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This lesson covers Ohm's law and the equation V=IR, as required by the Edexcel GCSE Combined Science specification (1SC0). You will learn what Ohm's law states, how to use and rearrange the equation, and the difference between ohmic and non-ohmic conductors.
The relationship between potential difference, current and resistance is:
V=IR
where:
This equation applies to any component in a circuit — it is the definition of resistance.
You will frequently need to rearrange this equation:
| To find... | Rearranged form |
|---|---|
| Voltage (V) | V=IR |
| Current (I) | I=RV |
| Resistance (R) | R=IV |
Exam Tip: Use a formula triangle to help you rearrange. Place V at the top, I at the bottom left and R at the bottom right. Cover the quantity you want to find — the remaining two show you the calculation.
Ohm's law states:
The current through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant.
This means:
An ohmic conductor is a component where the current is directly proportional to the voltage at constant temperature. The I-V graph is a straight line through the origin.
| Examples of Ohmic Conductors |
|---|
| A wire at constant temperature |
| A fixed resistor (at constant temperature) |
For an ohmic conductor, the resistance is constant — it does not change as the current or voltage changes.
A non-ohmic conductor does not obey Ohm's law — its resistance changes as conditions change.
| Component | Behaviour |
|---|---|
| Filament lamp | Resistance increases as the filament heats up |
| Diode | Very high resistance in one direction; low resistance in the other |
| Thermistor | Resistance changes with temperature |
| LDR | Resistance changes with light intensity |
These components have curved I-V graphs (filament lamp) or strongly asymmetric graphs (diode).
A 12 V battery is connected to a 4 Ω resistor. Calculate the current.
I=RV=412=3 A
A current of 2.5 A flows through a heater connected to a 230 V supply. Calculate the resistance.
R=IV=2.5230=92Ω
A current of 0.5 A flows through a 20 Ω resistor. Calculate the potential difference across it.
V=IR=0.5×20=10 V
Exam Tip: Always show your working clearly: write the equation, substitute the values, and give the final answer with the correct unit. Even if you make an arithmetic error, you can still gain marks for method.
A 9 V battery is connected in series with a 3 Ω resistor and a 6 Ω resistor.
Step 1 — Total resistance:
Rtotal=3+6=9Ω
Step 2 — Current:
I=RtotalV=99=1 A
Step 3 — Voltage across each resistor:
V1=1×3=3 V
V2=1×6=6 V
Check: 3+6=9 V (correct).
A 12 V battery is connected to two resistors in parallel: 4 Ω and 6 Ω.
Current through each branch:
I1=412=3 A
I2=612=2 A
Itotal=3+2=5 A
For metallic conductors:
This is why a filament lamp does not obey Ohm's law — the filament gets very hot, so its resistance changes as the current increases.
A student records the following readings from a component. Is it an ohmic conductor?
| V / V | I / A |
|---|---|
| 1.0 | 0.05 |
| 2.0 | 0.10 |
| 3.0 | 0.15 |
| 4.0 | 0.20 |
Compute R=V/I at each point:
| V / V | I / A | R / Ω |
|---|---|---|
| 1.0 | 0.05 | 20 |
| 2.0 | 0.10 | 20 |
| 3.0 | 0.15 | 20 |
| 4.0 | 0.20 | 20 |
Resistance is constant, so the component is an ohmic conductor. The I-V graph would be a straight line through the origin with gradient 1/R=0.05 A/V.
A filament lamp gives the following readings.
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