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 or variable resistor. Its function is to control the current flow in the circuit and help measure potential differences accurately.

The potential difference was measured using a voltmeter connected in parallel to the copper sulfate solution, allowing for accurate readings of the voltage across the solution.

To draw the graph:

1. Label the horizontal axis as 1/A and the vertical axis as V/V.
2. Plot the points from the table:
   • (0, 0)
   • (0.4, 1.0)
   • (0.8, 2.0)
   • (1.2, 3.0)
   • (1.6, 4.0)
   • (2.0, 5.0)
   • (2.4, 6.0)
3. Connect the points with a straight line.
4. Ensure the graph is neat, and all points are accurately represented.

The slope of the graph can be calculated using two points. For example, using the points (0, 0) and (2.0, 5.0):

Slope = rac{ΔV}{Δ(1/A)} = rac{5.0 - 0.0}{2.0 - 0.0} = rac{5.0}{2.0} = 2.5.

Thus, the slope of the graph is 2.5.

Using Ohm's law, R = rac{V}{I}, we know from the graph that the slope represents rac{V}{I} = 2.5. Therefore, the resistance of the copper sulfate solution is:

$R = 2.5 ext{ ohms}$.