How to calculate resistance from points using linear regression?

Question

I have a lists of current and voltage of one device and I would like to calculate the resistance. There are some errors with the coordinate, so I have made a script where I calculate the line by linear regression.

from scipy.stats import linregress
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd

voltage = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
current = np.array([3, 6, 6, 10, 9, 15, 14, 16, 20, 25])
slope, intercept, r_value, p_value, std_err = linregress(voltage, current)
print(linregress(voltage, current))

plt.plot(voltage, current, 'o', label='original data')
plt.plot(voltage, intercept + slope * voltage, 'r', label='fitted line')
plt.xlabel('Voltage')
plt.ylabel('Current')
plt.show()

Which plots me:

and prints:

LinregressResult(slope=2.206060606060606, intercept=0.2666666666666675, rvalue=0.9703665463597563, pvalue=3.2552108142876276e-06, stderr=0.1942232770783282, intercept_stderr=1.2051237414984548)

What is the value of the resistance for this device, how to calculate it from line?

Answer 1

Look at Ohm's law for resistors:

V = I x R

Then solve it for R:

R = V/I

Looking at your graph, the slope of the line is I/V, which is the reciprocal of R.

So R = 1/m

Where m is the slope of your graph.

Answer 2

Ohms law tells us \$V=IR\$, so \$R=\frac{V}{I}\$, so \$\frac{1}{\text{slope}}\$ is your resistance.

I suspect your current number are actually milliamps and not amps, so you need to correct for that.

A more meaningful way to do this is to force your line fit to have a zero intercept.

Answer 3

First make sure both units have the same magnitude (V and A, mV and mA etc).

Second, if you force the intercept to zero and switch V with A, so the slope is directly the resistance, the code becomes simply:

import numpy as np

voltage = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
current = np.array([3, 6, 6, 10, 9, 15, 14, 16, 20, 25])
resistance = np.sum(voltage*current)/np.sum(current**2)
print(resistance)

Answer 4

Using Mathematica I make the following picture:

The equation of the line of found using the least square method and is given by:

$$\text{Current}\left(\text{Voltage}\right)=\frac{364}{165}\cdot\text{Voltage}+\frac{4}{15}\tag1$$

We know Ohm's law:

$$\text{R}\space\left[\Omega\right]=\frac{\text{V}\space\left[\text{Volt}\right]}{\text{I}\space\left[\text{Ampere}\right]}\tag2$$

So, we can write:

$$\text{R}=\frac{1}{\text{Current}'\left(\text{Voltage}\right)}=\left(\frac{\text{d}\text{Current}\left(\text{Voltage}\right)}{\text{dVoltage}}\right)^{-1}=\frac{165}{364}\approx0.453297\space\Omega\tag3$$

---

The Mathematica-code I just is:

y = Fit[{{1, 3}, {2, 6}, {3, 6}, {4, 10}, {5, 9}, {6, 15}, {7, 
     14}, {8, 16}, {9, 20}, {10, 25}}, {1, x}, x];
Show[ListPlot[{{1, 3}, {2, 6}, {3, 6}, {4, 10}, {5, 9}, {6, 15}, {7, 
    14}, {8, 16}, {9, 20}, {10, 25}}], 
 Plot[y, {x, 1, 10}, PlotStyle -> Red], 
 AxesLabel -> {HoldForm[Voltage], HoldForm[Current]}, 
 PlotLabel -> None, LabelStyle -> {GrayLevel[0]}, ImageSize -> Large]