PhysicsCore22 min read

Practical Electricity

Power, energy, the cost of running an appliance, and mains safety

This topic appears in:

01

Electrical power

Definition

Electrical power — The rate at which a component transfers energy, P = E/t. One watt is one joule per second.

Every coulomb passing through a component gives up V joules, and I coulombs pass every second. Multiply the two and you have the energy transferred per second, which is the power. That is the whole derivation of P = VI, and it is worth understanding rather than memorising, because the other two forms follow from it in one line each.

Substituting V = IR gives P = I²R. Substituting I = V/R gives P = V²/R. All three are the same statement; you choose whichever one matches the quantities the question gives you.

The form matters more than it looks. Because power loss depends on the square of the current, halving the current cuts the heating to a quarter. That single fact is why electricity is transmitted across the country at hundreds of thousands of volts.

P = V IP = I² RP = V² / RE = P tpick the form that matches what the question gives you
P
powerW
V
potential differenceV
I
currentA
R
resistanceΩ
E
energyJ
Worked example 14 marks

A 2.0 kW electric kettle runs from the 230 V mains. Calculate the current it draws and the resistance of its element.

  1. Convert: 2.0 kW = 2000 W.Watts go with volts and amps; kilowatts belong with the kWh.
  2. Rearrange P = VI to I = P/V.
  3. I = 2000 / 230 = 8.7 A.
  4. R = V/I = 230 / 8.7 = 26 Ω.Or use P = V²/R directly, which gives the same value.

I = 8.7 A, R = 26 Ω

Change the supply voltage and watch the power reading. Doubling the voltage across a fixed resistance quadruples the power, because P = V²/R — the relationship is square, not proportional.

02

Paying for electricity: the kilowatt-hour

Definition

Kilowatt-hour — The energy transferred by a 1 kW appliance running for 1 hour. It is a unit of energy, not power, and equals 3.6 million joules.

The joule is far too small a unit for a household bill. A single kettle boiling uses about half a million of them. So energy suppliers bill in kilowatt-hours, and the arithmetic becomes easy: power in kilowatts, multiplied by time in hours, multiplied by the price per unit.

The name causes trouble because it sounds like a rate. It is not. A kilowatt is a rate; a kilowatt-hour is a rate multiplied by a time, which is a quantity of energy. Examiners ask you to state this, and "a unit of energy" is often a mark on its own.

The practical lesson from any bill is that it is the heating appliances that cost money. Anything designed to warm something — kettle, iron, immersion heater, air conditioner — runs at kilowatts. Anything designed to process information runs at watts. A phone charger left plugged in all year costs less than one hot bath.

energy (kWh) = power (kW) × time (h)cost = energy (kWh) × price per kWhkeep power in kilowatts and time in hours, or the numbers are meaningless
Worked example 25 marks

A 2.0 kW heater is used for 3.0 hours a day. Electricity costs 22 rupees per kWh. Find the cost of running it for 30 days.

  1. Daily energy = 2.0 × 3.0 = 6.0 kWh.Power already in kW, time already in hours — nothing to convert.
  2. Monthly energy = 6.0 × 30 = 180 kWh.
  3. Cost = 180 × 22.
  4. = 3960 rupees.
  5. Sense check: one heater, one month, roughly four thousand rupees. Plausible for a 2 kW load.An answer in the tens or the millions would signal a conversion error.

3960 rupees

03

The three-pin plug and why each wire is there

Mains wiring in a plug is not arbitrary. Each of the three wires does one job, and exam questions almost always ask you to explain that job rather than recite a colour.

The live wire carries the alternating supply voltage and is the dangerous one. The neutral completes the circuit and sits at roughly earth potential. The earth wire normally carries no current at all: it is a safety path, connected to the metal case of the appliance.

If a fault lets the live wire touch a metal case, the case becomes live. Without an earth wire, the next person to touch it becomes the path to ground. With an earth wire, a very large current flows instantly to earth instead, which blows the fuse and disconnects the appliance before anyone touches it.

WireColour (international)Job
Livebrowncarries the alternating supply voltage — the dangerous one
Neutralbluecompletes the circuit, near earth potential
Earthgreen and yellowsafety path from the metal case to the ground

The fuse goes in the live wire

A fuse must break the connection to the dangerous side of the supply. Put it in the neutral instead and it will still blow, but the appliance stays connected to the live wire and remains lethal to touch. Examiners award the mark for saying exactly that.

04

Fuses, circuit breakers and double insulation

A fuse is a deliberately weak link: a thin wire that melts when the current through it exceeds its rating, breaking the circuit. It is chosen to be just above the appliance's normal working current — near enough to react to a fault, far enough above to avoid blowing every time the appliance switches on.

Choosing a fuse is a two-step calculation that appears constantly. Work out the normal current from I = P/V, then pick the next standard fuse above it. The common ratings are 3 A, 5 A and 13 A.

A circuit breaker does the same job with an electromagnetic switch instead of a melting wire. It trips faster and, more usefully, can simply be reset instead of replaced.

Some appliances have no earth wire at all and are still safe. These are double insulated: the casing is plastic, and there is a second layer of insulation between the electrical parts and anything you can touch. With no metal case there is nothing to become live, so there is nothing to earth.

Worked example 33 marks

A 920 W hairdryer runs on 230 V. Calculate the normal operating current and state a suitable fuse.

  1. I = P/V = 920 / 230.Both already in base units.
  2. I = 4.0 A.
  3. The next standard rating above 4.0 A is 5 A.A 3 A fuse would blow in normal use; a 13 A fuse would allow a dangerous fault current before reacting.

4.0 A, so a 5 A fuse

Key points

  1. Fuse rating is chosen just above the normal working current.
  2. The fuse always goes in the live wire.
  3. A circuit breaker does the same job and can be reset.
  4. Double-insulated appliances need no earth wire because they have no metal case.
  5. Earth plus fuse is the pair that makes a metal-cased appliance safe.

Practice questions

6 questions · 28 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain why a fuse must be fitted in the live wire and not the neutral wire.
Model answer

When the fuse blows it must disconnect the appliance from the dangerous side of the supply. A fuse in the neutral would leave the appliance connected to the live wire and still dangerous.

Examiner tip. Both halves — what a live fuse achieves, and what a neutral fuse fails to achieve.

SQ2[2 marks]
State what is meant by the kilowatt-hour.
Model answer

The energy transferred by an appliance of power 1 kilowatt operating for 1 hour. It is a unit of energy, not power.

Examiner tip. Adding "a unit of energy" is often the second mark, because the name sounds like a power unit.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A 60 W lamp is left on for 8.0 hours. Calculate the energy used in kilowatt-hours and in joules.

Given. P = 60 W, t = 8.0 h

Full working
  1. Converts to kilowatts: 60 W = 0.060 kW[1]
  2. E = 0.060 × 8.0 = 0.48 kWh[1]
  3. Converts hours to seconds: 8.0 × 3600 = 28 800 s[1]
  4. E = 60 × 28 800 = 1.73 × 10⁶ J[1]

0.48 kWh, or 1.7 × 10⁶ J

Examiner tip. The two answers describe the same energy. If you get one right and the other wrong, the error is always in a unit conversion rather than in the physics.

Long questions

1 · 9 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[9 marks]
A household uses a 3.0 kW immersion heater for 2.5 hours a day and five 12 W LED lamps for 6.0 hours a day. Electricity costs 22 rupees per kWh.
  1. Calculate the daily energy used by the immersion heater, in kWh. [2]
  2. Calculate the daily energy used by the five lamps, in kWh. [3]
  3. Calculate the total daily cost. [2]
  4. The immersion heater runs on 230 V. Calculate the current it draws and state a suitable fuse. [2]
Mark scheme
  1. Uses E = Pt with power in kW and time in hours[1]
  2. E = 3.0 × 2.5 = 7.5 kWh[1]
  3. Total lamp power = 5 × 12 = 60 W = 0.060 kW[1]
  4. Uses E = 0.060 × 6.0[1]
  5. = 0.36 kWh[1]
  6. Total = 7.5 + 0.36 = 7.86 kWh[1]
  7. Cost = 7.86 × 22 = 173 rupees[1]
  8. I = P/V = 3000 / 230 = 13.0 A[1]
  9. A 13 A fuse is marginal — a higher-rated fuse or a dedicated circuit is neededaccept 13 A with a comment, or a stated higher rating[1]

(a) 7.5 kWh (b) 0.36 kWh (c) about 173 rupees (d) 13 A

Examiner tip. Multiply the lamp power by five before converting to kilowatts, not after. Doing it in the wrong order is where this question usually goes wrong.

Exam questions

2 · 11 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
An electric kettle is rated 2.3 kW and is used on a 230 V mains supply. It is used for a total of 15 minutes each day. Electricity costs 25 rupees per kWh.
  1. Calculate the current drawn by the kettle.
  2. State the most suitable fuse from 3 A, 5 A and 13 A, and justify your choice.
  3. Calculate the daily cost of running the kettle.
Mark scheme
  1. Uses I = P/V = 2300 / 230converting kW to W[1]
  2. I = 10 A[1]
  3. 13 A fuse[1]
  4. It is the next standard rating above the normal working current of 10 A — a 5 A fuse would blow in normal usethe justification is a separate mark[1]
  5. Energy = 2.3 kW × 0.25 h = 0.575 kWh15 minutes is 0.25 hours[1]
  6. Cost = 0.575 × 25 = 14.4 rupees[1]

(a) 10 A (b) 13 A, the next rating above 10 A (c) about 14 rupees per day

Examiner tip. Fifteen minutes is 0.25 hours, not 15. Converting time to hours before touching the kilowatt-hour formula prevents the commonest error in this whole topic.

Q2[5 marks]
A metal-cased electric drill is connected to the mains with a three-core cable.
  1. State the colour of the earth wire and where it is connected.
  2. Explain how the earth wire and fuse together protect the user if the live wire touches the metal case.
Mark scheme
  1. Green and yellow[1]
  2. Connected to the metal case[1]
  3. A fault would send a very large current from the live wire through the case to earth[1]
  4. This large current melts the fuse[1]
  5. Which disconnects the live supply, so the case cannot give a shockthe fuse must be in the live wire for this to work[1]

Examiner tip. The chain is: fault → large current to earth → fuse melts → live disconnected. Four links, and each is a mark.