Second full Unit Exam — all-new questions, same scope and difficulty as Unit Exam 1. 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 charge of 180 C passes a point in a circuit in 2.0 minutes. What is the current?
2. Which statement defines the p.d. across a component?
3. A wire is replaced by a wire of the same material that is half as long with twice the diameter. Its resistance becomes:
4. A p.d. of 6.0 V is applied across a 15 Ω resistor. What is the current?
5. Three 6.0 Ω resistors are connected in parallel. What is their combined resistance?
6. A current of 3.0 A enters a junction. One branch carries 1.2 A away. What current flows in the other branch?
7. A 9.0 V supply is connected across a 3.0 kΩ resistor in series with a 6.0 kΩ resistor. What is the p.d. across the 6.0 kΩ resistor?
8. In a potential divider, Vout is taken across an LDR. The light gets brighter. What happens to Vout?
9. A resistor is connected across a supply of fixed p.d. Its resistance is then halved. The power dissipated:
10. An 800 W appliance runs for 45 minutes. How much energy does it transfer, in kWh?
11. A 2.4 kW water heater runs 50 minutes each day for 30 days. Electricity costs 15 cents per kWh. What is the total cost?
12. A 1500 W heater is switched on for 2.0 minutes. How much energy does it transfer, in joules?
13. A lamp unit is rated 230 V, 690 W. Fuses rated 3 A, 5 A and 13 A are available. Which should be fitted?
14. The live wire comes loose inside an earthed metal-cased appliance and touches the case. What happens, in order?
15. An appliance is labelled double-insulated. What does this mean?
16. A balloon rubbed on dry hair gains electrons. Which statement is correct?
17. Which describes the electric field between two oppositely charged parallel plates (away from the edges)?
18. A 6.0 V battery of negligible internal resistance is connected to a 1.0 Ω resistor in series with a parallel pair of two 4.0 Ω resistors. What is the power dissipated in ONE of the 4.0 Ω resistors?
19. Which describes the current–voltage graph of a resistor kept at constant temperature?
20. A battery of e.m.f. 4.5 V drives 20 C of charge around a complete circuit. How much energy does the battery transfer?
Question 1 -- Temperature-Controlled Fan Circuit
Section B — Structured · Total: 8 marks
A cooling fan for a computer uses the potential-divider circuit shown below. A 9.0 V battery is connected across a fixed resistor of resistance 3.0 kΩ in series with a thermistor. The output voltage Vout is taken across the thermistor and is connected to a control unit that switches the fan on.
(a)[2]
At 20 °C the resistance of the thermistor is 6.0 kΩ. Calculate Vout at 20 °C.
Model Answer -- 1(a)
M1 -- correct potential-divider ratio used:
V_out = 9.0 × 6000 / (6000 + 3000)
A1 -- Vout = 6.0 V
⚠ If you missed marks here: The commonest error is putting the WRONG resistance on top of the fraction. Vout is across the thermistor, so the thermistor's 6.0 kΩ goes on top. If you got 3.0 V you found the p.d. across the fixed resistor instead. The two shares must add to 9.0 V: 6.0 + 3.0 = 9.0. Always check that.
M1 -- potential divider set up with 6.0 kΩ on top: 9.0 × 6/(6+3) (1 mark)
A1 -- 6.0 V (1 mark)
(b)[2]
At 80 °C the resistance of the thermistor falls to 1.0 kΩ. Calculate Vout at 80 °C.
Model Answer -- 1(b)
M1 -- new ratio with 1.0 kΩ:
V_out = 9.0 × 1000 / (1000 + 3000)
A1 -- Vout = 2.25 V (accept 2.3 V)
⚠ If you missed marks here: Watch the units — both resistances must be in the SAME unit before you add them. Mixing "1000" with "3" gives the nonsense answer 8.97 V. Work in kΩ throughout if you prefer: 9.0 × 1.0/(1.0 + 3.0). Sanity check: the thermistor now has the smaller resistance, so it must take the smaller share of the 9.0 V.
Using your answers to (a) and (b), explain in words why Vout decreases as the temperature increases.
Model Answer -- 1(c)
B1 -- as temperature increases, the resistance of the thermistor decreases
B1 -- the 9.0 V is shared in proportion to the resistances, so the thermistor takes a smaller share of the total p.d. and Vout falls
⚠ If you missed marks here: "The thermistor 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 -- thermistor resistance decreases as temperature increases (1 mark)
B1 -- p.d. shared in ratio of resistances, so thermistor takes smaller share of 9.0 V (1 mark)
(d)[2]
The control unit switches the fan on when Vout falls below 3.0 V. Calculate the resistance of the thermistor at the moment the fan switches on.
Model Answer -- 1(d)
M1 -- divider equation set equal to 3.0 V and rearranged (either route):
3.0 = 9.0 × R / (R + 3000) → 3.0R + 9000 = 9.0R
OR: p.d. across fixed resistor = 9.0 - 3.0 = 6.0 V, so I = 6.0/3000 = 2.0 mA, then R = 3.0/0.0020
A1 -- R = 1500 Ω = 1.5 kΩ
⚠ If you missed marks here: This is the reverse of (a) — you are given the voltage and asked for the resistance. If you got 6.0 kΩ you put the fixed resistor's 3.0 kΩ on top of the fraction. The slick route: the fixed resistor has the remaining 6.0 V across it, so the series current is 6.0 V / 3000 Ω = 2.0 mA everywhere, and the thermistor is 3.0 V / 2.0 mA = 1.5 kΩ. Ratio check: 3.0 V : 6.0 V is 1 : 2, and 1.5 kΩ : 3.0 kΩ is also 1 : 2.
M1 -- valid method: divider equation with 3.0 V, or current route via 6.0 V across fixed resistor (1 mark)
A1 -- R = 1.5 kΩ (1500 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. 20 V and negligible internal resistance is connected to a 4.0 Ω resistor in series with a parallel combination of a 10 Ω resistor and a 15 Ω resistor.
(a)[2]
Calculate the combined resistance of the parallel combination of the 10 Ω and 15 Ω resistors.
Model Answer -- 2(a)
M1 -- reciprocal formula used:
1/R = 1/10 + 1/15 = 5/30 = 1/6.0
A1 -- R = 6.0 Ω
⚠ If you missed marks here: If your answer was 25 Ω you added them as if in series. A parallel combination is always SMALLER than the smallest branch (here smaller than 10 Ω). The other classic slip is stopping at 1/R = 1/6 and writing "0.17 Ω" — flip the reciprocal at the end: R = 6.0 Ω.
M1 -- 1/R = 1/10 + 1/15 used (1 mark)
A1 -- 6.0 Ω (1 mark)
(b)[1]
Calculate the total resistance of the circuit.
Model Answer -- 2(b)
B1 -- Rtotal = 4.0 + 6.0 = 10 Ω
⚠ If you missed marks here: Once the parallel pair is reduced to a single 6.0 Ω block, the circuit is simply two resistors in series: 4.0 + 6.0. If you got 29 Ω you added the raw 10 and 15 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. If you got 5.0 A you used 20/4.0 — only one part of the circuit. This is an error-carried-forward chain: a wrong (b) gives a wrong (c), so double-check the total before moving on.
B1 -- I = 20/10 = 2.0 A (1 mark)
(d)[2]
Calculate the current in the 15 Ω resistor. Show your working clearly.
Model Answer -- 2(d)
M1 -- p.d. across the parallel pair found: V = 2.0 × 6.0 = 12 V (or 20 - 2.0 × 4.0 = 12 V)
A1 -- I = 12 V / 15 Ω = 0.80 A
⚠ If you missed marks here: The trap is 20 V / 15 Ω = 1.3 A — the full 20 V is NOT across the parallel pair, because the 4.0 Ω series resistor takes 8.0 V first. Find the p.d. at the junction (12 V), then apply V = IR to the one branch. Consistency check: 1.2 A (through 10 Ω) + 0.80 A (through 15 Ω) = 2.0 A total, matching (c).
M1 -- p.d. across parallel pair = 12 V found first (1 mark)
A1 -- current in 15 Ω branch = 0.80 A (1 mark)
(e)[2]
Calculate the charge that flows through the battery in 4.0 minutes.
Model Answer -- 2(e)
M1 -- Q = It with time converted to seconds: Q = 2.0 × (4.0 × 60)
A1 -- Q = 480 C
⚠ If you missed marks here: If you wrote 8.0 C you left the time in minutes. Q = It only works with time in SECONDS: 4.0 min = 240 s, so Q = 2.0 × 240 = 480 C. Unit conversion before substitution — every time.
M1 -- Q = It with t = 240 s (1 mark)
A1 -- 480 C (1 mark)
Question 3 -- Mains Electricity: the Electric Iron
Section B — Structured · Total: 8 marks
An electric iron is marked 230 V, 1840 W. It has a metal soleplate 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 iron is switched on. Show that the current in the heating element is 8.0 A.
Model Answer -- 3(a)
M1 -- P = IV rearranged for current:
I = P / V = 1840 / 230
A1 -- I = 8.0 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 1840/230, and the result. Writing "8.0 A" alone earns nothing in a show-that question. Dividing the wrong way round gives 0.125 A — check the reverse: 8.0 × 230 = 1840 W, matching the rating plate.
M1 -- I = P/V = 1840/230 clearly shown (1 mark)
A1 -- 8.0 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 8.0 A working current, so the iron runs normally but a fault current still melts it
B1 -- the 3 A and 5 A fuses are both below 8.0 A, so they would melt as soon as the iron 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 8.0 A. And the explanation for rejecting 3 A and 5 A must be by CALCULATION comparison: the normal 8.0 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 8.0 A working current (1 mark)
B1 -- 3 A and 5 A rejected: normal current 8.0 A exceeds both, so they melt in normal use (1 mark)
(c)[2]
The iron is used for a total of 15 minutes each day. Electricity costs 14 cents per kilowatt-hour (1 unit). Calculate the cost, in cents, of using the iron every day for 30 days.
Model Answer -- 3(c)
M1 -- energy in kWh with BOTH conversions done (W to kW, minutes to hours):
⚠ If you missed marks here: The kilowatt-hour needs kilowatts and HOURS. Two conversions before anything else: 1840 W = 1.84 kW, and 15 min = 15/60 h = 0.25 h. Using 1840 with 15 gives the absurd 11 592 000 — an iron cannot cost a hundred thousand dollars a month, so always sanity-check the size of a cost answer. Each day is 1.84 × 0.25 = 0.46 kWh; thirty days is 13.8 kWh.
M1 -- E = 1.84 kW × (15/60) h × 30 = 13.8 kWh, both unit conversions correct (1 mark)
Explain why the fuse and the switch are both 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 iron and touches the metal soleplate.
Model Answer -- 3(d)
B1 -- the live wire is the one at high voltage; with the fuse (or switch open) in the live wire the iron is disconnected from the high-voltage side, so no part of it remains at 230 V (in the neutral, the iron would stay live)
B1 -- the earth wire gives a very low-resistance path from the soleplate; if the live touches it a very large current flows through the earth wire, melting the fuse and cutting off the supply, so the soleplate 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: a blown fuse or open switch 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 and switch in live so the iron is disconnected from the 230 V side; in neutral it would stay live (1 mark)
B1 -- earth = low-resistance path; fault gives large current, melts fuse, supply cut, soleplate 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. 6.0 V and negligible internal resistance is connected in series with an ammeter, a resistor R and a buzzer. The ammeter reads 0.25 A. A voltmeter connected across the resistor reads 4.2 V.
(a)[2]
The e.m.f. of the battery and the p.d. across the resistor 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 the buzzer, and explain, using conservation of energy, why it must have this value.
Model Answer -- 4(b)
B1 -- p.d. across buzzer = 6.0 - 4.2 = 1.8 V
B1 -- each coulomb gains 6.0 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 — and if you got 10.2 V you ADDED instead of subtracting; the shares cannot exceed the e.m.f. The explanation mark is the ENERGY argument: 6.0 joules go into every coulomb, 4.2 of them come out in the resistor, so the remaining 1.8 must come out in the buzzer — energy cannot vanish or appear. SAYING WHY (conservation of energy per coulomb) earns the mark.
B1 -- p.d. across buzzer = 1.8 V (1 mark)
B1 -- energy per coulomb conserved: 6.0 J supplied = 4.2 J + 1.8 J transferred, so series p.d.s sum to e.m.f. (1 mark)
(c)[2]
Calculate the charge that passes through the buzzer in 8.0 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 = 0.25 × (8.0 × 60) = 120 C
A1 -- number of electrons = 120 / (1.6 × 10−19) = 7.5 × 1020
⚠ If you missed marks here: Two traps. First, Q = It needs SECONDS: 8.0 min = 480 s, so Q = 0.25 × 480 = 120 C (writing 2.0 C means you left minutes in). Second, the electron count DIVIDES the total charge by the charge of one electron — 120 is the big number and 1.6 × 10−19 the tiny one, so the answer must be enormous. If you got 10−21-something, you divided the wrong way round.
One of the copper connecting wires is replaced by a copper wire that is three times as long and has twice 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: tripling the length trebles the resistance, and doubling the area halves it
A1 -- the resistance increases by a factor of 3 × ½ = 1.5
⚠ If you missed marks here: Treat the two changes SEPARATELY, then multiply. Longer wire = more metal for the charge to get through, so R trebles with length. Fatter wire = a wider path, so DOUBLING the area HALVES the resistance (inverse proportion). If you got ×6 you multiplied by 2 for the area instead of dividing. 3 × ½ = 1.5, not 3 + ½.
M1 -- R proportional to L, inversely proportional to A: ×3 from length, ×½ from area (1 mark)
A1 -- resistance becomes 1.5 times larger (1 mark)
Question 5 -- Static Charge, Fields and Circuit Power
Section B — Structured · Total: 8 marks
(a)[2]
An acetate rod becomes positively charged when rubbed with a dry 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 rod to the cloth (only electrons move, never the positive charges)
B1 -- the cloth is left with an equal negative charge (it has gained electrons)
⚠ If you missed marks here: The particle must be named: ELECTRONS move, and you must give the direction. A POSITIVE rod has LOST electrons, so they went rod → cloth. Writing "positive charge moves onto the rod" loses the mark — in solids the positive nuclei are fixed. Charge is conserved, so whatever the rod lost, the cloth gained: equal and opposite.
B1 -- electrons transferred from rod to cloth (1 mark)
B1 -- cloth left with equal negative charge (1 mark)
(b)[2]
Two parallel metal plates are connected to a high-voltage supply so that one plate is positive and the other negative. Describe the electric field pattern between the plates, including the direction of the field lines, and state what the direction of an electric field represents.
Model Answer -- 5(b)
B1 -- a uniform field: straight, parallel, evenly spaced field lines running from the positive plate to the negative plate
B1 -- the field direction is the direction of the force on a positive (test) charge placed in the field
⚠ If you missed marks here: Three words carry the first mark: PARALLEL, EVENLY SPACED, and from + to −. Drawing radial lines is the point-charge pattern, not the plate pattern. 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 -- uniform: parallel, evenly spaced lines from positive plate to negative plate (1 mark)
B1 -- field direction = direction of force on a positive charge (1 mark)
(c)[3]
In the circuit below, a 9.0 V battery of 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 12 Ω resistor. Calculate the current in the 6.0 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: (6.0 × 12)/(6.0 + 12) = 4.0 Ω; total R = 2.0 + 4.0 = 6.0 Ω; I = 9.0/6.0 = 1.5 A; V_pair = 1.5 × 4.0 = 6.0 V
A1 -- current in the 6.0 Ω resistor = 6.0 / 6.0 = 1.0 A
A1 -- power = VI = 6.0 × 1.0 = 6.0 W (or I²R = 1.0² × 6.0)
⚠ If you missed marks here: The trap is 9.0 V / 6.0 Ω = 1.5 A — the full 9.0 V is NOT across the branch, because the 2.0 Ω series resistor takes 3.0 V first. Reduce the pair (4.0 Ω), find the battery current (1.5 A), then the p.d. at the junction (6.0 V), and only then use the ONE branch. Checks: 3.0 + 6.0 = 9.0 V, and branch currents 1.0 + 0.5 = 1.5 A. For the power, both routes must agree: 6.0 × 1.0 and 1.0² × 6.0 are both 6.0 W.
M1 -- p.d. across parallel pair = 6.0 V (via pair 4.0 Ω, total 6.0 Ω, battery current 1.5 A) (1 mark)
A1 -- current in 6.0 Ω resistor = 1.0 A (1 mark)
A1 -- power dissipated = 6.0 W (1 mark)
(d)[1]
In the circuit of (c), the conventional current flows clockwise around the external circuit. State the direction in which the electrons actually move, and explain why the two directions differ.
Model Answer -- 5(d)
B1 -- the electrons move anticlockwise (the opposite way to the conventional current), because they carry negative charge and are attracted towards the positive terminal; conventional current was defined as the direction of positive charge flow
⚠ If you missed marks here: Both halves are needed: the direction (anticlockwise — the OPPOSITE way) AND the reason (electrons carry NEGATIVE charge, while conventional current follows positive charge). The convention was fixed before the electron was discovered, which is why every circuit arrow points against the electron drift.
B1 -- electrons move anticlockwise, opposite to conventional current, because their charge is negative (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.