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Resistance (R = V / I)

Key idea: Define resistance and Ohm’s law, use R = V/I and V = IR, and apply the length and cross-sectional area proportionalities for wires (O Level).

  • SEC G3 Physics 2027
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Learning objectives

  • State current as rate of charge flow measured in amperes
  • Distinguish conventional current from electron flow
  • Apply charge equals current multiplied by time
  • Define source e.m.f. as work done per unit charge around a circuit
  • Calculate total e.m.f. for sources in series
  • Define component potential difference as work done per unit charge
  • State resistance as potential difference divided by current
  • Apply resistance equals potential difference divided by current
  • Apply wire-resistance proportionalities for length and cross-sectional area
  • Describe the effect of temperature on metallic resistance
  • Sketch and interpret required current–voltage characteristics

1. Definition

A. Resistance

Resistance, R (Ω), is a measure of how much a component opposes the flow of current.

B. Resistance equation and Ohm’s law

Resistance at an operating point is defined by:

R = V/I or V = IR

Ohm’s law is the more specific statement that current is directly proportional to potential difference for a metallic conductor when temperature and other physical conditions remain constant. Its resistance is therefore constant and its I–V graph is a straight line through the origin.

What you need for this course

You should be able to use R = V/I and the proportional relationships between a wire’s resistance, length and cross-sectional area.

2. Key Ideas

  • Unit: 1 Ω = 1 V A⁻¹.
  • Rearrangements: V = IR, I = V/R.
  • For a wire of the same material (and at the same temperature):
    • R ∝ L (longer wire → larger resistance)
    • R ∝ 1/A (thicker wire → smaller resistance)
  • For a metallic conductor, resistance increases when temperature increases.
Resistance and wire dimensionsComparison of a long thin wire and a short thick wire to show how resistance depends on length and cross-sectional area.Long, thin wireL largeR highShort, thick wireL smallA largeR low
For the same material and temperature: longer wire means higher resistance, thicker wire means lower resistance.

3. Detailed Explanations

A. What does resistance do in a circuit?

A larger resistance means it is harder for charges to flow, so for the same voltage you get a smaller current:

I = V/R

B. Factors affecting resistance of a wire (qualitative)

For the same material:

  • longer wire → more opposition to current → bigger R
  • thicker wire (larger cross-sectional area) → more “paths” for charge → smaller R

C. Temperature effect (metals)

In metals, higher temperature makes the lattice ions vibrate more. Electrons collide more often, so resistance increases.

Link

How resistance appears on I–V graphs is covered in: Metallic Conductor (Ohmic), Filament Lamp, and Diode.

4. Common Mistakes

  • Mixing up the unit: resistance is in Ω, not V or A.
  • Calling every use of R = V/I “Ohm’s law”. The equation defines resistance at a point; Ohm’s law requires constant resistance and an I ∝ V relationship.
  • Treating “thicker wire” as “bigger diameter → area doubles” (area depends on diameter²).

5. Exam Tips

  1. Write the relationship you need: V = IR / R = V/I / I = V/R.
  2. Substitute with units (V, A, Ω), then compute and state the unit.
  3. For wire comparisons (same material): use R ∝ L and R ∝ 1/A.
  4. For metals and temperature: “temperature increases → more collisions → R increases”.

6. Worked Examples

Modelled example 1

Finding resistance

Core

Problem

A resistor has 6.0 V across it and a current of 0.50 A through it. Find its resistance.
Study the worked solution

  1. Use the resistance definition

    Method

    Divide potential difference by current.

    Reason

    Resistance at an operating point is R = V/I.

    Working

    R = 6.0/0.50 = 12 Ω

  2. Interpret the result

    Method

    State that each ampere corresponds to 12 V across this resistor at the stated point.

    Reason

    1 Ω = 1 V A⁻¹.

    Working

    12 Ω = 12 V A⁻¹.

Guided practice 2

Finding current

About 4 min

Problem

A 12 Ω resistor is connected across a 6.0 V supply. Find the current.

Make current the subject

Unit: A

Hints

Hint 1: start from V = IR
Make I the subject.
Hint 2: divide voltage by resistance
Calculate 6.0/12.
View solution step by step

  1. Calculate current

    Method

    Divide the p.d. by resistance.

    Reason

    At fixed p.d., greater resistance permits less current.

    Working

    I = V/R = 6.0/12 = 0.50 A

Common misconception 3

Finding voltage

Find and correct the mistake

Learner response

A current of 0.20 A flows through a 15 Ω resistor. A learner calculates 0.20/15 for voltage. Diagnose the rearrangement and find the p.d.

Use units to check the rearrangement

Unit: V

View solution step by step

  1. Make voltage the subject

    Method

    Multiply current by resistance.

    Reason

    R = V/I rearranges to V = IR.

    Working

    V = (0.20)(15) = 3.0 V

  2. Check dimensions

    Method

    Confirm amperes times ohms gives volts.

    Reason

    Unit consistency helps expose the incorrect division.

    Working

    A · Ω = V

Examiner practice 4

Comparing wires (same material)

3 marks

Examination question

Wires A and B have the same material, temperature and cross-sectional area. A is twice as long as B. Compare their resistances and explain microscopically. [3 marks]

State proportionality, ratio and physical reason

View solution step by step

  1. Apply the proportionality

    2 marks

    Method

    Use R ∝ L to state R_A = 2R_B.

    Reason

    Material, temperature and area are held constant.

    Working

    R_A/R_B = L_A/L_B = 2

  2. Explain the larger resistance

    1 mark

    Method

    State that electrons travel through more lattice and undergo more collisions in the longer wire.

    Reason

    More scattering opposes charge flow.

    Working

    Longer path → more collisions → larger R.

Challenge 5

Unit conversion (mA to A)

Minimal support

Prefix-conversion transfer

A resistor has 12 V across it and a current of 200 mA. Find its resistance.

Convert current before substitution

Hints

Hint 1: convert the prefix
1000 mA = 1 A.
Hint 2: then divide
Use R = V/I with current in amperes.
View solution step by step

  1. Convert current

    Method

    Express 200 mA as 0.200 A.

    Reason

    Ohms follow from volts divided by amperes.

    Working

    200 mA = 0.200 A

  2. Calculate resistance

    Method

    Divide p.d. by the converted current.

    Reason

    Resistance at the operating point is V/I.

    Working

    R = 12/0.200 = 60 Ω

7. Mind Stretchers

Mind stretcher 1: Diameter changeExtension

Two wires are the same material and same length. Wire A has twice the diameter of Wire B. Compare their resistances.

Show Answer

Cross-sectional area A ∝ d², so doubling diameter makes area 4 × bigger. Since R ∝ 1/A, R_A = (1/4)R_B.

Mind stretcher 2: Stretching a wire (constant volume)Extension

A wire is stretched so its length doubles, but its volume stays constant. Compare the new resistance to the original resistance.

Show Answer

Constant volume means AL is constant, so if L doubles, A halves.

Since R ∝ L/A, the resistance becomes:

R' ∝ 2L/(A/2) = 4L/A

So the resistance becomes 4× the original.

8. Practice and next step

Complete the resistance and wire-comparison questions in Structured Current Electricity. Then continue to the metallic conductor I–V graph to see how constant resistance appears graphically.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027