IGCSE Physics Practical Electricity: Paper 2 Worked Solutions

www.megalecture.com · www.youtube.com/@megalecture

Online tutoring for Cambridge IGCSE, O Level & A Level Physics

Fahad H. Ahmad

Contact: +92 323 509 4443

M.Eng NUS Singapore · B.Eng NUST · 15+ years of teaching experience in Singapore, UAE & Pakistan · 60,000+ subscribers and 20 million+ video lecture views.

www.Megalecture.com · +92 323 509 4443

IGCSEPHYSICS0625·TOPICALPASTPAPERS

Practical Electricity — Paper 2

Worked Solutions (Theory / Structured)

Practical Electricity — Paper 2 · Worked Solutions

Megalecture worked solutions — model answers with working; please verify before classroom use.

These structured-question solutions for Practical Electricity (IGCSE Physics 0625, Paper 2) were solved from first principles by the Megalecture team using the data in each question. Key relations used throughout: P = VI, P = I²R = V²/R, E = Pt, energy in kWh = power(kW) × time(h), cost = energy(kWh) × price per kWh, and fuse rating chosen just above the normal operating current. Bold values are the final answers with units.

Questions 1 – 5 Question 1 — Lamps in a house (parallel)

5054/02/M/J/03 Q6

• Each lamp must be drawn with its own switch, and every lamp+switch branch connected in parallel across the live and neutral lines (so each lamp gets the full mains voltage and can be switched independently). The single fuse stays in the live wire.
• Current in one 240 V, 30 W lamp.

I = P / V = 30 / 240 = 0.125 A

• Lamps in parallel, so currents add: Imains = 3 × 0.125 = 0.375 A.

Question 2 — Electric grill

5054/02 M/J/04 Q9

• The metal case is connected to earth (the earth pin / earth wire of the plug). If a fault makes the live wire touch the case, a large current flows straight to earth through the low-resistance earth wire; this blows the fuse (or trips the breaker) and disconnects the supply, so the case cannot stay live and give a shock.

(c)(i) Equation linking current, power and voltage:

P = V × I (power = voltage × current)

(c)(ii) Normal current = 8.3 A, so choose the standard fuse rated just above this. The next standard value above 8.3 A is 13 A (a 13 A fuse carries 8.3 A normally but blows on a fault).

(c)(iii) The student is wrong. For a fixed resistance heating element, P = V²/R, so power depends on the square of the voltage.

Halving V (240 → ~half): P ∝ V² → P becomes (½)² = ¼ of the original, not ½. So at 115 V the power is about one quarter, not half.

www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443

Question 3 — Two household appliances

5054/02/M/J/05 Q6

• Water heater (240 V, 12.6 A, 0.50 h): P = VI = 240 × 12.6 = 3024 W = 3.024 kW; energy/day = P × t =

3.024 × 0.50 = 1.51 kW h.

appliance power / W power / kW energy / kW h water heater 3.024 1.51

• At the same mains voltage the water heater takes a much larger current (12.6 A vs 1.20 A); since P =

VI, the larger current means more power.

• The normal working current of the water heater is 12.6 A, which is greater than 3 A. A 3 A fuse would blow immediately in normal use, so the heater could never operate. A 3 A fuse is too small (a 13 A fuse is needed).

Question 4 — Two lamps A and B

5054/02 M/J/06 Q?

• Component C is a (variable) resistor / rheostat — its purpose is to control (limit/vary) the current in the circuit. Component D is a switch — its purpose is to turn the circuit on and off.
• Lamp A is 240 V, 60 W; total current at C = 0.42 A.

(b)(i) Current in lamp A: I = P/V = 60/240 = 0.25 A.

(b)(ii) A and B are in parallel, so IB = Itotal − IA = 0.42 − 0.25 = 0.17 A.

(b)(iii) Resistance of lamp A: R = V/I = 240 / 0.25 = 960 Ω.

• Two lamps wired in parallel: each lamp connected directly across the supply on its own branch, so both receive the full 240 V (correct domestic wiring).
• Same material and length but half the cross-sectional area. Since R ∝ 1/area, halving the area doubles the resistance: Rnew = 2 × 960 = 1920 Ω.

Question 5 — Mains extension lead

5054/02/M/J/06 Q6

(a)(i) Live: the wire that carries the alternating supply voltage (alternates above and below 0 V); it delivers current to the appliance and is the dangerous wire.

(a)(ii) Neutral: the return wire that completes the circuit; it stays at (or near) 0 V / earth potential.

(a)(iii) Earth: a safety wire connected to the metal case (and to the ground); it carries current safely away to earth if a fault occurs.

(b)(i) As more lamps are switched on, the total current drawn through the single cable increases (so the cable gets hotter).

(b)(ii) A fuse of too high a value will not blow when the current becomes dangerously large; the cable/ appliance can then overheat and start a fire.

• Water conducts electricity, so wet hands lower the resistance of the path through the body; if you touch a live part a larger current flows and gives a worse (possibly fatal) shock.

www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443

Questions 6 – 10 Question 6 — Lamp power & parallel/series circuits

5054/02 M/J/09 Q9 · M/J/11 Q9

• Experiment to check the power of a 24 V, 100 W lamp at 24 V: connect the lamp to a 24 V supply with an ammeter in series and a voltmeter across the lamp; adjust so the voltmeter reads 24 V; record the current I and voltage V; then power = V × I and compare this value with the marked 100 W.

(b)(i) Lamps in parallel across 240 V; lamp A = 190 Ω (branch R), lamp B = 380 Ω (branch Q).

At R (lamp A): I = V/R = 240/190 = 1.26 A | At Q (lamp B): I = 240/380 = 0.63 A | At P (main): I = 1.26 +

0.63 = 1.89 A

(b)(ii) Total resistance of the parallel combination:

1/R = 1/190 + 1/380 → R = (190×380)/(190+380) = 127 Ω (or 240/1.89 ≈ 127 Ω)

(b)(iv) Drill of power 1000 W on a 240 V supply; choose a fuse from 1 A, 3 A, 4 A, 13 A.

Normal current I = P/V = 1000/240 = 4.17 A. The fuse must be rated just above this, so a 4 A fuse would blow. The correct choice is the 13 A fuse (the smallest standard value above 4.17 A).

(c)(i) Same two lamps now in series (190 Ω + 380 Ω) across 240 V; current at S:

I = V/R = 240 / (190 + 380) = 240/570 = 0.42 A

(c)(ii) P.d. across lamp B: VB = I × RB = 0.42 × 380 = 160 V.

• The lamps are brighter in parallel. In parallel each lamp gets the full 240 V (so each dissipates more power, P = V²/R), whereas in series they share the 240 V and each carries a smaller current, dissipating less power.

Question 7 — Hairdryer & electric heater (mains)

5054/22 M/J/15 Q7

• The earth wire provides a low-resistance path to earth for the current if a fault makes the live wire touch the metal case, keeping the case safe.
• If the live wire touches the metal case, a large fault current flows through the earth wire to earth;

this large current blows the fuse, which breaks the live wire and disconnects the appliance, so the case is no longer live and cannot give a shock.

• The hairdryer has a plastic (insulating) case and double insulation, so a fault cannot make any touchable metal part live; therefore no earth wire is needed.
• A circuit breaker can be reset (switched back on) after it trips rather than being replaced — it is reusable, faster to restore, and trips more quickly/accurately than a fuse melts.

www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443

Questions 11 – 16 Question 8 — Washing machine (earth wire & energy)

5054/22 M/J/11 Q6

• If a fault connects the live wire to the metal case, the earth wire gives a low-resistance route to earth;

a large current flows and blows the fuse, disconnecting the supply so the case cannot remain live and shock the user.

• Input power 500 W = 0.500 kW; time 45 min = 0.75 h.

Energy = P × t = 0.500 × 0.75 = 0.375 kW h

Question 9 — Room heater (two switched elements)

5054/21 M/J/12 Q9

(a)(i) A fuse is drawn as a small rectangle (box) in the live wire, in series with switch A (the fuse symbol is a line through a rectangle).

(a)(ii) The earth wire is connected to the metal case / metal body of the heater.

(a)(iii)1 A fault could make the live wire touch the metal case, so the case becomes live; if a person touches it, current passes through them to earth → electric shock.

(a)(iii)2 The earth connection gives a low-resistance path to earth, so a large current flows and blows the fuse, cutting off the supply before a person can receive a dangerous shock.

(c)(i) Both switches give 2100 W; switch A alone gives 600 W. The elements are in parallel so powers add:

switch B alone = 2100 − 600 = 1500 W.

(c)(ii)1 Both switches: 2100 W = 2.1 kW for 2.5 h.

Energy = 2.1 × 2.5 = 5.25 kW h

(c)(ii)2 In joules: E = P × t = 2100 × (2.5 × 3600) = 2100 × 9000 = 1.89 × 107 J (18 900 000 J).

Question 10 — Freezer (kW h and cost)

5054/21 M/J/13 Q10

(a)(i) One kilowatt-hour is the energy transferred (used) by a device of power 1 kW operating for 1 hour

(= 3.6 × 106 J). It is the "unit" used for electricity bills.

(a)(ii) Power 80 W = 0.080 kW; one week = 7 × 24 = 168 h.

Energy = 0.080 × 168 = 13.44 kW h

Cost = 13.44 × 25 = 336 cents = $3.36 (336 cents) www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443

Question 11 — Heater in a cup of water

5054/22 M/J/13 Q11

• Circuit to measure electrical power of the heater: heater connected to the power supply with an ammeter in series with the heater and a voltmeter connected in parallel across the heater (so V and I can be read together; power = V × I).

(b)(i) Supply 12 V, current 4.2 A.

P = V × I = 12 × 4.2 = 50.4 W

(b)(ii) Energy input in 8.0 min (= 480 s), then convert to kW h.

E = P × t = 50.4 × 480 = 24 192 J

In kW h: 24 192 ÷ 3 600 000 = 6.72 × 10-3 kW h (0.00672 kW h)

(Equivalently: 0.0504 kW × (8/60) h = 0.00672 kW h.)

Question 12 — Hairdryer & electric heater (safety)

5054/22 M/J/15 Q7

• The neutral wire completes the circuit, providing the return path for the current back to the supply (it is at about 0 V).
• If a fault makes the live wire touch the metal case, the earth wire carries a large current to earth;

this large current blows the fuse, breaking the live connection so the case is no longer live — protecting the user from a shock.

• The hairdryer is double-insulated with a plastic (non-conducting) case, so no exposed metal part can become live; an earth wire is therefore unnecessary and it is still safe.
• A circuit breaker can simply be reset (switched back on) after a fault — it is reusable and acts faster, unlike a fuse which must be replaced each time it blows.

Question 13 — Electrical generator supplying a factory

5054/21 M/J/16 Q11

(b)(i)1 Power 500 kW delivered at 33 kV; current in the supply wires.

I = P / V = (500 000 W) / (33 000 V) = 15.2 A

(b)(i)2 Electrical energy supplied in 1 hour:

E = P × t = 500 000 × (1 × 3600) = 1.8 × 109 J (= 500 kW h = 1800 MJ)

Question 14 — Television and two lamps

5054/21 M/J/17 Q?

(a)(i) Total power = television + 2 lamps = 120 + 40 + 40 = 200 W.

(a)(ii) Energy in 3.0 h: E = P × t = 0.200 kW × 3.0 h = 0.60 kW h.

(a)(iii) Each lamp is 40 W across 240 V (parallel): I = P/V = 40/240 = 0.167 A (0.17 A).

• A switch is placed in the live wire so that when it is off the appliance is disconnected from the dangerous live (high-voltage) supply. If the switch were in the neutral wire, the appliance would still be connected to the live wire — it could remain live and give a shock even when "switched off".

www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443

Note from Megalecture. These are original Megalecture worked solutions prepared for revision use; they were solved independently from the question papers and are not copied from any official mark scheme. Final numerical answers are shown in bold with units; please verify before classroom use.

Question 15 — Three resistors (3.3 Ω, 3 W each)

5054/21 M/J/18 Q5

• Method: connect the three resistors in the circuit with a battery, a variable resistor (rheostat), a voltmeter across one resistor and an ammeter in series; vary the current, record V and I, and find resistance from R = V / I (taking several readings / a graph improves accuracy).

(b)(i) P.d. 4.2 V across the resistor, current 1.2 A.

R = V / I = 4.2 / 1.2 = 3.5 Ω (close to the marked 3.3 Ω)

(b)(ii) The resistor overheats because the power dissipated exceeds its 3 W rating.

P = V × I = 4.2 × 1.2 = 5.04 W (> 3 W rating → overheats and smokes)

(b)(iii) With one resistor removed, the other two (3.3 Ω each) are in series across 4.2 V.

I = 4.2 / (3.3 + 3.3) = 0.64 A; power in each = I²R = 0.64² × 3.3 ≈ 1.3 W

Each resistor now dissipates only ≈ 1.3 W, which is below the 3 W rating, so they do not overheat — they are safe.

Question 16 — Fuses and a lighting circuit

5054/22 M/J/18 Q8

• Two other forms of protection in household circuits: 1. earth wire (connects metal cases to earth); 2.

circuit breaker / trip switch (cuts the supply on excess current). (Other accepted: insulation of wires, double insulation.)

(b)(i) Wire W is the live wire because it is the wire that contains the fuse and the switches (protection and switching are always placed in the live wire).

(b)(ii) A switch that controls the lamps must be placed at the position in the live wire before the lamps branch (point s), so that opening it isolates the lamps from the live supply.

(b)(iii) A 5 A fuse means the fuse wire melts and breaks the circuit if the current exceeds 5 A, disconnecting the lamps to prevent the wires overheating (the lamps' normal total current is below 5 A).

www.Megalecture.com Fahad H. Ahmad · +92 323 509 4443