Voltage, Energy and EMF

Current tells you how much charge is moving; voltage tells you how hard it is being pushed, and how much energy each unit of charge is carrying. By the end of this lesson you should be able to say what one volt actually means, calculate the energy a supply gives to a known amount of charge and run that backwards, work out the voltage of several cells connected together whichever way round they face, and calculate electrical energy from power and time.

Voltage is energy per unit of charge

A cell does not make charge and it does not make current. What it does is give energy to the charge that passes through it, using a chemical reaction inside itself as the source, and the charge then carries that energy round the circuit and hands it over to whatever it passes through. So the natural way to measure what a supply is doing is to ask how much energy it gives to each coulomb — energy transferred per unit of charge. That quantity is the potential difference, usually just called the voltage, and its unit is the volt, written V. One volt means one joule of energy for every coulomb of charge. From the energy module, a joule is the unit of energy and of work, and one joule is the work done by a force of one newton moving something one metre.

Rearranging that gives the form you will use most often. Multiply both sides by the charge: on the right the charges cancel, leaving energy = charge × voltage. So a 12-volt supply hands 12 joules to every coulomb, and if five coulombs pass through it the supply has delivered 60 joules. Divide both sides of that version by the voltage instead and the voltages cancel, leaving $\text{charge} = \frac{\text{energy}}{\text{voltage}}$ — the arrangement for the rarer question that gives you an energy and a voltage and wants the charge.

Why it is called a difference

The full name, potential difference, is not decoration: a voltage is always a comparison between two points, and quoting one without saying which two points is meaningless. A charge arriving at a lamp carries a certain amount of energy per coulomb, and a charge leaving it carries less; the difference between those two, the energy given up per coulomb inside the lamp, is the potential difference across the lamp. The same idea applies to the supply, where the charge gains rather than loses. This is why voltage is described as being across a component, while current is described as going through it — and it is why a voltmeter is connected to two points either side of a component while an ammeter has to be inserted into the path itself.

The useful mental picture is height. Lifting water to the top of a tower gives every litre of it a certain amount of energy, and it does not matter whether you lift one litre or a thousand — the energy per litre depends only on the height. Voltage is the electrical version of that height, and it is why nothing about a voltage tells you how much current is flowing: a 9-volt battery with nothing connected to it still has 9 volts across its terminals, and no current at all. It also explains the single commonest error in the vocabulary. Voltage does not flow. Charge flows, current is the rate at which it flows, and voltage is the difference in energy per coulomb between two points. An answer describing the voltage as flowing through something has confused it with the current.

EMF, and cells in series

The voltage a supply produces when it is doing the driving has its own name: the electromotive force, almost always shortened to EMF. The name is unfortunate, because it is not a force at all and is not measured in newtons — it is the energy the supply gives to each coulomb passing through it, measured in volts like any other voltage. A single ordinary cell has an EMF of about 1.5 volts, which means it gives 1.5 joules to each coulomb. A car battery is 12 volts, and mains electricity in Europe is about 230 volts.

Connect two cells one after the other, so that the charge passes through both, and each of them adds its own energy to every coulomb going by. The energies per coulomb therefore add, and so do the voltages — provided the cells are the same way round, meaning the positive terminal of one connects to the negative terminal of the next, so that both are pushing the charge in the same direction.

Turn one of them round, though, and it pushes the charge the other way, so its contribution subtracts instead of adding. Two identical cells in opposition cancel exactly and the combination drives nothing at all, which is why a torch with one battery in backwards is dead rather than dim. Orientation, not count, is what decides the total.

Why a real cell sags under load

A cell is not a perfect supply, and the reason matters for reading questions correctly. The chemical reaction happens inside a physical object, and the charge has to travel through that object's own material to get from one terminal to the other. That material opposes the flow just as the rest of the circuit does — this is the cell's internal resistance, and resistance gets its own treatment shortly. What it costs you is some of the energy per coulomb: the charge gives up part of what it gained before it even leaves the cell, so the voltage actually available at the terminals is less than the EMF.

The amount lost inside depends on the current: the more charge per second the cell is pushing, the more it loses internally, so a lightly loaded cell delivers nearly its full EMF while a heavily loaded one sags noticeably. This is why a car's headlights dim while the starter motor turns — the starter draws an enormous current, the battery's internal loss becomes large, and the terminal voltage available to everything else drops. It is also why an old or cold battery can read a healthy 1.5 volts on a meter and still fail to run anything: an unloaded measurement draws almost no current, so it hides exactly the defect it was meant to find.

Electrical energy over time

There is a second route to the energy, and it is the one most questions take. From the energy module, power is the rate of doing work — the energy transferred each second, measured in watts, where one watt is one joule per second. That module gave power as work divided by time; rearranging it by multiplying both sides by the time gives the energy directly, and the relationship holds for electrical energy exactly as it does for any other kind.

The units are the whole reason this works: watts are joules per second, so watts multiplied by seconds gives joules. That also tells you which conversions matter. A power in kilowatts, where the prefix kilo- means a thousand, must be multiplied by a thousand to give watts before you use it; a time in minutes must be multiplied by sixty to give seconds. Get either wrong and the answer is out by exactly that factor, which is where two of the four options usually come from.

Why it matters in aviation

An aircraft distributes power at a fixed voltage precisely because voltage is energy per unit of charge. Hold a busbar at a known number of joules per coulomb and every service connected to it can draw whatever current its own job requires, with no service needing to know about the others. Transport aircraft typically work with 28 V DC and 115 V AC, and each of those figures is a statement about energy per coulomb rather than about how much current is flowing.

In flight the engine-driven generators maintain that voltage; on the ground, and as a reserve if generation is lost, the battery maintains it instead. A battery's internal resistance is why the difference between EMF and terminal voltage matters here rather than being a technicality. Under a heavy load such as an engine start the loss inside the battery becomes large, its terminal voltage sags, and everything else fed from it sees the reduced figure — the same mechanism as headlights dimming while a starter turns, and the reason a battery that reads its full voltage unloaded can still be unfit for the job.

What the test asks

The reliable distractor here is language: an option describing voltage as something that flows through a component, or as being used up along a wire the way water pressure is lost in a pipe. Voltage is a difference between two points, so the correct answer will always be phrased as across something. Expect an item on two cells where the trap is counting them rather than reading their orientation, and the safe habit is to check which way each one faces before adding anything. Expect an E = Pt calculation whose only difficulty is a minute or a kilowatt left unconverted. And expect one item on a battery that measures its full voltage but cannot run a load, where the correct reasoning is internal resistance: the measurement drew no current, and the sag only appears when current flows.