Aim
To investigate the relationship between the length of a wire and its resistance.
Variables
- Manipulated: Length of the wire, l
- Responding: Resistance of the wire, R
- Constant: The cross-sectional area and material of the wire, and its temperature
Apparatus & materials
- Constantan wire fixed along a metre rule
- Ammeter
- Voltmeter
- Battery
- Rheostat
- Switch
- Crocodile clips
- Connecting wires
Procedure
- Fix the constantan wire straight along a metre rule and connect the circuit with crocodile clips so that length l of wire is included.
- Set the length in the circuit to l = 20.0 cm and record the ammeter reading I and the voltmeter reading V.
- Calculate the resistance R = V / I.
- Repeat for l = 40.0, 60.0, 80.0 and 100.0 cm, taking each reading quickly to avoid heating the wire.
- Record l, V, I and calculate R for each length.
Tabulating results
Record the length l in cm, the voltmeter reading V in V, the ammeter reading I in A, and the calculated resistance R = V / I in Ω.
The graph
Plot R (y-axis) against l (x-axis). A straight line through the origin shows the resistance is directly proportional to the length.
Analysis
Since R = ρl / A, the resistance is directly proportional to the length when the area and material are unchanged. The gradient of the R against l graph equals ρ / A.
Precautions
- Use the same wire throughout so the area and material stay the same.
- Take each reading quickly and switch off between readings to avoid heating.
- Make sure the crocodile clips make firm contact and the wire is straight.
Sample results and what they show
Example data for a constantan wire, with the resistance found from R = V / I at each length:
- l = 20.0 cm: V = 0.40 V, I = 0.20 A, R = 2.0 Ω
- l = 40.0 cm: V = 0.80 V, I = 0.20 A, R = 4.0 Ω
- l = 60.0 cm: V = 1.20 V, I = 0.20 A, R = 6.0 Ω
- l = 80.0 cm: V = 1.60 V, I = 0.20 A, R = 8.0 Ω
- l = 100.0 cm: V = 2.00 V, I = 0.20 A, R = 10.0 Ω
Each time the length doubles, the resistance doubles, so the resistance is a constant multiple of the length. This shows that the resistance of a wire is directly proportional to its length when the cross-sectional area, material and temperature are unchanged.
A longer wire opposes the current more, so a larger potential difference is needed to keep the same current.
Reading the graph and finding the answer
Plot resistance R (y-axis) against length l (x-axis). The points give a straight line through the origin, confirming R is directly proportional to l.
The gradient equals ρ / A, the resistivity divided by the cross-sectional area. Take a large triangle using (0.20 m, 2.0 Ω) and (1.00 m, 10.0 Ω), converting the length to metres:
gradient = ΔR ÷ Δl = (10.0 − 2.0) Ω ÷ (1.00 − 0.20) m = 8.0 Ω ÷ 0.80 m = 10 Ω m⁻¹
So the resistance rises by 10 Ω for every 1 m of wire. If the cross-sectional area A is known, the resistivity can be found from ρ = gradient × A. A line through the origin means a wire of zero length would have zero resistance, as expected.
Marks examiners look for
For the practical marks: use the same wire throughout so the area and material stay the same, take each reading quickly and switch off between readings so the wire does not heat up, and make sure the crocodile clips make firm contact and the wire is pulled straight along the rule.
For the Paper 3 science-process-skill marks, state the variables: the manipulated variable is the length l, the responding variable is the resistance R, and the fixed variables are the cross-sectional area, the material and the temperature. Tabulate l in cm, V in V, I in A and R in Ω, with a unit in every heading and consistent decimal places.
Plot R against l on even scales, draw one thin best-fit line, and show the gradient triangle with coordinates and unit. The conclusion should link the straight line through the origin to the aim: resistance is directly proportional to length.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)