Year 9 Science Unit 3 — Energy Block 3: Electrical Energy SC5-EGY-01 ⏱ ~35 min Lesson 19 of 24

Ohm's Law Investigation

In 1827, German physicist Georg Ohm discovered a simple but powerful relationship: the current through a conductor is directly proportional to the voltage across it. This relationship — V = IR — is the foundation of circuit design. From the wiring in your walls to the microchips in your phone, Ohm's Law governs how electricity behaves. In this lesson, you will investigate this law interactively, calculate unknown values, and discover why not all materials obey Ohm's simple rule.

📐

Think First

🤔

Before you begin, predict:

If you double the voltage across a resistor, what happens to the current? And if you double the resistance while keeping voltage constant, what happens to the current? Make two predictions with reasoning, then use the calculator below to test them.

Core Concepts

Core Concept

Ohm's Law: V = IR

Ohm's Law states that for many conductors (called ohmic conductors), the voltage across the conductor is directly proportional to the current flowing through it:

V = I × R

Voltage (volts) = Current (amps) × Resistance (ohms)

This means:

If you increase voltage across a fixed resistor, current increases proportionally
If you increase resistance at a fixed voltage, current decreases proportionally
If you double both voltage and resistance, current stays the same

The magic triangle helps you rearrange the formula:
Cover the variable you want to find:
• Cover V → V = I × R
• Cover I → I = V ÷ R
• Cover R → R = V ÷ I

🧮 Ohm's Law Calculator

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

Voltage (V): volts

Current (I): amps

Resistance (R): ohms (Ω)

Enter two values to calculate the third.

Working Scientifically

Investigating Ohm's Law

To verify Ohm's Law in the laboratory, you would set up a circuit like this:

Method:

Set up the circuit with a variable power supply, a resistor of known value, an ammeter in series, and a voltmeter in parallel across the resistor.
Start at a low voltage (e.g., 1 V) and record the current.
Increase the voltage in steps (1 V, 2 V, 3 V, 4 V, 5 V) and record the current at each step.
Plot a graph of voltage (y-axis) vs current (x-axis).
For an ohmic conductor, the graph should be a straight line through the origin. The gradient equals the resistance.

Safety: Always start with the power supply at minimum voltage. Never exceed the resistor's power rating (P = V × I). Use heat-resistant mats if resistors become warm.

📊 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.

Extension

When Ohm's Law Breaks Down

Not all conductors obey Ohm's Law. Materials where V is not proportional to I are called non-ohmic conductors.

Light bulb: As current increases, the filament heats up. Hot metal has higher resistance than cold metal. So the V-I graph curves — the gradient (resistance) increases with voltage.

Diode / LED: A diode only allows current in one direction. Below a threshold voltage (~0.7 V for silicon, ~2 V for LEDs), almost no current flows. Above the threshold, current increases rapidly. The graph shows a sharp "knee" rather than a straight line.

Thermistor: A temperature-sensitive resistor. Its resistance decreases as temperature rises — the opposite of a light bulb. Used in digital thermometers and engine temperature sensors.

🇦🇺

Australian Context

Ohm's Law in the NEM

The National Electricity Market (NEM) that powers eastern Australia operates on principles that extend Ohm's Law to grid scale. When demand spikes on a hot summer afternoon — millions of air conditioners turning on simultaneously — the effective "resistance" of the grid drops (more parallel paths drawing current) and the system must supply more current to maintain voltage.

Grid voltage must stay within 230–250 V. If voltage drops too low (a "brownout"), motors in refrigerators and air conditioners draw more current to compensate — which can overheat and damage them. This is why AEMO dispatches peaking power stations and batteries within minutes when demand surges: they inject extra "push" (voltage) to keep the grid stable.

Fact: During the January 2024 heatwave, grid demand in NSW peaked at 14,500 MW. AEMO called on the Tallawarra B gas power station and grid batteries to inject extra current and stabilise voltage across Sydney. Ohm's Law at grid scale: V = IR, where V is 240 V, I is millions of amps summed across the network, and R is the combined resistance of every device switched on.

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.

Quick Questions

1. According to Ohm's Law, if the voltage across a resistor is doubled while the resistance stays the same, the current:

AStays the same BDoubles CHalves DBecomes zero

2. A circuit has a voltage of 9 V and a current of 0.5 A. What is the resistance?

A4.5 Ω B9.5 Ω C18 Ω D0.056 Ω

3. Which of the following is a non-ohmic conductor?

AA light bulb filament BA copper wire at constant temperature CA carbon resistor DA nichrome heating element at low current

4. In an Ohm's Law investigation, the ammeter must be connected:

AIn parallel across the resistor BIn series with the resistor COnly to the positive terminal of the battery DBetween the voltmeter and the resistor

5. The gradient of a V-I graph for an ohmic conductor represents:

AThe power dissipated BThe current at 1 volt CThe resistance of the conductor DThe energy transformed per second

Short Answer Questions

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.

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.

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

▼

MCQ Answers

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

2. C — R = V/I = 9/0.5 = 18 Ω.

3. A — Light bulb filaments heat up, changing resistance. Non-ohmic.

4. B — Ammeters must be in series to measure current through the component.

5. C — Gradient = 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.

Game: Doodle Jump

🔄 Revisit These Concepts

L17: Circuit Basics L18: Series & Parallel L14: The Grid L15: Storage L16: Checkpoint 2

🦘

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 MW — enough 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.

🏉

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.

Lesson Complete

Mark this lesson as finished to unlock the next one.

← Lesson 18: Series & Parallel Lesson 20: Checkpoint 3 →