Determine the intercept of l/l axis

The relationship between $\frac{1}{I}$ and $R$ for a simple circuit with internal resistance is linear and can be expressed as:

$$\frac{1}{I} = \frac{1}{E}R + \frac{r}{E}$$

This is in the form of a straight line equation $y = mx + c$, where $y = \frac{1}{I}$, $x = R$, $m = \frac{1}{E}$ (slope), and $c = \frac{r}{E}$ (y-intercept). The intercept of the $\frac{1}{I}$ axis is the value of $c$.

We use the given data points $(R, \frac{1}{I})$:
$(2 \, \Omega, 5.00 \, \text{A}^{-1})$, $(4 \, \Omega, 5.60 \, \text{A}^{-1})$, $(6 \, \Omega, 10.00 \, \text{A}^{-1})$, $(7 \, \Omega, 12.50 \, \text{A}^{-1})$, $(9 \, \Omega, 16.70 \, \text{A}^{-1})$.
Let $x = R$ and $y = \frac{1}{I}$.
We calculate the necessary sums for the least squares method:
$\sum x = 2+4+6+7+9 = 28$
$\sum y = 5.00+5.60+10.00+12.50+16.70 = 49.80$
$\sum x^2 = 2^2+4^2+6^2+7^2+9^2 = 4+16+36+49+81 = 186$
$\sum xy = (2 \times 5.00) + (4 \times 5.60) + (6 \times 10.00) + (7 \times 12.50) + (9 \times 16.70)$
$\sum xy = 10.00 + 22.40 + 60.00 + 87.50 + 150.30 = 330.20$
Number of data points $n = 5$.

The slope $m$ is given by:

$$m = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2}$$ $$m = \frac{5(330.20) - (28)(49.80)}{5(186) - (28)^2}$$ $$m = \frac{1651 - 1394.4}{930 - 784}$$ $$m = \frac{256.6}{146}$$ $$m \approx 1.7575 \, \text{A}^{-1}\Omega^{-1}$$

The y-intercept $c$ is given by:

$$c = \frac{\sum y - m \sum x}{n}$$

Using the calculated slope $m \approx 1.7575$:

$$c = \frac{49.80 - (1.7575)(28)}{5}$$ $$c = \frac{49.80 - 49.21}{5}$$ $$c = \frac{0.59}{5}$$ $$c = 0.118 \, \text{A}^{-1}$$

The intercept of the $\frac{1}{I}$ axis is $\boxed{\mathbf{0.118 \, A^{-1}}}$.