Resistance and Ohm's Law - Physics for Pilot Assessments

Two quantities are now in place — the current through a component and the voltage across it — and this lesson is about the relationship between them, which is the single most useful piece of arithmetic in the whole module. By the end of it you should be able to say what resistance physically is and what an ohm means, calculate any one of voltage, current and resistance from the other two, show the rearrangement you used rather than reciting a mnemonic, and say what happens to a current when the resistance in its path doubles.

What resistance is

Push a current through a conductor and it does not flow as freely as it would in empty space. The free electrons drifting through the metal keep colliding with the vibrating atoms of the lattice they move through, and each collision takes energy out of the flow and leaves it in the metal, which is why a wire carrying a current warms up. Resistance is the name for that opposition — a measure of how hard the material makes it for a given push to produce a flow. It is a property of the particular component, not of the circuit it sits in: a coil of wire has the resistance it has whether you connect it to a cell or to nothing at all.

The unit follows from what resistance compares. If it takes a large voltage to drive a small current, the opposition is large; if a small voltage produces a large current, it is small. So resistance is defined as the voltage across a component divided by the current through it, and the unit is the ohm, written with the Greek capital omega, Ω. One ohm is one volt per ampere: a component has a resistance of one ohm if one volt across it drives one amp through it. Values in circuits range from small fractions of an ohm for a length of connecting wire, through a few ohms for a lamp filament, to thousands of ohms — kilohms, where kilo- again means a thousand — for the resistors inside electronic equipment.

The relationship, and why it takes this form

For a metal conductor at a steady temperature, something clean happens when you measure it: double the voltage across it and the current through it doubles, treble the voltage and the current trebles. The current is proportional to the voltage, which means their ratio never changes, and that unchanging ratio is what the resistance is. Draw current against voltage and you get a straight line through the origin, and both of those features carry meaning. Straight means the ratio is constant; through the origin means no voltage gives no current, with no threshold to overcome first.

A straight line through the origin: double the voltage and you double the current, so the ratio of voltage to current is the same at every point on the line. A component with more resistance gives a shallower line — the same voltage now drives less current.

Written as a formula this is Ohm's law, and it is usually arranged so that the voltage stands alone, because that is the form most easily read aloud: the voltage across a component equals the current through it multiplied by its resistance.

Ohm's law

V = I R

$V$: the potential difference across the component, in volts (V)
$I$: the current through the component, in amperes (A)
$R$: the resistance of the component, in ohms (Ω)

That single line covers all three of the questions you can be asked, provided you can rearrange it, so do the rearranging now rather than in the exam. Divide both sides by the resistance: on the right the resistances cancel and leave the current, so $\text{current} = \frac{\text{voltage}}{\text{resistance}}$ . Go back to the original and divide both sides by the current instead: the currents on the right cancel and leave the resistance, so $\text{resistance} = \frac{\text{voltage}}{\text{current}}$ . That last form is the definition the lesson started with, which is a good sign — the law and the definition of the ohm are the same statement.

Reading the relationship the right way round

It is worth spending a moment on what the formula does not say, because half the wrong answers in this topic come from turning it round. Resistance is not something that grows when you put more voltage across a component: the resistance is a property of the component, and it is what stays the same while the voltage and the current change together. So if an item doubles the voltage across a fixed resistor, the resistance is unchanged and the current doubles. And if an item doubles the resistance in a circuit while leaving the supply alone, the voltage is unchanged and the current halves — the current is inversely proportional to the resistance, meaning that whatever you multiply the resistance by, you divide the current by.

The triangle mnemonic that gets taught alongside this is a way of remembering which letter goes where, and it is worth being blunt about its limits: it tells you nothing about what any of the three quantities means, so it cannot help you notice that a question has given you a resistance where you expected a voltage, or that an answer of 3000 amps through a torch bulb is absurd. Read the question for which quantity is missing, write down the arrangement that isolates it, and then put the numbers in.

Example 1 — finding a current, twice

A 12 V supply is connected across a resistance of 4 Ω. What current flows? The 4 Ω resistor is then swapped for a 2 Ω one, with the same supply — what current flows now?

The current is the unknown, so use the arrangement that isolates it: $\text{current} = \frac{\text{voltage}}{\text{resistance}}$
With 4 Ω: $I = \frac{12}{4} = 3\ \text{A}$
With 2 Ω: $I = \frac{12}{2} = 6\ \text{A}$
Halving the resistance doubled the current, which is what inversely proportional means — and notice the voltage never changed, because the supply decides the voltage, not the resistor.
Check the first answer by going forwards: $V = I \times R = 3 \times 4 = 12\ \text{V}$ That is the supply we started with.

Example 3 — finding the resistance

A component has 9 V across it and 1.5 A flowing through it. What is its resistance? And what current would the same component carry at 18 V?

The resistance is the unknown, so use $\text{resistance} = \frac{\text{voltage}}{\text{current}}$ .
That is $\frac{9}{1.5}$ . Multiply both numbers by 10 to clear the decimal, which leaves the answer unchanged because the top and bottom of the fraction were scaled equally: $R = \frac{90}{15} = 6\ \Omega$
For the second part, the resistance is a property of the component, so it is still 6 Ω at the higher voltage.
$\text{Current} = \frac{\text{voltage}}{\text{resistance}} = \frac{18}{6} = 3$ amperes.
The voltage doubled from 9 V to 18 V and the current doubled from 1.5 A to 3 A, exactly as the straight line through the origin says it must.

Getting a resistance out of the graph

The straight line is not just an illustration; it is how a resistance is actually determined in a laboratory, and reading one is a fair question. Pick any point on the line, read off the voltage on the horizontal axis and the current on the vertical axis, and divide the first by the second — that is the resistance, and because the line is straight and passes through the origin it does not matter which point you pick. A line passing through the point where the voltage is 6 V and the current is 2 A belongs to a resistance of $\frac{6}{2} = 3$ Ω, and the point at 12 V and 4 A on the same line gives $\frac{12}{4} = 3$ Ω again. Two consequences are worth naming. A steeper line means more current for the same voltage, so a steeper line is a smaller resistance — the intuition runs the opposite way to most graphs, and items exploit that. And a line that curves has a resistance that changes as you move along it, which is the case dealt with at the end of this lesson.

Measuring the two quantities

The two instruments involved are worth understanding rather than just naming, because how each one has to be connected follows directly from what it measures. An ammeter measures a current, and a current is a flow through something, so the meter has to be inserted into the path itself: you break the circuit at the point of interest and put the meter in the gap, so that the charge you are counting actually goes through it. A good ammeter is therefore built to have as little resistance as possible, because whatever resistance it adds becomes part of the circuit and reduces the very current it is trying to report.

A voltmeter measures a potential difference, and a potential difference is a difference between two points, so the meter is connected to the two ends of whatever you are interested in without breaking anything. Nothing is cut and nothing is rearranged — you simply touch its two leads to the two points. A good voltmeter is built to have as high a resistance as possible, for the mirror-image reason: it is offering the charge an alternative route, and if that route were easy the meter would divert current away from the component and change the reading it was supposed to take. Between them these two facts explain a common practical mistake. Connecting a voltmeter into a circuit as though it were an ammeter puts an enormous resistance in the path and nearly stops the current; connecting an ammeter as though it were a voltmeter offers the charge a near-resistanceless bypass, and a very large current flows through it instead of through the component.

Where the straight line stops being straight

One line of honesty, and no more than one, because the batteries do not go further. Not every component gives a straight line: a filament lamp gets appreciably hotter as the current through it rises, and its resistance rises with it, so its graph bends over instead of staying straight. Components that behave that way are called non-ohmic, and for them the ratio of voltage to current is still a resistance at any given point — it just is not the same number at every point. Every calculation in this module assumes ohmic behaviour at a steady temperature, which is what the questions assume too.

Why it matters in aviation

Ohm's law decides how much copper an aircraft has to carry. Current flowing along a wire develops a voltage across the wire's own resistance, so a service mounted at the far end of the airframe sees less than the full busbar voltage, and the shortfall grows with both the current drawn and the length of the run.

The cure is thicker wire, since a larger cross-section means less resistance and therefore less voltage lost on the way, but thicker wire is heavier and on an aircraft weight is never free. Choosing a conductor is therefore a routine trade: work out the current the service will draw, calculate the drop the run would produce, and accept the lightest wire that still leaves the voltage at the load inside tolerance. The same relationship sets the ratings of the protective devices fitted in each circuit, because what such a device has to be matched to is the current the wire behind it can safely carry, and that current follows from the voltage driving it and the resistance of the path.

What the test asks

Two traps, and both are about the arrangement rather than the arithmetic. The first is inverting the relationship: an item gives a voltage and a resistance and offers $\frac{\text{resistance}}{\text{voltage}}$ among the options, which for the numbers usually chosen produces a plausible-looking small number. Guard against it by checking the units of your answer — coulombs per second is a current, and volts per amp is a resistance, so a current worked out as $\frac{4}{12}$ rather than $\frac{12}{4}$ is answering a different question. The second is the direction of the effect: if the resistance in a circuit is doubled at a fixed supply voltage, the current halves, and the distractor offers a doubled current on the reasoning that a bigger number in the formula must give a bigger answer. And expect at least one item where the given resistance is in kilohms, with the trap being to use the number as though it were ohms.