⚡ Spark Academy53 lessons

Power & Energy

Watts, joules, kilowatt-hours — where the energy goes, how fast it goes, and why components have ratings.

lesson 4 of 5 in this unit

Builds on: 0.2 Voltage: Energy per Charge0.3 Current: Charge in Motion1.2 Ohm's Law

Power is the rate of energy flow

Put the last lessons together. Voltage is joules per coulomb; current is coulombs per second. Multiply them and the coulombs cancel — leaving joules per second, which is watts: the rate at which electrical energy is being converted into light, heat or motion.

P = V × Iwatts = volts × amps · with Ohm’s law: P = I²R = V²/R

The two derived forms come free by substituting V=IR or I=V/R, and each has its moment: P = I²R when you know the current through a part, P = V²/R when you know the voltage across it. Sense of scale: an LED runs at ~0.04 W, a phone charger ~10 W, a bright old-style bulb 60 W, a kettle 2000 W.

Energy is power × time — and it costs money

Your electricity meter counts energy in kilowatt-hours: one kWh is 1000 W flowing for one hour (3.6 million joules). At a typical $0.30/kWh, a 60 W bulb burning 4 hours a day costs about $2.20 a month, while an LED bulb doing the same job at 8 W costs $0.29. Multiply by every lamp in a country and you see why lighting technology mattered so much.

Ratings: why parts burn

Every real component can only shed heat so fast. A standard small resistor is rated ¼ watt — ask it to dissipate more and it cooks, drifts, smokes, and eventually opens. Checking is a one-liner with P = V²/R. Put 330 Ω straight across 9 V: P = 81/330 ≈ 0.245 W — that is 98% of the rating, technically survivable but bad practice. The same resistor across 12 V: 0.44 W — it will burn. This tiny calculation is a professional habit worth building now.

Rule of thumb

Keep parts below about half their rated power for a long, cool life. If the math says more, use a higher-value resistor, a beefier part, or rethink the circuit.

⚡ Lab — Watts, Heat & the Bill

Left: a bulb whose brightness is pure P = V×I. Right: a ¼ W resistor being heat-checked at your chosen voltage.

  • Find three different V–I combinations that give exactly 6 W.
  • Put 330 Ω across 9 V and read the rating bar — then try 12 V. 🔥
  • Set your local electricity price and see the bulb’s monthly cost.
6.0 V
500 mA
330 Ω
4 h
$0.30/kWh
Power P = V × I
3 W
Energy per day
43.2 kJ (0.012 kWh)
Cost per month
$0.11
Resistor verdict
cool — plenty of margin

Check your understanding

Q1. A component drops 5 V while 2 A flows through it. Its power dissipation is…

Q2. Which formula gives a resistor's power directly from the voltage across it?

Q3. A 100 W device runs 10 hours. Energy used?

Q4. A ¼ W, 330 Ω resistor is placed straight across 12 V. What happens?