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📖 Lesson 19 ⏱ ~30 min Year 9 · Unit 3 ⚡ +100 XP

Ohm's Law Investigation

In 1827, Georg Ohm measured 13 different wire lengths and found that doubling voltage exactly doubled current, but only for some materials.

Today's hook: Georg Ohm published his famous law in 1827 and was ridiculed by scientists who thought it was too simple to be true. Nearly 200 years later, his $V = IR$ sits at the centre of every circuit investigation on Earth. Today you will test it yourself, and find out whether it actually holds for every material.
0/5QUESTS
Warm-up
Think First
+5 XP each

Q1 · Ohm's Law states that doubling the voltage doubles the current, but a student asks "does that always work for every material?" Before reading, predict: do you think all conductors obey this rule, or might some behave differently?

Q2 · In an investigation, how would you test whether a component obeys Ohm's Law? What would you measure, what would you vary, and what result would you expect to see if it does obey the law?

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Core Concept
Ohm's Law: V = IR
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Set up a simple circuit with a battery, a resistor, and a voltmeter and ammeter. Connect a 1.5 V battery, the ammeter reads, say, 0.15 A. Swap in a 3 V battery, the reading doubles to 0.30 A. Swap in a 6 V battery, it doubles again to 0.60 A. Every time you double the voltage, the current doubles, in a perfectly constant ratio. That ratio is resistance, and the relationship is Ohm's Law: V = I × R, where V is voltage in volts, I is current in amps, and R is resistance in ohms. If you know any two of these quantities, you can calculate the third.

For many materials, called ohmic conductorsthis relationship is linear. Double the voltage across a resistor and the current doubles. Triple the voltage and the current triples. This proportionality makes circuits predictable and designable. However, not all materials are ohmic. Diodes, LEDs and transistors have non-linear behaviour.

Example

A 12 V car battery connected to a 4 Ω starter motor draws I = V/R = 12/4 = 3 A. If the battery voltage drops to 10 V during cranking, the current falls to 2.5 A and the motor turns more slowly.

What to write in your book
  • Ohm's Law: V = I × R
  • For ohmic conductors, current is directly proportional to voltage
  • Not all materials are ohmic, resistance can change with temperature
Move the sliders+5 XP

Drag the sliders to explore Ohm's Law. Watch the current update live.

Current I
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From the lesson
Calculator

🧮 Ohm's Law Calculator

Enter any two values to calculate the third. Use the magic triangle above if you get stuck.

volts
amps
ohms (Ω)
Enter two values to calculate the third.
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Working Scientifically
Investigating Ohm's Law
+5 XP

Ohm's Law lets you predict circuit behaviour before you build it. Engineers use it to select appropriate wire thicknesses, calculate power dissipation, and design safety systems. The law is not universal, it fails for superconductors (zero resistance), semiconductors, and components like light bulbs whose resistance changes with temperature.

The power dissipated in a resistor can also be calculated from Ohm's Law. Combining P = V × I with V = I × R gives P = I²R and P = V²/R. These forms are essential for checking whether a component will overheat.

Example

A phone charger labelled 5 V, 2 A delivers P = 5 × 2 = 10 W. The internal resistance of the charger circuit must be low enough that most of this power reaches the phone rather than being wasted as heat in the charger itself.

Watch out

Students often think Ohm's Law applies to everything electrical. It does not. A light bulb filament has much higher resistance when hot than when cold, so its current is not proportional to voltage. Batteries have internal resistance that changes with charge level. Real circuits are more complex than ideal resistors.

What to write in your book
  • Always state Ohm's Law before substituting numbers
  • Power in a resistor: P = I²R = V²/R
  • Check whether the material is approximately ohmic
Fill the blanks+4 XP

Complete this Ohm's Law calculation.

A resistor has V = V across it and R = Ω. Using Ohm's Law, I = V ÷ R = ÷ = A. If the voltage doubles to V, the current becomes A, showing that for an ohmic conductor, current is proportional to voltage.
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From the lesson
Graph

📊 V vs I Graph, Ohmic vs Non-Ohmic

Click buttons to see how different conductors behave. Ohmic conductors show a straight line; non-ohmic conductors curve.

Current I (A) Voltage V (V) 0.5 1.0 1.5 2.0 3 6 9 12 Ohmic conductor (R = 6 Ω)
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Extension
When Ohm's Law Breaks Down
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Ohmic conductors are materials where resistance stays constant regardless of voltage or current. Copper wire, carbon resistors, and nichrome heating elements are approximately ohmic at constant temperature. This predictability makes them essential for circuit design.

Non-ohmic materials have uses too. Diodes allow current in one direction only. Light-dependent resistors change resistance with light level. Thermistors change resistance with temperature. These non-linear behaviours are the basis of sensors and electronic control systems.

Example

A dimmer switch uses a semiconductor device called a triac to chop the AC waveform. The bulb receives less average voltage and dims. This is non-ohmic behaviour that could not be achieved with a simple resistor.

What to write in your book
  • Ohmic conductors have constant resistance at constant temperature
  • Copper wire and carbon resistors are approximately ohmic
  • Diodes, LEDs and light bulb filaments are non-ohmic
Which material would most likely obey Ohm's Law over a wide range of voltages?
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Australian Context
Ohm's Law in the NEM
+5 XP

The assumption that all conductors are ohmic is one of the most common errors in introductory physics. While Ohm's Law is an excellent approximation for metals at constant temperature, it breaks down for many important materials and devices.

Even a simple torch bulb is non-ohmic. When cold, its filament has low resistance and draws a large inrush current. As it heats to over 2,000°C, its resistance increases by a factor of ten or more. The steady-state current is much lower than the initial surge. This is why bulbs often burn out when switched on, the initial current spike stresses the filament.

Example

A 60 W, 240 V bulb has a hot resistance of R = V²/P = 240²/60 = 960 Ω. But when cold, the same filament might measure only 100 Ω. The initial switch-on current is nearly 2.4 A, ten times the normal operating current of 0.25 A.

What to write in your book
  • Not all conductors are ohmic, resistance often changes with temperature
  • A cold light bulb filament has much lower resistance than a hot one
  • Real circuits require more complex analysis than ideal resistors
True or false?
For all materials, doubling the voltage always doubles the current.
Reflect
Revisit your thinking
reflect

At the start of this lesson you read that Georg Ohm published his famous law in 1827 and was ridiculed for it being too simple, yet nearly 200 years later, V = IR sits at the centre of every circuit investigation on Earth. You were challenged to test it yourself and find out whether it holds for every material.

Now that you've completed the investigation, does Ohm's Law hold up for all materials? What did your evidence show, and why did Ohm's critics turn out to be wrong?

Interactive Tool, Ohm's Law Lab Open fullscreen ↗
Ohm's Law states that V = IR. If voltage doubles and resistance stays the same, current:
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From the lesson
Think First
🤔

Before you begin, estimate:

A 12 V car battery is connected to a resistor. If the current is 3 A, what is the resistance? And if you replace the resistor with one of double the resistance but keep the same battery, what is the new current? Use V = IR. Record your estimates, then verify with the calculator.

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From the lesson
MCQ 1
1. According to Ohm's Law, if the voltage across a resistor is doubled while the resistance stays the same, the current:
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From the lesson
MCQ 2
2. A circuit has a voltage of 9 V and a current of 0.5 A. What is the resistance?
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From the lesson
MCQ 3
3. Which of the following is a non-ohmic conductor?
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From the lesson
MCQ 4
4. In an Ohm's Law investigation, the ammeter must be connected:
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MCQ 5
5. The gradient of a V-I graph for an ohmic conductor represents:
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From the lesson
SAQ 1
1. A student measures 2.0 V across a resistor and records a current of 0.4 A. Calculate the resistance. The student then increases the voltage to 6.0 V. Predict the new current and explain your reasoning using Ohm's Law. (3 marks)
💡 Hint: Step 1: R = V/I = 2.0/0.4 = 5 Ω. Step 2: New I = V/R = 6.0/5 = 1.2 A. Step 3: Explain proportionality, voltage tripled, so current triples (R constant).
✏️ Answer in your exercise book.
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From the lesson
SAQ 2
2. Describe how you would set up a practical investigation to test whether a metal resistor is ohmic. Include the circuit diagram, the variables (independent, dependent, controlled), and how you would process and represent your data. (4 marks)
💡 Hint: Independent = voltage (changed by variable PSU). Dependent = current (measured by ammeter). Controlled = resistor, temperature, same equipment. Data: table of V and I, then V-I graph. Straight line through origin = ohmic.
✏️ Answer in your exercise book.
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From the lesson
SAQ 3
3. A light bulb does not obey Ohm's Law. Explain why, using the concepts of temperature and resistance. Then explain why this non-ohmic behaviour makes light bulbs less efficient than they might appear from their power rating alone. (5 marks)
💡 Hint: Filament heats up → resistance increases → V-I graph curves. At high current, more energy goes to heat than light. Cold resistance is low, so inrush current is high when first switched on. Efficiency = useful light output / total electrical input.
✏️ Answer in your exercise book.
Model answers (click to reveal)

📖 Model Answers

MCQ Answers

1. BI = V/R; double V with constant R → double I.

2. CR = V/I = 9/0.5 = 18 Ω.

3. ALight bulb filaments heat up, changing resistance. Non-ohmic.

4. BAmmeters must be in series to measure current through the component.

5. CGradient = ΔV/ΔI = R for ohmic conductors.

SAQ 1, Ohm's Law Calculation (3 marks)

Marking Criteria: 1 mark, correct resistance calculation with units. 1 mark, correct new current with working. 1 mark, explains proportional relationship using Ohm's Law.

Model answer: Using Ohm's Law: R = V / I = 2.0 V / 0.4 A = 5 Ω. When the voltage increases to 6.0 V, the new current is I = V / R = 6.0 V / 5 Ω = 1.2 A. The current has tripled because the voltage has tripled while resistance stayed constant. This demonstrates the direct proportionality in Ohm's Law: for an ohmic conductor, current is directly proportional to voltage when resistance is unchanged.

SAQ 2, Investigation Design (4 marks)

Marking Criteria: 1 mark, valid circuit diagram description (variable PSU, resistor, ammeter in series, voltmeter in parallel). 1 mark, correct identification of all three variable types. 1 mark, describes range of measurements (multiple V-I pairs). 1 mark, describes data processing (table and V-I graph) and conclusion criteria.

Model answer: To test whether a metal resistor is ohmic, I would set up a circuit containing a variable DC power supply, the test resistor, an ammeter connected in series, and a voltmeter connected in parallel across the resistor.

The independent variable is the voltage across the resistor, which I change using the variable power supply in steps (e.g., 1 V, 2 V, 3 V, 4 V, 5 V). The dependent variable is the current through the resistor, measured by the ammeter. Controlled variables include the same resistor (to ensure consistent resistance), the same equipment, and allowing the resistor to cool between measurements (to control temperature).

I would record my results in a table with columns for Voltage (V) and Current (A). Then I would plot a graph of Voltage (y-axis) versus Current (x-axis). If the resistor is ohmic, the graph will be a straight line passing through the origin, and the gradient will equal the resistance. If the graph curves, the resistor is non-ohmic.

SAQ 3, Light Bulb Non-Ohmic Behaviour (5 marks)

Marking Criteria: 1 mark, explains heating of filament increases temperature. 1 mark, links increased temperature to increased resistance in metals. 1 mark, explains curved V-I graph / non-ohmic behaviour. 1 mark, explains energy distribution (more heat, less light at high current). 1 mark, explains efficiency implications or inrush current.

Model answer: A light bulb does not obey Ohm's Law because its filament heats up dramatically as current increases. The tungsten filament in an incandescent bulb reaches temperatures of 2,500°C. In metals, resistance increases with temperature because the vibrating metal ions scatter electrons more effectively. At low current (cold filament), resistance is low. At high current (hot filament), resistance is much higher. This means the V-I relationship is not proportional, the graph curves upward with a decreasing gradient.

This non-ohmic behaviour has important consequences for efficiency. A 60 W incandescent bulb transforms only about 5% of its electrical energy into visible light; the remaining 95% becomes heat. When the bulb first turns on, the cold filament has low resistance, causing a brief inrush current up to 10× the normal operating current. This is why bulbs often burn out at the moment of switching on, the sudden current surge stresses the filament. Modern LED bulbs avoid this problem entirely: they are non-ohmic in a different way (diode behaviour) but operate at much lower temperatures, achieving 80–90% efficiency in converting electricity to light.

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From the lesson
Additional content
1
Quick check
According to Ohm's Law, if the voltage across a resistor is doubled while the resistance stays the same, the current: A Stays the same B Doubles C Halves D Becomes zero Answer: B, V = IR, so I = V/R. If V doubles and R is constant, I doubles proportionally.
+10 XP
2
Quick check
A circuit has a voltage of 9 V and a current of 0.5 A. What is the resistance? A 4.5 Ω B 9.5 Ω C 18 Ω D 0.056 Ω Answer: C, R = V/I = 9/0.5 = 18 Ω.
+10 XP
3
Quick check
Which of the following is a non-ohmic conductor? A A light bulb filament B A copper wire at constant temperature C A carbon resistor D A nichrome heating element at low current Answer: A, A light bulb filament heats up as current increases, changing its resistance. The V-I graph is curved, not straight.
+10 XP
4
Quick check
In an Ohm's Law investigation, the ammeter must be connected: A In parallel across the resistor B In series with the resistor C Only to the positive terminal of the battery D Between the voltmeter and the resistor Answer: B, Current is the same in series, so the ammeter must be in series to measure the current flowing through the resistor.
+10 XP
5
Quick check
The gradient of a V-I graph for an ohmic conductor represents: A The power dissipated B The current at 1 volt C The resistance of the conductor D The energy transformed per second Answer: C, V = IR, so V/I = R. The gradient of V vs I is ΔV/ΔI = R.
+10 XP
Quick-fire challenge
Game time
+25 XP
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From the lesson
Revisit

🔄 Revisit These Concepts

L17: Circuit Basics L18: Series & Parallel L14: The Grid L15: Storage L16: Checkpoint 2
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From the lesson
Fun Fact
🦘
Australian Fun Fact

The Silverton Wind Farm's 200 Turbines

The Silverton Wind Farm near Broken Hill, NSW, is home to 58 giant turbines, each 200 metres tall, taller than the Sydney Harbour Bridge's pylons. Each turbine generates electricity through electromagnetic induction: wind spins the blades, which turn a shaft inside a generator. The generator contains coils of wire rotating within a magnetic field, inducing a voltage via Faraday's Law. At maximum output, Silverton produces 200 MWenough to power 137,000 Australian homes. The farm connects to the grid through a 132 kV transmission line, stepping up voltage to minimise resistive losses over the 300 km journey to consumers.

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From the lesson
Sports Science
🏉
Sports Science

Catapult Power in AFL

AFL players wear GPS tracking devices made by Australian company Catapult Sports (founded in Melbourne, now global). These small units contain accelerometers, gyroscopes, and GPS receivers, all powered by tiny lithium-polymer batteries. The devices measure player speed, distance covered, heart rate, and collision forces. A typical AFL midfielder runs 12–15 km per game, with peak speeds over 35 km/h. The battery must supply consistent voltage to sensors for 3+ hours in all weather conditions. Catapult's engineers use Ohm's Law to design power circuits that maximise battery life while maintaining data accuracy, a real-world optimisation problem where P = VI and every milliwatt counts.

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