3.6 · Generating and distributing electricity and household use

Generating and distributing electricity

Move a magnet, move a current — a p.d. appears. OxfordAQA's sixth topic runs the full chain from the generator effect through the national grid's transformers to the appliances and safety systems of the home. All statements in this topic are Paper-2 (P).

OxfordAQA International GCSE Physics 9203, topic 3.6 (Version 5.1) — entirely (P) content.

Generating electricity (P)

The generator effect 发电机效应 induces a p.d. across a conductor that moves through a magnetic field, or sits in a changing field. Reverse the motion or the polarity and the induced p.d. reverses; the induced current's own field opposes the change. The p.d. grows with faster movement, stronger field, more turns and larger coil area.

An alternator spins a coil in a field to make a.c.; a dynamo gives d.c. — know the p.d.–time graphs. Power stations spin turbines to turn coils between magnets.

Transmission, transformers and the home (P)

From the station to the socket, every watt travels the same chain of coils and cables.

For a given power, high transmission voltage means low current and so little cable heating. A transformer 变压器 is two coils on a soft-iron core: alternating current in the primary makes a changing field that induces a p.d. in the secondary.

Step-up before the lines, step-down after; switch-mode transformers (50–200 kHz) are small and light — phone chargers.

Worked check. A transformer steps 25 kV up to 400 kV with 1000 primary turns: $n_s = 1000 \times 400/25 = 16\,000$ turns. At 100 MW the line current is $I = P/V = 10^8/(4 \times 10^5) = 250$ A.

Worked example (P) (a step-down transformer). A laptop transformer changes 230 V to 11.5 V; the primary has 2000 turns. $n_s = 2000 \times 11.5/230 = 100$ turns. If it gives 2.0 A to the laptop and is 100% efficient, the primary current is $I_p = V_s I_s / V_p = 11.5 \times 2.0 / 230 = 0.10$ A. Lower voltage on one side always means higher current on that side.

Worked example (P) (reading a trace). One complete cycle of the mains trace on an oscilloscope covers 4 squares at 5 ms per square. Period $= 4 \times 5 = 20$ ms $= 0.020$ s, so frequency $= 1 \div 0.020 = 50$ Hz.

Cells give d.c.; mains is a.c. with fixed frequency and voltage — read period and frequency from oscilloscope-style traces. Safety: earthing 接地 drains fault current from metal casings; a fuse 保险丝 melts or a circuit breaker trips on excess current — the breaker is resettable, and its trip time depends on its design and the fault current; double-insulated appliances need no earth.

The motor effect (P)

A current-carrying wire across a magnetic field feels a force (zero if parallel to it). Force grows with stronger field, bigger current and more wire in the field; reverse either to reverse the force — Fleming's left-hand rule 左手定则 predicts the direction.

Worked example (P) (using the left-hand rule). A wire sits between a N pole on the left and a S pole on the right, so the field points to the right; the current flows into the page. First finger to the right, second finger into the page: the thumb points down, so the wire is pushed downwards. Reverse the current (or the poles) and the force is up.

A current-carrying coil in a field rotates — the electric motor.

Transferring electrical energy (P)

Appliance labels give power; compute the current to choose a fuse, and the energy in kilowatt-hours for billing (E = Pt; cost = kWh × price per unit, read from meter readings).

Worked check. A 2.0 kW heater for 3 h uses 6.0 kWh; at 0.9 per kWh that costs 5.4. A 230 V, 1.1 kW kettle draws I = 1100/230 ≈ 4.8 A — fit a 5 A or 13 A fuse (next rating above).

Worked example (P) (meter readings). An electricity meter reads 12 345 kWh at the start of a month and 12 465 kWh at the end. Energy used $= 12\,465 - 12\,345 = 120$ kWh; at 0.9 per kWh the bill is $120 \times 0.9 = 108$. A 1500 W heater left on for 30 minutes uses $1.5\ \text{kW} \times 0.5\ \text{h} = 0.75$ kWh: change W to kW and minutes to hours first.

Common mistakes

• (P) Saying a transformer works on d.c. It needs a.c., because only a changing current makes a changing field in the core.
• (P) Saying a step-up transformer increases the power. It raises the voltage and lowers the current; at best the power stays the same.
• (P) Saying high voltage is used because it travels faster. It means low current, so less energy is lost heating the cables.
• (P) Using watts and seconds for kWh. Use kilowatts and hours: $E\,(\text{kWh}) = P\,(\text{kW}) \times t\,(\text{h})$.
• (P) Choosing a fuse below the working current. Use the next rating above the current.
• (P) Saying the motor force is biggest when the wire is parallel to the field. There is no force then; it is biggest at right angles.
• (P) Saying a voltage is induced in a still wire in a steady field. The wire must move, or the field must change.

Check yourself

1. State two ways to induce a p.d. in a conductor.
2. Give three factors that increase the size of an induced p.d.
3. Sketch the p.d.–time graphs of an alternator and a dynamo.
4. Why does the grid transmit at high voltage?
5. A transformer has 200 primary and 3000 secondary turns on 12 V a.c. Calculate the secondary p.d.
6. State the turns-ratio and power equations for a 100 % efficient transformer.
7. Explain how earthing and a fuse protect a user together.
8. State Fleming's left-hand rule.
9. Calculate the current of a 230 V, 2.3 kW heater, and a sensible fuse.
10. Calculate the energy in kWh used by a 1.5 kW kettle in 20 minutes.

Answers: 1 move the conductor through a field; place it in a changing magnetic field; 2 faster motion; stronger field; more turns (also larger coil area); 3 alternator — a sine wave about zero; dynamo — humps all above (or all below) the axis; 4 for the same power the current is smaller, so I²R heating losses in the cables are far lower; 5 $V_s = 12 \times 3000/200 = 180$ V; 6 $V_p/V_s = n_p/n_s$ and $V_p I_p = V_s I_s$; 7 the earth wire gives the fault current a path to earth, and the surge blows the fuse, disconnecting the live wire; 8 first finger Field, second finger Current, thumb Motion — mutually at right angles; 9 I = 2300/230 = 10 A — use a 13 A fuse; 10 t = 1/3 h, so E = 1.5 × 1/3 = 0.5 kWh.