In 1827 Georg Ohm found a beautifully simple rule for the metal wires he was testing: double the voltage and the current doubles too. That single straight line, current proportional to voltage, is the backbone of every circuit calculation you will ever do. But the rule has limits, and the most interesting components break it on purpose.
You connect a battery to a metal wire and measure the current. Then you swap in a fresh battery with twice the voltage.
Before reading: what do you predict happens to the current, does it stay the same, double, or quadruple? Write your prediction and one sentence saying why.
One equation links the push, the flow and the opposition
Voltage is the push, current is the flow it produces, and resistance is whatever gets in the way, and the three are locked together by a single equation.
For any component, resistance is defined as the ratio of the voltage across it to the current through it. Rearranged, this gives Ohm's law in its familiar form:
$$V = IR \qquad\Longleftrightarrow\qquad R = \frac{V}{I} \qquad\Longleftrightarrow\qquad I = \frac{V}{R}$$
where $V$ is in volts, $I$ in amperes and $R$ in ohms. For a fixed resistance, increasing the voltage increases the current in direct proportion: double $V$ and you double $I$.
$V = IR$ links voltage, current and resistance. For a constant resistance, current is directly proportional to voltage. $R = V/I$ always defines the resistance at that instant, even when $R$ is not constant.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: for a resistor of fixed resistance, doubling the voltage across it doubles the current through it.
Picking the right rearrangement for the unknown you want
We just saw that $V$, $I$ and $R$ are tied together. That raises a question: if you know any two of them, how do you find the third? This card answers it → rearrange $V = IR$ to make the unknown the subject and substitute carefully.
Once you have the equation, every numerical circuit question becomes the same three steps: list what you know, choose the rearrangement, and substitute.
The key is to use consistent units (volts, amperes, ohms) and to be sure both measurements refer to the same component.
A resistor carries a current of $0.25\ \text{A}$ when the voltage across it is $6.0\ \text{V}$. Find its resistance.
To solve a circuit problem: list $V$, $I$, $R$ in consistent units, rearrange $V = IR$ for the unknown ($R = V/I$ or $I = V/R$), then substitute. Always confirm the two quantities refer to the same component.
Add the highlighted method to your notes before the check below.
Quick check: A $12\ \text{V}$ supply is connected across a $48\ \Omega$ resistor. What current flows?
An ohmic conductor gives a straight line through the origin; anything else does not
Use the simulation. Record the filament lamp at $2\ \text{V}$, $4\ \text{V}$ and $6\ \text{V}$. What happens to its resistance, and why?
Use the simulation. True or false: because $V = IR$ can be used to work out the lamp's resistance at every point, the lamp is an ohmic conductor.
Pattern A, Plan then justify
You are given a power supply, a rheostat (variable resistor), an ammeter, a voltmeter and a length of nichrome wire. In your book: (1) draw the circuit you would build, showing where the ammeter and voltmeter go; (2) describe how you would vary the voltage and what you would record each time; (3) sketch the V–I graph you expect and state how you would read the resistance off it.
Odd one out: three of these are correct rearrangements of $V = IR$. Which one is not?
Dot point: Analyse the relationships between voltage, current and resistance using V = IR for electric circuits having constant and variable resistance
When the straight line bends, the resistance is changing
We just learned to use $V = IR$ on a single component. That raises a question: does the resistance ever change while we test it? This card answers it → for ohmic conductors it stays constant, but for non-ohmic ones it does not.
A metal at fixed temperature gives the clean straight line Ohm found, but heat a filament until it glows and its resistance climbs, bending the graph.
An ohmic conductor (such as a metal wire held at constant temperature) has constant resistance, so its V–I graph is a straight line through the origin. A non-ohmic conductor has a resistance that changes with conditions, so its V–I graph curves. A filament lamp is the classic example: as current rises the filament heats up, the metal's resistance increases, and the graph bends so that less extra current flows for each extra volt.
An ohmic conductor has constant resistance and a straight V–I line through the origin. A non-ohmic conductor (e.g. a filament lamp) changes resistance with conditions, giving a curved V–I graph, the lamp's resistance rises as it heats.
Pause, copy the highlighted comparison into your book before the check below.
Last check: Decide if the statement is true or false.
A filament lamp is an ohmic conductor because its V–I graph passes through the origin.
Match the pairs: Match each experimental observation to what it tells you about the conductor.
Dot point: conduct a laboratory experiment to examine the relationships between voltage and current for ohmic conductors and non-ohmic conductors
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
UnderstandBand 4(4 marks) 1. A student tests a metal resistor and a filament lamp, plotting voltage against current for each. Explain, with reference to Ohm's law, why one graph is a straight line and the other curves.
1 mark: states $V = IR$ / resistance = $V/I$ · 1 mark: metal resistor has constant R → straight line · 1 mark: lamp heats up as current rises · 1 mark: rising temperature increases R → graph curves
ApplyBand 4(3 marks) 2. A resistor carries a current of $0.25\ \text{A}$ when connected across a $9.0\ \text{V}$ supply. Calculate its resistance, and determine the current if the supply voltage is increased to $15\ \text{V}$, assuming the resistance is unchanged.
1 mark: $R = V/I = 36\ \Omega$ · 1 mark: rearranges to $I = V/R$ · 1 mark: $I = 0.42\ \text{A}$ (2 s.f.)
Dot point: Analyse the relationships between voltage, current and resistance using V = IR for electric circuits having constant and variable resistance
AnalyseBand 5(3 marks) 3. A student plots $V$ against $I$ for a filament lamp and obtains a curve that becomes steeper at higher currents. Explain what this shows about the lamp, and contrast it with the graph expected for an ohmic conductor.
1 mark: the gradient of a $V$-$I$ graph is the resistance · 1 mark: a steepening curve means the resistance increases with current · 1 mark: the filament heats up, raising its resistance, so it is non-ohmic; an ohmic conductor gives a straight line through the origin
Dot point: conduct a laboratory experiment to examine the relationships between voltage and current for ohmic conductors and non-ohmic conductors
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Ohm's law, $V = IR$, means the gradient of a voltage–current graph (with $V$ on the vertical axis) equals the resistance $R$. The metal resistor is an ohmic conductor: held at roughly constant temperature its resistance stays constant, so $V$ and $I$ are directly proportional and the graph is a straight line through the origin. The filament lamp is non-ohmic: as the current increases, the filament heats up, and the resistance of the metal increases with temperature. Because $R$ is no longer constant, less additional current flows for each extra volt, so the graph curves upward rather than following a straight line.
Q2 (3 marks): From Ohm's law, $R = V/I = 9.0/0.25 = 36\ \Omega$. If the resistance stays constant and the supply is raised to $15\ \text{V}$, the new current is $I = V/R = 15/36 = 0.42\ \text{A}$ to two significant figures. The assumption that $R$ is unchanged is the assumption that the resistor is ohmic and that its temperature has not risen appreciably; if the resistor heated up, its resistance would increase and the current would be a little lower than this.
Q3 (3 marks): On a graph of $V$ against $I$, the gradient at any point is $V/I$, which is the resistance. A curve that becomes steeper at higher currents therefore shows that the resistance of the lamp is increasing as the current increases. The physical reason is heating: a larger current dissipates more power in the filament, its temperature rises sharply, the lattice ions vibrate more and collide with the drifting electrons more often, so the resistance rises. Because its resistance is not constant, the filament lamp is a non-ohmic conductor. An ohmic conductor held at constant temperature would instead give a straight line passing through the origin, because $V$ and $I$ are directly proportional and the gradient, the resistance, is the same everywhere on the line.
You can now use $V = IR$ to solve circuit problems and tell ohmic from non-ohmic conductors. Next lesson uses this to work out electrical power with $P = VI$ and $P = V^2/R$.
↩ Back to focus-area hubAt the start you predicted what doubling the voltage does to the current. For an ohmic conductor the answer is the current doubles, because $I = V/R$ with $R$ constant. If you predicted "stays the same" or "quadruples", revisit Card 1, current is directly proportional to voltage when resistance is fixed.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
Tell us what you would like help with. Register interest; support details will be confirmed before enrolment.
Register interest →