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Power transmission at high voltage

Key idea: Sending the same power at a higher voltage needs a smaller current. The heating loss in the cables is I²R, so halving the current cuts this loss to a quarter.

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Electricity is often generated far from the towns that use it, and it has to travel through many kilometres of cable. The cables have resistance, so some energy is always lost as heat on the way. Power companies keep this loss small with a simple trick: they send the electricity at a very high voltage. This lesson explains why that works.

Energy lost in the cables

A current in a cable heats it, because the cable has resistance. The power lost as heat in a cable of resistance R carrying a current I is

Pₗₒₛₛ = I²R

The loss depends on the square of the current. Halving the current cuts the heating to a quarter; reducing the current to a tenth cuts it to a hundredth. For the same transmitted power and cable resistance, reducing current reduces this loss. Voltage cannot be raised without limit: insulation and other engineering constraints still matter.

Cable heating increases with current squared

A rising curved graph of cable heating power against current, calculated from I squared times 0.50 ohm.

Scroll across the figure to read all labels.

A rising curved graph of cable heating power against current, calculated from I squared times 0.50 ohm.A rising curved graph of cable heating power against current, calculated from I squared times 0.50 ohm.
Illustrative values for a cable resistance held at 0.50 Ω. Doubling current quadruples heating power: 2.0 A gives 2.0 W and 4.0 A gives 8.0 W. These are model values, not measured cable data.
Open full-size graph
What to noticeConstructed resistive-heating model. Cable resistance is assumed constant; actual resistance can change with temperature.
View figure data
Figure data: Cable heating increases with current squared
Current (A)R = 0.50 Ω
0.00.000
0.50.125
1.00.500
1.51.125
2.02.000
2.53.125
3.04.500
3.56.125
4.08.000
4.510.125
5.012.500
5.515.125
6.018.000
6.521.125
7.024.500
7.528.125
8.032.000
8.536.125
9.040.500
9.545.125
10.050.000

Worked example 1

Doubling the current

Problem

A cable has a resistance of 0.50 Ω. Calculate the power lost as heat when it carries a current of 2.0 A, and when it carries 4.0 A. Compare the two.

Show full solution

  1. At 2.0 A

    Method

    Square the current, then multiply by the resistance.

    Reason

    The heating power in the cable is I²R.

    Working

    Answer

    P₁ = (2.0 A)² × 0.50 Ω = 2.0 W

  2. At 4.0 A

    Method

    Use the same resistance with the doubled current.

    Reason

    Keeping the resistance the same shows the effect of the current alone.

    Working

    Answer

    P₂ = (4.0 A)² × 0.50 Ω = 8.0 W

  3. Compare

    Working

    Answer

    8.0 W is four times 2.0 W: doubling the current makes the heating loss four times as large.

Why a high voltage means a small current

A power station supplies a certain power P to the cables. In this calculation model, voltage and current are r.m.s. values with power factor taken as one, so P = VI and the current needed is

I = P/V

For the same power, a higher voltage needs a smaller current. Multiply the voltage by 100 and the current falls to a hundredth, so the heating loss in the same cables falls to (1/100)² = 1/(10 000) of what it was.

A smaller current can reduce the conductor size needed for a given heating limit, although insulation, mechanical strength and cost also constrain cable design.

Check your understanding 1 Convert before dividing

You will compare 20 kW entering the cables at 200 V or 20 kV. Write the power in watts and the higher voltage in volts before using I = P/V.

Unit: W
Unit: V
Show answer

20 kW = 20 000 W and 20 kV = 20 000 V. Use the same unit system for both cases; the numerical equality here does not mean power and voltage are the same quantity.

Try it yourself 2

200 V or 20 kV?

Minimal support

Problem

A generator supplies 20 kW at the entrance to a cable pair whose total resistance is 0.50 Ω. Using the stated calculation model, compare the current and cable heating power at entry voltages (i) 200 V and (ii) 20 kV. Hold entry power and total cable resistance fixed, not the power delivered after the cable losses.

Find each current first, then square it for the loss

Calculate on paper or here. Check both currents and both cable losses; the complete calculation remains below.

Show the steps Optional working
Unit: W
Unit: A
Show the steps Optional working
Unit: W
Unit: V
Unit: A
Show the steps Optional working
Unit: A
Unit: W
Show the steps Optional working
Unit: A
Unit: W

Hints

Hint 1

current

Convert to watts and volts, then use I = P/V for each voltage.

Hint 2

loss

Use Pₗₒₛₛ = I²R with each current and the same 0.50 Ω.

Show solution step by step

  1. The currents

    Method

    Divide the power by each voltage.

    Reason

    The same power is sent each time, so I = P/V.

    Working

    Answer

    I₁ = (20 000 W)/(200 V) = 100 A I₂ = (20 000 W)/(20 000 V) = 1.0 A

  2. The losses

    Method

    Use I²R with each current.

    Reason

    The cables are the same, so R = 0.50 Ω both times.

    Working

    Answer

    P₁ = (100 A)² × 0.50 Ω = 5000 W P₂ = (1.0 A)² × 0.50 Ω = 0.50 W

  3. Compare

    Working

    Answer

    At 200 V the cables waste 5000 W, a quarter of the power sent. At 20 kV they waste only 0.50 W: the current is 1/100 as large, so the loss is 1/(10 000) as large.

Check your understanding 2 Account for the energy reaching the load

Use the losses you just found. From 20 000 W entering the cable pair, how much power reaches the load at 200 V entry, and at 20 kV entry? Assume these are the only losses. Keep one decimal place in this model subtraction to show the small loss at 20 kV.

Unit: W
Unit: W
Show answer

At 200 V entry, 20 000 − 5000 = 15 000 W reaches the load. At 20 kV entry, 20 000 − 0.50 = 19 999.5 W reaches it. Entry power is the same, while delivered power differs. In each case input equals delivered power plus cable heating.

Check your understanding 3

Two systems send the same power through cables of the same resistance. System B uses three times the voltage of system A. What fraction of A’s current, and of A’s cable loss, does B have? Enter each fraction as a decimal.

Show the steps Optional working
Show answer

For the same power, I = P/V, so B’s current is 1/3 of A’s. The loss is I²R, so B’s loss is (1/3)² = 1/9 of A’s.

From power station to home

  1. The generator at the power station produces A.C. at a moderate voltage.
  2. A step-up transformer raises the voltage to a very high value for the long-distance cables. For the same power, the current in the cables is now small, so little energy is lost as heat.
  3. Near the users, step-down transformers lower the voltage in stages to the mains value used by homes and appliances, giving the voltage needed by the appliances. The mains supply still presents a serious shock hazard.

In this A.C. transmission system, ordinary transformers provide the step-up and step-down stages because their operation needs a changing magnetic field; a steady D.C. supply does not give continuous transformer output. In an ideal transformer, power in equals power out, so the step-up and step-down stages change the voltage and current without losing power.

Check your understanding 4 Explain the complete transmission chain

Explain on paper or here why the system steps voltage up before a long cable run and down near users. Include current, cable heating and energy conservation, then compare the full chain below; a note is optional.

Show answer

For a given power entering the long cables, a step-up stage raises voltage and reduces current. The same cable resistance then gives a smaller I²R heating loss, so more of that entry power reaches users. Step-down stages near users provide the lower voltages required by local distribution and appliances; mains voltage remains hazardous. Ideal transformer stages conserve power, while real transformers and cables lose some energy as heat. Check both transformer roles, the current link, the squared-current loss and where energy is transferred. Step-up does not create power and step-down does not remove the need for electrical protection.

Check your understanding 5

A student says: “A higher voltage always means a bigger current, so transmitting at high voltage must increase the losses.” What has she missed?

Constraint in the comparison
Voltage across the cable resistance alone
Show answer

V = IR applies when a voltage is connected across a fixed resistance. In transmission, the power sent is fixed, and I = P/V, so raising the voltage reduces the current needed. A smaller current means a smaller I²R loss in the cables.

Check your understanding 6 A changed power requirement

System B sends four times A’s entry power at three times A’s entry voltage, through cables with the same resistance. Calculate B/A for current and cable heating power. Enter decimals.

Show the steps Optional working
Show answer

The current ratio is 4/3 ≈ 1.33. Cable heating is proportional to current squared, so its ratio is (4/3)² = 16/9 ≈ 1.78. Both exceed one: the higher power requirement makes B’s current and heating greater despite its higher voltage.

Check your understanding 7 When the required power also changes

System B sends four times the entry power of system A at three times the entry voltage, through the same cable resistance. Calculate the ratio of their currents and of their cable heating losses. Explain whether heating is lower in this new comparison. Write the full argument on paper or here, then compare below; notes are optional.

Show answer

The current ratio is

IB/IA = (PB/PA)/(VB/VA) = 4/3

and the heating-power ratio, at the same resistance, is

P_(loss,B)/P_(loss,A) = (4/3)² = 16/9

So B has larger current and greater cable heating despite its higher voltage. The power requirement has increased more than the voltage. Check both ratios, the squared-current step and the constraint that changed. The earlier lower-loss conclusion required the same entry power; it cannot be applied unchanged to this new comparison.

Common mistakes

  • Saying that a high voltage reduces the losses without mentioning the current. At fixed entry power and cable resistance, higher voltage reduces current and therefore the I²R loss.
  • Forgetting to square the current when comparing losses.
  • Using V = IR with the transmission voltage to find the cable current. Use I = P/V for the power being sent.
  • Saying transformers create power. An ideal transformer changes the voltage and current but not the power.
  • Forgetting the step-down transformers near the users.

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