Ohm's Law, Power and Energy

If you learn one equation for this exam, learn this one. It appears in Section A directly, it appears again disguised as a power question, and it is what tells you whether the 50-watt figure on your licence has been exceeded. Everything else in Module 1 is scaffolding around it.

The three quantities, one more time

The previous lesson introduced them separately. Ohm's law is the statement that they are not independent — fix any two and the third is decided for you.

The symbol for current is I, not C — it comes from the French intensité de courant. C is already taken by the coulomb and by capacitance, which is exactly the sort of collision the exam likes to test.

The law itself

Georg Ohm's finding, in 1827, was that for a metallic conductor at constant temperature the current through it is directly proportional to the voltage across it. Write the constant of proportionality as resistance and you get:

Two rearrangements follow, and you must be able to produce either without hesitating:

Try it before you are examined on it

Type any two of the four boxes below and the other two appear, together with the formula that produced them. Work through the three examples underneath by hand first, then check yourself here.

Worked example 1 — current from voltage and resistance

Your handheld draws its supply through a 5.5 Ω dummy load from a 13.8 V bench supply. What current flows?

I = V ÷ R = 13.8 ÷ 5.5 = 2.51 A

Worked example 2 — resistance from voltage and current

A panel lamp glows correctly on 12 V and draws 50 mA. What is its resistance when lit?

First the prefix: 50 mA = 0.05 A. Then R = V ÷ I = 12 ÷ 0.05 = 240 Ω.

Worked example 3 — voltage from current and resistance

150 mA flows through a 47 Ω resistor. What voltage is developed across it?

V = I × R = 0.15 × 47 = 7.05 V

What Ohm's law does not cover

Ohm's law describes ohmic or linear conductors — metals at a constant temperature. Say that qualification out loud, because the exam tests the exception:

• A diode is non-ohmic. Doubling the voltage across it does not double the current; below about 0.6 V almost nothing flows at all.
• A filament lamp is non-ohmic in practice, because as it heats its resistance rises — which is why a cold bulb draws a surge at switch-on.
• Temperature matters generally. In a metal, resistance rises with temperature (positive temperature coefficient). In carbon and in a semiconductor, resistance falls with temperature (negative temperature coefficient).

Power

Power is the rate at which energy is converted — joules per second. One joule per second is one watt. In an electrical circuit the rate is simply the push times the flow:

Substitute Ohm's law into that and you get the two forms the exam prefers, because they let you find power without measuring all three quantities:

Worked example 4

A 47 Ω resistor carries 150 mA. What must its power rating be?

P = I²R = 0.15² × 47 = 0.0225 × 47 = 1.06 W.

A quarter-watt resistor would burn. You would fit a 2 W part — the working rule in a real shack is to specify at least double the calculated dissipation, because a component run at exactly its rating runs hot enough to drift.

Power and energy are not the same word

The syllabus names both, and the distinction is a favourite one-mark question.

Power

A rate. Measured in watts. It tells you how hard something is working at this instant. Your rig draws 50 W.

Energy

An amount. Measured in joules, or — because a joule is tiny — in kilowatt-hours. It tells you how much was used in total. Your rig drew 50 W for two hours, so it used 0.1 kWh.

The electricity board bills you for energy — kilowatt-hours — not for power. A 1 kW heater run for one hour and a 2 kW heater run for half an hour cost the same.

Efficiency, briefly

No stage converts all of its input into useful output; the difference becomes heat. Efficiency is the ratio, expressed as a percentage:

A class-C amplifier might reach 70%; a class-A stage struggles past 30%. Lesson 14 returns to this. For now, note that the missing power is not lost — it is dissipated as heat, which is why power amplifiers wear heatsinks and small-signal stages do not.