Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Linked equation: V = IR
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Aim
To investigate the I–V characteristics of a fixed resistor (at constant temperature), a filament lamp and a diode.
Method
- Set up a circuit with the component, a variable resistor and an ammeter in series, and a voltmeter in parallel across the component.
- Adjust the variable resistor to change the current. Record the current and p.d. for a range of settings.
- Reverse the connections to the power supply and repeat, recording the readings as negative values.
- Repeat for each component.
- Plot current (y-axis) against potential difference (x-axis) for each component.
| Independent variable | Potential difference across the component |
| Dependent variable | Current through the component |
| Control variables | The component; temperature (for the fixed resistor) |
Questions
Q1 (F) What is the variable resistor used for?
Show answer
To change the current in the circuit (and so the p.d. across the component), so you can take a range of readings.
Q2 (F) Describe the I–V graph for a fixed resistor at constant temperature.
Show answer
A straight line through the origin: current is directly proportional to p.d., so the resistance is constant (an ohmic conductor).
Q3 (F) Describe and explain the I–V graph for a filament lamp.
Show answer
A curve through the origin that gets less steep as the p.d. increases (in both directions). As the current increases the filament gets hotter, so its resistance increases.
Q4 (F) Describe the I–V graph for a diode.
Show answer
In the forward direction, current flows once the p.d. passes a small value and then rises steeply. In the reverse direction the resistance is very high, so almost no current flows.
Q5 (F/H) Why are the power supply connections reversed during the practical?
Show answer
To get negative values of p.d. and current, showing how the component behaves when current flows the other way (important for the diode).
Q6 (F/H) A lamp has 2.0 V across it and 0.25 A through it. At 6.0 V the current is 0.50 A. Calculate the resistance at each p.d. and explain the difference.
Show answer
R = V ÷ I
At 2.0 V: 2.0 ÷ 0.25 = 8.0 Ω
At 6.0 V: 6.0 ÷ 0.50 = 12 Ω
At the higher current the filament is hotter, so its resistance is higher.
Q7 (F/H) Why must the fixed resistor be kept at a constant temperature?
Show answer
Its resistance changes with temperature. If it heats up, the graph would no longer be a straight line. Switching off between readings keeps it cool.
Q8 (H) A fixed resistor has a current of 0.15 A when the p.d. is 3.0 V. Calculate the current when the p.d. is 4.5 V, to 2 significant figures.
Show answer
Step 1 – resistance: R = V ÷ I = 3.0 ÷ 0.15 = 20 Ω
Step 2 – new current: I = V ÷ R = 4.5 ÷ 20 = 0.225 A
A: 0.23 A (2 s.f.)
