Primary & Middle School Physics · Grades 1–9

68Electric Power and Energy

The electricity bill does not charge for volts, nor for amperes — it charges for something your family’s appliances have been quietly multiplying together all month. This chapter weds the two electrical quantities into a third, gives fast and slow spending their proper names, and ends at the meter by the front door: physics you can audit against real money.

68.1 Power: the rate of delivery

Definition 68.1 (Power and the watt)

The power $P$ of a device is the rate at which it converts energy — joules per second. Its unit is the watt ( $\mathrm{W}$ ): one joule each second, honoring the engineer of the steam age; the kilowatt ( $1\,\mathrm{kW}$ $=$ $1000\,\mathrm{W}$ ) serves the kitchen’s heavyweights. Power is not an amount of energy but a pace of spending — a tap’s flow, not a bucket’s content.

Proposition 68.2 (Electric power)

A device fed voltage $U$ and drawing current $I$ receives electric power

P = U \times I ,

watts from volts times amperes. The formula computes in all three directions, and its second recipe runs the fuse box: $I = P/U$ tells what current an appliance’s rated power will pull from the mains.

Example 68.3 (The household ladder)

Read the plates on the family’s machines: an indicator light, $1\,\mathrm{W}$ ; a modern lamp, $5\,\mathrm{W}$ (its glowing ancestor spent sixty for the same light — the difference was heat); a laptop, $50\,\mathrm{W}$ ; a television, $100\,\mathrm{W}$ ; a washing machine heating its water, $2000\,\mathrm{W}$ ; the kettle, $2300\,\mathrm{W}$ ; an instant water heater, up to $9000\,\mathrm{W}$ . The pattern of the ladder: anything whose job is heating climbs to the kilowatts — warmth is the costliest form of energy to supply.

Example 68.4 (The fuse box arithmetic)

The kettle on the $230\,\mathrm{V}$ mains: $I = 2300 \div 230 = 10\,\mathrm{A}$ — the ten-ampere appliance of the old chapters, now derived. The $9000\,\mathrm{W}$ water heater: nearly $40\,\mathrm{A}$ , demanding its own stout branch and breaker. And the overloaded multi-socket of the fire investigation: kettle, heater and fryer summing their $P/U$ currents past the branch’s rating — three innocents, one arithmetic crime, exactly as charged.

68.2 Energy: the amount delivered

Proposition 68.5 (Energy from power and time)

A device of power $P$ running for a time $t$ converts the energy

E = P \times t ,

joules from watts times seconds. Fast tap or long trickle, the bucket fills by the product: a $2300\,\mathrm{W}$ kettle’s three minutes ( $2300 \times 180 \approx 4.1 \times 10^{5}\,\mathrm{J}$ ) spends more than a $5\,\mathrm{W}$ lamp’s whole day ( $5 \times 86400 = 4.3 \times 10^{5}\,\mathrm{J}$ ) — almost exactly a tie, in fact: one boiling matches one day of good light.

Definition 68.6 (The kilowatt-hour)

Household energies make ungainly joule-counts, so the meter speaks kilowatt-hours ( $\mathrm{kWh}$ ): the energy of one kilowatt sustained for one hour,

1\,\mathrm{kWh} = 1000 \times 3600 = 3.6 \times 10^{6}\,\mathrm{J}

— three point six million joules per unit on the bill. With $P$ in kilowatts and $t$ in hours, $E = P \times t$ delivers kilowatt-hours directly: the formula the meter lives by.

Method 68.7 (Auditing an appliance)

For any machine and month:

read its power $P$ on the plate (or compute $U \times I$ );
honestly estimate its running time $t$ — per day, then per month;
$E = P \times t$ in kilowatt-hours ( $P$ in $\mathrm{kW}$ , $t$ in hours);
multiply by the tariff (take about $0.25$ currency units per $\mathrm{kWh}$ ) for the month’s cost.

The kettle, six minutes daily: $2.3 \times 0.1 \times 30 = 6.9\,\mathrm{kWh}$ a month. The television, three hours daily: $0.1 \times 3 \times 30 = 9\,\mathrm{kWh}$ . Small powers with long hours rival big powers with short ones — the audit’s recurring lesson.

Energy as area: power times time. The kettle’s tall sliver and the television evening’s long strip can enclose comparable areas — comparable kilowatt-hours.

68.3 The meter and the bill

Example 68.8 (Reading the bill)

The meter by the front door counts every kilowatt-hour that enters the house; the bill is its month’s difference times the tariff. A typical family’s $250$ monthly units decompose by audit: water heating and radiators first (the kilowatt club), then the cold appliances — modest watts, but running always — then cooking, washing, light and electronics. The audit’s power: it finds the real levers. Replacing ten old $60\,\mathrm{W}$ bulbs with $5\,\mathrm{W}$ ones saves more than unplugging every charger in the house a thousand times over.

Example 68.9 (The standby vampires)

Small print, long hours: a television’s standby light sips $0.5\,\mathrm{W}$ — but for all $720$ hours of the month: $0.0005 \times 720 = 0.36\,\mathrm{kWh}$ . One vampire is harmless; a house with twenty sippers at a watt each pays $0.02 \times 720 \approx 14\,\mathrm{kWh}$ monthly for nothing. Power times time forgives no term: the second factor bites when the first one hides.

Remark 68.10 (Power in the wider world)

The watt, born electric here, measures every energy pace. Your resting body idles near $100\,\mathrm{W}$ — a glowing ancestor-bulb of warmth (recall the crowded classroom’s stuffiness: thirty pupils, three kilowatts of biology). A cyclist sustains $200\,\mathrm{W}$ ; a car’s engine unleashes tens of kilowatts; the town’s power station, hundreds of millions of watts; the sunlight falling on the town, comfortably more — the ceiling under which all the other paces work. Next chapter follows the grid’s watts to their spinning source.

68.4 Exercises

Exercise 68.1 ★

Define power and its unit — tap or bucket? Write both formulas of the chapter.

Solution

Solution of Exercise 68.1.

Power is the pace of energy conversion — joules per second, in watts: the tap’s flow, not the bucket. $P = U \times I$ and $E = P \times t$ .

Exercise 68.2 ★

Compute powers: a lamp at $230\,\mathrm{V}$ drawing $0.022\,\mathrm{A}$ ; a starter motor at $12\,\mathrm{V}$ drawing $100\,\mathrm{A}$ .

Solution

Solution of Exercise 68.2.

Lamp: $230 \times 0.022 \approx 5\,\mathrm{W}$ . Starter: $12 \times 100 = 1200\,\mathrm{W}$ .

Exercise 68.3 ★

The fuse-box recipe: what currents do a $2300\,\mathrm{W}$ kettle and a $700\,\mathrm{W}$ microwave pull at $230\,\mathrm{V}$ ?

Solution

Solution of Exercise 68.3.

Kettle: $2300 \div 230 = 10\,\mathrm{A}$ ; microwave: $700 \div 230 \approx 3\,\mathrm{A}$ .

Exercise 68.4 ★

Why do heating appliances crowd the top of the household power ladder?

Solution

Solution of Exercise 68.4.

Because warmth is the costliest form of energy to supply: raising water’s or a room’s temperature devours joules at a pace lamps and electronics never approach — so every heater’s plate reads in kilowatts.

Exercise 68.5 ★

Convert: $1\,\mathrm{kWh}$ to joules (show the two factors). Why did the meter’s makers abandon the joule?

Solution

Solution of Exercise 68.5.

$1\,\mathrm{kWh} = 1000 \, \text{W} \times 3600 \, \text{s} = 3.6 \times 10^{6}\,\mathrm{J}$ . A month of joules runs to ten digits; the meter’s makers chose a unit that keeps household months in comfortable hundreds.

Exercise 68.6 ★

Energy check: which spends more, a $2\,\mathrm{kW}$ heater for half an hour or a $100\,\mathrm{W}$ television for a whole evening of five hours?

Solution

Solution of Exercise 68.6.

Heater: $2 \times 0.5 = 1\,\mathrm{kWh}$ . Television: $0.1 \times 5 = 0.5\,\mathrm{kWh}$ . The half-hour heater outspends the whole television evening, twofold.

Exercise 68.7 ★

Audit one machine: a $50\,\mathrm{W}$ laptop used four hours daily, for a thirty-day month, at $0.25$ per $\mathrm{kWh}$ . Energy and cost?

Solution

Solution of Exercise 68.7.

$E = 0.05 \times 4 \times 30 = 6\,\mathrm{kWh}$ ; cost $6 \times 0.25 = 1.5$ currency units — a month of homework for the price of a pastry.

Exercise 68.8 ★

What does the front-door meter count, and how does the bill turn its count into money?

Solution

Solution of Exercise 68.8.

Every kilowatt-hour entering the house. The bill subtracts last month’s reading from this month’s and multiplies the difference by the tariff.

Exercise 68.9 ★★

The bulb swap: ten $60\,\mathrm{W}$ ancestors versus ten $5\,\mathrm{W}$ moderns, four hours nightly, thirty nights. Compute both monthly energies and the saving — in kilowatt-hours and at $0.25$ each.

Solution

Solution of Exercise 68.9.

Ancestors: $0.6 \times 4 \times 30 = 72\,\mathrm{kWh}$ . Moderns: $0.05 \times 4 \times 30 = 6\,\mathrm{kWh}$ . Saving: $66\,\mathrm{kWh}$ — $16.5$ currency units monthly, from one afternoon on a stepladder.

Exercise 68.10 ★★

A hair dryer’s plate reads “230 V, 2000 W”. Its current? May it share a $16\,\mathrm{A}$ branch with the $2300\,\mathrm{W}$ kettle, both running? Show the sum.

Solution

Solution of Exercise 68.10.

Dryer: $2000 \div 230 \approx 8.7\,\mathrm{A}$ . With the kettle’s $10\,\mathrm{A}$ : $18.7$ — past the $16\,\mathrm{A}$ branch’s rating: the breaker will (rightly) end the experiment.

Exercise 68.11 ★★

An electric car’s battery stores $60\,\mathrm{kWh}$ . How long could that run the $2.3\,\mathrm{kW}$ kettle? The $5\,\mathrm{W}$ lamp? And what does the pair of answers teach about the size of transport’s energy appetite?

Solution

Solution of Exercise 68.11.

Kettle: $60 \div 2.3 \approx 26$ hours. Lamp: $60 \div 0.005 = 12000$ hours — a year and four months. Yet the same store drives the car only a few hundred kilometres: moving a tonne of machine devours energy on a scale beside which household lighting is a rounding error.

Exercise 68.12 ★★★

The kettle test, end to end: $1.5\,\mathrm{L}$ of water from $20\,{}^{\circ}\mathrm{C}$ to the boil needs about $5.0 \times 10^{5}\,\mathrm{J}$ of heat. Time the family kettle ( $2300\,\mathrm{W}$ ): predict its boiling time from $E = P t$ , then explain why the real kettle takes a little longer — where do the missing joules go, by the oldest law of the energy chapters?

Solution

Solution of Exercise 68.12.

Prediction: $t = E/P = 5.0 \times 10^{5} \div 2300 \approx 217\,\mathrm{s}$ — about three and a half minutes. The real kettle runs longer: part of its joules leak past the water — warming the kettle’s own body, the air, escaping as early steam. Energy is never destroyed, but neither is it all delivered where intended: the chain’s leak, timed at last in your own kitchen.

68.5 Problem: The Home Energy Audit

Problem 68.1

Weekend problem — the family energy audit; plates, meters and the month’s bill; finding the real levers

Armed with the two formulas and the tariff ( $0.25$ per $\mathrm{kWh}$ ), the family audits a month (30 days). The inventory: kettle $2300\,\mathrm{W}$ , $6$ minutes daily; television $100\,\mathrm{W}$ , $4$ hours daily; refrigerator — special case below; ten lamps $5\,\mathrm{W}$ each, $5$ hours daily; washing machine $2\,\mathrm{kW}$ , one $1$ -hour hot cycle every other day; water heater $2.5\,\mathrm{kW}$ , $2$ hours daily; standby sippers totaling $10\,\mathrm{W}$ , always on.

Part I — Machine by machine.

The kettle’s month, in $\mathrm{kWh}$ .
The television’s and the ten lamps’ months.
The washing machine’s ( $15$ cycles) and the water heater’s months.
The standby sippers’ month — the second factor at work.

Part II — The refrigerator and the sum. The refrigerator’s compressor runs at $120\,\mathrm{W}$ , but only a third of the time, cycling day and night.

Its effective average power, and its month in $\mathrm{kWh}$ .
Total the audit’s seven lines. Which three lines dominate?
The month’s bill at the tariff — and the water heater’s share of it in currency.
The meter read $2150$ units at the month’s start. Predict its end-of-month reading.

Part III — The levers.

Proposal one: shorter showers — the heater drops to $1.5\,\mathrm{h}$ daily. Monthly saving, energy and money?
Proposal two: a public campaign urges unplugging phone chargers ( $0.2\,\mathrm{W}$ each when idle). Audit one charger’s idle month and compare with proposal one; what does the contrast teach about audits versus slogans?
Proposal three: grandmother suggests the old habit of filling the kettle only as needed — half the water, half the energy. New kettle line, and the month’s saving?
Write the audit’s conclusions: three sentences — where the kilowatt-hours truly live, which formula factor the standby case teaches, and which single household change pays best.

Solution

Solution of Problem 68.1.

1. $2.3 \times 0.1 \times 30 = 6.9\,\mathrm{kWh}$ . 2. Television: $0.1 \times 4 \times 30 = 12\,\mathrm{kWh}$ ; lamps: $0.05 \times 5 \times 30 = 7.5\,\mathrm{kWh}$ . 3. Washer: $2 \times 1 \times 15 = 30\,\mathrm{kWh}$ ; heater: $2.5 \times 2 \times 30 = 150\,\mathrm{kWh}$ . 4. $0.010 \times 720 = 7.2\,\mathrm{kWh}$ — ten hidden watts, a kettle’s whole month. 5. Average $120 \div 3 = 40\,\mathrm{W}$ ; month: $0.04 \times 720 = 28.8\,\mathrm{kWh}$ . 6. Total $\approx 242\,\mathrm{kWh}$ . Dominant: the water heater ( $150\,$ ), the washer ( $30\,$ ), the refrigerator ( $28.8\,\mathrm{kWh}$ ) — heating and always-on. 7. $242 \times 0.25 \approx 60.6$ units; the heater’s share: $150 \times 0.25 = 37.5$ — well over half the bill. 8. $2150 + 242 = 2392$ units. 9. New heater line: $2.5 \times 1.5 \times 30 = 112.5\,\mathrm{kWh}$ : saving $37.5\,\mathrm{kWh}$ , about $9.4$ units of currency monthly. 10. One idle charger: $0.0002 \times 720 = 0.14\,\mathrm{kWh}$ — about four hundredths of a currency unit: proposal one outsaves it two-hundredfold per charger. Audits find levers; slogans find chargers. 11. Kettle line halves to $3.45\,\mathrm{kWh}$ : saving $3.45\,\mathrm{kWh}$ , under one currency unit — grandmother’s habit is right, and modest. 12. For example: “The kilowatt-hours live where water gets hot and where machines never sleep. The standby line teaches the time factor: tiny powers grow bills when $t$ is always. And the best-paying change is the shorter shower — the heater owns the bill, so its hours own the savings.”

Terms defined in this chapter

See all 393 terms in the glossary