Mains electricity, electrical power and the National Grid
Answer on paper first, then reveal the mark scheme one mark at a time.
1Grade 1–3Foundation3 marks
(a) State the frequency of the mains electricity supply in the UK. [1]
(b) State the potential difference of the mains supply in the UK. [1]
(c) Mains electricity is an alternating current (ac) supply. Describe the difference between an alternating current and a direct current. [1]
(a) 50 Hz(1)
(b) 230 V(1)
(c) ac repeatedly / constantly changes direction, dc flows in one direction only both ideas needed; allow answers in terms of potential difference; allow 'ac changes direction but dc does not'(1)
Answer: (a) 50 Hz (b) 230 V (c) ac changes direction; dc is one direction only
Examiner tip: Give the unit with each value — '50' alone does not score.
2Grade 4–54 marks
Most electrical appliances are connected to the mains using a three-core cable.
(a) Give the colour of the insulation on the earth wire. [1]
(b) State the potential of the earth wire. [1]
(c) Explain why touching the live wire is dangerous, even if a switch in the circuit is open. [2]
(a) green and yellow (stripes) both colours needed(1)
(b) 0 V allow zero / earth potential(1)
(c) the live wire is still at a high potential / 230 V (even when the switch is open) allow there is a large pd between the live wire and the body / earth; allow the body is at 0 V so there is a large pd(1)
(c) so a current would flow through the body (to earth), causing an electric shock allow 'you would be electrocuted' only if linked to current flowing through you(1)
Answer: (a) green and yellow (b) 0 V (c) large pd between live wire and body, so current flows through you
Examiner tip: The danger is the potential difference between the wire and your body — say both potentials.
3Grade 4–54 marks
A hairdryer is connected to the 230 V mains supply. The current in the hairdryer is 8.0 A.
(a) Calculate the power of the hairdryer. [2]
(b) Calculate the energy transferred by the hairdryer in 5.0 minutes. [2]
(a) $P = 230 \times 8.0$(1)
(a) 1840 W allow 1.84 kW(1)
(b) $E = 1840 \times 300$ allow ecf from (a); time in seconds(1)
(b) 552 000 J allow 5.52 × 10⁵ J or 552 kJ; allow ecf: their (a) × 300(1)
Answer: (a) 1840 W (b) 552 000 J
Examiner tip: Convert 5.0 minutes to 300 s; $1840 \times 5$ gives a value 60 times too small.
4Grade 6–7Higher5 marks
(a) A cable has a resistance of 0.50 Ω. The current in the cable is 12 A.
Calculate the power dissipated in the cable. [2]
(b) The national grid transmits electrical power at a very high potential difference.
Explain why this makes the national grid an efficient way to transfer energy. [3]
(a) $P = 12^2 \times 0.50$(1)
(a) 72 W(1)
(b) for the same power transmitted, a higher pd means a lower current (since $P = VI$) reject 'high pd reduces the resistance'(1)
(b) lower current means less heating of the cables / less power lost in the cables ($P = I^2R$)(1)
(b) so less energy is wasted / dissipated to the surroundings (so a greater proportion of the energy reaches the consumer)(1)
Answer: (a) 72 W (b) high pd → low current → less heating in cables → less energy wasted
Examiner tip: Link three steps in order — high pd, low current, less heating. 'Less energy is lost' alone scores 1.
5Grade 6–7Higher4 marks
A charge of 600 C flows through an electric motor in 2.0 minutes. The potential difference across the motor is 12 V.
Calculate the power of the motor. [4]
$E = 600 \times 12$ OR $I = \dfrac{600}{120}$ (either route; time in seconds)(1)
$E = 7200$ (J) OR $I = 5.0$ (A)(1)
$P = \dfrac{7200}{120}$ OR $P = 5.0 \times 12$ allow ecf from their E or I(1)
60 W correct answer with no working scores 4 marks(1)
Answer: 60 W
Examiner tip: Two routes both work — find energy then divide by time, or find current then use $P = VI$.
6Grade 4–55 marks
An electric shower has a power rating of 9.5 kW and is connected to the 230 V mains.
(a) Calculate the current in the shower. Give your answer to 2 significant figures. [3]
(b) The shower is connected using a much thicker cable than a table lamp.
(a) 41 A (2 sf) a correct answer of 41 A with no working scores 3 marks; an answer of 0.041 (from 9.5 kW not converted) scores 0 for marks 1–2 but may score mark 3 if correctly given to 2 sf(1)
(b) the shower has a (much) larger current allow the shower has a higher power so draws a larger current; ignore 'more electricity'(1)
(b) a thicker cable has a lower resistance, so less heating / it does not overheat allow a thin cable would get too hot / could melt the insulation / cause a fire(1)
Answer: (a) 41 A (b) larger current; thicker cable has lower resistance so heats less
Examiner tip: Convert 9.5 kW to 9500 W first — $9.5 \div 230$ gives 0.041 A.
7Grade 4–53 marks
The diagram shows part of the national grid.
(a) Name the device at P. [1]
(b) What does device P do to the potential difference? [1]
(c) Explain why device Q is needed. [1]
(a) (step-up) transformer(1)
(b) increases it allow steps it up / makes it higher(1)
(c) to decrease / step down the pd (to 230 V) so that it is safe for use in homes both the decrease and the safety / homes idea needed; allow 'voltage' for pd; allow 'to reduce the pd to 230 V' / 'to the level used by appliances in homes' as both ideas(1)
Answer: (a) (step-up) transformer (b) increases the pd (c) decreases the pd (to 230 V) so it is safe for use in homes
Examiner tip: P and Q are both transformers — say which one steps up and which steps down.
8Grade 6–77 marks
An electric bike has a rechargeable battery.
The battery charger supplies a current of 1.5 A at a potential difference of 42 V. The battery is charged for 4.0 hours.
(a) Calculate the charge that flows into the battery. [3]
(b) Calculate the energy transferred to the battery while it is charging. [2]
(c) When the bike is ridden, the motor has an average power of 250 W.
Assume that all the energy transferred to the battery in part (b) is transferred to the motor.
Calculate how long the motor can run before the battery needs recharging. Give your answer in minutes. [2]
(a) $t = 4.0 \times 60 \times 60 = 14\,400$ s conversion of time to seconds(1)
(a) $Q = 1.5 \times 14\,400$(1)
(a) 21 600 C allow $2.16 \times 10^4$ C; allow 22 000 C; 6 (C) from using 4.0 h unconverted scores 1 mark (substitution); 360 C (using minutes) scores 1 mark(1)
(b) $E = 21\,600 \times 42$ allow their (a) × 42; allow $E = Pt$ route: $P = 42 \times 1.5 = 63$ W and $E = 63 \times 14\,400$(1)
(b) 907 200 J allow $9.1 \times 10^5$ J; allow correct evaluation of their (a) × 42 (ecf)(1)
(c) $t = \dfrac{907\,200}{250} = 3629$ s allow 3628.8 s; allow ecf from (b)(1)
(c) 60 (minutes) allow 60.5 / 60.48 min; allow ecf: their time in s ÷ 60 correctly evaluated; an answer of 60 min scores 2 marks(1)
Examiner tip: Convert hours to seconds before using $Q = It$ or $E = Pt$ — 4.0 h is 14 400 s, not 240 s.
9Grade 8–9Higher6 marks
A power station generates electricity at a potential difference of 25 000 V. The table shows two ways the power could be sent through the cables of the National Grid.
Option 1: no step-up transformer
Option 2: step-up transformer used
power transmitted
100 MW
100 MW
pd of transmission cables
25 000 V
400 000 V
total resistance of cables
2.0 Ω
2.0 Ω
A student says:
“Using a step-up transformer is pointless. The pd has to be reduced to 230 V by a step-down transformer before it can be used in homes anyway.”
Evaluate the student’s statement.
Use calculations based on the data in the table in your answer. [6]
current using $I = P/V$: option 1, $I = 100\,000\,000 \div 25\,000 = 4000$ A(1)
power lost in cables using $P = I^2R$: option 1, $4000^2 \times 2.0 = 32\,000\,000$ W (32 MW)(1)
option 2, $250^2 \times 2.0 = 125\,000$ W (0.125 MW)(1)
option 1 wastes 32% of the power transmitted; option 2 wastes about 0.13% or power lost is 256 times smaller (current 16 times smaller, loss depends on current squared)(1)
the energy is wasted by heating the cables (and dissipated to the surroundings)(1)
so using the step-up transformer makes the National Grid much more efficient / far more of the energy reaches consumers(1)
the student is correct that the pd must be reduced (to 230 V) before it reaches homes, for safety(1)
but the high pd is only needed in the transmission cables, so stepping down at the end does not undo the saving(1)
conclusion: the student is wrong / the step-up transformer is not pointless(1)
Level
Marks
What it looks like
Level 3
5–6
Correct currents and power losses calculated for both options and used to make a clear comparison. Links high pd to low current and to reduced heating of the cables, and reaches a reasoned judgement on the statement that also deals with why the pd is stepped down for homes.
Level 2
3–4
At least one option's current and power loss calculated correctly, or both currents correct with a valid comparison. Some correct linking of pd, current and energy wasted in the cables, with a judgement that may be only partly supported.
Level 1
1–2
Isolated correct points, e.g. one current calculated, or a statement that higher pd means less energy is wasted, or that the pd is reduced for safety in homes. Little linking and no supported judgement.
No relevant content
0
Nothing worthy of credit.
Answer: Option 1: I = 4000 A, loss 32 MW (32%). Option 2: I = 250 A, loss 125 kW (0.125%) — 256 times less. Stepping down to 230 V is needed for safety but only at the end, so the step-up transformer is far from pointless: the student is wrong.
Examiner tip: Top-band answers back up the judgement with numbers: calculate both currents with $P = VI$, then both losses with $P = I^2R$ — and say why the pd is still stepped down for homes.
10Grade 1–3Foundation6 marks
A student connects two different power supplies, one at a time, to an oscilloscope. The graph shows how the potential difference (pd) of each supply changes with time.
(a) Which trace, A or B, shows an alternating potential difference?
Give a reason for your answer. [2]
(b) Use the graph to determine the maximum pd of supply A. [1]
(c) Use the graph to determine the frequency of supply A. [2]
(d) The UK mains supply is also alternating.
Give one way in which the mains supply is different from supply A. [1]
(a) A(1)
(a) the pd keeps changing direction / goes from positive to negative (and back) allow it changes direction; allow the line goes above and below zero; ignore 'it is a wave' / 'it goes up and down' unless direction or negative values are mentioned; reason only scores if A is chosen(1)
(b) 6 V allow 6.0 V; unit needed if no unit already printed(1)
(c) evidence of using the time for one cycle (20 ms) or the number of cycles in a known time (e.g. 3 cycles in 60 ms / 0.060 s) allow $f = \dfrac{1}{0.020}$(1)
(c) 50 Hz allow 50 cycles per second; 50 alone with no working scores 2 marks; 0.05 (from 1/20 with ms) scores 1 mark(1)
(d) the mains pd is much larger / about 230 V (supply A has a maximum of only 6 V) allow the mains is a higher voltage; do not accept 'the mains has a different frequency' (both are 50 Hz); ignore 'mains is more dangerous' alone(1)
Answer: (a) A — the pd changes direction (positive then negative) (b) 6 V (c) 50 Hz (d) mains pd is much larger (about 230 V)
Examiner tip: For frequency, read the time for one complete cycle (up, down and back to the start) — many students read only half a cycle (10 ms) and get 100 Hz.
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