Ohm's law and V-I graphs

Ohm's law states that, at constant temperature and with all other physical conditions unchanged, the current $I$ through a metallic conductor is directly proportional to the potential difference $V$ across its ends. In symbols:

where: $V$ is the potential difference in volts (V), $I$ is the current in amperes (A), and $R$ is the resistance of the conductor in ohms ($\Omega$), the constant of proportionality. Rearranging gives $R = V/I$: resistance is simply the ratio of potential difference to current, and it measures how strongly the conductor opposes the flow of charge. One ohm is defined as the resistance of a conductor across which 1 volt drives a current of exactly 1 ampere. A conductor that obeys this proportionality is called ohmic — a pure metal wire such as nichrome or copper, held at constant temperature, is the standard example. Because $V$ and $I$ grow in lockstep, plotting $V$ against $I$ gives a straight line through the origin, and the slope of that line is $R$ itself. A conductor whose $V$-$I$ graph is not a straight line through the origin — whose $R = V/I$ changes from point to point — is non-ohmic; a filament bulb, whose tungsten heats and resists more as current rises, is the classroom example. The figure below puts both behaviours on one plot: the straight ohmic line and the flattening non-ohmic curve.

The goal is to show, from a page of raw voltmeter and ammeter readings taken across an unknown conductor, exactly how to establish whether that conductor obeys Ohm's law and, if it does, extract its single resistance value — the everyday task any V-I graph question in this chapter is really testing.

Step 1. Take several readings of $V$ and the corresponding $I$ across the same conductor, keeping the temperature constant (a rheostat lets you vary $V$ without letting the wire heat appreciably). Each pair $(I, V)$ is one data point.

Step 2. Compute the ratio $V/I$ for every pair. If Ohm's law holds, this ratio is the same number every time — it does not depend on which pair you pick, because $V$ and $I$ are scaling together.

Step 3. Plot $V$ on the y-axis against $I$ on the x-axis, following the ICSE convention. If the ratio $V/I$ was constant in Step 2, the points must lie on a straight line through the origin, since $V = IR$ with fixed $R$ is exactly the equation of such a line.

Step 4. Read off the resistance as the slope of that line, $R = \Delta V/\Delta I$ (rise over run). This single number then predicts $V$ or $I$ for any future setting of the circuit, via $V = IR$, without needing to take another reading.

A student sets up a circuit with a nichrome wire, an ammeter, a voltmeter, a rheostat and a battery, and records: $V$ (volts): 1.0, 2.0, 3.0, 4.0 — $I$ (amperes): 0.20, 0.40, 0.60, 0.80.

Step 1. Check proportionality by computing $V/I$ at each point: $1.0/0.20 = 5\ \Omega$, $2.0/0.40 = 5\ \Omega$, $3.0/0.60 = 5\ \Omega$, $4.0/0.80 = 5\ \Omega$. The ratio is the same every time.

Step 2. Since $V/I$ is constant, the wire obeys Ohm's law: it is an ohmic conductor with resistance $R = 5\ \Omega$, and a plot of these four points would sit on a straight line through the origin with slope $5\ \Omega$.

Step 3. Use $R$ to predict beyond the table. If the potential difference is raised to $6.0\ \text{V}$, the current should become $I = V/R = 6.0/5 = 1.2\ \text{A}$ — no new measurement needed, only the equation.

Step 4 (check against a non-ohmic conductor). A torch bulb tested the same way gives $V/I$ ratios of $5.0\ \Omega$, $6.25\ \Omega$, $7.5\ \Omega$ and $8.7\ \Omega$ at rising voltages of $0.5$-$2.0\ \text{V}$ — the ratio climbs steadily rather than staying fixed, because the tungsten filament heats up and its resistance rises with temperature. The nichrome wire is ohmic with $R = 5\ \Omega$; the bulb is non-ohmic, and no single $R$ describes it everywhere on its curve.

Competitive stretch

If the same nichrome wire (still $R = 5\ \Omega$) is instead connected so that the ammeter reads $I = 2.5\ \text{A}$, what voltage must the source supply, and would the wire still behave ohmically at this higher current? Since resistance is a property of the material and geometry at fixed temperature, not of $I$ itself, $V = IR = 2.5 \times 5 = 12.5\ \text{V}$ — provided the wire has not heated up enough to change its resistance. In practice, at large enough currents even nichrome would start to warm and drift from perfectly ohmic behaviour, which is exactly why the "constant temperature" clause in Ohm's law matters.

• Treating Ohm's law as a universal law true for every conductor — it is a specific property of ohmic materials under constant physical conditions, not a law every circuit element must obey.
• Confusing "the graph's slope gives R" with "R is the slope everywhere" — this only works when the graph is a straight line through the origin. For a curved (non-ohmic) graph, $V/I$ at one point gives the resistance only at that operating point.
• Forgetting that temperature is the hidden variable behind Ohm's law: the same nichrome wire that behaves ohmically at room temperature would stop doing so once current made it hot enough to glow.
• Swapping the axes — by ICSE convention $V$ goes on the y-axis and $I$ on the x-axis, so resistance is $\Delta V/\Delta I$ (rise over run), not the reciprocal.

Ask the tutor: Want to practise reading an unfamiliar V-I table and deciding ohmic or non-ohmic in under a minute? Ask the AI tutor to drill you with a few more examples.