3. Practical Electricity

a. Electrical power

6091 16(a), (b) · IGCSE 4.2.5

Lesson flow

Begin with appliances in the flat; ask students to rank by current before switching on. The Predict question exposes over-reliance on P = I²R when V is fixed. Mains is modelled by its r.m.s. value (230 V), so P = VI gives average power.

Predict-then-test

Prompt: A 2200 W kettle element (24 Ω) is replaced by a 12 Ω element on the same 230 V supply. Predict the new power.
Outcome: Power doubles to 4400 W.

Challenge (auto-checked)

Task: Choose the kettle element so that it heats 1.0 kg of water from 30 °C to 100 °C in 2.0 minutes (294 000 J needed). Assume no energy is lost.

Answer: P = 294 000 J ÷ 120 s = 2450 W, so R = 230² ÷ 2450 = 21.6 Ω (accepted within ±3 s: about 21.1–22.1 Ω).

Exam-style questions and answers

1. A rice cooker draws 3.0 A from the 230 V mains. Calculate its power. [2]
   690 W.
2. A 1000 W iron is connected to 230 V mains. Calculate the current in it. [2]
   4.35 A. I = P/V = 1000/230 = 4.35 A.
3. A heating element of resistance 50 Ω carries a current of 4.0 A. Calculate the power transferred. [2]
   800 W. P = I²R = 16 × 50 = 800 W.
4. Which appliance relies mainly on the heating effect of a current? [1]
   A  Ceiling fan
   B  Electric kettle
   C  Phone charger
   D  Television
   B (Electric kettle).
5. An electric water heater is rated 230 V, 3000 W. Calculate the resistance of its element and explain why the element is made of a high-resistance alloy rather than copper. [3]
   R = V²/P = 230²/3000 = 17.6 Ω. A thin copper wire with this resistance would need to be impractically long; an alloy such as nichrome has a high resistivity, so a short coil gives the resistance needed, and it does not oxidise easily when red hot.

b. Energy and cost

6091 16(b), (c) · IGCSE 4.2.5

Lesson flow

Tariff stored as an editable value: SP Group 28.59 ¢/kWh before GST (31.16 ¢ with GST) for 1 October – 31 December 2026, checked 13 October 2026. Update at the start of each quarter. In Generic mode students enter their own tariff and currency.

Predict-then-test

Prompt: Which costs more per month: 1.0 kW air-con for 8 h/day, or 3.0 kW water heater for 0.5 h/day?
Outcome: Air-conditioner: 240 kWh/month against 45 kWh for the heater.

Challenge (auto-checked)

Task: Change the hours of use so that the household’s monthly bill (30 days, before GST) is below S$100.00 (generic mode: the equivalent of 350 kWh). The refrigerator must stay on 24 hours a day.

Answer: Default use is about 451 kWh/month (S$129). Cutting the air-conditioner from 7 h to 3 h a day saves 120 kWh, bringing the bill to about S$94. Other combinations are accepted.

Exam-style questions and answers

1. A 2.2 kW kettle is used for 15 minutes each day. Calculate the energy it uses in 30 days, in kWh. [2]
   16.5 kWh. 2.2 kW × 0.25 h × 30 = 16.5 kWh.
2. Electricity costs 28.59 cents per kWh. Calculate the cost of running a 150 W refrigerator for 24 hours a day for 30 days, in dollars. [3]
   30.88 $. 0.15 kW × 24 h × 30 = 108 kWh; 108 × $0.2859 = $30.88.
3. How many joules are there in 1 kWh? [1]
   3600000 J.
4. Which uses the most energy? [1]
   A  A 2000 W heater for 15 minutes
   B  A 100 W TV for 4 hours
   C  A 10 W charger for 24 hours
   D  A 1000 W iron for 20 minutes
   A (A 2000 W heater for 15 minutes). 0.50 kWh, 0.40 kWh, 0.24 kWh and 0.33 kWh.
5. A family wants to reduce its electricity bill. Suggest two changes with the largest effect, and explain using E = Pt. [3]
   Energy used depends on both power and time. Reduce the use of high-power appliances that are on for long periods, e.g. run the air-conditioner for fewer hours or at a higher set temperature, and shorten water-heater use. Cutting a low-power device (a charger) saves little, since its P × t is small.

c. Mains wiring and hazards

6091 16(d), (g), (h) · IGCSE 4.4

Lesson flow

Use Real view for the plug. Colours are always shown with letters (L, N, E) so students with colour-vision deficiency can follow. The Predict step reads node potentials from the circuit solver: live 230 V, neutral 0 V relative to earth (the neutral is bonded to earth at the substation).

Predict-then-test

Prompt: Which single wire is most dangerous to touch: live, neutral or earth? Why?
Outcome: Live: 230 V; neutral and earth: 0 V.

Challenge (auto-checked)

Task: Wire the plug for a 2200 W kettle correctly and fit the correct fuse. (Use Real view to wire the plug.)

Answer: Brown → L (fuse side), blue → N, green/yellow → E; 13 A fuse (normal current 9.6 A).

Exam-style questions and answers

1. What is the colour of the insulation on the earth wire? [1]
   A  Brown
   B  Blue
   C  Green and yellow stripes
   D  Red
   C (Green and yellow stripes).
2. Which wire is at a high alternating p.d. relative to the ground? [1]
   A  Live
   B  Neutral
   C  Earth
   D  None of them
   A (Live).
3. Why is it dangerous to use a hair dryer with wet hands? [1]
   A  Water increases the mains p.d.
   B  Wet skin has a lower resistance, so a larger current can flow through the body
   C  Water makes the fuse blow
   D  The hair dryer uses more power
   B (Wet skin has a lower resistance, so a larger current can flow through the body).
4. State two hazards of using one extension adaptor for a 2200 W kettle, a 1000 W iron and a 2000 W heater, all switched on at once on 230 V mains, and explain one of them. [3]
   The total current is very large (about 9.6 + 4.3 + 8.7 ≈ 23 A), so the cable and adaptor overheat; the insulation can melt, exposing live wires (shock) or causing a fire. The fuse may also blow.
5. Describe how a three-pin plug is wired. [3]
   Brown (live) wire to the bottom-right terminal through the fuse; blue (neutral) to the bottom-left terminal; green and yellow (earth) to the top terminal. The cable grip holds the outer insulation firmly; no bare wire is exposed beyond the terminals.

d. Fuses and circuit breakers

6091 16(e), (i) · IGCSE 4.4

Lesson flow

The fuse is modelled as an ideal zero-resistance link that opens when the current exceeds its rating (the solver re-solves after each blow). Use the neutral-wire option to show why fuses belong in the live wire: after the blow the element is still at 230 V relative to earth.

Predict-then-test

Prompt: 2200 W kettle on a 13 A fused extension; an 1800 W hair dryer is added. Predict what happens.
Outcome: 17.4 A exceeds 13 A, so the fuse blows and both stop.

Challenge (auto-checked)

Task: Choose the correct plug fuse for each appliance (230 V mains).

Answer: Kettle 9.57 A → 13 A; lamp 0.26 A → 3 A; TV 0.43 A → 3 A; iron 4.35 A → 5 A.

Exam-style questions and answers

1. A 230 V hair dryer is rated 1800 W. Calculate its normal current. [2]
   7.83 A.
2. Which fuse should be fitted to the plug of the 1800 W hair dryer? [1]
   A  3 A
   B  5 A
   C  13 A
   D  30 A
   C (13 A).
3. Why are fuses fitted in the live wire rather than the neutral wire? [1]
   A  The live wire carries more current
   B  So that when the fuse blows the appliance is disconnected from the high p.d.
   C  The neutral wire is too thin
   D  So that the fuse blows more easily
   B (So that when the fuse blows the appliance is disconnected from the high p.d.).
4. What is one advantage of a circuit breaker over a fuse? [1]
   A  It works without any current
   B  It can be reset after it trips
   C  It increases the mains p.d.
   D  It never needs to be in the live wire
   B (It can be reset after it trips).
5. Explain how a fuse protects a circuit when too much current flows. [3]
   The current through the thin fuse wire heats it. If the current exceeds the fuse rating, the wire becomes hot enough to melt, which breaks the circuit and stops the current. This prevents the cables from overheating, which could melt the insulation and start a fire.

e. Earthing and faults

6091 16(e), (f) · IGCSE 4.4

Lesson flow

Misconception focus: fuse vs earth. Model: 230 V supply, 0.4 Ω live cable, 0.5 Ω earth path, 2 kΩ body, 24 Ω element, 13 A fuse. Run Fault 1 then Fault 2 and keep both predictions on screen. The no-fuse option shows that an earth wire alone leaves a huge current and a casing at about 128 V.

Misconception focus. The Predict step locks in each student’s prediction and shows it next to the measured result.

Predict-then-test

Prompt: Fault 1: earthed metal kettle + 13 A fuse, live touches casing. Fault 2: same but earth wire missing. Predict what happens in each.
Outcome: Fault 1: fuse blows, casing safe. Fault 2: fuse does not blow; casing at 230 V; person ≈ 115 mA.

Challenge (auto-checked)

Task: With the fault switched on, find two different arrangements that keep the person safe (no current through them) and leave nothing overheating.

Answer: Two safe arrangements: (1) earth connected + 13 A fuse (fuse blows, casing 0 V); (2) double-insulated plastic casing (fault cannot reach the user). Earth without fuse fails (≈ 260 A, cable overheats, casing ≈ 128 V); fuse without earth fails (casing 230 V, person ≈ 115 mA).

Exam-style questions and answers

1. What is the main purpose of the earth wire? [1]
   A  To carry the normal current back to the supply
   B  To provide a low-resistance path to the ground for a fault current
   C  To make the fuse blow during normal use
   D  To reduce the p.d. of the mains
   B (To provide a low-resistance path to the ground for a fault current).
2. A hair dryer has a plastic casing and the symbol ⧈. Why does it not need an earth wire? [1]
   A  It uses very little current
   B  It is double insulated: the casing is an insulator so it cannot become live
   C  Its fuse is in the neutral wire
   D  Plastic conducts the current to earth
   B (It is double insulated: the casing is an insulator so it cannot become live).
3. In a fault, the live wire touches an earthed metal casing. The total resistance of the fault path is 0.80 Ω. Calculate the fault current from 230 V mains. [2]
   287.5 A.
4. An unearthed metal kettle has a fault: its casing is live. A person touching it carries 0.12 A. Why does the 13 A fuse not blow? [1]
   A  The fuse is in the live wire
   B  0.12 A is far below the fuse rating
   C  The current is a.c.
   D  The person’s resistance is zero
   B (0.12 A is far below the fuse rating).
5. Explain how the earth wire and the fuse together protect a person using a metal-cased appliance when a live wire touches the casing. [4]
   The earth wire connects the casing to the ground with a very low resistance, so a very large current flows from the live wire through the casing to earth (rather than through a person). This current is far above the fuse rating, so the fuse in the live wire melts and breaks the circuit. The appliance, and its casing, are then disconnected from the live supply and are safe to touch.