Built for the September Unit Assessment. Every question is multi-step, top-of-grade-8/9 difficulty — harder than a normal mock. 60%+ here is a genuinely strong result. Section A: 20 multiple-choice marks in about 25 minutes. Section B: 40 structured marks. Show ALL working in Section B.
Instructions
Section A: 20 multiple-choice questions, 1 mark each — about 25 minutes. Choose ONE option per question.
Section B: 5 structured questions, 40 marks — about 65 minutes. Answer in the spaces provided.
Answer all questions.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 5.
Your answers will be automatically graded when you submit. For every Section A question you will be shown why the option YOU chose is right or wrong. Section B model answers appear for self-marking.
Question Navigation
Section A — Multiple Choice
20 questions · 1 mark each · about 25 minutes · choose ONE option per question. After you submit, every question explains why the option you chose is right or wrong.
1. A current of 0.25 A flows in a wire for 4.0 minutes. How much charge passes a point in the wire?
2. Which statement defines the e.m.f. of a battery?
3. A copper wire is replaced by another copper wire that is three times as long and has twice the cross-sectional area. By what factor does the resistance change?
4. The p.d. across a lamp is 6.0 V and the current in it is 0.50 A. What is the resistance of the lamp?
5. A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel. What is their combined resistance?
6. Lamp X has a higher resistance than lamp Y. They are connected in SERIES with a battery. Which statement is correct?
7. A 12 V supply is connected across an 8.0 kΩ resistor in series with a 4.0 kΩ resistor. What is the p.d. across the 4.0 kΩ resistor?
8. A potential divider consists of a thermistor in series with a fixed resistor. Vout is taken across the FIXED resistor. The temperature rises. What happens to Vout?
9. The current in a fixed resistor is doubled. The power dissipated in the resistor becomes:
10. A 2.0 kW heater runs for 90 minutes. How much energy does it transfer, in kWh?
11. A 1500 W heater is used for 4.0 hours every day for 7 days. Electricity costs 12 cents per kWh. What is the total cost?
12. A 60 W lamp is switched on for 5.0 minutes. How much energy does it transfer, in joules?
13. A kettle is rated 230 V, 1380 W. Fuses rated 3 A, 5 A and 13 A are available. Which should be fitted in the plug?
14. A phone charger has a plastic case and a two-core cable with no earth wire. Why is it still safe?
15. Why must the switch of a mains appliance be connected in the live wire rather than the neutral wire?
16. A polythene rod becomes negatively charged when rubbed with a cloth. This is because:
17. Which best describes the electric field around an isolated negatively charged sphere?
18. A 12 V battery of negligible internal resistance is connected to a 4.0 Ω resistor in series with a parallel combination of a 12 Ω resistor and a 6.0 Ω resistor. What is the current in the 6.0 Ω resistor?
19. As the p.d. across a filament lamp increases, its resistance increases. Why?
20. Two identical resistors are connected across the same battery, first in series and then in parallel. Total power dissipated in parallel = k × total power in series. What is k?
Question 1 -- Light-Controlled Switching Circuit
Section B — Structured · Total: 8 marks
An automatic garden lamp uses the potential-divider circuit shown below. A 6.0 V battery is connected across a fixed resistor of resistance 2.0 kΩ in series with a light-dependent resistor (LDR). The output voltage Vout is taken across the LDR and is connected to a switching unit that turns the lamp on.
(a)[2]
In darkness the resistance of the LDR is 10 kΩ. Calculate Vout in darkness.
Model Answer -- 1(a)
M1 -- correct potential-divider ratio used:
V_out = 6.0 × 10 000 / (10 000 + 2000)
A1 -- Vout = 5.0 V
⚠ If you missed marks here: The commonest error is using the WRONG resistance on top of the fraction. Vout is across the LDR, so the LDR's 10 kΩ goes on top. If you got 1.0 V you found the voltage across the fixed resistor instead — that is the share of the OTHER component. The two shares must add to 6.0 V: 5.0 + 1.0 = 6.0. Always check that.
M1 -- potential divider set up with 10 kΩ on top: 6.0 × 10/(10+2) (1 mark)
A1 -- 5.0 V (1 mark)
(b)[2]
In bright sunlight the resistance of the LDR falls to 500 Ω. Calculate Vout in bright sunlight.
Model Answer -- 1(b)
M1 -- new ratio with 500 Ω:
V_out = 6.0 × 500 / (500 + 2000)
A1 -- Vout = 1.2 V
⚠ If you missed marks here: Watch the units — 500 Ω and 2.0 kΩ must be in the SAME unit before you add them. Mixing "500" with "2" gives nonsense. Convert 2.0 kΩ to 2000 Ω first. A sanity check: the LDR now has the smaller resistance, so it must take the smaller share of the 6.0 V.
Using your answers to (a) and (b), explain in words why Vout decreases as the light intensity increases.
Model Answer -- 1(c)
B1 -- as light intensity increases, the resistance of the LDR decreases
B1 -- the 6.0 V is shared in proportion to the resistances, so the LDR takes a smaller share of the total p.d. and Vout falls
⚠ If you missed marks here: "The LDR resistance goes down" alone is only 1 of 2. The second mark is the DIVIDER idea: p.d. is shared in the ratio of the resistances, so a smaller resistance takes a smaller share. Without linking resistance share to voltage share you have described the component, not explained the circuit.
B1 -- p.d. shared in ratio of resistances, so LDR takes smaller share of 6.0 V (1 mark)
(d)[2]
The switching unit turns the lamp on when Vout rises above 4.0 V. Calculate the resistance of the LDR at the moment the lamp switches on.
Model Answer -- 1(d)
M1 -- divider equation set equal to 4.0 V and rearranged (either route):
4.0 = 6.0 × R / (R + 2000) → 4.0R + 8000 = 6.0R
OR: p.d. across fixed resistor = 6.0 - 4.0 = 2.0 V, so I = 2.0/2000 = 1.0 mA, then R = 4.0/0.0010
A1 -- R = 4000 Ω = 4.0 kΩ
⚠ If you missed marks here: This is the reverse of (a) — you are given the voltage and asked for the resistance. The slick route: the fixed resistor must have the remaining 2.0 V across it, so the series current is 2.0 V / 2000 Ω = 1.0 mA everywhere, and the LDR is 4.0 V / 1.0 mA = 4.0 kΩ. Notice the ratio check: 4.0 V : 2.0 V is 2 : 1, and 4.0 kΩ : 2.0 kΩ is also 2 : 1.
M1 -- valid method: divider equation with 4.0 V, or current route via 2.0 V across fixed resistor (1 mark)
A1 -- R = 4.0 kΩ (4000 Ω) (1 mark)
Question 2 -- Series-Parallel Circuit Analysis
Section B — Structured · Total: 8 marks
In the circuit below, a battery of e.m.f. 12 V and negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 3.0 Ω resistor.
(a)[2]
Calculate the combined resistance of the parallel combination of the 6.0 Ω and 3.0 Ω resistors.
Model Answer -- 2(a)
M1 -- reciprocal formula used:
1/R = 1/6.0 + 1/3.0 = 3/6.0
A1 -- R = 2.0 Ω
⚠ If you missed marks here: The classic slip is stopping at 1/R = 0.5 and writing "0.5 Ω" — you must flip the reciprocal at the end: R = 1/0.5 = 2.0 Ω. Sanity check: a parallel combination is always SMALLER than the smallest branch (here smaller than 3.0 Ω). If your answer was 9.0 Ω you added them as if in series.
M1 -- 1/R = 1/6.0 + 1/3.0 used (1 mark)
A1 -- 2.0 Ω (1 mark)
(b)[1]
Calculate the total resistance of the circuit.
Model Answer -- 2(b)
B1 -- Rtotal = 2.0 + 2.0 = 4.0 Ω
⚠ If you missed marks here: Once the parallel pair is reduced to a single 2.0 Ω block, the circuit is simply two resistors in series: 2.0 + 2.0. Never add the raw 6.0 and 3.0 into a series total — the parallel pair must be combined FIRST.
⚠ If you missed marks here: The battery sees the TOTAL resistance from (b), not any single resistor. Using 12/2.0 = 6.0 A means you divided by only one part of the circuit. This is an "error carried forward" chain — a wrong (b) gives a wrong (c), so always double-check the total before moving on.
B1 -- I = 12/4.0 = 3.0 A (1 mark)
(d)[2]
Calculate the current in the 6.0 Ω resistor. Show your working clearly.
Model Answer -- 2(d)
M1 -- p.d. across the parallel pair found: V = 3.0 × 2.0 = 6.0 V (or 12 - 3.0 × 2.0 = 6.0 V)
A1 -- I = 6.0 V / 6.0 Ω = 1.0 A
⚠ If you missed marks here: The trap is 12 V / 6.0 Ω = 2.0 A — the full 12 V is NOT across the parallel pair, because the 2.0 Ω series resistor takes 6.0 V first. Find the p.d. across the junction (6.0 V), then apply V = IR to the one branch. Consistency check: 1.0 A (through 6.0 Ω) + 2.0 A (through 3.0 Ω) = 3.0 A total, matching (c).
M1 -- p.d. across parallel pair = 6.0 V found first (1 mark)
A1 -- current in 6.0 Ω branch = 1.0 A (1 mark)
(e)[2]
Calculate the charge that flows through the battery in 5.0 minutes.
Model Answer -- 2(e)
M1 -- Q = It with time converted to seconds: Q = 3.0 × (5.0 × 60)
A1 -- Q = 900 C
⚠ If you missed marks here: If you wrote 15 C you left the time in minutes. Q = It only works with time in SECONDS: 5.0 min = 300 s, so Q = 3.0 × 300 = 900 C. Unit conversion before substitution — every time.
M1 -- Q = It with t = 300 s (1 mark)
A1 -- 900 C (1 mark)
Question 3 -- Mains Electricity: the Kitchen Kettle
Section B — Structured · Total: 8 marks
An electric kettle is marked 230 V, 2760 W. It has a metal case and is connected to the a.c. mains by a cable containing live, neutral and earth wires. A fuse is fitted in the plug, as shown below.
(a)[2]
The kettle is switched on. Show that the current in the heating element is 12 A.
Model Answer -- 3(a)
M1 -- P = IV rearranged for current:
I = P / V = 2760 / 230
A1 -- I = 12 A
⚠ If you missed marks here: "Show that" means every step must be written out — the equation P = IV, the rearrangement I = P/V, the substitution 2760/230, and the result. Writing "12 A" alone earns nothing in a show-that question. Check the reverse: 12 × 230 = 2760 W, which matches the rating plate.
M1 -- I = P/V = 2760/230 clearly shown (1 mark)
A1 -- 12 A obtained with working (1 mark)
(b)[2]
Fuses rated 3 A, 5 A and 13 A are available. State which fuse should be fitted in the plug, and use the current from (a) to explain why each of the other two ratings must not be used.
Model Answer -- 3(b)
B1 -- the 13 A fuse: it is the smallest rating above the 12 A working current, so the kettle runs normally but a fault current still melts it
B1 -- the 3 A and 5 A fuses are both below 12 A, so they would melt as soon as the kettle is switched on, even with no fault
⚠ If you missed marks here: The rule is JUST ABOVE the normal working current. "13 A because it is the biggest" earns nothing — the reason is that 13 A sits just above 12 A. And the explanation for rejecting 3 A and 5 A must be by CALCULATION comparison: the normal 12 A already exceeds both ratings, so they blow during normal use, not just during a fault.
B1 -- 13 A chosen because it is the smallest rating above the 12 A working current (1 mark)
B1 -- 3 A and 5 A rejected: normal current 12 A exceeds both, so they melt in normal use (1 mark)
(c)[2]
The kettle is used for a total of 20 minutes each day. Electricity costs 10 cents per kilowatt-hour (1 unit). Calculate the cost, in cents, of using the kettle every day for 30 days.
Model Answer -- 3(c)
M1 -- energy in kWh with BOTH conversions done (W to kW, minutes to hours):
E = 2.76 kW × (20/60) h × 30 = 27.6 kWh
A1 -- cost = 27.6 × 10 = 276 cents (= $2.76)
⚠ If you missed marks here: The kilowatt-hour needs kilowatts and HOURS. Two conversions before anything else: 2760 W = 2.76 kW, and 20 min = 20/60 h = 0.333 h. Using 2760 with 20 gives an absurd number — a kettle cannot cost thousands of dollars a month, so always sanity-check the size of a cost answer. Each day is 2.76 × 1/3 = 0.92 kWh; thirty days is 27.6 kWh.
M1 -- E = 2.76 kW × (20/60) h × 30 = 27.6 kWh, both unit conversions correct (1 mark)
A1 -- cost = 276 cents ($2.76) (1 mark)
(d)[2]
Explain why the fuse is connected in the live wire and not in the neutral wire, and explain how the earth wire protects the user if the live wire comes loose inside the kettle and touches the metal case.
Model Answer -- 3(d)
B1 -- the live wire is the one at high voltage; with the fuse in the live wire, when the fuse melts the kettle is disconnected from the high-voltage side, so no part of it remains at 230 V (a fuse in the neutral would blow but leave the whole kettle still live)
B1 -- the earth wire gives a very low-resistance path from the case; if the live touches the case a very large current flows through the earth wire, melting the fuse and cutting off the supply, so the case cannot stay at mains voltage and the user is not electrocuted
⚠ If you missed marks here: These two answers form a CHAIN, and each mark needs the whole chain. Live-wire mark: blown fuse must DISCONNECT the appliance from the live (high-voltage) side — "the live is dangerous" alone is not an explanation. Earth-wire mark: low resistance → LARGE current → fuse melts → supply cut. The earth wire does not "absorb" the electricity; it deliberately blows the fuse.
B1 -- fuse in live so a blown fuse disconnects the kettle from the 230 V side; in neutral the kettle would stay live (1 mark)
B1 -- earth = low-resistance path; fault gives large current, melts fuse, supply cut, case cannot stay at mains voltage (1 mark)
Question 4 -- e.m.f., p.d. and Electric Charge
Section B — Structured · Total: 8 marks
A battery of e.m.f. 9.0 V and negligible internal resistance is connected in series with an ammeter, a filament lamp and a resistor R. The ammeter reads 0.30 A. A voltmeter connected across the lamp reads 5.4 V.
(a)[2]
The e.m.f. of the battery and the p.d. across the lamp are both measured in volts, but they are defined differently. Define e.m.f. and p.d. in terms of energy and charge, making the difference between them clear.
Model Answer -- 4(a)
B1 -- e.m.f. is the electrical work done (energy transferred) BY the source per unit charge driven around the complete circuit
B1 -- p.d. is the work done (energy transferred) per unit charge passing THROUGH the component — energy given out in one part of the circuit, where e.m.f. is energy put in by the source
⚠ If you missed marks here: "Voltage is energy" earns nothing — both definitions are energy PER UNIT CHARGE (joules per coulomb), and the words "per unit charge" must appear. The difference the examiner wants: e.m.f. is the energy the SOURCE gives each coulomb (whole circuit); p.d. is the energy each coulomb delivers to ONE component. Same unit, opposite ends of the energy story.
B1 -- e.m.f.: work done by the source per unit charge around the complete circuit (1 mark)
B1 -- p.d.: work done per unit charge through the component, i.e. energy delivered not supplied (1 mark)
(b)[2]
State the p.d. across resistor R, and explain, using conservation of energy, why it must have this value.
Model Answer -- 4(b)
B1 -- p.d. across R = 9.0 - 5.4 = 3.6 V
B1 -- each coulomb gains 9.0 J from the battery and must transfer all of it in the circuit, so the p.d.s across the components in series add up to the e.m.f. (energy is conserved)
⚠ If you missed marks here: The number is the easy half. The explanation mark is the ENERGY argument: 9.0 joules go into every coulomb, 5.4 of them come out in the lamp, so the remaining 3.6 must come out in R — energy cannot vanish or appear. "The voltages add up in series" restates the rule; SAYING WHY (conservation of energy per coulomb) earns the mark.
B1 -- p.d. across R = 3.6 V (1 mark)
B1 -- energy per coulomb conserved: 9.0 J supplied = 5.4 J + 3.6 J transferred, so series p.d.s sum to e.m.f. (1 mark)
(c)[2]
Calculate the charge that passes through the lamp in 2.0 minutes, and the number of electrons that carry this charge. (charge on one electron = 1.6 × 10−19 C)
Model Answer -- 4(c)
M1 -- Q = It with time in seconds:
Q = 0.30 × (2.0 × 60) = 36 C
A1 -- number of electrons = 36 / (1.6 × 10−19) = 2.25 × 1020 (accept 2.2–2.3 × 1020)
⚠ If you missed marks here: Two traps. First, Q = It needs SECONDS: 2.0 min = 120 s, so Q = 0.30 × 120 = 36 C (writing 0.6 C means you left minutes in). Second, the electron count DIVIDES the total charge by the charge of one electron — 36 is the big number and 1.6 × 10−19 the tiny one, so the answer must be enormous. If you got 10−20-something, you divided the wrong way round.
One of the copper connecting wires is replaced by a copper wire that is twice as long and has half the cross-sectional area. Explain, in terms of how resistance depends on length and cross-sectional area, what happens to the resistance of that wire, and state the factor by which it changes.
Model Answer -- 4(d)
M1 -- resistance is proportional to length and inversely proportional to cross-sectional area: doubling the length doubles the resistance, and halving the area doubles it again
A1 -- the resistance increases by a factor of 2 × 2 = 4
⚠ If you missed marks here: Treat the two changes SEPARATELY, then multiply. Longer wire = more metal for the charge to get through, so R doubles with length. Thinner wire = a narrower path, so HALVING the area DOUBLES the resistance (inverse proportion — the commonest error is halving it instead). 2 × 2 = 4, not 2 + 2. A factor answer of 2 almost always means one of the two effects was dropped.
M1 -- R proportional to L, inversely proportional to A: each change doubles R (1 mark)
A1 -- resistance becomes 4 times larger (1 mark)
Question 5 -- Static Charge, Fields and Circuit Power
Section B — Structured · Total: 8 marks
(a)[2]
A polythene rod becomes negatively charged when rubbed with a wool cloth. Explain, in terms of the movement of charged particles, how the rod becomes charged, and state the charge left on the cloth.
Model Answer -- 5(a)
B1 -- electrons are transferred from the cloth to the rod (only electrons move, never the positive charges)
B1 -- the cloth is left with an equal positive charge (it has lost electrons)
⚠ If you missed marks here: The particle must be named: ELECTRONS move, and you must give the direction (cloth → rod). Writing "positive charge moves to the cloth" loses the mark — in solids the positive nuclei are fixed. Charge is conserved, so whatever the rod gained, the cloth lost: equal and opposite.
B1 -- electrons transferred from cloth to rod (1 mark)
B1 -- cloth left with equal positive charge (1 mark)
(b)[2]
The charged rod is used to charge a small isolated conducting sphere negatively. Describe the electric field pattern around the sphere, including the direction of the field lines, and state what the direction of an electric field represents.
Model Answer -- 5(b)
B1 -- radial field lines, evenly spaced around the sphere, pointing inwards towards the sphere (because it is negative)
B1 -- the field direction is the direction of the force on a positive (test) charge placed in the field
⚠ If you missed marks here: Two words carry the first mark: RADIAL and INWARDS. Arrows point in for a negative charge, out for a positive one. The definition mark is exact wording territory: field direction = direction of the force on a POSITIVE charge. Leave out "positive" and the definition is wrong.
B1 -- radial field lines pointing inwards towards the negative sphere (1 mark)
B1 -- field direction = direction of force on a positive charge (1 mark)
(c)[3]
In the circuit below, a 12 V battery of negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel combination of a 12 Ω resistor and a 4.0 Ω resistor. Calculate the current in the 12 Ω resistor and the power dissipated in it. Show your working clearly.
Model Answer -- 5(c)
M1 -- p.d. across the parallel pair found first:
pair: (12 × 4.0)/(12 + 4.0) = 3.0 Ω; total R = 2.0 + 3.0 = 5.0 Ω; I = 12/5.0 = 2.4 A; V_pair = 2.4 × 3.0 = 7.2 V
A1 -- current in the 12 Ω resistor = 7.2 / 12 = 0.60 A
A1 -- power = VI = 7.2 × 0.60 = 4.3 W (accept 4.32 W; or I²R = 0.60² × 12)
⚠ If you missed marks here: The trap is 12 V / 12 Ω = 1.0 A — the full 12 V is NOT across the branch, because the 2.0 Ω series resistor takes 4.8 V first. Reduce the pair (3.0 Ω), find the battery current (2.4 A), then the p.d. at the junction (7.2 V), and only then use the ONE branch. Checks: 4.8 + 7.2 = 12 V, and branch currents 0.60 + 1.8 = 2.4 A. For the power, both routes must agree: 7.2 × 0.60 and 0.60² × 12 are both 4.32 W.
M1 -- p.d. across parallel pair = 7.2 V (via pair 3.0 Ω, total 5.0 Ω, battery current 2.4 A) (1 mark)
A1 -- current in 12 Ω resistor = 0.60 A (1 mark)
A1 -- power dissipated = 4.3 W (accept 4.32 W) (1 mark)
(d)[1]
In the circuit of (c), state the direction in which the conventional current flows around the external circuit, and how the direction of the electron flow compares with it.
Model Answer -- 5(d)
B1 -- conventional current flows from the positive terminal of the battery through the external circuit to the negative terminal; the electrons, being negatively charged, flow the opposite way round (from negative to positive)
⚠ If you missed marks here: Both halves are needed: the direction (positive terminal → external circuit → negative terminal) AND the comparison (electrons go the OPPOSITE way, because they carry negative charge). Conventional current is a historical convention fixed before the electron was discovered — every circuit rule uses it, which is why 4(c)'s electrons and this arrow point in opposite directions.
B1 -- conventional current + terminal to - terminal externally; electrons flow the opposite way (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
Section A (20 marks) is counted automatically from your answers. For Section B, tick all mark checkboxes you have earned, then click "Calculate Grade" below.