Ohm's law and resistance — V equals IR revisited — Physics

Before this

Electric potential and potential energy
Voltage, current and resistance · coming soon
Electric current — charge flowing · coming soon

This leads to

A lamp that changes its resistance

A student connects a small filament lamp to a variable power supply. At a potential difference (p.d.) of 4.0 V, the current is 0.20 A. At 8.0 V, the current is 0.30 A. The student works out the resistance both times, using $R = V \div I$ :

\frac{4.0}{0.20} = 20 Ω \frac{8.0}{0.30} \approx 26.7 Ω

The same lamp has two different resistances. Doubling the p.d. did not double the current. So is $V = I R$ wrong for a lamp? Many students have been told that $V = I R$ is "Ohm's law", and that it is always true. Both of those statements need care.

The answer is that two different ideas usually hide inside one equation. One is a definition, and it is true for every component. The other is a claim about some materials, and it is often false. This lecture separates them. After that, the lamp stops being a puzzle.

What current, p.d. and resistance measure

Start with the two quantities you can measure. Current, $I$ , is the rate at which charge flows past a point. One ampere is one coulomb per second. Potential difference, $V$ , is the energy transferred per coulomb of charge between two points. One volt is one joule per coulomb. Some books call the p.d. the "voltage".

The last lecture ended with a promise about circuits: every point in a wire has a potential, and a voltmeter reads the difference between two points. Here is that picture at work. In a circuit with a 12 V battery and one resistor, the end of the resistor joined to the positive terminal is at about 12 V. The other end is at about 0 V. Charge "falls" through 12 V of potential as it passes through the resistor, and each coulomb transfers 12 J there. A voltmeter connected across the resistor reads the gap: 12 V.

Now the third quantity. Different components let different currents through for the same p.d. A component that lets only a small current through has a large resistance. We measure this by asking how many volts are needed for each ampere.

The definition of resistance

$R = \frac{V}{I}$ , so $V = I R$ , where $V$ is the p.d. across the component and $I$ is the current through it.

The unit is the ohm, Ω. One ohm is one volt per ampere.

This is a definition. It is true for any component, at any p.d., because it is simply how resistance is measured. For the lamp, it gave 20 Ω at 4.0 V and 26.7 Ω at 8.0 V. Both values are correct. They describe the lamp at two different moments.

Ohm's law is a different claim

In 1827, the German physicist Georg Ohm published his results for metal wires. He found that, for a wire kept at a steady temperature, the current is proportional to the p.d. Double the p.d. and the current doubles. Triple it and the current triples. This means the ratio $V \div I$ stays the same. The resistance is a constant.

Ohm's law

For a metal conductor at constant temperature, the current is proportional to the potential difference: $I \propto V$ .

A component that obeys this is called ohmic. Its resistance does not change when the p.d. or the current changes.

That is it. That is the whole idea. $V = I R$ is the definition of resistance, and it always holds. Ohm's law is the claim that $R$ stays constant, and it holds only for some components. The lamp obeys the definition at every point. It does not obey Ohm's law, because its resistance changes. So the lamp is non-ohmic.

Test the idea before you read on.

Predict — does V = IR work for a lamp?

A filament lamp has 6.0 V across it and 0.26 A through it. Can you use $V = I R$ to find its resistance at that moment? Choose one, then read the explanation.

This box asks whether $V = I R$ can be used for a filament lamp, and answers it.

Yes. $R = V \div I$ is the definition of resistance, so it works for any component at any moment: $6.0 \div 0.26 \approx 23$ Ω. What a lamp breaks is Ohm's law. Its resistance is different at a different p.d., so you cannot use this value to predict the current at another p.d.

Reading an I–V graph

The clearest way to see how a component behaves is to plot its I–V graph ("characteristic"): current up the side, p.d. along the bottom. The p.d. is made negative by turning the component round, so that current flows the other way.

For an ohmic resistor, the graph is a straight line through the origin. Its gradient is $I \div V$ , which is $1 \div R$ . A steeper line means more current for each volt, so a smaller resistance.

Now the hard part of the lecture. Spend time here. For a curved graph, the resistance at a point P is still $V \div I$ at P. On the graph, $I \div V$ is the gradient of the straight line from the origin to P. It is not the gradient of the tangent at P. The tangent tells you how fast the current changes for a small change in p.d. That is a different quantity, and it is not the resistance.

The lamp at P (8.0 V, 0.30 A). The dashed line from the origin has gradient 0.30 ÷ 8.0, so R = 8.0 ÷ 0.30 ≈ 26.7 Ω. The tangent at P is much less steep. One divided by its gradient is about 53 Ω, which is not the resistance.

Stop and think

On the lamp's graph, the curve becomes less steep as the p.d. rises. A student says: "So the resistance falls." What is wrong?

Show the answer

The resistance rises, and the student has read the gradient upside down. The current is on the vertical axis, so a gradient is current per volt, which is $1 \div R$ . A line from the origin that becomes less steep means less current per volt, which is more resistance. Always check which quantity is on which axis. In some books the axes are swapped, with $V$ up the side. Then a steeper line from the origin means a larger resistance.

Try the three components yourself. For each point, compare the dashed line from the origin with the tangent.

Graph builder — the resistance at a point

Choose a component. Move the p.d. and watch the point move along the I–V graph. The dashed line from the origin gives the resistance. The grey tangent does not.

Component:

Potential difference: 8.0 V

These three I–V graphs show a 30 Ω resistor, a filament lamp and a diode. The values in the table are worked out from the same models.

The resistor's graph is straight through the origin. The lamp's curve becomes less steep at higher p.d. in both directions. The diode passes almost no current until about 0.6 V in the forward direction.

Component	V	I	R = V ÷ I
Resistor	6.0 V	0.20 A	30 Ω
Resistor	12 V	0.40 A	30 Ω
Filament lamp	4.0 V	0.20 A	20 Ω
Filament lamp	8.0 V	0.30 A	26.7 Ω
Diode	0.60 V	1.0 mA	600 Ω
Diode	0.70 V	21.5 mA	32.5 Ω

You should have seen three different behaviours. For the resistor, the dashed line and the tangent are the same line, and $R$ is 30 Ω everywhere. For the lamp, the two lines separate as the p.d. rises, and the resistance grows. For the diode, almost nothing happens until about 0.6 V. Then the current rises very steeply, and the resistance falls fast.

The filament lamp: hotter means harder to pass

A metal is a lattice of positive ions with free electrons moving between them. A p.d. pushes the free electrons along the wire. As they move, they collide with the ions and transfer energy to them. That is why a wire carrying a current gets warm.

The ions are never still. They vibrate about their positions. When the metal is hotter, they vibrate with larger amplitude. This means the moving electrons collide with them more often. More collisions mean less current for the same p.d., which is a larger resistance.

In a lamp, this effect is huge. The filament is a very thin wire, and the current heats it until it glows white-hot. A larger p.d. gives a larger current, which makes the filament hotter, which raises its resistance. So the current grows more slowly than the p.d. The graph bends over in both directions, because heating does not depend on which way the current flows.

The diode: a one-way valve

A diode is made from a semiconductor such as silicon. In one direction, called forward bias, it passes almost no current until the p.d. reaches about 0.6–0.7 V for a silicon diode. Above that, a tiny extra p.d. gives a large extra current. In the other direction, called reverse bias, it passes almost no current at all, as long as the p.d. is not too large.

So a diode lets current through in one direction only. Circuits use this to protect components from a battery connected the wrong way round, and to turn alternating current into current in one direction. A light-emitting diode (LED) is a diode that gives out light when current flows forward.

The thermistor: hotter means easier to pass

A thermistor is also made from a semiconductor, and it does the opposite of a metal. In the common type, its resistance falls as its temperature rises. This type is called an NTC thermistor, for "negative temperature coefficient".

The reason is the number of charge carriers. In a metal, the number of free electrons is fixed, so only the extra vibration matters. In a semiconductor, very few electrons are free at room temperature. Heating gives some bound electrons enough energy to break free and carry current. This increase in carriers is much larger than the effect of the extra vibration. So the resistance falls. One thermistor design, for example, has 10 kΩ at 25 °C and about 3.6 kΩ at 50 °C. (These values come from a typical design, given as an example.) A circuit can measure this change and use it as a thermometer.

Worked examples

Worked Example 1 — An ohmic resistor

A resistor carries a current of 0.40 A when the p.d. across it is 12 V. (a) Find its resistance. (b) It is ohmic. Find the current when the p.d. is 9.0 V.

Think first. Part (a) uses the definition. Part (b) uses Ohm's law: an ohmic resistance stays the same.

(a) $R = \frac{V}{I} = \frac{12}{0.40} = 30$ Ω.

(b) $I = \frac{V}{R} = \frac{9.0}{30} = 0.30$ A.

Check. 9.0 V is three quarters of 12 V, and 0.30 A is three quarters of 0.40 A. For an ohmic component, current and p.d. change by the same factor. ✓

Worked Example 2 — The lamp from the start

A filament lamp carries 0.20 A at 4.0 V and 0.30 A at 8.0 V. Find its resistance at each p.d. Is it ohmic?

At 4.0 V: $R = 4.0 \div 0.20 = 20$ Ω. At 8.0 V: $R = 8.0 \div 0.30 \approx 26.7$ Ω (3 s.f.).

The resistance changes, so the lamp is not ohmic. A student who used 20 Ω to predict the current at 8.0 V would get $8.0 \div 20 = 0.40$ A, which is a third too high.

Check. The p.d. doubled but the current rose only by half. If the lamp were ohmic, the current would have doubled. ✓ The resistance rose, which fits a filament getting hotter. ✓

Worked Example 3 — A diode in both directions

A model silicon diode carries 1.0 mA at 0.60 V and 21.5 mA at 0.70 V in the forward direction. In reverse, at −5.0 V, it carries about $1 \times 1 0^{- 11}$ A. (These values come from a simple model of a diode.) Find its resistance at each point.

At 0.60 V: $R = 0.60 \div 0.0010 = 600$ Ω. At 0.70 V, using the unrounded current of 21.54 mA: $R = 0.70 \div 0.02154 \approx 32.5$ Ω (3 s.f.). In reverse: $R \approx 5.0 \div (1 \times 1 0^{- 11}) = 5 \times 1 0^{11}$ Ω.

Check. A rise of only 0.10 V made the current about 20 times larger, so the resistance fell to about one twentieth. ✓ In reverse, the resistance is so large that the diode acts as a break in the circuit. ✓

Worked Example 4 — Potential along a wire

A 12 V battery is connected to a 30 Ω resistor by two copper leads. Each lead has a resistance of 0.050 Ω. Assume the battery itself has no resistance. Find the current, and the p.d. across each lead and across the resistor.

Think first. The same current passes through every part of this single loop. The total resistance is $30 + 0.050 + 0.050 = 30.1$ Ω.

Current: $I = 12 \div 30.1 \approx 0.399$ A. Across each lead: $V = I R \approx 0.399 \times 0.050 \approx 0.020$ V. Across the resistor: $V \approx 0.399 \times 30 \approx 11.96$ V.

Check. $11.96 + 0.020 + 0.020 = 12.00$ V, the whole battery p.d. ✓ This shows the promised picture. Along a copper lead, the potential hardly changes: both ends are at almost the same potential. Nearly all the 12 V "drop" happens across the resistor, which is where a voltmeter finds it.

Three traps

Trap 1 — "V = IR only works for ohmic components"

$R = V \div I$ is the definition of resistance, so it works for every component at every point. What only works for ohmic components is using one value of $R$ to predict the current at a different p.d.

Trap 2 — Using the tangent on a curve

On a curved I–V graph, the resistance at P is $V \div I$ at P: the line from the origin, not the tangent. For the lamp at 8.0 V, the tangent suggests about 53 Ω, but the resistance is 26.7 Ω.

Trap 3 — Reading the gradient upside down

With current up the side, a steeper line from the origin means a smaller resistance. With p.d. up the side, it means a larger one. Check the axes before you say anything about resistance.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A p.d. of 2.4 V across a resistor gives a current of 0.16 A. The resistor is ohmic. Find its resistance, and the current at 6.0 V.

Hint 1
Use the definition first: $R = V \div I$ .

Hint 2
"Ohmic" means the same $R$ works at 6.0 V. Rearrange to $I = V \div R$ .

Full answer
$R = 2.4 \div 0.16 = 15$ Ω. At 6.0 V, $I = 6.0 \div 15 = 0.40$ A. Check: 6.0 V is 2.5 times 2.4 V, and 0.40 A is 2.5 times 0.16 A. ✓
A lamp gives these readings: 1.0 V, 0.067 A; 2.0 V, 0.12 A; 4.0 V, 0.20 A; 6.0 V, 0.26 A. Find the resistance at each p.d., and say whether the lamp is ohmic.

Hint 1
Divide each p.d. by its current.

Hint 2
An ohmic component gives the same answer every time.

Full answer
$1.0 \div 0.067 \approx 15$ Ω, $2.0 \div 0.12 \approx 17$ Ω, $4.0 \div 0.20 = 20$ Ω, $6.0 \div 0.26 \approx 23$ Ω. The resistance rises, so the lamp is not ohmic. Check: at small p.d. the values change slowly, so a cold lamp is nearly ohmic. The change grows as the filament heats up. ✓
The thermistor described above (10 kΩ at 25 °C, about 3.6 kΩ at 50 °C) is connected to a steady 6.0 V supply. Find the current at each temperature, and explain why it changes.

Hint 1
Change kΩ to Ω first: 10 kΩ = 10 000 Ω.

Hint 2
$I = V \div R$ at each temperature. Then think about charge carriers.

Full answer
At 25 °C: $I = 6.0 \div 10000 = 6.0 \times 1 0^{- 4}$ A = 0.60 mA. At 50 °C: $I \approx 6.0 \div 3600 \approx 1.7 \times 1 0^{- 3}$ A = 1.7 mA. Warming frees more electrons in the semiconductor, so more charge carriers carry the current, and the resistance falls. Check: the resistance fell to about a third, and the current rose to about three times as much. ✓
A stretch. The lamp in this lecture follows the rule $I = \frac{3 V}{40 + 5 V}$ for positive $V$ , with $V$ in volts and $I$ in amperes. (a) Show that its resistance is $R = \frac{40 + 5 V}{3}$ . (b) At what p.d. is its resistance 30 Ω? (c) What is its resistance when it is almost cold, at very small p.d.?

Hint 1
$R = V \div I$ . Divide $V$ by the fraction, which is the same as multiplying by it upside down.

Hint 2
For (b), set $\frac{40 + 5 V}{3} = 30$ and solve. For (c), let $V$ become very small.

Full answer
(a) $R = V \div \frac{3 V}{40 + 5 V} = V \times \frac{40 + 5 V}{3 V} = \frac{40 + 5 V}{3}$ . (b) $40 + 5 V = 90$ , so $V = 10$ V. Check: $I = 30 \div 90 \approx 0.333$ A, and $10 \div 0.333 = 30$ Ω. ✓ (c) As $V$ becomes very small, $R$ approaches $40 \div 3 \approx 13.3$ Ω. A cold filament has a much lower resistance than a hot one. This is why a lamp draws its largest current in the moment it is switched on, before the filament heats up.

The bigger idea

Ohm published his law in 1827, in a book called The Galvanic Circuit Investigated Mathematically. It was received so coldly that he gave up his teaching post in Cologne. Recognition came slowly. In 1841 the Royal Society in London gave him its Copley Medal, and the unit of resistance now carries his name. (Sources: Britannica, "Georg Ohm"; Engineering and Technology History Wiki, "Ohm's Law".)

His law is a property of some materials, not a law of nature like the conservation of energy. And much of modern technology depends on components that break it. Diodes turn the alternating current from the mains into current in one direction for a phone charger. Thermistors measure temperature. Transistors, the switches inside every computer chip, are non-ohmic too. Knowing exactly where Ohm's law stops is what makes those devices possible. The next lecture uses resistance to derive the rules for series and parallel circuits.

Key concepts in this lesson

Resistance is defined by $R = V \div I$ , in ohms, and this definition holds for every component at every point.
Ohm's law says that, for a metal at constant temperature, current is proportional to p.d., so the resistance stays constant.
On an I–V graph, the resistance at a point is $V \div I$ there: the line from the origin, never the tangent.
A filament lamp's resistance rises as it heats up, because the vibrating ions cause more collisions with the free electrons.
A diode conducts in one direction only, and an NTC thermistor's resistance falls as its temperature rises, because heating frees more charge carriers.