Show all working for calculation questions — marks are awarded for method.
Useful constants: g = 10 N/kg (or 10 m/s²), speed of light c = 3.0 × 10⁸ m/s.
After completing all questions, click Submit Exam at the end of Question 7 to enter marking mode.
After submitting, answers are auto-graded using keyword matching. Review each model answer to verify your marks.
Question 1
Magnetism & Electromagnets — 12 marks
(a)[2]
State two properties of magnets.
Model Answer — 1(a)
Like poles repel [1]
Unlike poles attract [1]
⚠ If you missed marks here: Naming the poles ("magnets have a north and a south pole") describes a magnet rather than stating a property, and "magnets attract all metals" is wrong — copper and aluminium are not attracted, only iron, steel, cobalt and nickel. The two marks are for two separate statements: like poles repel, unlike poles attract.
Mark 1 — Like poles repel
Mark 2 — Unlike poles attract
(b)[3]
Draw the magnetic field pattern around a bar magnet. Include the direction of the field lines.
Model Answer — 1(b)
Your diagram should look like this:
Correct shape showing field lines from N to S [1]
Lines not crossing, correct spacing (closer together at poles) [1]
Direction arrows from N to S [1]
⚠ If you missed marks here: Arrows pointing from S to N outside the magnet score nothing — outside the magnet the field runs N → S, and only inside does it return. Two other frequent losses: field lines that cross or touch (they never do), and lines drawn evenly spaced all round, when they must be closest together at the poles where the field is strongest. Every line must start and end on a pole, not stop in mid-air.
Mark 1 — Correct shape showing field lines from N to S
Mark 2 — Lines not crossing, correct spacing
Mark 3 — Direction arrows from N to S
(c) (i)[2]
Describe how an electromagnet is made.
Model Answer — 1(c)(i)
Coil of insulated wire [1]
Wound around a soft iron core [1]
⚠ If you missed marks here: Answering "wrap wire around a magnet" gains nothing — the core starts unmagnetised, and the whole point is that the current magnetises it. The word insulated carries a mark: with bare wire the turns would touch and the current would take a short cut instead of going round the coil. Soft iron must be named as the core material.
Mark 1 — Coil of insulated wire
Mark 2 — Wound around a soft iron core
(c) (ii)[2]
State two ways to increase the strength of an electromagnet.
Model Answer — 1(c)(ii)
Increase the current [1]
Increase the number of turns on the coil [1]
⚠ If you missed marks here: Vague answers like "use a bigger battery" or "make it more powerful" score nothing; the marks are for increase the current and increase the number of turns on the coil. Two common wrong answers here are "use a stronger magnet" (there is no magnet in an electromagnet) and "use a steel core", which would stay magnetised after switching off.
Mark 1 — Increase the current
Mark 2 — Increase the number of turns on the coil
(d)[3]
Explain why soft iron is used as the core of an electromagnet rather than steel.
Model Answer — 1(d)
Soft iron is easily magnetised [1]
Soft iron is easily demagnetised [1]
So the electromagnet can be switched on and off / steel would remain permanently magnetised [1]
⚠ If you missed marks here: Writing "soft iron is a better magnet" or "soft iron is softer" earns no marks. The answer needs both halves of the contrast — soft iron magnetises easily and demagnetises easily — and the third mark is the consequence: the electromagnet can be switched off, whereas steel would keep its magnetism permanently. Stopping after "easily magnetised" scores 1 of 3.
Mark 1 — Soft iron is easily magnetised
Mark 2 — Soft iron is easily demagnetised
Mark 3 — Electromagnet can be switched on and off, steel remains permanently magnetised
Question 2
Electrostatics & Electric Fields — 10 marks
(a)[2]
Describe how a polythene rod can be charged by friction.
Model Answer — 2(a)
Rub the rod with a dry cloth or duster [1]
Electrons transfer from cloth to rod giving it a negative charge [1]
⚠ If you missed marks here: Reversing the electron transfer is the standard error: if electrons moved from the rod to the cloth the rod would end up positive, but polythene gains electrons and becomes negative. Never write that protons or positive charges move — only electrons transfer. The first mark also needs the method named: rub with a dry cloth or duster.
Mark 1 — Rub the rod with a dry cloth or duster
Mark 2 — Electrons transfer from cloth to rod giving it a negative charge
(b)[2]
A negatively charged rod is brought close to a small piece of uncharged paper. Explain why the paper is attracted to the rod.
Model Answer — 2(b)
The rod repels electrons in the paper to the far side, inducing positive charge on the near surface [1]
The attraction between the rod and the nearer positive charges is greater than the repulsion with the more distant negative charges [1]
⚠ If you missed marks here: Quoting "opposite charges attract" alone scores nothing, because the paper is uncharged to begin with. Mark 1 needs induction described: the negative rod repels electrons in the paper to the far side, leaving the near surface positive. Mark 2 is the comparison — the near positive charges are closer, so their attraction beats the repulsion of the more distant negative charges. The paper never gains overall charge.
Mark 1 — Rod repels electrons in the paper to the far side, inducing positive charge on near surface
Mark 2 — Attraction between rod and nearer positive charges is greater than repulsion with distant negative charges
(c)[3]
Draw the electric field pattern between two parallel plates, one positive and one negative.
Model Answer — 2(c)
Your diagram should look like this:
Straight parallel lines between the plates [1]
Lines perpendicular to the plates [1]
Direction from positive plate to negative plate, arrows shown correctly [1]
⚠ If you missed marks here: Curved lines fanning out like a bar magnet’s pattern lose two of the three marks. Between parallel plates the field is uniform: straight, evenly spaced lines meeting both plates at 90°. The direction mark is lost by arrows drawn from − to +; field arrows always point from the positive plate towards the negative plate.
Mark 1 — Straight parallel lines between the plates
Mark 2 — Lines perpendicular to the plates
Mark 3 — Direction from positive plate to negative plate with arrows
(d)[3]
A Van de Graaff generator is used to demonstrate electrostatic effects. A student's hair stands on end when they touch the dome. Explain why.
Model Answer — 2(d)
Charge transfers from dome to student and hair [1]
Each strand of hair receives the same charge (positive) [1]
Like charges repel, so the strands repel each other and stand on end [1]
⚠ If you missed marks here: Saying "the hair is attracted to the dome" is the wrong direction — the strands are pushed apart. The explanation needs all three links: charge passes from the dome onto the student and hair, every strand receives the same sign of charge, and like charges repel each other. "Static electricity makes it stand up" restates the observation instead of explaining it.
Mark 1 — Charge transfers from dome to student and hair
Mark 2 — Each strand receives the same charge (positive)
Mark 3 — Like charges repel so strands repel each other and stand on end
Question 3
Current, Voltage, Resistance Calculations — 12 marks
(a) (i)[1]
State the equation linking current, charge and time.
Model Answer — 3(a)(i)
I = Q/t or Q = It [1]
I = Q / t
⚠ If you missed marks here: The mark is for the equation with charge, current and time only. Writing it upside down, Q = I/t, scores nothing, and neither does quoting a different relation such as V = IR or P = IV. Words are acceptable — charge = current × time — provided the three quantities are the right way round.
Mark 1 — Correct equation I = Q/t or Q = It stated
(a) (ii)[3]
A current of 2.5 A flows through a lamp for 3 minutes. Calculate the charge that flows through the lamp.
Model Answer — 3(a)(ii)
Convert time: t = 3 × 60 = 180 s [1]
Q = It = 2.5 × 180 [1]
Q = 450 C [1]
Q = 2.5 × 180 = 450 C
⚠ If you missed marks here: Leaving the time in minutes is the mark-losing slip: Q = 2.5 × 3 gives 7.5 C instead of 450 C. Convert first, 3 × 60 = 180 s, and show that line because it carries a mark of its own. The unit is the coulomb (C), not the amp.
Mark 1 — Convert time to seconds: 3 × 60 = 180 s
Mark 2 — Correct substitution Q = 2.5 × 180
Mark 3 — Correct answer Q = 450 C
(b) (i)[1]
State the equation linking voltage, current and resistance.
Model Answer — 3(b)(i)
V = IR or R = V/I [1]
V = I × R
⚠ If you missed marks here: This part wants voltage, current and resistance linked, so V = IR (or R = V/I, or I = V/R) all earn the mark, but V = I/R does not. Do not answer with the charge equation from part (a) — each "state the equation" part is marked on its own.
Mark 1 — Correct equation V = IR or R = V/I stated
(b) (ii)[3]
A resistor has a resistance of 47 Ω. The potential difference across it is 9.4 V. Calculate the current flowing through the resistor.
Model Answer — 3(b)(ii)
I = V/R [1]
I = 9.4 / 47 [1]
I = 0.20 A [1]
I = 9.4 / 47 = 0.20 A
⚠ If you missed marks here: Dividing the wrong way round gives 47 / 9.4 = 5.0 A instead of 0.20 A; sanity-check it — a large resistance with a small p.d. must give a small current. The rearrangement I = V/R has to appear before the numbers to earn mark 1, and the answer should be quoted to 2 significant figures with the unit A.
Mark 1 — Correct rearrangement I = V/R
Mark 2 — Correct substitution I = 9.4 / 47
Mark 3 — Correct answer I = 0.20 A
(c)[4]
The I–V characteristic of a filament lamp is non-linear. Explain why the resistance of a filament lamp increases as the current increases.
Model Answer — 3(c)
As current increases, the filament gets hotter [1]
The metal atoms vibrate more [1]
This causes more frequent collisions with the charge carriers (electrons) [1]
Therefore the resistance increases (V increases more than proportionally to I) [1]
⚠ If you missed marks here: Writing "the lamp gets hot so the resistance increases" is worth one mark of four, because it skips the mechanism. The full chain is: larger current → filament hotter → metal ions vibrate more → more frequent collisions with the electrons → resistance rises. Saying "more electrons flow so they collide more" describes the wrong cause, and "resistance increases because the voltage increases" reverses the argument.
Mark 1 — As current increases the filament gets hotter
Mark 2 — The metal atoms vibrate more
Mark 3 — More frequent collisions with charge carriers or electrons
Mark 4 — Therefore resistance increases, V increases more than proportionally to I
Question 4
Series and Parallel Circuit Analysis — 12 marks
A circuit contains a 12 V battery connected to two resistors: R1 = 4.0 Ω and R2 = 8.0 Ω.
(a) (i)[2]
Calculate the total resistance of the circuit when the resistors are connected in series.
Model Answer — 4(a)(i)
R_total = R1 + R2 [1]
R_total = 4.0 + 8.0 = 12 Ω [1]
R = 4.0 + 8.0 = 12 Ω
⚠ If you missed marks here: Applying the parallel formula here gives 2.67 Ω instead of 12 Ω. In series the resistances add, so R = 4.0 + 8.0 = 12 Ω, and the formula line R = R1 + R2 carries a mark even when the arithmetic is obvious.
Mark 1 — Correct formula R_total = R1 + R2
Mark 2 — Correct answer R_total = 12 ohm
(a) (ii)[2]
Calculate the current flowing through the circuit.
Model Answer — 4(a)(ii)
I = V/R = 12/12 [1]
I = 1.0 A [1]
I = 12 / 12 = 1.0 A
⚠ If you missed marks here: Using one resistor instead of the total is the usual error: 12 / 4.0 = 3.0 A rather than 12 / 12 = 1.0 A. The battery drives current through the whole 12 Ω, and in a series circuit that same 1.0 A flows through both resistors and the battery — the current is not shared out.
Mark 1 — Correct substitution I = V/R = 12/12
Mark 2 — Correct answer I = 1.0 A
(a) (iii)[2]
Calculate the potential difference across R2.
Model Answer — 4(a)(iii)
V = IR = 1.0 × 8.0 [1]
V = 8.0 V [1]
V = 1.0 × 8.0 = 8.0 V
⚠ If you missed marks here: Splitting the supply equally, 6.0 V across each resistor, is wrong — the p.d. divides in the ratio of the resistances, so the 8.0 Ω takes twice as much as the 4.0 Ω. Use V = IR with the circuit current: 1.0 × 8.0 = 8.0 V. Quoting the battery voltage, 12 V, gains nothing.
Mark 1 — Correct substitution V = IR = 1.0 × 8.0
Mark 2 — Correct answer V = 8.0 V
The same two resistors (4.0 Ω and 8.0 Ω) are now connected in parallel across the same 12 V battery.
(b) (i)[3]
Calculate the total resistance of the parallel combination.
Model Answer — 4(b)(i)
1/R = 1/R1 + 1/R2 = 1/4.0 + 1/8.0 [1]
1/R = 0.25 + 0.125 = 0.375 [1]
R = 1/0.375 = 2.67 Ω (accept 2.7 Ω or 8/3 Ω) [1]
1/R = 1/4.0 + 1/8.0 = 0.375 → R = 2.67 Ω
⚠ If you missed marks here: The commonest loss is stopping at 1/R = 0.375 and writing R = 0.375 Ω — you must invert to get 2.67 Ω. Adding the resistors (12 Ω) is the series method. Use this as a check: a parallel total is always smaller than the smallest resistor, so any answer above 4.0 Ω is wrong before you look at the working.
Mark 1 — Correct formula 1/R = 1/R1 + 1/R2 with substitution
⚠ If you missed marks here: Sharing the 12 V between the branches (6 V each) is the error to avoid — in parallel both resistors have the full 12 V across them, so 3.0 A and 1.5 A. Mark 3 needs the total stated, and it is worth checking against 12 / 2.67 = 4.5 A from part (b)(i); if those two disagree, one of the answers is wrong.
Mark 1 — Current through R1: I = 12/4.0 = 3.0 A
Mark 2 — Current through R2: I = 12/8.0 = 1.5 A
Mark 3 — Total current = 3.0 + 1.5 = 4.5 A
Question 5
Electrical Power and Energy Costs — 12 marks
(a) (i)[1]
State the equation linking power, current and voltage.
Model Answer — 5(a)(i)
P = IV [1]
P = I × V
⚠ If you missed marks here: Only P = IV answers this part, because the question names power, current and voltage. P = I2R brings in resistance and P = E/t brings in energy and time, so neither earns the mark here even though both are correct equations.
Mark 1 — Correct equation P = IV stated
(a) (ii)[2]
An electric kettle operates at 230 V and draws a current of 10 A. Calculate the power of the kettle.
Model Answer — 5(a)(ii)
P = IV = 10 × 230 [1]
P = 2300 W (or 2.3 kW) [1]
P = 10 × 230 = 2300 W
⚠ If you missed marks here: Dividing instead of multiplying gives 230 / 10 = 23 W rather than 2300 W. Watch the unit as well: 2300 W is 2.3 kW, and writing "2.3 W" or "2300 kW" throws away the answer mark after the substitution was right.
Mark 1 — Correct substitution P = 10 × 230
Mark 2 — Correct answer P = 2300 W or 2.3 kW
(a) (iii)[3]
The kettle takes 4 minutes to boil water. Calculate the energy transferred in kilowatt-hours and the cost if electricity costs 15p per kWh.
Model Answer — 5(a)(iii)
Time = 4/60 = 1/15 hours (or 0.0667 h) [1]
Energy = 2.3 × (4/60) = 0.153 kWh [1]
Cost = 0.153 × 15 = 2.3p [1]
E = 2.3 kW × (4/60) h = 0.153 kWh; Cost = 0.153 × 15 = 2.3p
⚠ If you missed marks here: A kilowatt-hour needs the power in kW and the time in hours. Leaving 4 minutes as it stands gives 2.3 × 4 = 9.2 kWh and a cost of 138p instead of 0.153 kWh and 2.3p. Note that 4/60 = 0.0667 h, not 0.4 h. The cost mark is for multiplying the kWh by 15, giving pence — do not convert into pounds unless asked.
Mark 1 — Convert time: 4/60 = 0.0667 hours
Mark 2 — Energy = 2.3 × (4/60) = 0.153 kWh
Mark 3 — Cost = 0.153 × 15 = 2.3p
(b) (i)[2]
The kettle has a three-pin plug. State the colours of the wires connected to the live, neutral and earth terminals.
Model Answer — 5(b)(i)
Live = brown, Neutral = blue [1]
Earth = green and yellow (striped) [1]
⚠ If you missed marks here: Red, black and green is the pre-1970 code and scores nothing on a modern plug: live is brown, neutral is blue. Swapping brown and blue loses mark 1. Earth must be described as green and yellow striped — "green" alone is not enough for mark 2.
Mark 1 — Live is brown, Neutral is blue
Mark 2 — Earth is green and yellow striped
(b) (ii)[2]
Explain why a fuse is connected in the live wire.
Model Answer — 5(b)(ii)
If the current exceeds the rated value, the fuse melts/blows and breaks the circuit [1]
This must be in the live wire so that the appliance is disconnected from the high voltage supply when the fuse blows / prevents the case becoming live [1]
⚠ If you missed marks here: Answering "so the fuse can protect the appliance" is too vague for either mark. Mark 1 is the action: if the current exceeds the rated value the fuse wire melts and breaks the circuit. Mark 2 is the reason for the position — in the live wire the appliance is cut off from the high-voltage side, whereas a fuse in the neutral would leave the appliance and its case still connected to the live supply after blowing.
Mark 1 — Fuse melts and breaks the circuit if current exceeds rated value
Mark 2 — Must be in live wire to disconnect appliance from high voltage supply
(c)[2]
The kettle is rated at 2300 W and operates at 230 V. Determine the most appropriate fuse rating. Choose from 3 A, 5 A, or 13 A.
Model Answer — 5(c)
I = P/V = 2300/230 = 10 A [1]
13 A fuse (must be higher than 10 A but closest available rating) [1]
I = 2300 / 230 = 10 A → 13 A fuse
⚠ If you missed marks here: Answering "10 A" is not one of the three choices and scores the second mark nothing; a 5 A fuse would melt straight away because the kettle draws 10 A in normal use. Work out I = P/V = 2300 / 230 = 10 A first, then pick the smallest listed rating above the working current, which is 13 A.
Mark 1 — Correct calculation I = P/V = 2300/230 = 10 A
Mark 2 — 13 A fuse selected (higher than 10 A, closest available)
Question 6
Electromagnetic Induction & Transformers — 12 marks
(a) (i)[2]
State two ways of inducing an e.m.f. in a coil of wire.
Model Answer — 6(a)(i)
Move a magnet into or out of the coil [1]
Change the magnetic field through the coil / vary the current in a nearby coil [1]
⚠ If you missed marks here: Connecting the coil to a battery produces a current but no induced e.m.f., so it earns nothing. Something has to change: move a magnet into or out of the coil, or vary the current in a nearby coil so the field through this one changes. A magnet held still inside the coil, however strong, induces no e.m.f. at all.
Mark 1 — Move a magnet into or out of the coil
Mark 2 — Change the magnetic field through the coil or vary current in a nearby coil
(a) (ii)[2]
State two ways of increasing the magnitude of the induced e.m.f.
Model Answer — 6(a)(ii)
Move the magnet faster / increase the rate of change of flux [1]
Use more turns on the coil / use a stronger magnet [1]
⚠ If you missed marks here: The two answers must be genuinely different physics — "move the magnet quickly" and "move the magnet faster" count once. Accepted: increase the speed (rate of change of the field), use more turns on the coil, use a stronger magnet. Answers such as "leave the magnet in the coil longer" or "use a bigger resistor" score nothing.
Mark 1 — Move the magnet faster or increase rate of change of flux
Mark 2 — Use more turns on the coil or use a stronger magnet
(b)[1]
State the equation linking the voltages and number of turns in a transformer.
Model Answer — 6(b)
Vp/Vs = Np/Ns [1]
Vp / Vs = Np / Ns
⚠ If you missed marks here: Writing the ratio upside down, Vp/Vs = Ns/Np, loses the mark: the two primary quantities must sit on the same side of the equation. The current relation IpVp = IsVs is a different equation and does not answer this part.
Mark 1 — Correct transformer equation Vp/Vs = Np/Ns
(c)[3]
A transformer has 400 turns on its primary coil and 2000 turns on its secondary coil. The input voltage is 25 V. Calculate the output voltage and state whether this is a step-up or step-down transformer.
Model Answer — 6(c)
Vs = Vp × Ns/Np = 25 × 2000/400 [1]
Vs = 125 V [1]
Step-up transformer (because Vs > Vp or Ns > Np) [1]
Vs = 25 × (2000/400) = 25 × 5 = 125 V
⚠ If you missed marks here: Inverting the turns ratio gives 25 × 400/2000 = 5.0 V and a step-down answer instead of 125 V. More turns on the secondary always means a higher secondary voltage, so use that as a check. Mark 3 is separate from the calculation — the words "step-up" must actually be written, with the reason (Ns > Np).
Mark 1 — Correct substitution Vs = 25 × 2000/400
Mark 2 — Correct answer Vs = 125 V
Mark 3 — Step-up transformer correctly identified
(d)[4]
Explain why transformers are used in the transmission of electrical power over long distances. Include relevant equations in your answer.
Model Answer — 6(d)
Power loss in cables = I²R (or P = I²R) [1]
Transformers step up the voltage for transmission [1]
For the same power transmitted, P = IV means higher V gives lower I [1]
Lower current means less power wasted as heat in the cables (I²R losses are reduced) [1]
P_loss = I²R; P = IV → higher V, lower I, less loss
⚠ If you missed marks here: One sentence such as "high voltage loses less energy" covers at most one of the four marks. Build the chain and quote both equations: for a fixed power P = IV, so raising V lowers I; the power wasted in the cables is I2R, and because the current is squared, cutting the current to a tenth cuts the loss to a hundredth. Saying the electricity "travels faster" or that the voltage "pushes it further" is wrong physics.
Mark 1 — Power loss in cables equals I squared R
Mark 2 — Transformers step up the voltage for transmission
Mark 3 — P = IV so higher voltage gives lower current for same power
Mark 4 — Lower current means less power wasted as heat in cables
Question 7
DC Motor and Generator — 10 marks
(a)[2]
State the rule used to find the direction of force on a current-carrying conductor in a magnetic field. Describe what each finger represents.
Model Answer — 7(a)
Fleming's left-hand rule [1]
First finger = Field, seCond finger = Current, thuMb = Motion/Force [1]
⚠ If you missed marks here: Naming the right-hand rule scores nothing — that is the generator effect; the force on a current-carrying wire uses the left hand. All three fingers must be assigned to earn mark 2: First finger = Field (N → S), seCond finger = Current (conventional, + to −), thuMb = Motion or force. The three must be held at right angles to each other.
Mark 1 — Fleming's left-hand rule named
Mark 2 — First finger = Field, Second finger = Current, Thumb = Motion or Force
(b)[3]
Describe the structure of a simple DC motor. Include the purpose of the commutator.
Model Answer — 7(b)
A coil of wire (armature) placed between the poles of a magnet [1]
The coil is connected to a power supply via a split-ring commutator and carbon brushes [1]
The commutator reverses the current direction every half turn to maintain continuous rotation in the same direction [1]
⚠ If you missed marks here: Saying "the coil spins in the magnetic field" is not a description of the structure. Mark 1 needs the coil or armature placed between the poles of a magnet, mark 2 the split-ring commutator and carbon brushes, and mark 3 the commutator’s job: it reverses the current in the coil every half turn so the turning effect keeps acting the same way round. Writing "slip rings" describes an a.c. generator, and "the commutator changes a.c. into d.c." is the wrong way round for a motor.
Mark 1 — Coil of wire (armature) between poles of a magnet
Mark 2 — Connected via split-ring commutator and carbon brushes
Mark 3 — Commutator reverses current every half turn for continuous rotation
(c)[2]
State two ways to increase the speed of rotation of a DC motor.
Model Answer — 7(c)
Increase the current [1]
Use a stronger magnet / increase the number of turns on the coil [1]
⚠ If you missed marks here: Loose answers such as "give it more power" or "use a bigger battery" do not earn the current mark — state increase the current. The alternatives are a stronger magnet or more turns of wire on the coil, all of which increase the force on the sides of the coil. Answers about lighter bearings, thinner wire or a longer axle are not the physics being tested.
Mark 1 — Increase the current
Mark 2 — Use a stronger magnet or increase number of turns on coil
(d)[3]
Describe the differences between a DC motor and an AC generator.
Model Answer — 7(d)
A motor converts electrical energy to kinetic energy; a generator converts kinetic energy to electrical energy [1]
A motor has a split-ring commutator; a generator has slip rings [1]
A motor requires an input current; a generator produces an output e.m.f./current [1]
⚠ If you missed marks here: Each mark needs both machines compared in the same sentence; describing only what a motor does leaves the comparison mark unearned. The three contrasts are: electrical → kinetic against kinetic → electrical, split-ring commutator against slip rings, and current supplied in against e.m.f. produced out. Giving a generator a commutator is the usual error — an a.c. generator has slip rings, which is exactly why its output stays alternating.
Mark 1 — Motor converts electrical to kinetic energy; generator converts kinetic to electrical energy
Mark 2 — Motor has split-ring commutator; generator has slip rings
Mark 3 — Motor requires input current; generator produces output e.m.f. or current
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