\Delta V}}{\Delta V})^2 +(x)^2 } Where x is the error of one division in the graph paper used to measure the potential difference. This will give you a more accurate estimate of the error on the evaluation of the resistance.