## EMF and Internal Resistance

Apparatus:

1.5V cell, Mounted and taped unknown resistor labelled 'S', push to make switch, 0 to 1000 OHM resistance box, two digital multimeters, wires.

Diagram: The internal resistance, of a cell is typically about To prevent damage to the cell, you are going to pretend that the resistor labelled 'S' is also inside the cell.

The dashed lines in the above diagram will represent the boundary of the 'modified' cell.

The resistor box will act as the external resistance, of this circuit.

Procedure:

1. Adjust the resistor box so that it has a resistance of 2. Set up the circuit as above.

3. Measure the current through the cell (I) and the potential difference across the cell (V)

for values of and IN THIS ORDER

DO NOT PRESS DOWN THE SWITCH LONGER THAN NECESSARY

4. Tabulate the values of &amp; in a suitable table.

5. Draw a graph of / volts ( - axis) against / A ( - axis). It should be a straight line of negative gradient.

(NOTE: Plot current in amperes!)

6. Measure the intercept. This should equal the EMF of the cell.

7. Measure the gradient of your graph. This will equal the negative of the combined resistance of resistor 'S' and the actual internal resistance r of the cell.

8. equals the power being transferred by the cell to the external resistance Draw a graph of / watts (y - axis) against / ( - axis). It should be a curve with a maximum.

Approximately what value of yields the maximum power transfer?

How does this compare with the 'modified' internal resistance?

THEORY:

For a complete circuit where The voltmeter reading, is equal to the external potential difference, Therefore and so rearranged is The equation is now in the form is equivalent to to to the gradient and to the y-axis intercept.

Hence the gradient of the against graph is equal to and the intercept with the y-axis to  