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Power Distribution

Key idea: Use P = VI and Ploss = I²R to explain high-voltage transmission and solve questions on transmission current, power loss and efficiency (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use magnetic flux and flux-linkage relationships.
  • Apply Faraday's and Lenz's laws to induced e.m.f. and direction.
  • Explain simple applications of electromagnetic induction, including motional e.m.f. and eddy currents.
  • Explain simple iron-core transformer operation and apply ideal transformer ratios.

1. Definitions (Must Know)

A. Transmission loss (cable heating)

Power lost as heat in transmission cables is:

Pₗₒₛₛ = I²R

where R is the total cable resistance and I is the transmission current.

2. Key Ideas (What Earns Marks)

  • For a given transmitted power P: P = VI ⇒ I = P/V
  • Substitute into cable loss: Pₗₒₛₛ = I²R = (P/V)²R so increasing V reduces loss strongly (loss ∝ 1/V²).
  • Transformers make it practical to:
    • step up voltage for transmission (reduce current),
    • step down voltage for safe use.

Why higher transmission voltage reduces losses (scaled)

Two curves showing transmission current decreasing as 1/V and cable loss decreasing as 1/V^2 for fixed transmitted power and cable resistance.

Scroll across the graph to read all labels.

Two curves showing transmission current decreasing as 1/V and cable loss decreasing as 1/V^2 for fixed transmitted power and cable resistance.Two curves showing transmission current decreasing as 1/V and cable loss decreasing as 1/V^2 for fixed transmitted power and cable resistance.
Doubling transmission voltage halves current, but quarters I²R loss. That squared effect is why step-up transformers are so valuable.
Open full-size graph
View figure data
Values for Why higher transmission voltage reduces losses (scaled)
Transmission voltage (V/V₀)Current fraction, I/I₀ = 1/(V/V₀)Loss fraction, Ploss/Ploss₀ = 1/(V/V₀)²
111
20.50.25
30.3330.111
50.20.04
100.10.01
Scope note

The ideal transformer is core outcome 18g. Grid transmission and I²R loss are valuable applications, but they are treated here as extension context rather than a separately named 9478 outcome.

3. Detailed Explanations

A. Why we don’t just use thick cables

Reducing R needs very thick conductors, which are heavy and expensive.

Reducing I by increasing V is usually the more efficient engineering choice.

B. Why a.c. matters

Transformers need changing flux, so they work with a.c. This is why a.c. is used in large-scale power distribution.

4. Common Mistakes

  • Using Pₗₒₛₛ = IV (that is power delivered/absorbed at a potential difference, not specifically cable heating).
  • Forgetting that Pₗₒₛₛ depends on I² (current matters a lot).

5. Exam Tips

  • If the question keeps power fixed and changes voltage, use the ratio method: Pₗₒₛₛ ∝ 1/V²
  • Always state “high voltage → low current → low I²R loss”.

6. Worked Examples

Modelled example 1

Compare losses at two transmission voltages

Core

Problem

Power of 1.0 MW is transmitted through cables of total resistance 2.0 Ω. Find the cable loss at 10 kV and at 100 kV.
Study the worked solution

  1. Find current at 10 kV

    Method

    I = 100 A.

    Reason

    At fixed transmitted power, current is power divided by transmission voltage.

    Working

    I = (1.0 × 10⁶)/(1.0 × 10⁴) = 100 A

  2. Find loss at 10 kV

    Method

    Pₗₒₛₛ = 2.0 × 10⁴ W.

    Reason

    Cable heating depends on current squared.

    Working

    Pₗₒₛₛ = (100)²(2.0) = 2.0 × 10⁴ W

  3. Repeat at 100 kV

    Method

    I = 10 A and Pₗₒₛₛ = 2.0 × 10² W.

    Reason

    A tenfold voltage increase makes current one tenth and I²R loss one hundredth.

    Working

    I = (1.0 × 10⁶)/(1.0 × 10⁵) = 10 A, Pₗₒₛₛ = 10²(2.0) = 200 W

Guided practice 2

Find the transmission current and loss

About 5 min

Problem

Power of 2.0 MW is transmitted at 50 kV through cables of total resistance 1.5 Ω. Find the transmission current and cable-loss power.

Try this before viewing the solution

Unit: A
Unit: W

Hints

Hint 1: calculate delivered current first
Use I = P/V before applying Pₗₒₛₛ = I²R.
View solution step by step

  1. Calculate transmission current

    Method

    I = 40 A.

    Reason

    The line carries the stated transmitted power at the stated voltage.

    Working

    I = (2.0 × 10⁶)/(5.0 × 10⁴) = 40 A

  2. Calculate cable loss

    Method

    Pₗₒₛₛ = 2.4 × 10³ W.

    Reason

    Resistive cable heating scales as current squared.

    Working

    Pₗₒₛₛ = 40²(1.5) = 2400 W

Common misconception 3

Required voltage for a maximum allowed loss

Find and correct the mistake

Learner claim

Power of 1.0 MW is transmitted through 4.0 Ω cables, with cable loss limited to at most 1.0 × 10⁴ W. A learner says the required voltage must equal exactly 20 kV. Find the boundary value and diagnose the conclusion.

Try this before viewing the solution

Unit: kV

View solution step by step

  1. Turn the loss limit into a current limit

    Method

    I ≤ 50 A.

    Reason

    For fixed cable resistance, loss increases with current squared.

    Working

    I ≤ square root of ((1.0 × 10⁴)/4.0) = 50 A

  2. Find the boundary voltage

    Method

    V = 20 kV when the loss is exactly at its maximum.

    Reason

    At the largest permitted current, V = P/I gives the smallest permitted voltage.

    Working

    Vₘᵢₙ = (1.0 × 10⁶)/50 = 2.0 × 10⁴ V = 20 kV

  3. Interpret the inequality

    Method

    The requirement is V ≥ 20 kV, not exactly 20 kV.

    Reason

    Any higher voltage gives lower current and therefore a loss below the cap.

    Working

    Pₗₒₛₛ ≤ 10 kW ⇒ V ≥ 20 kV

Examiner practice 4

Transmission efficiency

5 marks

Examination question

A station transmits 500 kW at 25 kV through cables of resistance 3.0 Ω. Find the cable-loss power and transmission efficiency. [5 marks]

Try this before viewing the solution

Unit: W
Unit: %

View solution step by step

  1. Find line current

    1 mark

    Method

    I = 20 A.

    Reason

    The transmitted power and voltage set the current.

    Working

    I = (5.0 × 10⁵)/(2.5 × 10⁴) = 20 A

  2. Find cable loss

    2 marks

    Method

    Pₗₒₛₛ = 1.2 × 10³ W.

    Reason

    Heating is I²R in the total cable resistance.

    Working

    Pₗₒₛₛ = 20²(3.0) = 1200 W

  3. Find received power

    1 mark

    Method

    P_received = 498,800 W.

    Reason

    Subtract cable heating from the power entering the transmission line.

    Working

    P_received = 500,000-1200 = 498,800 W

  4. Calculate efficiency

    1 mark

    Method

    η = 99.76%, about 99.8%.

    Reason

    Efficiency is received output divided by transmitted input.

    Working

    η = 498,800/500,000 × 100% = 99.76%

Challenge 5

Ratio method

Minimal support

Independent transfer

Transmission voltage rises from 15 kV to 60 kV while transmitted power and cable resistance stay unchanged. By what factor does cable-loss power change?

Try this before viewing the solution

Hints

Hint 1: square the inverse voltage ratio
Use P_(loss,2)/P_(loss,1) = (V₁/V₂)².
View solution step by step

  1. Find the voltage factor

    Method

    Voltage increases by a factor of 4.

    Reason

    60/15 = 4.

    Working

    V₂/V₁ = 4

  2. Apply inverse-square scaling

    Method

    The new loss is 1/16 of the original.

    Reason

    Current falls inversely with voltage, and resistive loss depends on current squared.

    Working

    P_(loss,2)/P_(loss,1) = (15/60)² = 1/16 = 0.0625

7. Mind Stretchers

Mind stretcher 1: Ratio shortcutExtension

If transmission voltage is increased by a factor of 20 (same power, same cable), what happens to Pₗₒₛₛ?

Show Answer

Pₗₒₛₛ ∝ 1/V², so increasing V by 20 reduces loss by 20² = 400.

Mind stretcher 2: Why not “infinite voltage”?Extension

If higher voltage reduces cable losses, why don’t power companies transmit at arbitrarily high voltages?

Show Answer

Very high voltages increase insulation requirements, cost and size of equipment, and risk of electrical breakdown/arcing.

They also raise safety and clearance requirements. In practice there is an optimal engineering/economic trade-off.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027