AC Electrical Power Distribution (DP IB Physics: HL)

• 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:
  • P ∝ I²
  • 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

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

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