S4 Physics Practice Paper: Current, Voltage, Resistance and Electrical Power (100 marks)

2026-11-10

S4 Physics – Current and circuits

S4 Physics paper: 6 questions, 30 parts, 100 marks, 75 min. Covers current, voltage, resistance, series and parallel circuits, power and household electricity.

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Section A – Current, charge and electrical power

Show the formula, substitution and unit in each calculation.

#	Question	Answer and marking notes
1	A resistor of resistance 6 Ω is connected across a 12 V battery. Calculate the current through the resistor.	Use I = V ÷ R: I = 12 ÷ 6 = 2.0 A. (2 marks: formula 1; answer with unit 1)
2	The circuit is switched on for 5 minutes. Calculate the charge that passes any cross-section of the resistor in this time.	Q = I t = 2.0 × (5 × 60) = 600 C. (3 marks: time in seconds 1; formula 1; answer with unit 1)
3	Calculate the electrical power dissipated by the resistor.	P = V I = 12 × 2.0 = 24 W, or P = I²R = 2.0² × 6 = 24 W. (3 marks: formula 1; substitution 1; answer with unit 1)
4	Calculate the energy dissipated by the resistor in 5 minutes. Give the answer in kJ.	E = P t = 24 × 300 = 7200 J = 7.2 kJ. (4 marks: time 1; formula 1; 7200 J 1; conversion to kJ 1)
5	A second resistor of 3 Ω is now connected in series with the 6 Ω resistor, and the battery voltage stays at 12 V. Find the current and the power of the 6 Ω resistor, then compare with part 3 and explain the difference.	Total resistance = 6 + 3 = 9 Ω; I = 12 ÷ 9 = 1.3 A (1.33 A); P = I²R = 1.33² × 6 = 10.7 W. Power falls from 24 W to 10.7 W because the total resistance rises, so the current is smaller and P is proportional to I². (6 marks: total resistance 1; current 1; power 2; comparison 1; explanation 1)

Section B – Series and parallel circuits

Circuit described in words: a 20 V battery is in series with a 6 Ω resistor, followed by a parallel combination of a 12 Ω and a 6 Ω resistor, which returns to the battery. Show all working.

#	Question	Answer and marking notes
6	Find the equivalent resistance of the parallel combination.	1 / R = 1 / 12 + 1 / 6 = 1 / 12 + 2 / 12 = 3 / 12, so R = 4.0 Ω. (3 marks: formula 1; substitution 1; answer 1)
7	Find the total resistance of the circuit.	6 + 4.0 = 10 Ω. (2 marks: method 1; answer with unit 1)
8	Find the total current delivered by the battery.	I = V / R = 20 / 10 = 2.0 A. (3 marks: formula 1; substitution 1; answer 1)
9	Find the voltage across the parallel combination and the currents in the 12 Ω and 6 Ω resistors, then check their sum.	V = I R = 2.0 × 4.0 = 8.0 V; I(12 Ω) = 8.0 / 12 = 0.67 A; I(6 Ω) = 8.0 / 6 = 1.33 A; 0.67 + 1.33 = 2.0 A, equal to the total current. (4 marks: voltage 1; each branch current 1; check 1)
10	Find the power dissipated by the 6 Ω resistor in series.	P = I²R = 2.0² × 6 = 24 W. (5 marks: formula 1; substitution 2; answer 1; unit 1)

Section C – Electrical power and electricity bills

An air conditioner is rated at 1500 W and runs 6 hours a day. Take 30 days in a month and $1.2 per kWh.

#	Question	Answer and marking notes
11	Find the energy used per day in kWh.	Power = 1500 W = 1.5 kW; E = 1.5 × 6 = 9.0 kWh. (3 marks: unit conversion 1; formula 1; answer 1)
12	Find the total energy used in a month of 30 days.	9.0 × 30 = 270 kWh. (3 marks: method 1; answer 1; unit 1)
13	Find the monthly electricity cost of the air conditioner.	270 × 1.2 = $324. (4 marks: formula 1; substitution 1; answer 1; currency unit 1)
14	If it runs only 4 hours a day with all other conditions unchanged, find the monthly cost and the money saved.	1.5 × 4 × 30 = 180 kWh; 180 × 1.2 = $216; saving 324 – 216 = $108. (4 marks: energy 1; cost 1; saving 2)
15	If a 1000 W model is used 5 hours a day for 30 days, how much is saved compared with part 3?	1.0 × 5 × 30 = 150 kWh; 150 × 1.2 = $180; saving 324 – 180 = $144. (2 marks: calculation 1; difference 1)

Section D – Household electricity and safety

A household circuit is supplied at 220 V and protected by a fuse.

#	Question	Answer and marking notes
16	An appliance rated at 2000 W is connected to a 220 V supply. Find its working current.	I = P / V = 2000 / 220 = 9.1 A. (3 marks: formula 1; substitution 1; answer with unit 1)
17	Should a 5 A or a 13 A fuse be used for this appliance? Explain your choice.	Use the 13 A fuse. The working current 9.1 A exceeds 5 A, so a 5 A fuse would blow; 9.1 A is below 13 A, so a 13 A fuse allows normal operation and still cuts the circuit on overload. (3 marks: choice 1; comparison with 5 A 1; comparison with 13 A 1)
18	What happens if a 3000 W appliance is connected to a 13 A fuse? Support your answer with a calculation.	I = 3000 / 220 = 13.6 A, which is larger than 13 A, so the fuse melts and breaks the circuit; the appliance cannot work. (4 marks: current 2; fuse melts 1; consequence 1)
19	State the function of the earth wire.	When the metal casing becomes live accidentally, the earth wire carries the current safely to the ground and prevents the user from being part of the conducting path. (2 marks)
20	Explain why a copper wire must not be used to replace a fuse.	Copper has low resistance and a high melting point, so it does not melt even under excessive current and cannot break the circuit; the wiring then overheats and may start a fire. (5 marks: high melting point or low resistance 2; no melting 1; loss of protection 1; overheating and fire 1)

Section E – Experimental data and V-I characteristics

A student measures the voltage and current of a metal wire: 1.0, 2.0, 3.0, 4.0, 5.0 V with currents 0.20, 0.40, 0.60, 0.80, 1.00 A.

#	Question	Answer and marking notes
21	Use the data to calculate the resistance for each data point.	R = V / I: 1.0 / 0.20 = 5.0 Ω, and every other point also gives 5.0 Ω. (3 marks: formula 1; two evaluations 1; conclusion 1)
22	What physical quantity does the slope of the V-I graph represent? Find the slope from the data.	The slope represents the resistance of the wire; slope = (5.0 – 1.0) / (1.00 – 0.20) = 4.0 / 0.80 = 5.0 Ω. (3 marks: quantity 1; calculation 1; answer 1)
23	Is the metal wire an ohmic conductor? Explain your answer.	Yes. The resistance stays at 5.0 Ω for every voltage, so current is proportional to voltage and the V-I graph is a straight line through the origin. (3 marks: judgement 1; evidence 1; explanation 1)
24	If the wire is replaced by a filament lamp, the V-I curve is no longer a straight line. Explain why.	As the current increases the filament gets hotter, and the resistance of a metal rises with temperature, so the current grows less than in proportion and the curve bends upwards. (4 marks: temperature rises 1; resistance increases 1; no longer proportional 2)
25	Suggest two ways to reduce experimental error in this experiment.	For example: take readings quickly so the wire does not heat up and change resistance; check clean tight contacts, zero the meters and repeat readings to average. (3 marks: 1 per valid method, plus 1 for explanation)

Section F – Integrated application

Two filament lamps are both marked 220 V, 60 W.

#	Question	Answer and marking notes
26	Find the resistance of each lamp.	R = V² / P = 220² / 60 = 48400 / 60 = 807 Ω. (3 marks: formula 1; substitution 1; answer 1)
27	Find the current when a lamp works normally.	I = P / V = 60 / 220 = 0.27 A. (3 marks: formula 1; answer 1; unit 1)
28	The two lamps are connected in parallel to a 220 V supply. Find the total power and total current.	Total power = 60 + 60 = 120 W; total current = 120 / 220 = 0.55 A. (4 marks: power 2; current 1; unit 1)
29	The lamps are now connected in series to the same 220 V supply. Find the power of each lamp.	Total resistance = 2 × 807 = 1613 Ω; I = 220 / 1613 = 0.136 A; P = I²R = 0.136² × 807 = 15 W. (4 marks: total resistance 1; current 1; power 2)
30	Explain why the lamps are dimmer in series, and why household appliances are connected in parallel.	In series the current is smaller and each lamp takes only half the voltage, so its power is about 15 W instead of 60 W and it is dimmer. In parallel each appliance receives the full 220 V, can be switched independently and does not affect the others. (2 marks: series 1; parallel 1)

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