IGCSE Physics · Unit 4: Electricity and Magnetism · Part 2 of 2 · Core + Extended

Electricity and Magnetism II: Circuits, Safety & Electromagnetic Effects CoreExtended adds Kirchhoff’s laws, potential dividers, generators, motors & transformer equations

This is Part 2 of the Electricity and Magnetism note. It covers how to read and build circuits (series, parallel, and potential dividers), the safety features built into mains wiring, and the electromagnetic effects that connect electricity and magnetism together — induction, generators, motors, and transformers.

Why this matters Electromagnetic induction and its “reverse,” the motor effect, are really the same underlying physics viewed from two directions: a changing magnetic field near a conductor creates a current (induction, used in generators and transformers); a current in a magnetic field creates a force (the motor effect, used in motors and loudspeakers). Spotting that symmetry makes the second half of this note far easier to remember.

1. Circuit Diagrams and Components

You should be able to draw and interpret circuit diagrams containing standard symbols: cells, batteries, power supplies, generators, potential dividers, switches, fixed and variable resistors, heaters, thermistors (NTC), light-dependent resistors (LDRs), lamps, motors, bells, ammeters, voltmeters, magnetising coils, transformers, fuses, and relays — and know how each behaves in a circuit. Extended adds diodes and light-emitting diodes (LEDs), which conduct current in one direction only.

2. Series and Parallel Circuits

SeriesParallel
Currentthe same at every point in the circuitsplits between branches; total from the source is larger than in any single branch
Combining sourcese.m.f.s add together—
Combining resistorsresistances add together (total is larger than any one resistor)combined resistance is smaller than any individual resistor

Lamps connected in parallel in a lighting circuit have an important advantage: each lamp gets the full source voltage, and if one lamp fails (breaking its branch), the others stay lit, since their branches are unaffected.

Worked example — combining e.m.f.s and resistors in series Three 1.5 V cells are connected in series with two resistors of 4 Ω and 6 Ω, also in series. Find the total e.m.f. and total resistance.
Total e.m.f. = 1.5 + 1.5 + 1.5 = 4.5 V
Total resistance = 4 + 6 = 10 Ω
Kirchhoff’s laws in a circuit Extended At any junction in a parallel circuit, the sum of currents entering equals the sum of currents leaving — charge cannot build up or disappear at a junction, so whatever flows in must flow out.
In a series circuit, the total p.d. across all components equals the sum of the individual p.d.s across each one.
In a parallel arrangement, the p.d. across the whole arrangement is the same as the p.d. across any one branch of it.
Combined resistance of resistors in parallel Extended 1/R = 1/R₁ + 1/R₂
Find the combined resistance of a 6 Ω resistor and a 3 Ω resistor connected in parallel.
1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
R = 2 Ω (notice this is smaller than either individual resistor, as expected for parallel resistors)
Try it — Series and Parallel Circuits
  1. Two resistors of 5 Ω and 7 Ω are connected in series. Find the combined resistance.
  2. At a junction in a parallel circuit, 3 A and 2 A flow in. Find the current flowing out of the junction.
  3. Find the combined resistance of a 4 Ω and a 12 Ω resistor connected in parallel.
Show answers
  1. R = 5 + 7 = 12 Ω
  2. By Kirchhoff’s current law, current in = current out: 3 + 2 = 5 A
  3. 1/R = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 → R = 3 Ω

3. Action and Use of Circuit Components

For a constant current, the p.d. across a conductor increases as its resistance increases (directly from R = V/I).

Potential dividers Extended A potential divider uses two resistors in series across a supply to split the total voltage between them in a chosen ratio — useful for supplying a smaller, adjustable voltage to another part of a circuit (e.g. a variable resistor used as a “volume control”).
R₁/R₂ = V₁/V₂
Two resistors, 20 Ω and 30 Ω, are connected in series across a 10 V supply as a potential divider. Find the voltage across the 30 Ω resistor.
R₁/R₂ = V₁/V₂ → 20/30 = V₁/V₂, and V₁ + V₂ = 10 V. Since the resistors share voltage in the same ratio as their resistances, the 30 Ω resistor (being larger) takes the larger share: V₂ = (30/50) × 10 = 6 V

4. Electrical Safety

Hazards when using a mains supply include: damaged insulation (exposing live wires), overheating cables (risk of fire), damp conditions (water conducts electricity, increasing shock risk), and excess current from overloading plugs, extension leads, or sockets.

A UK-style mains circuit has three wires: the live wire (carries the alternating current, at high voltage relative to earth), the neutral wire (completes the circuit, close to earth potential), and the earth wire (a safety wire, normally carrying no current). A switch must always be connected in the live wire, so that when it is switched off, the appliance is fully disconnected from the dangerous high-voltage supply — switching off the neutral wire alone would leave the live wire (and the appliance) still live.

Fuses, trip switches, and earthing A fuse contains a thin wire that melts and breaks the circuit if the current exceeds its rated value, protecting the circuit and cabling from overheating. A trip switch (circuit breaker) does the same job but can simply be reset (rather than replaced) after switching off automatically.
The outer casing of an electrical appliance must be either non-conducting (double-insulated) — so there’s no exposed conducting surface a fault could make live — or earthed, so that if a fault causes the live wire to touch the casing, a large current flows safely to earth, blowing the fuse instantly and disconnecting the supply, rather than the user receiving a shock.
A double-insulated appliance has no earth wire; its fuse still protects the circuit and cabling from excess current, but earthing is unnecessary since the casing itself cannot become live.
Try it — Electrical Safety
  1. Explain why a switch must always be connected in the live wire rather than the neutral wire.
  2. Explain how earthing an appliance’s metal casing protects the user in the event of a fault.
  3. State one hazard associated with damp conditions when using electrical appliances.
Show answers
  1. The live wire carries the high-voltage supply. Connecting the switch here ensures that, when switched off, the entire appliance is disconnected from this dangerous supply. If the switch were on the neutral wire instead, the appliance would still be connected to the live wire even when “off,” posing a shock risk.
  2. If a fault causes the live wire to touch the casing, the earth wire (connected to the casing) provides a low-resistance path to earth, allowing a very large current to flow suddenly. This large current blows the fuse (or trips the circuit breaker) almost instantly, disconnecting the supply before a user touching the casing could receive a dangerous shock.
  3. Water conducts electricity, so damp conditions increase the risk of electric shock, for example if water provides an unintended conducting path between a live part of a circuit and a person.

5. Electromagnetic Induction

A conductor moving across a magnetic field, or a changing magnetic field linking with a stationary conductor, can induce an e.m.f. in that conductor — this is electromagnetic induction. This can be demonstrated by moving a bar magnet in and out of a coil connected to a sensitive ammeter (galvanometer): the needle deflects only while the magnet is moving, showing current is only induced while the field linking the coil is changing.

The magnitude of an induced e.m.f. increases with: a faster relative motion (or faster-changing field), a stronger magnetic field, and a coil with more turns.

Direction of the induced e.m.f. Extended The direction of an induced e.m.f. always opposes the change that is causing it — for example, if a magnet is pushed toward a coil, the induced current flows in a direction that creates a magnetic field opposing the magnet’s approach (as if trying to push it back out). This means work must always be done to induce a current, consistent with the conservation of energy. The relative directions of the force, the magnetic field, and the induced current can be found using the same left-hand-type relationship used for the force on a current-carrying conductor (see Section 7).
Try it — Electromagnetic Induction
  1. Describe an experiment that demonstrates electromagnetic induction.
  2. State two factors that increase the magnitude of an induced e.m.f.
  3. Explain why work must be done to push a magnet into a coil connected in a complete circuit.
Show answers
  1. Connect a coil to a sensitive ammeter (galvanometer) to form a complete circuit. Move a bar magnet in and out of the coil — the ammeter needle deflects while the magnet is moving (showing an induced current), and returns to zero once the magnet stops moving.
  2. Moving the magnet/conductor faster (or changing the field more quickly); using a stronger magnet/magnetic field; using a coil with more turns. (Any two.)
  3. The induced current’s direction always opposes the change causing it — so as the magnet is pushed toward the coil, the induced current creates a magnetic field that pushes back against the magnet’s motion. Overcoming this opposing force requires work to be done, which is consistent with energy being conserved (the electrical energy produced comes from the work done pushing the magnet in).

6. The A.C. Generator Extended only

A simple a.c. generator uses electromagnetic induction to produce electricity: a coil is rotated within a magnetic field (or a magnet is rotated near a stationary coil). As the coil turns, the magnetic field linking it constantly changes, inducing an alternating e.m.f. Slip rings and brushes maintain a continuous electrical connection to the rotating coil.

The e.m.f.-time graph of an a.c. generator As the coil rotates: the e.m.f. is zero when the coil is moving parallel to the field lines (cutting through none of them at that instant); the e.m.f. is at its peak (maximum) when the coil is moving perpendicular to the field lines (cutting through the most field lines per second at that instant). Over one full rotation, the e.m.f.-time graph is a smooth wave, alternating between positive and negative peaks and passing through zero twice per rotation — this shape is why the output is called “alternating.”

7. Magnetic Effect of a Current

A current flowing through a wire creates a magnetic field around it. Around a straight wire, the field forms concentric circles centred on the wire. Around a solenoid (a coil of wire), the field pattern resembles that of a bar magnet, with clear north and south poles at the ends. The field’s direction can be found experimentally using a plotting compass placed at various points near the current-carrying wire/solenoid.

This magnetic effect is used practically in: relays (a small current in an electromagnet switches a separate, often larger-current circuit) and loudspeakers (a varying current in a coil, within a magnetic field, produces a varying force that vibrates a cone, producing sound).

Field strength Extended The strength of the magnetic field around a straight wire or solenoid increases as the current increases (a larger current produces a stronger field), and the field is strongest closest to the wire, weakening with distance. Reversing the direction of the current reverses the direction of the magnetic field produced.
Try it — Magnetic Effect of a Current
  1. Describe the shape of the magnetic field pattern around a straight current-carrying wire.
  2. Explain how a relay allows a small current to control a much larger current safely.
  3. State the effect on the magnetic field strength around a solenoid if the current through it is increased.
Show answers
  1. The field forms a series of concentric circles centred on the wire, in planes perpendicular to the wire.
  2. A small current flows through an electromagnet (coil) inside the relay. When energised, this electromagnet attracts a switch (armature), which closes a separate circuit — allowing a small, safe control current to switch on a completely separate circuit that may carry a much larger current, without the two circuits’ currents ever mixing.
  3. The magnetic field strength around the solenoid increases as the current increases.

8. Force on a Current-Carrying Conductor

A current-carrying conductor placed within a magnetic field experiences a force (this is often called the “motor effect”). This can be demonstrated using a wire resting between the poles of a strong magnet, connected to a circuit — when current flows, the wire visibly moves. Reversing either the current direction or the magnetic field direction reverses the direction of the force; reversing both returns the force to its original direction.

Relative directions Extended The relative directions of the force, the magnetic field, and the current are all mutually perpendicular to each other. Given any two of these three directions, the third can be determined (commonly taught using the left-hand rule: thumb = force/motion, first finger = field, second finger = current). The same relationship determines the direction of the force on a beam of charged particles moving through a magnetic field, treating the beam’s motion as equivalent to a current.
Try it — Force on a Conductor
  1. A current-carrying wire in a magnetic field experiences a force. Describe what happens to this force if BOTH the current direction and the field direction are reversed.
  2. Explain why a current-carrying wire placed parallel to (along the same direction as) a magnetic field experiences no force.
Show answers
  1. The force stays in the same direction as before — reversing one of the two directions (current or field) reverses the force, but reversing both together cancels out, returning the force to its original direction.
  2. The three directions (force, field, current) must be mutually perpendicular for a force to be produced. If the current runs parallel to the field, there is no perpendicular relationship for a force to act along, so no force is produced on the wire.

9. The D.C. Motor

A current-carrying coil placed within a magnetic field experiences a turning effect (torque) — this is the basis of the electric motor. This turning effect is increased by increasing: the number of turns on the coil, the current flowing through it, and the strength of the magnetic field.

How a d.c. motor works Extended As the coil rotates within the magnetic field, the forces on its two sides (which carry current in opposite directions relative to each other) act in opposite directions, creating a turning effect. A split-ring commutator, together with fixed brushes that stay in contact with it, automatically reverses the direction of current in the coil every half-turn — this ensures the turning force always acts in the same rotational sense, keeping the motor spinning continuously in one direction rather than oscillating back and forth.

10. The Transformer

A simple transformer consists of a primary coil and a secondary coil, both wound around a shared soft-iron core, used to change (transform) an alternating voltage from one size to another. A step-up transformer increases voltage (secondary voltage > primary voltage); a step-down transformer decreases it.

Vp / Vs = Np / Ns
Worked example — transformer voltage ratio A transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. If the primary voltage is 12 V, find the secondary voltage. Is this a step-up or step-down transformer?
Vp/Vs = Np/Ns → 12/Vs = 200/1000
Vs = 12 × (1000/200) = 12 × 5 = 60 V
Since the secondary voltage is greater than the primary, this is a step-up transformer.
How a transformer works, and 100% efficiency Extended The alternating current in the primary coil creates a constantly changing magnetic field, which the soft-iron core carries efficiently to the secondary coil. This changing field induces an alternating e.m.f. in the secondary coil, by electromagnetic induction — this is why transformers only work with a.c., not d.c. (a constant d.c. current produces a constant, unchanging field, which induces nothing).
Assuming 100% efficiency (no energy losses): IpVp = IsVs — the electrical power supplied to the primary coil equals the electrical power delivered by the secondary coil, since power cannot be created, only transferred.
Worked example — using IpVp = IsVs A transformer has a primary current of 5 A and a primary voltage of 230 V. The secondary voltage is 2300 V. Assuming 100% efficiency, find the secondary current.
IpVp = IsVs → 5 × 230 = Is × 2300
1150 = 2300 × Is → Is = 0.5 A
High-voltage transmission and power losses Extended Electricity is transmitted across the country at very high voltage (using step-up transformers at power stations) and low current, then stepped back down (using step-down transformers) for use in homes and businesses. This matters because power lost as heat in transmission cables is given by:
P = I²R
For a fixed amount of power to be delivered, transmitting at a higher voltage means a correspondingly lower current is needed. Since power loss depends on the square of the current, even a modest reduction in current dramatically reduces the power wasted as heat in the cables — this is the entire reason high-voltage transmission is used.
Worked example — why higher voltage reduces losses A power station delivers 100 kW of power along a cable of resistance 2 Ω, once at 1000 V and once at 10000 V. Compare the power lost as heat in each case.
At 1000 V: I = P/V = 100000/1000 = 100 A. Power loss = I²R = 100² × 2 = 10000 × 2 = 20000 W (20 kW) — a fifth of the transmitted power is wasted!
At 10000 V: I = P/V = 100000/10000 = 10 A. Power loss = I²R = 10² × 2 = 100 × 2 = 200 W — a tiny fraction of the transmitted power.
Increasing the voltage by a factor of 10 reduced the current by a factor of 10, but reduced the power loss by a factor of 100 — because power loss depends on current squared.
Try it — Transformers
  1. A transformer has 100 turns on its primary coil and 25 turns on its secondary coil. If the primary voltage is 230 V, find the secondary voltage, and state whether this is step-up or step-down.
  2. A transformer’s primary coil draws a current of 2 A at 230 V. Assuming 100% efficiency, find the current in the secondary coil if the secondary voltage is 46 V.
  3. Explain why transformers only work with alternating current, not direct current.
Show answers
  1. Vp/Vs = Np/Ns → 230/Vs = 100/25 → Vs = 230 × (25/100) = 57.5 V. This is a step-down transformer, since Vs < Vp.
  2. IpVp = IsVs → 2 × 230 = Is × 46 → 460 = 46 × Is → Is = 10 A
  3. A transformer works by electromagnetic induction, which requires a CHANGING magnetic field to induce an e.m.f. in the secondary coil. Alternating current continuously changes direction, producing a constantly changing magnetic field. Direct current is constant, producing an unchanging (constant) magnetic field, which cannot induce an e.m.f. — so a transformer would not work with d.c.
Common mistakes to watch for
  • Adding resistances in parallel the same way as in series — remember 1/R = 1/R₁ + 1/R₂ for parallel, not R = R₁ + R₂.
  • Connecting a switch or fuse into the neutral wire instead of the live wire.
  • Forgetting that an induced current’s direction always opposes the change causing it (this is why work has to be done to induce a current).
  • Mixing up the roles of slip rings (a.c. generator — allow continuous connection while the coil reverses naturally) and a split-ring commutator (d.c. motor — actively reverses the current every half turn).
  • Using Vp/Vs = Np/Ns and IpVp = IsVs inconsistently — remember that as voltage steps up, current steps down by the same factor (for 100% efficiency), never both increasing or decreasing together.
  • Forgetting that power loss in transmission cables depends on current SQUARED (P = I²R) — a small change in current produces a much larger change in power loss.

11. Quick-Fire Challenge Round

True or False?
  1. The combined resistance of two resistors in parallel is always greater than either resistor alone.
  2. A fuse must always be connected in the live wire.
  3. A transformer works using direct current just as effectively as alternating current.
  4. Power loss in a transmission cable is proportional to the square of the current flowing through it.
Show answers
  1. False — the combined resistance of resistors in parallel is always LESS than either individual resistor, since parallel branches provide extra paths for current to flow.
  2. True — this ensures the circuit is fully disconnected from the dangerous supply when the fuse blows or the switch is off.
  3. False — a transformer relies on electromagnetic induction, which needs a changing magnetic field; only alternating current produces this. Direct current produces a constant field and induces nothing.
  4. True — P = I²R, so power loss depends on the square of the current, which is why high-voltage (low-current) transmission is used to minimise losses.
Spot the Error
  1. “Two resistors of 4 Ω and 4 Ω are connected in parallel, so the combined resistance is 4 + 4 = 8 Ω.”
  2. “A step-up transformer increases both the voltage and the current together, so more power comes out than went in.”
  3. “The switch in a lamp is connected in the neutral wire, so the lamp is completely safe to touch when switched off.”
Show answers
  1. Resistors in parallel are combined using 1/R = 1/R₁ + 1/R₂, not simple addition. Here: 1/R = 1/4 + 1/4 = 2/4 = 1/2 → R = 2 Ω, which is smaller than either individual resistor, not larger.
  2. A step-up transformer increases voltage but DECREASES current by the same factor (for 100% efficiency, IpVp = IsVs) — power in equals power out, it is never created. Voltage and current cannot both increase together.
  3. The switch must be connected in the LIVE wire, not the neutral wire, for the appliance to be safely disconnected from the dangerous supply when switched off. If it were in the neutral wire, the live wire (and hence the lamp) could remain live and dangerous even when “switched off.”

12. Mixed Practice — Bringing It All Together

  1. Two resistors, 8 Ω and 8 Ω, are connected in series. Find the combined resistance.
  2. Find the combined resistance of a 6 Ω and a 6 Ω resistor connected in parallel.
  3. Explain why an appliance’s outer casing must be either double-insulated or earthed.
  4. A transformer has 50 turns on its primary coil and 500 turns on its secondary coil, with a primary voltage of 20 V. Find the secondary voltage, and state whether it is step-up or step-down.
  5. Explain, using P = I²R, why electricity is transmitted at very high voltage over long distances.
Show answers
  1. R = 8 + 8 = 16 Ω
  2. 1/R = 1/6 + 1/6 = 2/6 = 1/3 → R = 3 Ω
  3. To protect the user from electric shock: double-insulation ensures there’s no exposed conducting surface that could become live, while earthing provides a safe path for current to flow to earth if a fault does occur, blowing the fuse instantly rather than leaving the casing dangerously live.
  4. Vp/Vs = Np/Ns → 20/Vs = 50/500 → Vs = 20 × (500/50) = 200 V. This is a step-up transformer.
  5. For a fixed amount of power delivered, using a higher transmission voltage requires a correspondingly lower current. Since power loss in the cables (P = I²R) depends on the SQUARE of the current, reducing the current even slightly produces a much larger reduction in wasted power — making high-voltage transmission far more efficient than transmitting at low voltage with high current.

Practice at Home

A very large topic — spread this across several sittings. Parts C, D and E include Extended-only content.

Part A — Circuits, series and parallel
  1. Three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series. Find the combined resistance.
  2. Explain the advantage of connecting lamps in parallel, rather than series, in household lighting.
  3. State how current behaves at every point in a series circuit.
Part B — Electrical safety
  1. Name the three wires in a standard mains circuit and state the purpose of each.
  2. Explain the difference between a fuse and a trip switch (circuit breaker).
  3. State two hazards that can arise from using a mains electrical supply.
Part C — Resistors in parallel and potential dividers (Extended)
  1. Find the combined resistance of a 10 Ω and a 15 Ω resistor connected in parallel.
  2. Two resistors, 40 Ω and 60 Ω, are connected in series as a potential divider across a 20 V supply. Find the voltage across the 60 Ω resistor.
Part D — Electromagnetic induction and generators (Extended)
  1. Describe how an a.c. generator produces an alternating e.m.f., referring to the coil and magnetic field.
  2. State the position of the coil (relative to the field) at which the induced e.m.f. is at a maximum.
  3. Explain what determines the direction of an induced current.
Part E — Motors and transformers (Extended)
  1. Explain the role of the split-ring commutator in a d.c. motor.
  2. A transformer has a primary voltage of 12 V and a primary current of 3 A. If the secondary voltage is 4 V, find the secondary current (assume 100% efficiency).
  3. Explain why increasing the transmission voltage reduces power losses in cables, even though the same total power is being delivered.

Key Vocabulary Recap

Potential divider — a circuit that splits a voltage between two resistors in the ratio of their resistances
Fuse — a safety device containing a wire that melts and breaks the circuit if current exceeds its rating
Electromagnetic induction — inducing an e.m.f. by a changing magnetic field linking a conductor
Split-ring commutator — reverses current in a d.c. motor’s coil every half turn, keeping it spinning one way
Step-up / step-down transformer — increases / decreases an a.c. voltage, using Vp/Vs = Np/Ns
P = I²R — the equation showing why high-voltage (low-current) transmission minimises power losses

Syllabus Reference

This note covers Cambridge IGCSE Physics (0625) section 4.3 (Electric circuits: 4.3.1 Circuit diagrams and components, 4.3.2 Series and parallel circuits, 4.3.3 Action and use of circuit components), section 4.4 (Electrical safety), and section 4.5 (Electromagnetic effects: 4.5.1 Electromagnetic induction, 4.5.2 The a.c. generator, 4.5.3 Magnetic effect of a current, 4.5.4 Force on a current-carrying conductor, 4.5.5 The d.c. motor, 4.5.6 The transformer). Core and Supplement content are both included, with Extended-only material (Kirchhoff’s laws, parallel resistance calculations, potential dividers, direction of induced e.m.f., the a.c. generator, field strength variation, force direction rules, motor operation, and the transformer equations) clearly flagged throughout. Part 1 of this note covers sections 4.1-4.2.

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