These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- The Electric Shower
Total: 12 marks
A house in Sheffield has an electric shower rated at 10.5 kW, 230 V. Unlike most appliances, the shower is not plugged into a wall socket: it is connected by a thick cable directly to the consumer unit (fuse box), through its own 50 A circuit breaker.
(a)[2]
Calculate the current in the shower when it is operating normally.
Model Answer -- 1(a)
I = P / V = 10500 / 230 [1]
I = 45.7 A (45.65 A) [1]
⚠ If you missed marks here: The trap is leaving the power in kilowatts: 10.5 ÷ 230 gives 0.046 A, which would not warm a teaspoon. Convert to watts first — 10500 W ÷ 230 V = 45.7 A. An answer of 45.6–45.7 A earns both marks with working.
Mark 1 -- current equation I equals P divided by V used with 10500 W (1 mark)
Mark 2 -- current 45.7 A or 45.65 A calculated (1 mark)
(b)[2]
Explain why the shower cannot be connected to the mains using an ordinary 3-pin plug with a 13 A fuse.
Model Answer -- 1(b)
The current in the shower (about 46 A) is far greater than 13 A [1]
The fuse would melt / blow immediately (or the plug and its wiring would overheat), so a dedicated high-current circuit is needed [1]
⚠ If you missed marks here: You must compare the numbers: the working current (≈46 A) is more than three times the 13 A fuse rating. Just saying "the shower is too powerful" without the current comparison is not a physics answer.
Mark 1 -- shower current about 46 A exceeds 13 A rating (1 mark)
Mark 2 -- fuse would blow or plug wiring would overheat so dedicated circuit needed (1 mark)
(c)[3]
A shower lasts 8.0 minutes. Calculate the energy transferred in kWh, and the cost at 30 p per kWh.
Model Answer -- 1(c)
E = P × t = 10.5 kW × (8.0/60) h [1]
E = 1.4 kWh [1]
cost = 1.4 × 30 = 42 p [1]
⚠ If you missed marks here: For kWh the power stays in kW and the time must be in HOURS: 8.0 min = 8/60 h = 0.133 h. If you used 8 directly you got 84 kWh and a £25 shower. 10.5 × 0.133 = 1.4 kWh, then 1.4 × 30p = 42p.
Mark 1 -- time converted to hours 0.133 h and E equals P t used (1 mark)
Mark 2 -- energy 1.4 kWh calculated (1 mark)
Mark 3 -- cost 42 p calculated (1 mark)
(d)[2]
Calculate the resistance of the shower heating element when operating.
Model Answer -- 1(d)
R = V² / P = 230² / 10500 (or R = V / I = 230 / 45.65) [1]
R = 5.0 Ω (5.04 Ω) [1]
⚠ If you missed marks here: Either route works: R = V²/P = 52900/10500 = 5.0 Ω, or R = V/I using your current from (a). If you got 5000 Ω you squared 230 but kept P in kW again — same unit trap as (a).
Mark 1 -- resistance route R equals V squared over P or V over I (1 mark)
Mark 2 -- resistance 5.0 ohm calculated (1 mark)
(e)[3]
The cable from the consumer unit to the shower is much thicker than the cable to the lighting circuit. Explain why, in terms of resistance and heating.
Model Answer -- 1(e)
A thicker wire has a lower resistance [1]
The shower current is very large, and the heating in the cable depends on I²R [1]
With a thin (higher-resistance) cable the energy dissipated would overheat the cable / melt insulation / risk fire; thick cable keeps the power wasted (and the p.d. dropped along the cable) small [1]
⚠ If you missed marks here: Three linked ideas are needed: thicker means lower resistance; heating goes as I²R so a 46 A current is punishing; and the consequence (overheating/fire or lost p.d.) if the resistance were high. "Thick wires carry more current" with no mention of resistance or heating earns nothing.
Mark 1 -- thicker wire has lower resistance (1 mark)
Mark 2 -- heating depends on current squared times R and shower current is large (1 mark)
Mark 3 -- thin cable would overheat or waste power or drop voltage (1 mark)
Question 2 -- Induction in the Physics Lab
Total: 11 marks
A student in a Manchester school investigates electromagnetic induction. She has a bar magnet, a coil of wire connected to a sensitive centre-zero meter, and a length of copper pipe.
(a)[3]
The student pushes the N pole of the magnet into the coil and watches the meter. Describe what the meter shows when the magnet is: (i) moving into the coil, (ii) held stationary inside the coil, (iii) pulled out of the coil twice as fast as it went in.
Model Answer -- 2(a)
Moving in: the meter deflects (to one side) [1]
Stationary: no deflection / reads zero [1]
Pulled out faster: deflects in the OPPOSITE direction and with a LARGER deflection [1]
⚠ If you missed marks here: Part (ii) is the classic trap: an e.m.f. is only induced while the magnetic field through the coil is CHANGING, so a stationary magnet gives zero. For (iii) you need both changes: opposite direction (motion reversed) and larger (faster change).
Mark 1 -- deflection while magnet moves in (1 mark)
Mark 2 -- zero deflection when magnet stationary (1 mark)
Mark 3 -- opposite direction and larger deflection when withdrawn faster (1 mark)
(b)[2]
State two changes the student could make to increase the size of the induced e.m.f.
Model Answer -- 2(b)
Any two of: move the magnet faster [1]
use a stronger magnet / use a coil with more turns (of wire) [1]
⚠ If you missed marks here: "Use a bigger battery" is the giveaway wrong answer — there is no battery; the coil itself is the source. The e.m.f. grows with the rate of change of the field: faster movement, stronger magnet, more turns.
Mark 1 -- one valid factor such as faster movement (1 mark)
Mark 2 -- second valid factor such as stronger magnet or more turns (1 mark)
(c)[4]
The student drops the magnet down the vertical copper pipe. It takes much longer to fall through than an identical unmagnetised steel bar, even though copper is not a magnetic material. Explain this observation.
Model Answer -- 2(c)
The falling magnet causes a changing magnetic field through the copper pipe [1]
This induces currents (eddy currents) in the pipe wall [1]
By Lenz’s law the induced currents create a magnetic field that opposes the change causing them / opposes the motion of the magnet [1]
So there is an upward (retarding) force on the magnet, its acceleration is reduced and it falls slowly; the unmagnetised bar induces no currents [1]
⚠ If you missed marks here: The chain must be complete: changing field → induced (eddy) currents → induced currents OPPOSE the change (Lenz’s law) → retarding force on the magnet. Saying "copper attracts the magnet" loses everything — copper is not magnetic, which is exactly why the effect needs induction to explain it.
Mark 1 -- falling magnet gives changing field through pipe (1 mark)
Mark 2 -- eddy currents induced in pipe wall (1 mark)
Mark 3 -- Lenz law induced currents oppose the change or motion (1 mark)
Mark 4 -- upward force reduces acceleration and plain bar induces nothing (1 mark)
(d)[2]
The magnet loses gravitational potential energy as it falls, but reaches the bottom moving slowly. Explain what happens to the energy.
Model Answer -- 2(d)
Work is done against the opposing (magnetic) force; energy is transferred electrically by the induced currents [1]
The currents dissipate the energy as heat in the copper pipe (its resistance) [1]
⚠ If you missed marks here: Energy is never "used up by the magnetic field". The g.p.e. drives the induced currents, and the currents heat the copper because it has resistance — the pipe ends up very slightly warmer.
Mark 1 -- work done against opposing force transfers energy to induced currents (1 mark)
Mark 2 -- energy dissipated as heat in the resistance of the pipe (1 mark)
Question 3 -- The Garden Lawnmower
Total: 12 marks
An electric lawnmower used in a garden in Surrey is rated 1800 W, 230 V. It is marked with the double-insulation symbol (a square inside a square) and is used with a 25 m extension lead. The extension lead has a total resistance of 0.60 Ω.
(a)[2]
The lawnmower has no earth wire. Explain what double insulation means and why an earth wire is not needed.
Model Answer -- 3(a)
The case (and all parts the user can touch) is made of an insulator / the live parts have two layers of insulation [1]
No exposed metal part can become live, so a fault cannot make the case dangerous — an earth wire would have nothing to protect [1]
⚠ If you missed marks here: The key phrase is that no touchable metal can ever become live. An earth wire exists to carry fault current away from a live metal case; with an all-plastic case there is no such path to protect.
Mark 1 -- case is an insulator or live parts doubly insulated (1 mark)
Mark 2 -- no exposed metal can become live so earth wire unnecessary (1 mark)
(b)[3]
The plug can be fitted with a 3 A, 5 A or 13 A fuse. Calculate the normal operating current and state which fuse should be fitted. Justify your choice.
Model Answer -- 3(b)
I = P / V = 1800 / 230 = 7.8 A [1]
13 A fuse chosen [1]
It is the smallest available fuse ABOVE the normal current; 3 A and 5 A would blow in normal use [1]
⚠ If you missed marks here: The fuse must sit just above the working current, never below it: 7.8 A rules out 3 A and 5 A. If you picked 5 A "to be safe", think about what happens the moment the mower starts — safe fuses protect the cable, they are not meant to stop the appliance working.
Mark 1 -- operating current 7.8 A calculated (1 mark)
Mark 2 -- 13 A fuse selected (1 mark)
Mark 3 -- justified as smallest rating above normal current (1 mark)
(c)[2]
Explain how a fuse protects the circuit when a fault causes too large a current.
Model Answer -- 3(c)
The large current heats the fuse wire until it melts [1]
This breaks the circuit, cutting off the current (before the cable overheats) [1]
⚠ If you missed marks here: A fuse does not "absorb" or "block" current — it is a thin wire that MELTS when the current is too large, breaking the circuit. Both steps (melts, breaks circuit) are needed.
Mark 1 -- excess current melts the fuse wire (1 mark)
Mark 2 -- circuit broken so current stops (1 mark)
(d)[3]
Calculate the p.d. across the extension lead when the mower draws its normal current, and the power wasted as heat in the lead.
Model Answer -- 3(d)
V = I R = 7.8 × 0.60 = 4.7 V [1]
P = I² R = 7.8² × 0.60 [1]
P = 37 W (36.7 W) [1]
⚠ If you missed marks here: The lead’s p.d. uses the CIRCUIT current through the lead’s own resistance: 7.8 × 0.60 = 4.7 V. For the wasted power, remember to square the current — 7.8 × 0.60 = 4.7 W is the classic slip; it must be 7.8² × 0.60 ≈ 37 W.
Mark 1 -- p.d. across lead 4.7 V from V equals I R (1 mark)
Mark 2 -- power formula I squared R substituted (1 mark)
Mark 3 -- wasted power about 37 W calculated (1 mark)
(e)[2]
The instructions say the extension lead must be fully uncoiled before use. Suggest and explain why a tightly coiled lead is dangerous when carrying a large current.
Model Answer -- 3(e)
The lead dissipates heat along its whole length; coiled up, the turns keep each other warm and the heat cannot escape to the air [1]
So the temperature rises until the insulation may soften / melt / catch fire [1]
⚠ If you missed marks here: The current (and so the heat generated) is the same coiled or straight — what changes is how well the heat ESCAPES. Trapped in a coil, the same 37 W raises the temperature far higher.
Mark 1 -- coiling traps the heat produced in the lead (1 mark)
Mark 2 -- temperature rises until insulation melts or fire risk (1 mark)
Question 4 -- Electrostatic Paint Spraying
Total: 11 marks
A car factory in Sunderland paints body panels electrostatically. The paint droplets are given a negative charge as they leave the spray gun, and the metal panel is given a positive charge. The spray gun is supplied with charge by a current of 15 μA.
(a)[2]
The droplets are charged by friction as they rub through the nozzle. Explain, in terms of particles, how an uncharged droplet becomes negatively charged.
Model Answer -- 4(a)
Electrons are transferred (by friction/rubbing) from the nozzle to the droplet [1]
The droplet gains (extra) electrons so has an overall negative charge; only electrons move, not positive charge [1]
⚠ If you missed marks here: Charging is always about ELECTRONS moving — positive charges (protons) are locked in nuclei and never transfer. Negative charge means the droplet has GAINED electrons.
Mark 1 -- electrons transferred to the droplet by friction (1 mark)
Mark 2 -- gaining electrons makes it negative and only electrons move (1 mark)
(b)[2]
Explain why giving all the droplets the same charge produces a fine, even cloud of paint rather than large drops.
Model Answer -- 4(b)
Like charges repel [1]
The droplets push apart / spread out, so they do not merge and the cloud spreads evenly [1]
⚠ If you missed marks here: One law does the work: like charges repel. The droplets cannot clump into big drops because each repels its neighbours, so the spray stays fine and spreads.
Mark 1 -- like charges repel stated (1 mark)
Mark 2 -- droplets spread apart giving fine even cloud (1 mark)
(c)[2]
Explain two advantages of giving the panel a charge opposite to that of the droplets.
Model Answer -- 4(c)
Unlike charges attract, so the droplets are pulled onto the panel — less paint misses / is wasted [1]
Field lines end on all faces, so paint is even attracted round onto the back/edges of the panel, giving an even coat [1]
⚠ If you missed marks here: Two separate advantages are being paid for: less wasted paint (droplets curve towards the panel instead of drifting past) and coverage of awkward faces (the attraction acts even on parts not in the direct line of spray).
Mark 1 -- opposite charges attract so less paint wasted (1 mark)
Mark 2 -- attraction covers edges and rear giving even coat (1 mark)
(d)[2]
An electric field exists between the negatively charged spray gun and the positively charged panel. State the direction of the electric field, and what is meant by the direction of an electric field.
Model Answer -- 4(d)
The direction of an electric field is the direction of the force on a positive charge (placed in the field) [1]
So the field points from the (positive) panel towards the (negative) gun [1]
⚠ If you missed marks here: Field direction is DEFINED by the force on a POSITIVE test charge, so field lines run from positive to negative — here from the panel to the gun, which surprises many students because the paint travels the other way (the droplets are negative).
Mark 1 -- field direction defined by force on positive charge (1 mark)
Mark 2 -- field points from positive panel to negative gun (1 mark)
(e)[3]
The gun is supplied with charge by a steady current of 15 μA. Calculate the charge sprayed onto the panels in 2.0 minutes.
Model Answer -- 4(e)
Q = I t [1]
Q = 15 × 10⁻⁶ × 120 [1]
Q = 1.8 × 10⁻³ C = 1.8 mC [1]
⚠ If you missed marks here: Two unit conversions before anything else: 15 μA = 15 × 10⁻⁶ A and 2.0 minutes = 120 s. Then Q = It = 1.8 × 10⁻³ C. If you got 30 C you used minutes; if 0.0018 looks "too small", remember a whole coulomb is an enormous charge.
Mark 1 -- charge equation Q equals I t stated (1 mark)
Mark 2 -- substitution with 15 microamp and 120 s (1 mark)
Mark 3 -- charge 1.8 millicoulomb calculated (1 mark)
Question 5 -- The Diode Investigation
Total: 12 marks
A student at a college in Nottingham investigates a semiconductor diode. She measures the current through the diode for p.d.s up to 0.70 V applied in the FORWARD direction, then reverses the connections. Her forward results are plotted below.
(a)[3]
Describe the shape of the graph, and state what happens when the p.d. is applied in the reverse direction.
Model Answer -- 5(a)
For small p.d.s (below about 0.5 V) the current is (almost) zero — the diode barely conducts [1]
Above this threshold the current rises steeply — the diode’s resistance falls rapidly [1]
With the p.d. reversed the current is (essentially) zero: a diode conducts in ONE direction only [1]
⚠ If you missed marks here: Three features: the dead region below the threshold (≈0.5 V), the steep rise beyond it, and one-way conduction — reverse-connected, it passes practically nothing. "The current increases with voltage" alone describes every component ever made; the threshold and the one-way behaviour are what make it a diode.
Mark 1 -- near zero current below threshold about 0.5 V (1 mark)
Mark 2 -- steep current rise above threshold (1 mark)
Mark 3 -- one-way conduction with reverse current zero (1 mark)
(b)[3]
Use the graph to find the resistance of the diode at 0.70 V and at 0.40 V. Comment on the change.
Model Answer -- 5(b)
At 0.70 V, I = 0.60 A: R = V / I = 0.70 / 0.60 = 1.2 Ω [1]
At 0.40 V, I = 0.04 A: R = 0.40 / 0.04 = 10 Ω [1]
The resistance falls dramatically (about 10× smaller) as the p.d. rises — the diode is strongly non-ohmic [1]
⚠ If you missed marks here: Resistance at a point on a curved graph is V ÷ I at that point, never the gradient: 0.70/0.60 = 1.2 Ω and 0.40/0.04 = 10 Ω. If your two values came out equal you have assumed Ohm’s law — the one thing a diode refuses to obey.
The diode is connected in series with a protective resistor across a 3.1 V battery. The diode then operates at exactly 0.70 V. Calculate the p.d. across the resistor and the resistance of the resistor.
Model Answer -- 5(c)
Series p.d.s add: VR = 3.1 − 0.70 = 2.4 V [1]
Series circuit: same current everywhere, I = 0.60 A (from the graph at 0.70 V) [1]
R = V / I = 2.4 / 0.60 = 4.0 Ω [1]
⚠ If you missed marks here: The resistor takes whatever the diode leaves: 3.1 − 0.7 = 2.4 V. The current is the SAME through both (series), 0.60 A off the graph, so R = 2.4/0.60 = 4.0 Ω. Dividing 3.1 by 0.60 uses the whole battery p.d. across only one component — the classic series slip.
Mark 1 -- resistor p.d. 2.4 V by subtraction (1 mark)
Mark 2 -- series current 0.60 A identified from graph (1 mark)
Mark 3 -- resistance 4.0 ohm calculated (1 mark)
(d)[3]
State the main use of diodes in power supplies, and explain what happens when an ALTERNATING p.d. is applied across a diode in series with a lamp.
Model Answer -- 5(d)
Diodes are used for RECTIFICATION — converting a.c. to d.c. [1]
The diode conducts only during the half-cycles in which it is forward-biased [1]
During the reverse half-cycles it blocks the current, so the lamp carries current in one direction only (pulsing on and off — half-wave rectified) [1]
⚠ If you missed marks here: Rectification is the word the examiner is fishing for. With a.c. applied, the diode is a one-way valve: forward half-cycles pass, reverse half-cycles are blocked, so the lamp sees a one-directional, interrupted current — half the wave, hence half-wave rectification.
Mark 1 -- rectification converting ac to dc named (1 mark)
Mark 2 -- conducts only in forward half-cycles (1 mark)
Mark 3 -- reverse half-cycles blocked giving one-way pulsed current (1 mark)
Question 6 -- Inside the Loudspeaker
Total: 12 marks
An electronics workshop in Birmingham repairs loudspeakers. In a moving-coil loudspeaker, a coil of wire is attached to a paper cone and sits in the radial magnetic field of a circular permanent magnet. The amplifier drives an alternating current through the coil.
(a)[4]
Explain how the alternating current makes the loudspeaker produce sound.
Model Answer -- 6(a)
The coil carries a current in a magnetic field, so it experiences a force (motor effect) [1]
When the current reverses, the force on the coil reverses direction [1]
The alternating current therefore makes the coil (and the attached cone) vibrate / move in and out [1]
The vibrating cone pushes on the air, producing a sound wave at the same frequency as the current [1]
⚠ If you missed marks here: Four links in the chain: force on a current-carrying conductor in a field; force reverses when current reverses; a.c. therefore vibrates the coil and cone; the cone’s vibration is what makes the sound, at the a.c. frequency. Most dropped mark: forgetting to say the force REVERSES with the current.
Mark 1 -- current in magnetic field experiences force (1 mark)
Mark 2 -- force reverses when current reverses (1 mark)
Mark 3 -- alternating current vibrates coil and cone (1 mark)
Mark 4 -- vibrating cone produces sound at same frequency (1 mark)
(b)[2]
State two changes that would increase the force on the coil.
Model Answer -- 6(b)
Increase the current (in the coil) [1]
Use a stronger magnet / increase the magnetic field strength (or more turns of wire in the field) [1]
⚠ If you missed marks here: The force on a conductor grows with the current and with the field strength (and with the length of wire in the field, i.e. more turns). "Turn up the volume" is the everyday version of "increase the current" — say the physics version.
Mark 1 -- larger current increases force (1 mark)
Mark 2 -- stronger field or more turns increases force (1 mark)
(c)[2]
In a test rig, a straight wire carries a current horizontally from west to east. The magnetic field points vertically downwards. State the direction of the force on the wire, and the rule used to find it.
Model Answer -- 6(c)
Fleming’s left-hand rule (First finger Field, seCond finger Current, thuMb Motion/force) [1]
Force is horizontal, towards the north [1]
⚠ If you missed marks here: Set your left hand: first finger down (field), second finger east (current) — the thumb points north. The three directions are mutually perpendicular; if your answer was "up" or "west" you likely used the right hand or swapped fingers.
Mark 1 -- Fleming left-hand rule named or described (1 mark)
Mark 2 -- force direction horizontal towards north (1 mark)
(d)[2]
State what happens to the direction of the force if BOTH the current and the magnetic field are reversed at the same time. Explain your answer.
Model Answer -- 6(d)
The force direction is unchanged [1]
Reversing one alone reverses the force; reversing both reverses it twice, back to the original direction [1]
⚠ If you missed marks here: Each reversal flips the force. Two flips bring it back: reversing current AND field leaves the force exactly as it was. Many students answer "reverses" — that is the one-change case.
Mark 1 -- force unchanged stated (1 mark)
Mark 2 -- two reversals cancel explanation given (1 mark)
(e)[2]
The magnet’s pole pieces are shaped so that the field is radial and the coil former contains soft iron. Explain why soft iron is used rather than steel.
Model Answer -- 6(e)
Soft iron is easily magnetised and demagnetised, and concentrates/strengthens the magnetic field through the coil [1]
Steel would stay permanently magnetised (hard magnetic material), which is not wanted here — the iron must simply carry the field [1]
⚠ If you missed marks here: "Soft" here means magnetically soft, not squishy: it magnetises strongly while the field is present and loses it instantly. Steel keeps its magnetism (that is why permanent magnets ARE steel/alloys) — the wrong property for a core.
Mark 1 -- soft iron easily magnetised and demagnetised concentrating field (1 mark)
Mark 2 -- steel would retain magnetism which is unwanted (1 mark)
Question 7 -- The Hill Farm Hydro Scheme
Total: 10 marks
A hill farm in Snowdonia installs a small hydroelectric generator beside a stream. The generator produces 2.0 kW of electrical power at 48 V a.c. A transformer steps this up to 240 V for transmission along a long cable (total resistance 2.0 Ω) to the farmhouse, where a second transformer steps it back down. Assume the transformers are 100% efficient.
(a)[2]
The step-up transformer’s primary coil has 120 turns. Calculate the number of turns on the secondary coil.
Model Answer -- 7(a)
Ns / Np = Vs / Vp = 240 / 48 = 5 [1]
Ns = 5 × 120 = 600 turns [1]
⚠ If you missed marks here: The turns ratio equals the voltage ratio: 240/48 = 5, so the secondary needs 5 × 120 = 600 turns. If you got 24 turns you inverted the ratio — a step-UP transformer must have MORE secondary turns.
Mark 1 -- turns ratio equals voltage ratio 5 (1 mark)
Mark 2 -- secondary turns 600 calculated (1 mark)
(b)[2]
Calculate the current in the transmission cable when the full 2.0 kW is being transmitted at 240 V.
Model Answer -- 7(b)
I = P / V = 2000 / 240 [1]
I = 8.3 A [1]
⚠ If you missed marks here: Use the transmission voltage, not the generator voltage: 2000 W ÷ 240 V = 8.3 A. Using 48 V gives 41.7 A — that is the current in the PRIMARY, before the step-up.
Mark 1 -- current from P over V at 240 V (1 mark)
Mark 2 -- cable current 8.3 A calculated (1 mark)
(c)[3]
Calculate the power wasted as heat in the cable, and express it as a percentage of the power generated.
Model Answer -- 7(c)
Ploss = I² R = 8.33² × 2.0 [1]
Ploss = 139 W [1]
139 / 2000 × 100 = 6.9% [1]
⚠ If you missed marks here: Square the current: 8.33² × 2.0 ≈ 139 W, which is 6.9% of the 2000 W generated. The unsquared slip (8.33 × 2.0 = 16.7 W) makes the scheme look better than it is.
Mark 1 -- loss formula I squared R substituted (1 mark)
Mark 2 -- power loss about 139 W calculated (1 mark)
Mark 3 -- percentage 6.9 percent calculated (1 mark)
(d)[2]
Calculate the current in the primary coil of the step-up transformer, assuming it is 100% efficient.
Model Answer -- 7(d)
100% efficient: input power = output power, so Ip = P / Vp = 2000 / 48 [1]
Ip = 42 A (41.7 A) [1]
⚠ If you missed marks here: For an ideal transformer the POWER is the same on both sides, so the low-voltage side must carry the big current: 2000/48 ≈ 42 A. Stepping voltage UP steps current DOWN by the same factor of 5 — check: 42/5 = 8.3 A, matching (b).
Mark 1 -- power equal both sides so I equals P over 48 (1 mark)
Mark 2 -- primary current about 42 A calculated (1 mark)
(e)[1]
The generator must produce alternating current for this scheme to work. State why a transformer cannot operate from a direct current.
Model Answer -- 7(e)
A steady (direct) current gives a constant magnetic field in the core; only a CHANGING field induces an e.m.f. in the secondary coil [1]
⚠ If you missed marks here: One idea, precisely put: no change, no induction. d.c. magnetises the core steadily, and a steady field induces nothing in the secondary.
Mark 1 -- constant field from d.c. induces no emf in secondary (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
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