S4 Physics Paper: Electricity - Electrostatics, Current, Resistance, Circuits And Power (100 Marks)

S4 Physics Paper: Electricity – Electrostatics, Current, Resistance, Circuits and Power (100 marks)

2026-11-11

S4 Physics – electricity and circuits

S4 physics electricity practice paper: static charge, current, resistance, series and parallel circuits, electrical power and energy, with full answers and mark schemes.

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Paper details　S4 physics electricity practice paper: electrostatics, current and charge, resistance and Ohm law, series and parallel circuits, electrical power, meter bills and household safety.

Section A – Electrostatics and electric charge

Answer in full sentences; show the working and the unit in any calculation.

#	Question	Answer and marking notes
1	Explain charging by friction in terms of particles and state why the total charge stays the same.	When two different materials are rubbed, one holds its outer electrons less tightly and electrons transfer to the other. The material losing electrons becomes positive and the one gaining them becomes negative. Only electrons move; protons do not, so the total charge stays zero, which is conservation of charge. (4 marks: electron transfer 2; gain and loss 1; conservation 1)
2	A small sphere carries a negative charge of 0.0 C. Find the number of excess electrons (elementary charge 1.6 × 10^-19 C).	n = Q / e = 0.0 / 1.6 x 10^-19 = 3.0 electrons. (4 marks: formula 1; substitution 1; answer 2)
3	Compare how conductors and insulators behave in electrostatic situations, giving one example of each.	A conductor has many free electrons, so charge spreads quickly over its surface, e.g. a brass sphere or a metal door handle. An insulator has very few free electrons, so charge stays where it was created and does not flow away, e.g. a plastic ruler or a glass rod. An insulator therefore keeps its charge for a long time, while a conductor needs an insulating handle to stay charged. (4 marks: conductor 2; insulator 2, with examples)
4	A lightning conductor is fixed at the highest point of a building and earthed by a thick copper strip. Explain how it protects the building.	The thundercloud induces a large charge on the top of the building; the sharp tip concentrates charge so the field is very strong, the air ionises and a conducting path forms, and the lightning current is carried safely to the ground through the thick copper strip instead of through the structure, avoiding intense heating or fire. The copper has low resistance and a large cross-section, so it carries the large current without melting. (4 marks: point discharge and guiding the strike 2; earthing the current 1; low resistance thick copper 1)

Section B – Electric current and charge

Take the elementary charge as 1.6 × 10^-19 C and give the unit of every answer.

#	Question	Answer and marking notes
5	Define electric current, give its formula and unit, and state how the conventional current direction relates to electron flow.	Current is the charge passing a cross-section of a conductor per unit time: I = Q / t, measured in amperes (A), where 1 A = 1 C/s. Conventional current is taken as the direction of positive charge flow, from the positive terminal through the external circuit to the negative terminal. In a metal the moving charges are electrons, so electron flow is opposite to the conventional current. (4 marks: definition 1; formula and unit 2; opposite direction 1)
6	A current of 0.5 A flows for 4 minutes. Find the charge that passes the cross-section.	t = 4 minutes x 60 = 240 s. Q = I t = 0.5 x 240 = 120 C. (4 marks: time conversion 1; formula 1; answer 1; unit 1)
7	Using the result above, find the number of electrons that pass the cross-section in that time.	n = Q / e = 120 / 1.6 x 10^-19 = 7.5 × 10^20 electrons. (4 marks: formula 1; substitution 1; answer 1; standard form 1)
8	Compare direct current with alternating current and state which type the local mains supply uses.	In direct current the direction is fixed and electrons drift one way, as from a battery. In alternating current the direction reverses periodically; the local mains supply is 50 Hz, so it reverses 100 times a second, at about 220 V. A domestic socket therefore supplies alternating current. (4 marks: direct current 1; alternating current and frequency 2; local supply is alternating 1)

Section C – Resistance and Ohm law

Show the formula, substitution and unit; give resistance in ohms.

#	Question	Answer and marking notes
9	Define resistance, write the formula of Ohm law and give the unit of resistance.	Resistance measures how strongly a conductor opposes current and equals the potential difference across it divided by the current: R = V / I. The unit is the ohm, 1 ohm = 1 V/A. If the temperature stays constant the resistance of a metal wire is nearly constant, so V is proportional to I. (4 marks: definition 2; formula 1; unit 1)
10	The potential difference across a resistor is 12 V and the current through it is 2.5 A. Find its resistance.	R = V / I = 12 / 2.5 = 4.8 ohms. (5 marks: formula 1; substitution 2; answer 1; unit 1)
11	Two resistors of 6 ohms and 12 ohms are connected first in series and then in parallel. Find the equivalent resistance in each case.	Series: R = 6 + 12 = 18 ohms. Parallel: 1/R = 1/6 + 1/12, so R = (6 x 12) / (6 + 12) = 4 ohms. (4 marks: series 2; parallel 2)
12	Explain how the resistance of a metal wire depends on its length, cross-sectional area and temperature.	A longer wire gives electrons more chance to collide with ions, so resistance is proportional to length. A larger cross-sectional area provides more paths for the electrons, so resistance is inversely proportional to area. At higher temperature the ions vibrate more strongly and collisions become more frequent, so resistance rises. (5 marks: length 2; area 2; temperature 1, reason needed for each)

Section D – Series and parallel circuit calculations

Neglect internal resistance of the battery and the resistance of the connecting wires; give units at every step.

#	Question	Answer and marking notes
13	A 12 V battery is connected in series with resistors of 4 ohms and 8 ohms. Find the current and the voltage across each resistor.	Total resistance = 4 + 8 = 12 ohms. I = V / R = 12 / 12 = 1 A. V1 = I R1 = 1 x 4 = 4 V; V2 = 1 x 8 = 8 V. The two voltages add to 12 V, matching the supply. (5 marks: total resistance 1; current 2; the two voltages 2)
14	A 6 V supply is connected to resistors of 12 ohms and 6 ohms in parallel. Find the equivalent resistance, the total current and the current in each branch.	Equivalent resistance = (12 x 6) / (12 + 6) = 4 ohms. Total current = 6 / 4 = 1.5 A. Branch currents: 6 / 12 = 0.5 A and 6 / 6 = 1.0 A, which add to 1.5 A as charge is conserved. (5 marks: equivalent resistance 2; total current 1; branch currents 2)
15	An appliance on a 12 V supply draws 1.5 A. Find its power.	P = V I = 12 x 1.5 = 18 W. (4 marks: formula 1; substitution 1; answer 1; unit 1)
16	An appliance rated 2000 W is used for 1.5 hours a day. Find the monthly (30 days) energy used and the cost at 1.2 dollars per kWh.	Daily energy = 2000 / 1000 x 1.5 = 3.0 kWh; monthly = 3.0 x 30 = 90.0 kWh. Cost = 90.0 x 1.2 = 108.0 dollars. (4 marks: kW conversion 1; daily energy 1; cost 1; unit 1)

Section E – Experiment: measuring resistance by the voltmeter-ammeter method

In experiment questions state how the meters are connected and where errors come from.

#	Question	Answer and marking notes
17	Describe the circuit used to measure resistance by the voltmeter-ammeter method and explain why each meter is connected that way.	The resistor, ammeter, rheostat and battery are connected in series as the main circuit, while the voltmeter is connected across the resistor. An ammeter has a very low resistance, so connecting it in series hardly changes the current while measuring the current through the resistor; a voltmeter has a very high resistance, so in parallel it takes almost no current while measuring the voltage across the resistor. Varying the rheostat gives several readings and R = V / I can be averaged. (6 marks: series main circuit 2; voltmeter in parallel 2; explanation of the two resistances 2)
18	In one trial the voltmeter reads 4.5 V and the ammeter reads 0.30 A. Find the resistance.	R = V / I = 4.5 / 0.30 = 15 ohms. (5 marks: formula 1; substitution 2; answer 1; unit 1)
19	State two reasons why the measured value may differ from the true value and give one improvement for each.	The current heats the resistor, so its resistance rises with temperature and the reading is too high; keep each current pulse short or use a smaller current. The ammeter itself has some resistance, which adds to the circuit and makes the measured voltage include the drop across the ammeter; use an ammeter of very low resistance. Poor contact at terminals or crocodile clips adds resistance; clean the contacts and clip firmly. (5 marks: 2 per reason, 1 per improvement, max 5)

Section F – Integrated application: household electricity and safety

Take the mains voltage as 220 V and the tariff as 1.2 dollars per kWh.

#	Question	Answer and marking notes
20	An appliance rated 2000 W is connected to the 220 V mains. Find its current and state whether a 3 A, 13 A or 30 A fuse should be used, with a reason.	I = P / V = 2000 / 220 = 9.1 A. A 13 A fuse is chosen: it is just above the working current so it will not blow in normal use, yet it melts and breaks the circuit when the current rises well above normal because of a short circuit or a fault. A 3 A fuse would keep blowing and a 30 A fuse would not protect the appliance. (6 marks: current 2; choice of 13 A 1; slightly above working current 1; overload protection 2)
21	An appliance rated 1500 W is used 2.0 hours a day for 30 days. Find the total energy used and the cost.	Power = 1500 / 1000 = 1.5 kW. Energy = 1.5 x 2.0 x 30 = 90 kWh. Cost = 90 x 1.2 = 108 dollars. (5 marks: kW conversion 1; energy 2; cost 2)
22	Explain how the earth wire, a metal casing and a circuit breaker protect the user, and why wet hands make touching an appliance dangerous.	The metal casing is joined to the earth wire so it stays at earth potential; if a live wire works loose and touches the casing, current flows away through the earth wire rather than through the user, and the large current blows the fuse or trips the breaker. A residual current device detects unequal currents in and out, that is current leaking through the user or the earth path, and cuts the supply in a very short time. Wet hands greatly lower the skin resistance, so a much larger current flows through the body at the same voltage, which is why plugs and appliances must not be touched with wet hands. (5 marks: earth wire and casing 2; breaker 2; wet hands lower resistance 1)

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