Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current

4.3.3 Action and Use of Circuit Components

Learning Objective

Students will be able to explain and use the relationship that, for a constant current, the potential difference (p.d.) across a conductor increases in direct proportion to its resistance (R). They will also be able to apply related formulas for resistance, power rating, and potential dividers, and to recognise the different behaviour of non‑ohmic components.

Key Concepts

• Potential difference (p.d.) – voltage (V): energy transferred per unit charge, measured in volts (V).
• Current (I): rate of flow of charge, measured in amperes (A).
• Resistance (R): opposition to the flow of current, measured in ohms (Ω).
• Resistivity (ρ): intrinsic property of a material, measured in Ω·m.
• Power rating (P₍rated₎): maximum power a resistor can dissipate safely, expressed in watts (W).

Ohm’s Law and Direct Proportionality

Ohm’s law links p.d., current and resistance:

V = I R

When the current is kept constant (I = const.), the equation simplifies to

V ∝ R

Thus, if the resistance of a conductor is increased while the current remains unchanged, the p.d. across it must increase in the same ratio.

Graphical illustration

Derivation of the Resistance Formula

For a uniform conductor the resistance depends on material, length and cross‑sectional area:

R = ρ L / A

• ρ (resistivity) – property of the material (Ω·m).
• L (length) – resistance is directly proportional to length (R ∝ L).
• A (cross‑sectional area) – resistance is inversely proportional to area (R ∝ 1/A).
• Temperature – for most metals, ρ (and therefore R) increases with temperature.

Resistivity of common materials

Variable Potential Divider

A potential divider consists of two resistors, R₁ and R₂, in series across a supply voltage Vₛ. The output voltage taken across R₂ is

V₂ = Vₛ · R₂ / (R₁ + R₂)

Re‑arranging gives the relationship explicitly required by the syllabus:

R₁ = V₁ R₂ / V₂

where V₁ is the voltage across R₁. This formula is used to calculate the required resistance when a particular output voltage is needed.

Example

A 12 V supply is connected to R₁ = 5 Ω and R₂ = 15 Ω. Find the voltage across R₂ and the value of R₁ that would give a 9 V output.

1. Voltage across R₂:

   V₂ = 12 V · 15 Ω / (5 Ω + 15 Ω) = 9 V

2. To obtain 9 V across a resistor R₂ = 15 Ω with the same supply, the required R₁ is

   R₁ = V₁ R₂ / V₂ = (12 V – 9 V) · 15 Ω / 9 V = 5 Ω

Power Dissipation and Power Rating of Resistors

The electrical power converted to heat in a resistor is

P = V I = I²R = V² / R

For a given current, a higher‑resistance resistor has a larger voltage drop but dissipates more power because P = I²R. Conversely, for a given voltage, a higher‑R resistor dissipates less power (P = V²/R).

Selecting a suitable resistor – the rated power must satisfy

P₍rated₎ ≥ I²R (or P₍rated₎ ≥ V² / R)

This is an explicit AO2 requirement in the syllabus.

Non‑linear (Ohm‑law‑violating) Components

Diodes and Light‑Emitting Diodes (LEDs) are covered in the supplementary section 4.3.1. Their V‑I characteristics are curved:

• In the forward‑bias region a small increase in p.d. can cause a large increase in current once the “knee” voltage is reached.
• In reverse‑bias only a very small leakage current flows (until breakdown).

Suggested activity: Sketch a typical diode V‑I curve, label the forward‑bias knee (≈0.7 V for a silicon diode) and the reverse‑bias region.

Practical Series‑Resistor Experiment

Goal: verify experimentally that, for a constant current, the p.d. across each resistor is proportional to its resistance.

1. Connect a 12 V battery to a series circuit containing an ammeter, a fixed resistor R₁, and a variable resistor R₂.
2. Adjust R₂ so that the ammeter reads I = 0.5 A. Record the reading on a voltmeter placed across each resistor.
3. Calculate the expected p.d. using V = I R and compare with the measured values.
4. Typical results:

   • R₁ = 24 Ω → expected V₁ = 0.5 A × 24 Ω = 12 V
   • Adding R₂ = 48 Ω (in series) gives V₂ = 0.5 A × 48 Ω = 24 V (the voltmeter will read 24 V across R₂ while the total supplied voltage is now 36 V, so a higher‑voltage source would be used in practice).

This hands‑on activity reinforces the proportional relationship V ∝ R for a constant current.

Safety Considerations

• Never touch a resistor that is heating under constant current – it can cause burns.
• Switch off the supply before changing resistor values or reconnecting the circuit.
• Choose a resistor whose power rating satisfies P₍rated₎ ≥ I²R (or V²/R) to avoid overheating.
• Use appropriate protective devices (e.g., a fuse) in circuits where a resistor could fail; this links to the broader safety topic in section 4.4.

Summary

• For a constant current, V = I R ⇒ p.d. and resistance are directly proportional.
• Resistance of a uniform conductor: R = ρ L / A (depends on material, length, area, temperature).
• Potential divider relationship required by the syllabus: R₁ = V₁ R₂ / V₂.
• Power dissipated: P = V I = I²R = V² / R. Select a resistor with a rated power ≥ calculated dissipation.
• Diodes/LEDs are non‑ohmic; their V‑I curves are curved, not straight lines.
• Practical verification: use a voltmeter across each resistor in a series circuit to confirm V ∝ R for a constant current.
• Safety: avoid hot components, switch off supplies, use correctly rated resistors, and incorporate protective devices where appropriate.

Practice Questions

1. A copper wire 2 m long has a cross‑sectional area of 1.0 × 10⁻⁶ m². Calculate its resistance. (ρ_Cu = 1.68 × 10⁻⁸ Ω·m.)
2. If a constant current of 3 A flows through the wire in (1), what p.d. appears across it?
3. Explain qualitatively what happens to the p.d. across a heating element if its temperature rises, assuming the current remains constant.
4. In a potential divider, a 12 V supply is connected to R₁ = 5 Ω and R₂ = 15 Ω. Find the voltage across R₂ and the value of R₁ that would give the same voltage if the required output is 9 V.
5. A resistor of 10 Ω carries a current of 0.2 A. Calculate the power dissipated and state whether a 0.5 W resistor is sufficient.
6. Sketch a typical silicon diode V‑I curve. Indicate the forward‑bias knee voltage and the region of reverse‑bias leakage.
7. In the series‑resistor experiment described above, the ammeter reads 0.5 A and the voltmeter across a 48 Ω resistor reads 24 V. Verify that the measured p.d. agrees with the theoretical value.