Back to course
1PH0 · CP10.17(a)Core practical investigationsCore

CP10.17(a): Resistor and Filament-Lamp Characteristics

Use an appropriate low-voltage supply and current-limiting control. The ammeter is in series with the component and the voltmeter is in parallel across it. Vary the operating potential difference within the approved ratings and record paired V and I readings. Check the circuit against a diagram before switching on; an ammeter directly across a supply can produce an unsafe short circuit.

35 min Video by Science School Core practical investigations
Edexcel Core Practical 9-1 GCSE P10.17 Electrical circuit for V=IR in a resistor & a filament lampCurated by Kepler, not produced by us. Watch it before the notes below.
Open on YouTube

Before you start

  • Use your teacher’s tier checklist: this overview can include Higher-only material, and is not a tier-specific revision list.
  • Pearson Edexcel separate GCSE Physics 1PH0. Use the original core-practical reference and retain every specified subpart; two guides can develop different parts of one core reference.
  • Read ammeter and voltmeter scales, including negative readings and milliampere conversion.
  • Use the centre’s approved supervised method and equipment limits. These notes prepare you to explain and evaluate practical work; they do not replace carrying it out and recording the evidence.

Watch, pause, try

Video by Science School. This video develops the resistor-and-filament-lamp part of CP10.17. The separate circuit guide develops the required series/parallel lamp comparison.

Viewing focus: Collect paired current and potential-difference readings

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Sketch the expected I–V shapes for a constant-temperature resistor and a filament lamp. Explain the difference.

Show answer and reasoning

The resistor has a straight proportional line. The lamp’s curve becomes shallower at larger magnitudes because heating increases resistance.

After watching 2: Why does a lamp’s tangent on an I–V curve not generally give its operating resistance directly?

Show answer and reasoning

The operating resistance is V/I at the point. A tangent gives a local change ratio, and on I-versus-V axes it is not even expressed in ohms.

What you'll be able to do

  • Collect paired current and potential-difference readings
  • Compare constant-temperature resistor, filament and diode behaviour
  • Distinguish operating resistance from a non-linear tangent gradient
1

Measure a component without confusing the meters

Use an appropriate low-voltage supply and current-limiting control. The ammeter is in series with the component and the voltmeter is in parallel across it. Vary the operating potential difference within the approved ratings and record paired V and I readings. Check the circuit against a diagram before switching on; an ammeter directly across a supply can produce an unsafe short circuit.

Where required, reverse polarity using the approved arrangement and record the signs consistently. A component or lead should not be reconnected while the circuit is live. Convert milliamperes to amperes before calculating resistance. Use meter ranges with suitable resolution and assess whether the meter loading or very small currents affect the inference.

2

Required resistor/lamp comparison and an optional diode extension

For an ohmic resistor at approximately constant temperature, current is proportional to potential difference and an I-against-V graph is a straight line through the origin. A filament lamp heats as the electrical input increases. Its resistance rises, so current increases less rapidly with voltage and the I–V curve becomes shallower at larger voltage magnitudes.

A semiconductor diode conducts appreciably in one direction once the applied forward conditions are sufficient; reverse current is very small within its approved operating range. Do not assume its forward characteristic is a straight line through the origin or that every diode has an identical sharp threshold. Protect it with the specified current-limiting arrangement.

CP10.17(a) requires the resistor and filament lamp: initially build the resistor circuit with ammeter and voltmeter, record current at different potential differences, then repeat with a filament lamp. Diode behaviour is included here as a supporting comparison, not as an additional named CP10.17 requirement.

3

Read a resistance without swapping the axes

At a stated operating point, resistance is V/I. On a linear I-against-V graph for an ohmic component, the gradient I/V is conductance, so resistance is its reciprocal. On a V-against-I graph the straight-line gradient is resistance. State the graph axes and units before interpreting the slope.

For a non-linear lamp or diode, V/I at a point is generally not the same as the local tangent gradient dV/dI. GCSE operating-resistance questions normally require the ratio of the point’s coordinates. Avoid calculating 0/0 at the origin; infer a suitable relationship from nonzero data where the model supports it. Explain temperature changes before claiming that disagreement proves an instrument fault.

4

Worked applications

Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.

Example 1An ohmic resistor carries 0.150 A at 3.00 V. Find R and the I-against-V straight-line gradient.
1R = 3.00/0.150 = 20.0 Ω.
2Gradient I/V = 0.150/3.00 = 0.0500 A V⁻¹.
Answer
Resistance 20.0 Ω; the plotted gradient is 0.0500 A V⁻¹, its reciprocal.
Example 2A lamp carries 0.200 A at 2.00 V and 0.300 A at 6.00 V. Compare its operating resistance.
1At the first point, R = 2.00/0.200 = 10.0 Ω.
2At the second, R = 6.00/0.300 = 20.0 Ω.
3The increased operating resistance is consistent with a hotter filament.
Answer
10.0 Ω then 20.0 Ω; the coordinate ratios differ.
Example 3A diode measurement is 0.700 V and 14.0 mA. Find V/I and explain why this does not establish a constant resistance for the entire diode curve.
114.0 mA = 0.0140 A.
2V/I = 0.700/0.0140 = 50.0 Ω.
3A single operating-point ratio cannot establish proportionality over other conditions.
Answer
50.0 Ω at this point; it is not a universal diode resistance.

Common mistakes to avoid

Connecting both meters in series.
Current is measured in the path and voltage across the component.
Calling every I–V slope resistance.
Check axes and whether the relationship is linear.

Key takeaways

  • You can collect paired current and potential-difference readings.
  • You can compare constant-temperature resistor, filament and diode behaviour.
  • You can distinguish operating resistance from a non-linear tangent gradient.

Apply it in an exam

Question 1: A diode measurement is 0.700 V and 14.0 mA. Find V/I and explain why this does not establish a constant resistance for the entire diode curve.

Show answer and reasoning

50.0 Ω at this point; it is not a universal diode resistance.

  • 14.0 mA = 0.0140 A.
  • V/I = 0.700/0.0140 = 50.0 Ω.
  • A single operating-point ratio cannot establish proportionality over other conditions.

Explore further

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

  • Why does a filament lamp’s resistance increase at higher input?
  • How should operating resistance be obtained from a non-linear I–V point?

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 10 lessons