Ohm's law and its limitations

Twist the volume knob on an old radio and, for one narrow slice of its travel, the sound rises in perfectly even steps — a rheostat behaving exactly as circuit theory promises. Push a diode into the same circuit the wrong way round and nothing happens at all, no matter how hard you turn; push it the right way past a threshold and it suddenly conducts as if a floodgate opened. Same wires, same battery, two utterly different personalities. The line separating them is one of the most quietly powerful ideas in Class 12 Physics: Ohm's law, $V = IR$ , and — just as important for the exam and for real engineering — knowing exactly where that law stops applying. Georg Simon Ohm found, by careful measurement on metal wires in 1827, that the potential difference across a conductor is directly proportional to the current through it, provided the conductor's physical state — above all its temperature — is held fixed. That single proportionality, $V \propto I$ , quietly underlies every resistor calculation you will ever do: series and parallel combinations, Kirchhoff's rules, power dissipation, the whole apparatus of circuit analysis assumes it. Yet Ohm's law is an empirical rule obeyed by some materials under some conditions, not a fundamental law of nature like conservation of charge. Filament bulbs heat up and drift off the straight line. Diodes and transistors are built specifically to defy it. Even the humble resistor stops being ohmic if you push enough current through it to warm it up. This article builds the law from its defining proportionality, reads the V–I graph the way an examiner wants it read, and then catalogues precisely how and why real devices fall off that straight line — the three signature failure patterns that CBSE loves to test with a graph and a one-line explanation.

Key takeaways

Ohm's law: for a conductor at constant physical conditions (mainly constant temperature), $V \propto I$ , so $V = IR$ with $R$ constant.
Resistance $R = V/I$ is measured in ohms ( $\Omega$ ); $1\ \Omega = 1\ \text{V/A}$ . This ratio can be computed for any device — it becomes "Ohm's law" only when it stays constant as $V$ and $I$ vary.
Ohmic conductors (metals like copper, nichrome, at constant temperature) give a straight-line V–I graph through the origin, of slope $R$ .
Non-ohmic devices (diodes, filament lamps run hot, and other non-linear elements) give a curved, sign-dependent, or even multi-valued V–I graph.
Three failure signatures: (i) $V$ not proportional to $I$ ; (ii) the $V$ – $I$ relation changes when the sign of $V$ is reversed; (iii) the relation is not single-valued — more than one current for the same voltage.
Resistance itself is not fixed for real materials; it depends on dimensions and resistivity ( $R = \rho L/A$ ) and, separately, on temperature — so comparing V–I data taken at different temperatures can wrongly look non-ohmic.
Key idea: Ohm's law says V is proportional to I at constant temperature, giving a straight V–I line of slope R through the origin, and a device is non-ohmic whenever that ratio V/I fails to stay constant, reverses asymmetrically with the sign of V, or becomes multi-valued — exactly what a forward-biased diode's steeply curving V–I graph shows.

The core insight

Ohm's law is the statement that V/I stays constant (equal to R) as V and I vary at fixed temperature, so its V–I graph is a straight line through the origin, and a device is non-ohmic the moment that ratio changes with the operating point, as it does dramatically for a diode.

Where it fits in the syllabus

This topic sits right after electric current and drift velocity in the Current Electricity chapter, and just before resistivity, temperature dependence of resistance, and the Wheatstone bridge. CBSE's 2025-26 prescription keeps Ohm's law and V–I characteristics (linear and non-linear, such as diode-like curves) squarely in scope; you are expected to state the law precisely, read and sketch V–I graphs, and give the physical reasoning for deviations — not to derive $V=IR$ from microscopic drift-velocity theory (that connection is built in the next node, on drift and resistivity). Nothing here touches potentiometers or colour codes, both explicitly out of this year's theory syllabus.

The physical picture

Ohm's law is a statement about a ratio staying fixed. When you connect a conductor across a battery and measure both the potential difference across it and the current flowing through it, you can always divide one by the other — that division is legitimate for literally any device, resistor or otherwise. What makes a material ohmic is the extra experimental fact that this ratio does not change as you vary the voltage, provided temperature and other physical conditions stay constant.

V \propto I \quad\Longrightarrow\quad V = IR \qquad (1)

where: $V$ is the potential difference (volts) across the conductor's ends, $I$ is the current (amperes) through it, and $R$ is the resistance (ohms, $\Omega$ ), a constant of proportionality fixed by the material, its geometry, and its temperature — but, for an ohmic conductor, not by $V$ or $I$ themselves. Read physically, equation (1) says: double the push (voltage), and you double the flow (current); the conductor's "difficulty," $R$ , does not change its mind depending on how hard you push it. A second, equivalent way to write the same idea uses resistivity, separating the material property from the object's shape:

R = \rho\,\dfrac{L}{A} \qquad (2)

where: $\rho$ is the resistivity (ohm-metre, $\Omega\,\text{m}$ ) — a property of the material alone — $L$ is the conductor's length, and $A$ is its cross-sectional area. Resistivity itself depends on temperature (a separate relation, $R_T = R_0(1+\alpha\Delta T)$ , covered in the next topic), which is precisely why "constant physical conditions" in the statement of Ohm's law really means, in practice, constant temperature: if the wire heats up while you take readings, $\rho$ drifts, $R$ drifts, and the V–I plot bends even though nothing "non-ohmic" is happening at the microscopic level — the material is simply at a different temperature at each point.

Figure

Two graphs of potential difference against current side by side, the left one a straight line through the origin for an ohmic resistor, the right one a curve that stays flat near the origin and then rises steeply for a forward biased diode

Ohm's law and the meaning of a constant slope

Derivation AWhy the ohmic V-I graph must be a straight line through the origin, and why the diode's cannot be

We want to show, starting from nothing more than the definition $R \equiv V/I$ and the physical statement of Ohm's law, exactly why an ohmic conductor's V–I plot is forced to be a straight line through the origin with slope $R$ , and why a genuinely non-linear device such as a diode cannot be described by any single constant $R$ at all — this is the reasoning an examiner expects when a question asks you to "justify" or "explain" a given V–I graph rather than just quote the law.

Step 1. Start from the defining ratio. For any two-terminal device, at any instant, we can always write $R_{\text{apparent}} = V/I$ — this is nothing more than a definition and is true even for a diode or a light bulb; it carries no physical content by itself.

Step 2. State the extra physical claim Ohm's law makes. Ohm's law asserts that for certain materials (metals, at constant temperature), $R_{\text{apparent}}$ computed at Step 1 comes out to be the same number no matter which $V$ (and corresponding $I$ ) you pick, as long as temperature and other physical conditions are unchanged. Call that fixed number $R$ .

Step 3. Turn the constant-ratio claim into an equation for the graph. If $V/I = R$ for every pair of readings, then $V = IR$ for all $I$ , which is the equation of a straight line in the $I$ – $V$ plane with slope $R$ and $V$ -intercept zero. Setting $I=0$ gives $V=0$ , so the line must pass through the origin — a resistor carrying no current has no potential difference across it, which is physically obvious and consistent.

Step 4. Reverse the current and check the required symmetry. Ohm's law for an ordinary resistor does not care about the direction of current flow: reversing $I$ to $-I$ simply reverses $V$ to $-V$ , since $R$ itself has no preferred sign. So the ohmic line is symmetric through the origin — it looks the same when rotated 180 degrees about the origin.

Step 5. Now examine why a diode violates Step 2 directly. Measuring $V/I$ for a diode at several forward voltages (see the worked example below) gives wildly different numbers — tens of kilo-ohms near zero volts, dropping to tens of ohms once conduction begins. Because the apparent resistance is not the same number twice, the device fails the very premise of Step 2, so no single $R$ and no straight line can describe it. Separately, a diode conducts easily in one direction and almost not at all in the other, breaking the symmetry required in Step 4. Both failures are independent — a device could break one without the other — but the diode breaks both, which is why its graph is both curved and asymmetric about the origin.

Step 6. Conclude with the general test. To decide whether any device is ohmic from data, compute $V/I$ at several points; if the value is constant across all of them (for both signs of $V$ ), the device is ohmic and its graph is the straight line of Step 3; if the value changes, or is asymmetric under $V \to -V$ , or the same $V$ ever gives more than one possible $I$ , Ohm's law fails, and only a curve — not the equation $V=IR$ with fixed $R$ — describes the device.

Worked examples

Board level - resistor obeying Ohm's law

A resistor is connected to a battery, and a potential difference of $12\ \text{V}$ across it drives a current of $0.4\ \text{A}$ through it. (a) Find its resistance. (b) If the voltage is increased to $18\ \text{V}$ and the resistor remains ohmic, find the new current.

Step 1. Apply $R = V/I$ at the given operating point: $R = \dfrac{12}{0.4} = 30\ \Omega$ .

Step 2. Because the resistor is stated to be ohmic, $R$ stays fixed at $30\ \Omega$ even after the voltage changes — that is exactly the content of Ohm's law being obeyed here.

Step 3. Find the new current from $I = V/R$ at the new voltage: $I = \dfrac{18}{30} = 0.6\ \text{A}$ .

The resistance is $\mathbf{30\ \Omega}$ , and the new current is $\mathbf{0.6\ A}$ . Notice the current increased in exactly the same proportion as the voltage ( $18/12 = 1.5 = 0.6/0.4$ ) — that proportionality, and nothing else, is the practical fingerprint of an ohmic device.

Competitive level - identifying non-ohmic behaviour from a diode's V-I table

The following V–I data were recorded for a p-n junction diode in forward bias. $V$ (volts): $0.5,\ 0.6,\ 0.65,\ 0.7$ . Corresponding $I$ (milliamperes): $0.01,\ 0.5,\ 5,\ 20$ . Show that the diode does not obey Ohm's law, and find the apparent resistance $V/I$ at each point.

Step 1. Convert every current to amperes before dividing, since $R$ must come out in ohms: $0.01\ \text{mA} = 1\times10^{-5}\ \text{A}$ , $0.5\ \text{mA} = 5\times10^{-4}\ \text{A}$ , $5\ \text{mA} = 5\times10^{-3}\ \text{A}$ , $20\ \text{mA} = 2\times10^{-2}\ \text{A}$ .

Step 2. Compute $R = V/I$ at each of the four points in turn: at $V=0.5\ \text{V}$ , $R = 0.5/(1\times10^{-5}) = 5\times10^{4}\ \Omega$ ; at $V=0.6\ \text{V}$ , $R = 0.6/(5\times10^{-4}) = 1.2\times10^{3}\ \Omega$ ; at $V=0.65\ \text{V}$ , $R = 0.65/(5\times10^{-3}) = 130\ \Omega$ ; at $V=0.7\ \text{V}$ , $R = 0.7/(2\times10^{-2}) = 35\ \Omega$ .

Step 3. Compare the four values against the ohmic test from the derivation above: a genuine resistor would return the same $R$ at every point; here $R$ falls from $50{,}000\ \Omega$ to $35\ \Omega$ — a factor of well over a thousand — as $V$ rises by less than a quarter of a volt.

The apparent resistance is $\mathbf{5\times10^{4}\ \Omega,\ 1.2\times10^{3}\ \Omega,\ 130\ \Omega,\ 35\ \Omega}$ at the four voltages, and since this ratio is nowhere near constant, the diode is confirmed strongly non-ohmic. Its V–I graph is the steep curve of the earlier figure, not a straight line through the origin. Timing note: in an exam, do not recompute all four ratios from scratch if pressed for time — compute the first and last only, note the order-of-magnitude change (here, roughly $10^3$ ), and state the conclusion; full marks for "shows non-ohmic behaviour" schemes usually credit the comparison, not every intermediate value, though showing the middle two costs under a minute and removes any doubt.

Common traps

Treating $R = V/I$ itself as the definition of Ohm's law. It is not — that ratio can be computed for any device. Ohm's law is the separate, stronger claim that the ratio stays constant as $V$ and $I$ vary.
Assuming every resistor is perfectly ohmic at every current. Even a metal filament deviates once self-heating becomes significant — a light bulb filament glowing white-hot has a much higher resistance than the same filament cold, because $\rho$ itself rose with temperature.
Forgetting the "constant physical conditions" qualifier and comparing V–I readings taken at different temperatures, which can make a genuinely ohmic material look non-ohmic purely because $R_T = R_0(1+\alpha\Delta T)$ shifted between readings.
Believing a locally straight-looking segment of a curved graph proves ohmic behaviour. A diode's curve can look nearly flat (or nearly vertical) over a narrow enough range, but the device is still non-ohmic overall — always check across the full range asked in the question, and check both signs of $V$ .
Mixing up units when computing $R = V/I$ from tabulated data in milliamperes or microamperes — a very common numerical slip that inflates or shrinks the answer by a factor of a thousand.

You should be able to…

State Ohm's law precisely, including the "constant physical conditions" (constant temperature) qualifier, and give its equation $V = IR$ .
Explain why $R = V/I$ is not by itself a statement of Ohm's law, and identify the extra condition that makes it one.
Sketch and interpret the V–I graph for an ohmic conductor (straight line through the origin) and for a forward-biased p-n junction diode (flat, then steep, asymmetric curve).
Compute resistance from tabulated V–I data, including correct unit conversion between milliamperes/microamperes and amperes.
State the three signature ways Ohm's law can fail: non-proportionality, sign-dependence, and non-single-valuedness.
Distinguish deviations caused by genuinely non-ohmic material behaviour from apparent deviations caused merely by a changing temperature during measurement.

Ask the tutor: Ask the AI tutor to hand you a fresh V–I data table and quiz you on deciding, with reasoning, whether the device it describes is ohmic or non-ohmic.

If R equals V over I always works out to some number, why do we even say some materials disobey Ohm's law?

Because computing $V/I$ is just arithmetic — it works for literally any two-terminal device, resistor or diode or light bulb. Ohm's law is the extra, non-trivial physical claim that this number stays the same no matter what voltage you apply, as long as temperature is fixed. For a copper wire that turns out to be true over a huge range; for a diode it is dramatically false, as the worked example shows — the "resistance" you calculate swings by a factor of over a thousand. So the arithmetic never fails, but the law can, and does, fail for many important devices.

Is a light bulb filament ohmic or not?

It is close to ohmic only over a narrow range near a fixed operating current, and clearly non-ohmic if you compare its cold resistance to its glowing-hot resistance. When you first switch it on, the filament is cool and its resistance is low, so a comparatively large current briefly flows; as it heats towards its working temperature (often over 2000 K), the resistivity of the tungsten filament rises substantially, resistance rises with it, and the current settles to a lower steady value. Plot V against I across this whole range and you get a curve, not a straight line — the filament's temperature is not staying constant, so the "constant physical conditions" clause of Ohm's law is being violated by the very act of turning it on.

Why does a diode conduct so easily one way and barely at all the other way?

That asymmetry comes from the internal structure of the p-n junction — a barrier that current can only easily cross once the forward voltage pushes charge carriers over a built-in potential step, roughly 0.6 to 0.7 volts for a typical silicon diode, which is why the table in the worked example shows almost no current below 0.5 V and then a rapid climb. In reverse bias the same junction widens the barrier instead of narrowing it, so only a tiny leakage current flows. CBSE only expects you to recognise and read this V–I shape, not derive it from junction physics — that construction is a Class 12 semiconductor-electronics topic, beyond what is tested here.

Do I need to memorise the diode's numerical resistance values, or just the shape of the graph?

Just the shape and the reasoning, not any specific numbers — the exact resistances in the worked example depend on the particular diode and the particular voltages chosen, and a question could easily hand you a different table. What is tested is whether you can compute $V/I$ correctly from given data, notice that it is not constant, and state clearly why that means the device does not obey Ohm's law — plus being able to sketch (or recognise) the characteristic near-flat-then-steep curve for a forward-biased diode.