Flashcards · 9203 · Topic 6

Generating and distributing electricity and household use flashcards

33 cards on Topic 6, subtopics 3.6.1 to 3.6.5. Pick a subtopic or study them all.

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3.6.1 Generating electricity

When is a potential difference induced across a conductor?: When it moves relative to a magnetic field (cutting field lines), or sits in a changing magnetic field. A current flows only if the circuit is complete.
A north pole is pushed into a coil. What pole forms at the near end, and why?: A north pole, so it repels the magnet: the induced current's field opposes the change causing it.
Four ways to increase an induced potential difference: Faster movement, a stronger magnetic field, more turns on the coil, a larger coil area.
Alternator vs dynamo: Alternator: ac output, a wave above and below the axis. Dynamo: dc output, humps all on one side of the axis, because the connection swaps every half turn.
Coil positions for maximum and zero induced p.d.: Maximum: plane of the coil parallel to the field lines. Zero: plane of the coil at right angles to the field lines.
A generator's output has a period of 0.10 s. Frequency and rotations per minute?: f = 1 ÷ 0.10 = 10 Hz, so a simple coil turns 10 times a second: 600 rotations per minute.
How does a power station generator produce electricity?: Fuel heats water to steam (or moving water or wind is used) to turn a turbine. The turbine turns coils between magnets, so the coils cut field lines and a p.d. is induced.

3.6.2 Electricity transmission and distribution

The chain from power station to consumer: Power station → step-up transformer → transmission cables → step-down transformer(s) → consumers.
Why is electricity transmitted at a high voltage?: P = V × I, so for the same power a higher voltage means a lower current. A lower current heats the cables less, so less energy is wasted and the system is more efficient.
How does a transformer work?: Alternating current in the primary coil gives a changing magnetic field in the soft iron core. The field passes through the secondary coil and induces a changing p.d. across it. It needs ac (a changing current).
Step-up vs step-down transformers: Step-up: Vs > Vp, more turns on the secondary; used before the transmission lines. Step-down: Vs < Vp, fewer secondary turns; used to give a safer voltage for local consumers.
Turns ratio. Vp = 1500 V, np = 600, Vs = 75 V. Find ns.: Vp/Vs = np/ns, so ns = 600 × 75 ÷ 1500 = 30 turns (step-down).
A 100% efficient transformer gives 6.0 V at 2.0 A from a 12 V primary. Primary current?: Vp × Ip = Vs × Is: power = 6.0 × 2.0 = 12 W, so Ip = 12 ÷ 12 = 1.0 A.
2.7 MW is sent through a 3.0 Ω cable at 90 kV, then at 270 kV. Power lost each time?: I = P/V: 30 A, then 10 A. Loss = I²R: 30² × 3.0 = 2700 W, then 10² × 3.0 = 300 W. One third of the current gives one ninth of the loss.
Switch mode transformers: Work at high frequency (often 50 kHz to 200 kHz), are much lighter and smaller than traditional 50 Hz transformers (e.g. phone chargers), and use very little power when on with no load.

3.6.3 Using electricity in the home

Direct vs alternating current: dc always passes in the same direction (cells and batteries). ac repeatedly changes direction. Mains is ac with a set frequency and voltage.
An ac trace peaks 2.5 divisions high at 4 V/div; one cycle is 4 divisions at 5 ms/div. Peak p.d. and frequency?: Peak p.d. = 2.5 × 4 = 10 V. Period = 4 × 5 = 20 ms = 0.020 s, so f = 1/T = 50 Hz.
The live wire touches the metal case of an earthed appliance. What happens?: A large current flows through the earth wire, the fuse melts (or circuit breaker trips) in the live wire, and the appliance is disconnected so the case is no longer live.
Fuse vs circuit breaker: Both go in the live wire and disconnect the circuit when the current is too great. A fuse overheats, melts and must be replaced. A circuit breaker trips, works much faster and can be reset.
Double-insulated appliance: Live parts are kept out of reach inside a non-conducting case, so it has no earth wire connection.
A 690 W iron runs on 230 V. Choose a fuse from 3 A, 5 A and 13 A.: I = P/V = 690 ÷ 230 = 3.0 A. A 3 A fuse could melt in normal use, so choose 5 A (the smallest rating above the current).

3.6.4 The motor effect

The motor effect, and when the force is zero: A current-carrying conductor in a magnetic field, cutting the field lines, experiences a force. There is no force when it is parallel to the field; the force is largest at 90°.
Three ways to increase the force on a current-carrying conductor in a field: A stronger magnetic field, a larger current, a longer length of conductor in the field.
How is the direction of the motor-effect force reversed?: Reverse either the current or the field. Reversing both leaves the force direction unchanged.
Fleming's left-hand rule: Thumb and first two fingers at right angles. First finger: field (north to south). Second finger: current (positive to negative). Thumb: force (motion).
The field points to the left and the current flows up the page. Direction of the force?: Out of the page, towards you (first finger left, second finger up, thumb out).
Why does a current-carrying coil in a magnetic field rotate?: The two long sides carry currents in opposite directions, so the forces on them are opposite, giving a turning effect. More current, a stronger magnet or more turns increase it.

3.6.5 Transferring electrical energy

Main energy transfer in a kettle, a motor, a loudspeaker and a battery charger: Kettle: electrical → thermal. Motor: electrical → kinetic. Loudspeaker: electrical → sound. Charger: electrical → chemical.
Power equations. A hob ring takes 4.0 A from 230 V:: P = E/t and P = I × V. P = 4.0 × 230 = 920 W.
The 920 W hob runs for 300 s at 4.0 A and 230 V. Energy, found two ways?: E = P × t = 920 × 300 = 276 000 J. Q = It = 1200 C, and E = Q × V = 1200 × 230 = 276 000 J.
Kilowatt-hour: The energy transferred by a 1 kW appliance running for 1 hour. kWh measure energy; watts measure power.
A 0.50 kW heater runs for 4.0 h at 31 p per kWh. Cost?: Energy = 0.50 × 4.0 = 2.0 kWh. Cost = 2.0 × 31 = 62 p.
A meter reads 10230, later 10265, at 30 p per kWh. Cost?: Units used = 10265 − 10230 = 35 kWh. Cost = 35 × 30 = 1050 p = £10.50.