OCR OCR 2021-style practice: A step-up transformer has a primary potential difference of 25\,kV and a secondary potential difference of 400\,kV. The power input is 200\,MW. Assuming the transformer is 100\% efficient, calculate the current in the secondary coil. (You are given V_p I_p = V_s I_s.)

A P8 Calculate question using the given transformer power relationship. For a 100\% efficient transformer the power output equals the power input, so the secondary power is 200\,MW = 200 × 10^6\,W. Power equals potential difference times current, so I_s = PV_s = 200 × 10^6400 × 10^3 = 500\,A (3 marks for the method and substitution). The answer is I_s = 500\,A (1 mark with the unit amperes). Markers reward using power equals voltage times current on the secondary side and the answer of 500\,A. A common error is to mix up the primary and secondary values, or to mishandle the kilo and mega prefixes.

EnglandPhysicsSyllabus dot point

How does the national grid transmit electricity efficiently, and what is mains electricity?

The national grid and the role of step-up and step-down transformers, why transmission is at high voltage and low current, the transformer turns and power relationships, and the nature of mains electricity as an alternating supply.

A focused answer to OCR Gateway GCSE Physics A topic P8 on the national grid and mains electricity, covering step-up and step-down transformers, why transmission is at high voltage and low current, the transformer power relationship, and the nature of mains electricity as an alternating supply.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
The national grid
Why transmit at high voltage and low current
Transformers and the power relationship
Mains electricity
Try this

What this topic is asking

OCR wants you to describe the national grid and the role of step-up and step-down transformers, explain why transmission is at high voltage and low current, use the transformer power relationship, and describe mains electricity. This is part of topic P8.2 of the OCR Gateway Physics A (J249) specification.

The national grid

Transmission over the grid is at very high voltage (typically hundreds of kilovolts) along the cables strung between pylons, then reduced in stages by step-down transformers before it reaches the local supply.

Why transmit at high voltage and low current

This is the key idea of the topic: it is the low current (not the high voltage directly) that cuts the heating loss, because the loss goes as the current squared.

Transformers and the power relationship

A step-up transformer has more turns on the secondary coil and raises the voltage; a step-down transformer has fewer turns on the secondary and lowers it. For an ideal (100% efficient) transformer, the power output equals the power input, so

V_p I_p = V_s I_s

where $V_p$ and $I_p$ are the primary potential difference and current, and $V_s$ and $I_s$ are the secondary values. This equation is given on the data sheet. It shows that when a transformer steps the voltage up, it steps the current down by the same factor (for the same power).

Mains electricity

A direct current (d.c.), such as from a battery, flows one way only. Mains is alternating current (a.c.) so that the national grid can use transformers to change the voltage efficiently.

Why high voltage cuts the transmission loss

A power line carries $10\,\text{MW}$ of power. Explain, using the equations, why stepping the voltage up before transmission reduces the energy wasted in the cables.

step 1: Link power, voltage and current

The power transmitted is $P = VI$ , so for a fixed power of $10\,\text{MW}$ , raising the voltage $V$ must lower the current $I$ (their product is fixed).

step 2: Link the heating loss to the current

The power wasted heating the cables is $P_{\text{loss}} = I^2 R$ , where $R$ is the cable resistance. The loss depends on the current squared.

step 3: Combine the ideas

Because a higher transmission voltage gives a smaller current, and the loss goes as the current squared, even a modest drop in current gives a large drop in wasted power.

step 4: State the conclusion

So stepping the voltage up (and the current down) before transmission greatly reduces the energy wasted as heat in the cables, making the grid efficient. A step-down transformer then lowers the voltage for safe use.

Try this

Q1. State why the current is made small for transmission across the national grid. [2 marks]

Cue. The power wasted heating the cables is proportional to the current squared, so a small current wastes much less energy as heat.

Q2. State the potential difference and frequency of UK mains electricity. [1 mark]

Cue. About $230\,\text{V}$ and $50\,\text{Hz}$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksExplain why electricity is transmitted across the national grid at very high voltage, and state the role of step-up and step-down transformers.

Show worked answer →

A P8 question worth four marks. For a given power transmitted, $P = VI$ , so a higher voltage means a lower current (1 mark). The power wasted heating the transmission cables is $P = I^2 R$ , so a smaller current wastes far less energy as heat, making transmission much more efficient over long distances (2 marks for low current giving low heating loss). A step-up transformer raises the voltage (and lowers the current) for transmission; a step-down transformer lowers the voltage again to safe values for homes and businesses (1 mark). Markers reward high voltage giving low current, the reduced heating loss, and the roles of the two transformers. A common error is to say high voltage directly reduces the heating; it is the lower current that does.

OCR 20214 marksA step-up transformer has a primary potential difference of

25\,\text{kV}

and a secondary potential difference of

400\,\text{kV}

. The power input is

200\,\text{MW}

. Assuming the transformer is

100\%

efficient, calculate the current in the secondary coil. (You are given

V_p I_p = V_s I_s

Show worked answer →

A P8 Calculate question using the given transformer power relationship. For a $100\%$ efficient transformer the power output equals the power input, so the secondary power is $200\,\text{MW} = 200 \times 10^{6}\,\text{W}$ . Power equals potential difference times current, so $I_s = \dfrac{P}{V_s} = \dfrac{200 \times 10^{6}}{400 \times 10^{3}} = 500\,\text{A}$ (3 marks for the method and substitution). The answer is $I_s = 500\,\text{A}$ (1 mark with the unit amperes). Markers reward using power equals voltage times current on the secondary side and the answer of $500\,\text{A}$ . A common error is to mix up the primary and secondary values, or to mishandle the kilo and mega prefixes.

Related dot points

Sources & how we know this

OCR GCSE (9-1) Physics A (Gateway Science) J249 specification — OCR (2016)