Electrical Power and Safety - Physics for Pilot Assessments

Power is the quantity that decides how bright a lamp is, how hot a heater gets and how thick a cable has to be, and for electricity it comes straight out of the two quantities already in hand. By the end of this lesson you should be able to calculate an electrical power from a voltage and a current, work out what happens to the power when the voltage across a fixed resistance changes, convert between joules and kilowatt-hours, and say what a fuse rating is actually protecting.

Power from voltage and current

From the energy module, power is the rate of transferring energy — joules per second, measured in watts. In an electrical component the energy arrives carried by charge, and the two quantities already defined say exactly how fast that happens. The voltage across the component is the energy given up by each coulomb passing through it, and the current is the number of coulombs passing each second. Multiply them and the coulombs cancel: joules per coulomb multiplied by coulombs per second leaves joules per second, which is watts. So the power is simply the voltage multiplied by the current, and the derivation is worth following once because it makes the formula unforgettable.

Electrical power

P = V I

$P$: the power, in watts (W) — one watt is one joule per second
$V$: the potential difference across the component, in volts (V)
$I$: the current through the component, in amperes (A)

Both rearrangements come up. Divide both sides by the voltage and the voltages on the right cancel, leaving $\text{current} = \frac{\text{power}}{\text{voltage}}$ — the arrangement you need when an appliance is labelled with its power and its supply voltage and you want to know what current it will draw, which is the single most common real calculation in this topic. Divide the original by the current instead and you get $\text{voltage} = \frac{\text{power}}{\text{current}}$ .

Power, resistance and the square

Very often the thing you know about a component is not its current but its resistance, and combining the two relationships already established handles that. Ohm's law says the current through a resistance is the voltage divided by that resistance; substitute that expression for the current in P = VI, and the power becomes the voltage multiplied by the voltage divided by the resistance — that is, the voltage squared, divided by the resistance. Squaring a number means multiplying it by itself.

Power dissipated in a resistance

P = \frac{V^2}{R}

$P$: the power, in watts (W)
$V$: the potential difference across the resistance, in volts (V)
$V^2$: the voltage multiplied by itself
$R$: the resistance, in ohms (Ω)

The square is the whole point of this form, and it is the most heavily examined single fact in the lesson. Double the voltage across a fixed resistance and the power does not double — it quadruples, because two multiplied by itself is four. Treble the voltage and the power goes up nine times. The reason is that raising the voltage raises the current in the same proportion, and the power depends on both, so the effect lands twice. Running the same substitution the other way, replacing the voltage in P = VI with the current multiplied by the resistance, gives a third useful form: the power is the current squared multiplied by the resistance. That version explains why a thin, high-resistance connection in a circuit is where the heat appears — the current through everything in a series path is the same, so the largest resistance dissipates the most power.

Power from current and resistance

P = I^2 R

$P$: the power dissipated, in watts (W)
$I$: the current, in amperes (A)
$I^2$: the current multiplied by itself
$R$: the resistance carrying that current, in ohms (Ω)

Example 2 — the square, in both directions

A heating element of resistance 20 Ω is connected to 20 V. What power does it dissipate? What happens to that power if the voltage is doubled to 40 V? And what voltage would be needed to get 80 W out of it?

Use $\text{power} = \frac{\text{voltage}^2}{\text{resistance}}$ . The voltage squared is 20 × 20 = 400, so the power is $\frac{400}{20} = 20$ watts.
At 40 V: the voltage squared is 40 × 40 = 1600, so the power is $\frac{1600}{20} = 80$ watts.
That is four times the original power for twice the voltage, not twice — which is the square doing its work.
The third part is the second one read backwards: 80 W is four times 20 W, and because power goes with the square of the voltage, four times the power needs twice the voltage, so 40 V. The arithmetic above already confirms it.
Worth checking by the other route: at 40 V the current through 20 Ω is $\frac{40}{20} = 2$ A, and power = voltage × current = 40 × 2 = 80 W. The two methods must agree, and they do.

Energy over time, and the kilowatt-hour

Power tells you the rate; multiply by how long it ran for and you have the energy, which is the relationship met in the voltage lesson: energy equals power multiplied by time, in joules when the power is in watts and the time in seconds. For anything domestic that unit is inconveniently small — a single electric heater gets through several million joules in an hour — so electricity is billed in a larger unit instead. A kilowatt-hour is the energy transferred by a power of one kilowatt running for one hour, and because it is a power multiplied by a time it is a unit of energy, not of power, despite the way it sounds.

Energy in kilowatt-hours

E_\text{kWh} = P_\text{kW} \times t_\text{hours}

$E_\text{kWh}$: the energy, in kilowatt-hours (kWh) — the unit an electricity bill counts
$P_\text{kW}$: the power, in kilowatts (kW) — watts divided by 1000
$t_\text{hours}$: the time, in hours

The only trap in that formula is the units: the power must be in kilowatts and the time in hours, which is the opposite habit from every other calculation in this module. A 2000-watt heater is 2 kilowatts, and running it for three hours uses 2 × 3 = 6 kilowatt-hours. If you ever need that in joules, a kilowatt is 1000 watts and an hour is 3600 seconds, so one kilowatt-hour is 1000 × 3600 = 3 600 000 joules.

Why a fuse has a rating

Everything above has a consequence for safety. A current in a wire dissipates power in that wire, at the rate given by the current squared multiplied by the wire's resistance, and that energy has nowhere to go but into the wire itself, which gets hotter. Every cable therefore has a current it can carry safely and a current at which it starts to be damaged, and the fault to worry about is not a lamp drawing slightly too much but a fault that lets a very large current flow. A fuse is the deliberate weak point that stops that: a short piece of thin wire, chosen so that it melts and breaks the circuit before the cable it protects gets anywhere near being damaged.

Why the thin part is the part that melts. The current is identical in the thick cable and the thin fuse wire — a series path allows nothing else — but the power dissipated in each part is I²R, and the resistance is concentrated in the thin section. On a fault current the fuse wire reaches melting point and opens the circuit while the cable either side is still cool.

The rating is the current at which it does that, and it is why the rating has to be chosen rather than picked at random. Fit a fuse rated well above what the circuit can survive and it will sit there intact while the cable overheats — the protection is nominally present and actually useless. Fit one rated below the appliance's normal working current and it will melt in ordinary use, which is merely annoying but leads people to replace it with something far too large. So the rating is chosen to sit just above the normal working current of the appliance and comfortably below what the cable can stand, and working out that normal current is exactly the $\text{current} = \frac{\text{power}}{\text{voltage}}$ calculation from Example 1. The detailed comparison of fuses with circuit breakers, and the wiring practice around them, belongs to the practical electrical exercises rather than here.

Why it matters in aviation

Power as voltage multiplied by current is the quantity an electrical system is actually managed by. With a busbar held at a fixed voltage, the demand of each service can be budgeted as a wattage and converted into the current it will draw, and adding those up gives the load a generator is being asked to carry, which is why generator loading is monitored against a rated output rather than assumed.

The heat side of the same physics is what the protective devices are there for. A current dissipates power in the conductor carrying it, in proportion to the square of that current, so a modest overload heats a wire considerably more than the overload figure suggests, and an overloaded circuit buried in a bundle behind a panel is a fire risk rather than an inconvenience. Lightning protection follows the same logic from the other end. A strike delivers an enormous current whatever anyone does, so the structure is bonded together to offer that current a low-resistance path through metal able to take the heating, instead of leaving it to find a route through something vulnerable.

What the test asks

The dominant distractor in this lesson is the missing square: an item doubles the voltage across a fixed resistance and offers double the power, which is the answer you get from P = VI if you forget that the current doubled as well. Whenever a question changes a voltage and asks about power, check whether the resistance was held fixed, and if it was, square the factor. Expect a current-from-power item on a 230 V supply, where the trap is multiplying instead of dividing. Expect a kilowatt-hour item where the power is given in watts or the time in minutes, with a distractor waiting at the unconverted value. And expect one fuse item whose wrong answer has a higher rating being the safer choice, when a rating too high simply fails to protect anything.