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First year Lesson 4 of 14 About 18 min

Ohm’s law and power

What you’ll learn

  • What Ohm’s law says, and why knowing any two of volts, amps and ohms gives you the third.
  • The power law, P = V × I, and the two squared forms that fall out of it.
  • How to pick the right rearrangement without memorising a wheel.
  • Why a loose connection gets hot, and why the wrong supply voltage changes power by a square.

Ohm’s law says the current in a circuit equals the voltage pushing it divided by the resistance in its way: I = V ÷ R. Written the usual way round, V = I × R. Add the power law, P = V × I, and you have the two formulas behind almost every calculation in your first two years. Everything on this page is those two, rearranged.

120 V source (V) 24 Ω heater (R) 600 W of heat I = 5 A
The whole idea on one drawing. The source’s 120 V pushes current through the 24 Ω heater: 120 ÷ 24 = 5 A, and 120 × 5 = 600 W comes out as heat.

The law, in plain words

Voltage is the push, current is the flow, resistance is what gets in the way (the previous lesson has the long version). Ohm’s law is the obvious relationship between them, made exact: more push, more flow; more in the way, less flow.

  • V = I × Rfind volts
  • I = V ÷ Rfind amps
  • R = V ÷ Ifind ohms

The units are what make it work: one volt pushes one amp through one ohm. Keep everything in plain volts, amps and ohms (convert kilo- and milli- first) and the arithmetic takes care of itself.

Power: how fast energy is used

Power is the rate the load turns electrical energy into heat, light or motion, measured in watts. It is the push times the flow: P = V × I. A 120 V circuit carrying 10 A is delivering 1,200 W. Swap in Ohm’s law and two more forms appear — you do not need to memorise them, just know they come from the first one:

  • P = V × Ivolts and amps known
  • P = I² × Ramps and ohms known
  • P = V² ÷ Rvolts and ohms known
The squared forms matter

Because current and voltage are squared in those forms, small changes make big differences. Double the current through a connection and it makes four times the heat. Halve the voltage on a heater and it gives a quarter of the heat.

The wheel, if you want it

Every rearrangement of the two laws, drawn in a circle. Pick the quantity you want in the middle, then the slice that uses the two you have.

V I × RP ÷ I√(P × R)volts I V ÷ RP ÷ V√(P ÷ R)amps R V ÷ IV² ÷ PP ÷ I²ohms P V × II² × RV² ÷ Rwatts

Five worked examples

1. A lamp: find the current and the hot resistance

  1. A 60 W lamp on 120 V. Want I: I = P ÷ V = 60 ÷ 120 = 0.5 A.
  2. Want R: R = V ÷ I = 120 ÷ 0.5 = 240 Ω — while it is lit.
  3. Measure a cold filament with a meter and you will read far less. The resistance climbs as the filament heats, which is why a lamp takes a surge the instant it is switched on.

2. A baseboard heater on the wrong voltage

  1. Plate: 1,500 W at 240 V. Current: 1,500 ÷ 240 = 6.25 A.
  2. Its resistance: R = V² ÷ P = 240 × 240 ÷ 1,500 = 57,600 ÷ 1,500 = 38.4 Ω.
  3. Fed 120 V by mistake: P = V² ÷ R = 14,400 ÷ 38.4 = 375 W. Half the voltage, a quarter of the heat — and a callback for “the heater doesn’t work”.

3. A loose connection

  1. A back-stabbed receptacle has worked loose and measures 0.05 Ω across the joint. The circuit is carrying 15 A.
  2. Voltage lost at the joint: V = I × R = 15 × 0.05 = 0.75 V — hardly noticeable at the load.
  3. Heat at the joint: P = I² × R = 15 × 15 × 0.05 = 11.25 W, all of it in a space the size of a fingertip. That is the scorch mark you find behind the plate.

4. From two meter readings

  1. Your meter reads 208 V across a load and your clamp reads 4 A on one of its conductors.
  2. R = V ÷ I = 208 ÷ 4 = 52 Ω (running).
  3. P = V × I = 208 × 4 = 832 W. That is true for a resistive load like a heater; a motor’s real watts also depend on its power factor — a second-year topic.

5. Milliamps and kilohms

  1. A 24 V control circuit with a 4.7 kΩ resistor. Convert first: 4.7 kΩ = 4,700 Ω.
  2. I = 24 ÷ 4,700 = 0.00511 A = 5.11 mA.
  3. P = V² ÷ R = 576 ÷ 4,700 = 0.123 W — a quarter-watt resistor will do, running at about half its rating; a half-watt part gives more margin.
Try one: a 1,200 W toaster on 120 V

Current: 1,200 ÷ 120 = 10 A. Hot resistance of its elements: 120 ÷ 10 = 12 Ω. Check: 10² × 12 = 1,200 W.

On the job

Tips from the job

  • Before you trust a number, sanity-check it: a 120 V circuit drawing 50 A through a 1,000 W load is a typo, not a reading.
  • Nameplates give you two of the four almost every time — usually volts and watts. Current is P ÷ V; that is the number your breaker and conductor care about.
  • A warm device plate or a discoloured terminal is I² × R made visible. Report it; do not just tighten it and walk away.
  • Carry the printable wheel in your code book until you stop needing it.

Common mistakes

  • Mixing kilo and plain units: 1.5 kW is 1,500 W. Divide 1.5 by 120 and you get 0.0125 A — obviously wrong.
  • Using the cold resistance of a lamp or heater to predict its running current. A lamp filament reads roughly a tenth of its hot resistance when cold; a heater element reads somewhat lower.
  • Thinking half the voltage gives half the power. With a fixed resistance it gives a quarter.
  • Measuring resistance on a live circuit. The meter’s ohms function supplies its own small voltage — isolate first.

Check yourself

8 questions. Tap an answer — it is marked straight away and the reason opens under it. Get 80% or better and the lesson is ticked off on the year page.

Not started yet.

  1. A 120 V circuit feeds a heater element of 24 Ω. What current does it draw?

    Answer and why

    B. I = V ÷ R = 120 ÷ 24 = 5 A.

  2. Same heater: 120 V and 5 A. How much power?

    Answer and why

    C. P = V × I = 120 × 5 = 600 W. Check with P = V² ÷ R: 14,400 ÷ 24 = 600 W.

  3. A clamp meter reads 8 A on a load, and the voltage across it is 240 V. What is its resistance while running?

    Answer and why

    B. R = V ÷ I = 240 ÷ 8 = 30 Ω.

  4. If the voltage across a fixed resistance doubles, the current…

    Answer and why

    C. I = V ÷ R. R has not changed, so I follows V: double the push, double the flow.

  5. If the voltage across a fixed resistance doubles, the power…

    Answer and why

    C. P = V² ÷ R. Double V and V² goes up by 2 × 2 = 4. The current doubled AND the voltage doubled, so power goes up four times.

  6. A 1,500 W heater built for 240 V is connected to 120 V by mistake. Roughly how much heat does it give?

    Answer and why

    D. Its resistance is fixed: R = 240² ÷ 1,500 = 38.4 Ω. On 120 V, P = 120² ÷ 38.4 = 375 W — a quarter, because half the voltage is a quarter of the power.

  7. A loose connection has 0.1 Ω of resistance and carries 20 A. How much heat is being made right at the connection?

    Answer and why

    C. P = I² × R = 20 × 20 × 0.1 = 40 W — about a small light bulb’s worth of heat inside a box. That is how a loose terminal starts a fire.

  8. Which formula finds current from power and voltage?

    Answer and why

    B. From P = V × I, divide both sides by V: I = P ÷ V.

Questions people ask

Is Ohm’s law always true?

It holds very well for ordinary conductors and resistors at a steady temperature. Some things do not follow it in a straight line — a lamp filament’s resistance climbs as it heats, and diodes and electronics have their own curves. For the wires, heaters and connections you meet in the first year, treat it as exact.

Why are there so many power formulas?

There is really one, P = V × I. The other two come from swapping V or I for its Ohm’s-law equivalent: put V = I × R in and you get P = I² × R; put I = V ÷ R in and you get P = V² ÷ R. Use whichever matches the two numbers you already have.

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