DP IB Physics: HL

Revision Notes

Syllabus Edition

First teaching 2014

Last exams 2024

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11.2.5 AC Electrical Power Distribution

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AC Electrical Power Distribution

  • Energy losses due to the heating of transmission lines in national power grids are significant
    • This is because the electrical energy is transmitted across long distances from power stations to buildings
  • Inefficiencies in a transformer appear not from just the core, but also in the wires

  • The coils of wire have resistance
    • This causes heat energy to be lost from the current flowing through the coils
    • The larger the current, the greater the amount of heat energy lost

  • In the core, the inefficiencies appear from:
    • Induced eddy currents
    • The reversal of magnetism
    • Poor insulation between the primary and secondary coil

  • Ways to reduce energy loss in a transformer are:
    • Making the core from soft iron or iron alloys to allow easy magnetisation and demagnetisation and reduce hysteresis loss
    • Laminating the core
    • Using thick wires, especially in the secondary coil of step-down transformers
    • Using a core that allows all the flux due to the primary coil to be linked to the secondary coil
  • Power losses from the current are calculated using the equation:

P = I2R

  • Where:
    • P = power (W)
    • I = current (A)
    • R = resistance (Ω)

  • The equation shows that:
    • PI2
    • This means doubling the current produces four times the power loss

  • Therefore, step-up transformers are used to increase the voltage which decreases the current through transmission lines
    • This reduces the overall heat energy lost in the wires during transmission

  • A step-down transformer is then used to decrease the voltage to that required in homes and buildings

Power Loss National Grid, downloadable AS & A Level Physics revision notes

The use of step-up and step-down transformers in the National Grid

Worked example

A current of 2500 A is transmitted through 150 km of cables. The resistance of the transmission cable is 0.15 Ω per km.

Calculate the power wasted.

Step 1: List the known quantities

    • Current, I = 2500 A
    • Length of cables, L = 150 km = 150 × 103 m
    • Resistance of the cables, R = 0.15 Ω km-1

Step 2: Write out the power equation

P = I2R

Step 3: Determine the total resistance, R

R = Resistance of the wires × Length of wires

R = 0.15 × 150

Step 4: Substitute values into the power equation

Power lost = (2500)2 × (0.15 × 150) = 141 MW

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Lindsay Gilmour

Author: Lindsay Gilmour

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.