12. A student uses the following circuit to investigate the internal resistance $r$ and EMF $E$ of a battery - Scottish Highers Physics - Question 12 - 2022

From the graph, we observe the y-intercept where the current $I$ is zero, which corresponds to the EMF. Thus, the EMF $E = 6.0 ext{ V}$.

To find the internal resistance, we use the gradient of the line in the graph, which is given by:

$m = \frac{\Delta V}{\Delta I} = \frac{4.0 - 2.0}{0.50 - 0.20} = \frac{2.0}{0.30} \approx 6.67 \, \Omega$

Using the relationship EMF = terminal voltage + current * internal resistance, we can derive:

$6.0 = 4.0 + I r$

If we take a point from the graph, for example when $I = 0.50 ext{ A}$, we get:

$6.0 = 4.0 + 0.50 r$

Solving gives us:

$r = 8.0 \, \Omega.$

The student should open the switch and take the reading from the voltmeter connected across the battery. This reading will reflect the EMF as there is no current and therefore no internal resistance to account for.

As the resistance of the variable resistor $R$ is decreased, the current through the circuit increases. This results in a greater voltage drop across the internal resistance of the battery, causing the terminal potential difference (the voltage across the terminals of the battery) to decrease.

The line representing the new battery will still have the same gradient reflecting the internal resistance, but will intersect the y-axis below 6.0 V corresponding to the smaller EMF. It will maintain a linear trend, demonstrating the same relationship between current and potential difference.