Experiment: Determining the e.m.f. and internal resistance of a dry cell

Aim

To determine the electromotive force (e.m.f.) and the internal resistance of a dry cell.

Variables

• Manipulated: Current drawn from the cell, I
• Responding: Terminal potential difference, V
• Constant: The same dry cell

Apparatus & materials

• Dry cell
• Rheostat
• Ammeter
• Voltmeter
• Switch
• Connecting wires

Procedure

1. Connect the dry cell in series with the rheostat, ammeter and switch, and connect the voltmeter across the cell terminals.
2. Set the rheostat for a small current, then close the switch and record the current I and the terminal voltage V.
3. Lower the rheostat resistance to increase the current a little and record the new I and V.
4. Take five or six pairs of readings over a range of currents, switching off between readings.
5. Record I and V for each setting.

Tabulating results

Record the current I in A and the terminal potential difference V in V for each setting. Keep the decimal places consistent.

The graph

Plot V (y-axis) against I (x-axis). The points give a straight line with a negative gradient; the intercept on the V-axis is the e.m.f. and the magnitude of the gradient is the internal resistance.

Analysis

The terminal voltage obeys V = ε − Ir. The intercept of the V against I graph on the V-axis gives the e.m.f. ε, and the magnitude of the gradient gives the internal resistance r.

Electricity · Graph skills

Sample results and what they show

Here is a set of example readings for one dry cell (label these clearly as example data on your script):

• I / A = 0.20, V / V = 1.48
• I / A = 0.40, V / V = 1.42
• I / A = 0.60, V / V = 1.36
• I / A = 0.80, V / V = 1.30
• I / A = 1.00, V / V = 1.24

As the current drawn from the cell rises, the terminal potential difference V falls steadily. The size of the drop is the lost volts, equal to Ir across the cell's own internal resistance.

This tells you that the cell cannot deliver its full e.m.f. to the external circuit once a current flows: some voltage is used up inside the cell.

Notice the pattern is even. Each 0.20 A rise in current gives about a 0.06 V fall in V, so the readings lie on a straight line.

A steady step like this is the sign of a constant internal resistance and a clean, usable result.

Reading the graph and finding the answer

Plot V on the y-axis against I on the x-axis. The expected shape is a straight line that slopes downward, because V = ε − Ir.

Two quantities fall out of that line: the intercept on the V-axis is the e.m.f. ε, and the magnitude of the gradient is the internal resistance r.

To find the gradient, draw a large triangle using two points far apart on the best-fit line, for example (0.20 A, 1.48 V) and (1.00 A, 1.24 V):

gradient = (1.24 − 1.48) V ÷ (1.00 − 0.20) A = −0.24 V ÷ 0.80 A = −0.30 V A⁻¹, so r = 0.30 Ω.

Extend the line back to I = 0 to read the intercept: ε = 1.48 V + (0.30 V A⁻¹ × 0.20 A) = 1.54 V. So this cell has an e.m.f. of about 1.54 V and an internal resistance of 0.30 Ω.

Marks examiners look for

Precautions that protect the result: open the switch between readings so the cell does not run down, avoid drawing very large currents that would warm the cell and change its behaviour, and check both meters for zero error before you start.

For the Paper-3 science-process skills, the marks come from good practice, not luck:

• Take five or six well-spread pairs of readings so the line is well defined.
• Draw a single best-fit straight line with the points scattered evenly on both sides.
• Use a large triangle for the gradient and quote its unit, V A⁻¹.
• Extrapolate the line neatly to I = 0 to read the e.m.f. as the intercept.
• State the final e.m.f. and internal resistance with correct units and sensible precision.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)