Question

1. open link to simulation
2. set the resistance constant to 500 ω. change the value of the voltage according to table 1 and record the corresponding value of the current.
3. attach the graph of voltage versus current and calculate the slope.
4. calculate the percentage error in slope (resistance), taking the real value = 500 ω

data analysis

voltage/ v	current/ a
4	8
5	10
6	12
8	16
9	18

slope =
% =

Explanation:

Step1: Recall slope - formula

The slope $m$ of a line for two variables $x$ and $y$ (in this case, voltage $V$ is $y$ and current $I$ is $x$) is given by $m=\frac{\Delta y}{\Delta x}=\frac{y_2 - y_1}{x_2 - x_1}$. We can also use the general formula $m=\frac{\sum_{i = 1}^{n}x_iy_i-\frac{1}{n}(\sum_{i = 1}^{n}x_i)(\sum_{i = 1}^{n}y_i)}{\sum_{i = 1}^{n}x_i^{2}-\frac{1}{n}(\sum_{i = 1}^{n}x_i)^{2}}$. But for a linear relationship $V = IR$ (Ohm's law), the slope of the $V - I$ graph is the resistance $R$. We can take any two - points from the table. Let's take the first two points $(I_1,V_1)=(4,2)$ and $(I_2,V_2)=(8,4)$. Then $m=\frac{V_2 - V_1}{I_2 - I_1}=\frac{4 - 2}{8 - 4}=\frac{2}{4}=0.5$. In general, for any two points $(I_i,V_i)$ and $(I_j,V_j)$ from the table, since $V=IR$, the slope of the $V - I$ graph is $R$. Using all the data points, we know that from Ohm's law $V = IR$, and the slope of the $V - I$ graph represents resistance. If we take two points, say $(I_1 = 4,A,V_1 = 2V)$ and $(I_2=8A,V_2 = 4V)$, the slope $m=\frac{V_2 - V_1}{I_2 - I_1}=\frac{4 - 2}{8 - 4}=0.5$. In fact, for a linear relationship $V=IR$, we can calculate the slope using the formula $m=\frac{\sum_{i = 1}^{6}V_iI_i-\frac{1}{6}(\sum_{i = 1}^{6}V_i)(\sum_{i = 1}^{6}I_i)}{\sum_{i = 1}^{6}I_i^{2}-\frac{1}{6}(\sum_{i = 1}^{6}I_i)^{2}}$.
$\sum_{i = 1}^{6}V_i=2 + 4+5 + 6+8 + 9=34$
$\sum_{i = 1}^{6}I_i=4 + 8+10 + 12+16+18=68$
$\sum_{i = 1}^{6}V_iI_i=2\times4+4\times8 + 5\times10+6\times12+8\times16+9\times18=8 + 32+50+72+128+162=452$
$\sum_{i = 1}^{6}I_i^{2}=4^{2}+8^{2}+10^{2}+12^{2}+16^{2}+18^{2}=16 + 64+100+144+256+324=904$
$m=\frac{452-\frac{1}{6}\times34\times68}{904-\frac{1}{6}\times68^{2}}=\frac{452-\frac{1}{6}\times2312}{904-\frac{1}{6}\times4624}=\frac{452 - 385.33}{904 - 770.67}=\frac{66.67}{133.33}=0.5$

Step2: Calculate percentage error

The percentage error formula is $\text{Percentage Error}=\frac{\vert\text{Measured Value}-\text{True Value}\vert}{\text{True Value}}\times100\%$. The measured value of the slope (resistance) is $R_{measured}=0.5\Omega$ and the true value $R_{true}=500\Omega$.
$\text{Percentage Error}=\frac{\vert0.5 - 500\vert}{500}\times100\%=\frac{499.5}{500}\times100\% = 99.9\%$

Answer:

Slope = $0.5\Omega$
Percentage Error = $99.9\%$