Describe the use of transformers in high-voltage transmission of electricity

4.5.6 The Transformer

Objective

Describe the use of transformers in high‑voltage transmission of electricity.

1. Why High‑Voltage Transmission?

• Power loss in a transmission line is given by
  \$P_{\text{loss}} = I^{2}R\$ where I is the current and R the resistance of the line.
• For a given transmitted power P, the current is
  \$I = \frac{P}{V}\$.
• Increasing the transmission voltage V reduces the current I, and therefore reduces the \$I^{2}R\$ losses.
• Higher voltage also allows thinner, cheaper conductors for the same power rating.

2. How a Transformer Works

A transformer transfers electrical energy between two circuits through electromagnetic induction. It consists of:

• Primary winding – connected to the source.
• Secondary winding – connected to the load.
• Core – usually laminated iron to guide the magnetic flux.

When an alternating current flows in the primary winding, a changing magnetic flux \$\Phi\$ is produced in the core. This flux links the secondary winding and induces an emf according to Faraday’s law.

3. Transformer Equations

For an ideal transformer (no losses):

\$\frac{V{\text{p}}}{V{\text{s}}} = \frac{N{\text{p}}}{N{\text{s}}}\$

where \$V{\text{p}}\$, \$V{\text{s}}\$ are the primary and secondary voltages, and \$N{\text{p}}\$, \$N{\text{s}}\$ are the numbers of turns in the primary and secondary windings respectively.

Power is conserved (ignoring losses):

\$P{\text{in}} = P{\text{out}} \quad\Longrightarrow\quad V{\text{p}} I{\text{p}} = V{\text{s}} I{\text{s}}\$

4. Types of Transformers Used in Transmission

5. Example Calculation

Suppose a power station generates 500 MW at 20 kV. To transmit this power efficiently, a step‑up transformer raises the voltage to 400 kV.

1. Calculate the current before stepping up:

   \$I{\text{low}} = \frac{P}{V{\text{low}}} = \frac{500\times10^{6}\,\text{W}}{20\times10^{3}\,\text{V}} = 25\,000\ \text{A}\$

2. Current after stepping up:

   \$I{\text{high}} = \frac{P}{V{\text{high}}} = \frac{500\times10^{6}}{400\times10^{3}} = 1\,250\ \text{A}\$

3. Assuming a line resistance of 0.05 Ω, compare losses:

   • Low‑voltage loss: \$P{\text{loss,low}} = I{\text{low}}^{2}R = (25\,000)^{2}\times0.05 = 31.25\ \text{MW}\$
   • High‑voltage loss: \$P_{\text{loss,high}} = (1\,250)^{2}\times0.05 = 0.078\ \text{MW}\$

The high‑voltage transmission reduces the power loss by a factor of about 400, illustrating why transformers are essential.

6. Efficiency and Real‑World Considerations

• Real transformers have losses: copper (I²R) loss, core (hysteresis and eddy‑current) loss.
• Typical efficiency for large power‑transmission transformers is 98–99 %.
• Cooling methods (oil‑filled, forced air) are used to keep temperatures low.
• Voltage regulation devices (tap changers) maintain constant output despite load variations.

7. Safety and Environmental Aspects

• High‑voltage equipment is insulated and grounded to protect personnel.
• Transformers are housed in substations with clear signage and barriers.
• Oil‑filled transformers require spill containment measures.
• Noise from cooling fans and humming can be mitigated with acoustic enclosures.

8. Summary

Transformers enable the efficient transmission of electrical power over long distances by stepping up voltage (reducing current and \$I^{2}R\$ losses) and stepping it down for safe distribution to consumers. Understanding the relationship between turns ratio, voltage, current, and power is fundamental for solving IGCSE physics problems on electricity transmission.