In an experiment to investigate the variation of current $I$ with potential difference $V$ for a copper sulfate solution, the following apparatus was used - Leaving Cert Physics - Question 4 - 2012

The apparatus Y is a potentiometer. Its function is to measure the potential difference across components accurately, allowing for control of the voltage applied in the experiment.

The potential difference was measured using a voltmeter connected in parallel with the copper sulfate solution. It displays the voltage readings directly.

To plot the graph:

1. Label the axes: The x-axis represents $1/A$ and the y-axis represents $V/V$.

2. Plot the points: Using the data provided, plot points for each pair of values in the table.

3. Draw the line: Connect the points to form a straight line, indicating a linear relationship.

4. Ensure clarity: Ensure that the graph is clearly visible and all points are marked accurately.

To calculate the slope (m) of the graph, use the formula: m = rac{ ext{Change in } V}{ ext{Change in } (1/A)} Using two points from the graph, for example:

• Point 1: (0.4, 1.0)
• Point 2: (2.0, 5.0)

The slope can be calculated as follows: m = rac{5.0 - 1.0}{2.0 - 0.4} = rac{4.0}{1.6} = 2.5

So, the slope is 2.5.

Using Ohm’s Law, which states that: R = rac{V}{I} Where $V$ is the potential difference and $I$ is the current, the resistance R can be inferred from the slope of the graph as: R = rac{V}{I} = rac{ ext{Slope}}{1} = 2.5 ext{ Ohms (if using SI units)}. Thus, the resistance of the copper sulfate solution is 2.5 Ohms.