Electric circuits · Quantitative
Resistance
Resistance is not something you measure directly. You measure the push, you measure the flow, and you divide one by the other. That ratio is the number.
Start here
Add a metre of thin wire. The bulb dims.
A torch bulb on two cells is bright. Break the loop and splice in a metre of thin nichrome wire — no extra components, nothing removed — and the bulb goes noticeably dimmer. The wire gets warm.
What has the thin wire done to the circuit?
The wire added resistance to the one loop. There is still only one current and it is still the same at every point — it is just smaller than it was, because the same push now has more to get through. The wire also takes a share of the p.d., which is the energy that comes out of it as warmth in your fingers.
Resistance is how hard a component makes it for charge to get through. It is measured in ohms, written with the Greek letter Ω. A short piece of copper wire is a few hundredths of an ohm; a thin nichrome wire is several; a resistor is whatever it says on the packet; a filament is tens of ohms once it is glowing.
There is no such thing as an ohm-meter that reads resistance off the component the way a ruler reads a length. Resistance is defined as a ratio: the potential difference across a component divided by the current through it. So you take two measurements and do one division. R = V ÷ I.
Read that ratio as a price. A component of 15 Ω charges you 15 volts for every amp you want through it. Something with a low resistance is cheap: a small push gets a big flow. Something with a high resistance is expensive, and if the price is high enough — a piece of plastic, a gap of air — no push you can safely arrange will buy you any current at all.
At the bench · one component under test, an ammeter and a voltmeter
Two readings. One division.
Change a control to begin
Clip a component between the terminals, set the supply, and read both meters. The resistance is whatever the division gives.
Commit first. A 10 Ω resistor is tested at 3.0 V, then at 6.0 V. What happens to the resistance you calculate?
The component under test
6.0 V
The voltmeter reads
—
across the component
The ammeter reads
—
through the component
So the resistance is
—
—
Turn the supply up and
—
The figure
The same division, four times over
Two components, each tested at four supply settings. Every row is one reading of the voltmeter, one reading of the ammeter, and one division. Watch the last column.
| Component | V | I | R = V ÷ I |
|---|---|---|---|
| 10 Ω resistor | 3.00 V | 0.300 A | 10.0 Ω |
| same resistor | 6.00 V | 0.600 A | 10.0 Ω |
| same resistor | 12.00 V | 1.200 A | 10.0 Ω |
| Filament lamp | 1.50 V | 0.259 A | 5.8 Ω |
| same lamp | 6.00 V | 0.536 A | 11.2 Ω |
| same lamp | 12.00 V | 0.652 A | 18.4 Ω |
The resistor gives the same answer every time, which is why one number on the packet is enough to describe it. The lamp does not: its filament gets hotter as you turn the supply up, and hot metal resists more. Every one of those readings is a correct resistance — of that lamp, at that moment, at that temperature.
Writing it down · the shape of this relationship
Potential difference = current × resistance
The triangle
Cover the one you want
V = I × R
I = V ÷ R
R = V ÷ I
Cover V and I and R are left side by side — multiply them.
Cover I on the triangle: V sits over R, so you divide.
Cover R on the triangle: V sits over I, so you divide.
Two things side by side means multiply. One thing over another means divide.
V · potential difference across the component · V
I · current through the same component · A
R · resistance of that component, at that moment · Ω
1 Ω is 1 V for each 1 A
Worked example · one step at a time
A voltmeter across a resistor reads 6.0 V. The ammeter in the loop reads 0.40 A. What is the resistance?
Step 0 of 5
Convert
6.0 V stays 6.0 V · 0.40 A stays 0.40 A
The p.d. is already in volts and the current is already in amps, so there is nothing to convert.
Formula
R = V ÷ I
Cover R on the triangle: V sits over I, so you divide.
Insert
R = 6.0 V ÷ 0.40 A
Both readings come from the same setting of the supply.
Fine-tune
6.0 ÷ 0.40 = 15
Volts divided by amps leaves ohms.
Answer
R = 15 Ω
It takes 15 V across this resistor to drive 1 A through it.
Worked example · one step at a time
A voltmeter across a torch lamp reads 3.0 V. The ammeter reads 250 mA. What is the resistance?
Step 0 of 5
Convert
250 mA ÷ 1000 = 0.250 A
The formula wants amps, and a milliamp is a thousandth of an amp, so divide by 1000.
Formula
R = V ÷ I
Cover R on the triangle: V sits over I, so you divide.
Insert
R = 3.0 V ÷ 0.250 A
The converted current goes in. The milliamp reading never does.
Fine-tune
3.0 ÷ 0.250 = 12
Volts divided by amps leaves ohms.
Answer
R = 12 Ω
Put 250 in instead of 0.250 and the answer comes out 1000 times too small.
Your turn · the same five steps
Your component: the voltmeter reads 6.00 V and the ammeter reads 0.600 A.
Write each line out yourself — starting by deciding whether anything needs converting. Then check your working and tick the lines you had.
The five lines, marked
Convert
6.00 V stays 6.00 V · 0.600 A stays 0.600 A
Both meters already read in the units the formula wants, so nothing changes.
Formula
R = V ÷ I
Cover R on the triangle: V sits over I, so you divide.
Insert
R = 6.00 V ÷ 0.600 A
Both readings come from the same setting of the supply.
Fine-tune
6.00 ÷ 0.600 = 10.0
Volts divided by amps leaves ohms.
Answer
R = 10.0 Ω
It takes 10.0 V across this component to drive 1 A through it.
The five lines give 10.0 Ω for the 10 ohm resistor at 6.00 V.
A torch bulb on a 4.5 V supply. The ammeter reads 150 mA.
This one needs the Convert line to do some work.
The five lines, marked
Convert
150 mA ÷ 1000 = 0.150 A
A milliamp is a thousandth of an amp, so divide by 1000 before you go any further.
Formula
R = V ÷ I
Cover R on the triangle: V sits over I, so you divide.
Insert
R = 4.5 V ÷ 0.150 A
The converted current goes in. The milliamp reading never does.
Fine-tune
4.5 ÷ 0.150 = 30
Volts divided by amps leaves ohms.
Answer
R = 30 Ω
Put 150 in instead of 0.150 and the answer comes out 0.03 Ω — 1000 times too small.
The five lines give 30 Ω. The whole question turned on the first one.
Key fact
Resistance is the ratio of potential difference to current: R = V ÷ I, measured in ohms (Ω). One ohm is one volt for each amp. You never measure it directly — you measure V across a component and I through it, and divide.
Think again
“Resistance is a force pushing back against the current.”
There is no push-back. Resistance is a ratio — one measurement divided by another — and a ratio is not a force. What actually happens inside the metal is that the drifting electrons keep colliding with the atoms of the lattice, which are themselves jiggling with heat. Each collision hands over a little energy, the metal warms up, and the drift is slower than the push alone would suggest. A thin wire has fewer routes and a long wire has more collisions, so both resist more.
“A component has one resistance, so it does not matter what supply you test it on.”
True for a resistor, and that is exactly why resistors are sold with a number printed on them. Not true for a filament lamp: turn the supply up, the filament gets hotter, the atoms jiggle harder, the collisions get worse and the resistance climbs — from about 6 Ω cold to about 18 Ω at full brightness. Both readings are right. Resistance belongs to a component in a state, and for a lamp the state is its temperature.
Mastery ladder
Not started yet.
Rungs 3 and 4 you mark yourself.
Rung 1 · Calculate
A voltmeter across a wire reads 4.5 V and the ammeter in the loop reads 0.90 A. What is the resistance of the wire?
Rung 2 · The one that catches people
A student tests a filament lamp at 2 V and gets 6.4 Ω, then at 10 V and gets 16.0 Ω, and decides one of the measurements must be a mistake. What is right?
Rung 3 · Explain
Describe how you would find the resistance of a length of nichrome wire using a battery, an ammeter and a voltmeter, and explain why you need both meters.
Rung 4 · Take it somewhere new
An electric kettle element is a coil of wire that must produce a lot of heat. A lighting flex must carry a similar current and stay cool. Explain what the resistance of each has to be, and how a manufacturer would achieve it with the same kind of metal.
Key note
Resistance measures how hard a component makes it for charge to get through, in ohms (Ω). It is defined as a ratio: the potential difference across the component divided by the current through it, R = V ÷ I, so finding it always means two measurements and one division. One ohm is one volt for each amp. A resistor gives the same ratio at every supply setting, which is why one number describes it; a filament lamp does not, because its resistance climbs as it heats.
Going further
Georg Ohm published the relationship in 1827 and was largely ignored for a decade; his own colleagues called the work a web of naked fancies. What he had found is now stated carefully as a special case: for a metal at constant temperature the ratio V ÷ I is constant. The words "at constant temperature" are doing real work — a filament lamp obeys no such rule, and neither does a diode, a thermistor or a light-dependent resistor. The equation R = V ÷ I is always true, because it is a definition. Ohm's law is the extra claim that R stays the same, and plenty of components refuse it.
Components that refuse it on purpose are the useful ones. A thermistor's resistance drops sharply as it warms, which is how an oven, a kettle and a car engine all know their own temperature: the circuit is not measuring heat, it is measuring a resistance and reading the temperature off it. A light-dependent resistor does the same trick with brightness, which is what switches a street light on at dusk.
Before this lesson
Connects to
At GCSE this becomes
- Ohm's law as a special case, current–p.d. graphs for a wire, a lamp and a diode, and resistances added in series and in parallel.
Where to next
Ask Mr Badmus AI
Got a voltmeter reading and an ammeter reading and want the resistance?
The bench is a teaching model. The four ohmic components are given fixed resistances of 2, 5, 10 and 30 ohms; real wires and resistors vary with temperature by a small amount, and a resistor's printed value carries a tolerance of a few per cent. The filament lamp is modelled by taking its resistance as rising with the p.d. across it, from about 6 ohms at 1.5 V to about 18 ohms at 12 V; the model fixes the two ends of that rise and makes no claim about its shape in between, and the values here are typical of a small lamp rather than measurements of a particular one. The supply, the wires and the ammeter are treated as having no resistance and the voltmeter as drawing no current. Currents are rounded to three decimal places and resistances to one.
Lesson content © MrBadmusAI.