Recall and use the equation P = I^2 R to explain why power losses in cables are smaller when the voltage is greater

4.5.6 The Transformer

Learning objective

Recall and use the equation P = I²R to explain why power losses in transmission cables are much smaller when the transmission voltage is high.

1. Construction & terminology

• Core – provides a low‑reluctance path for the magnetic flux.

  • Laminated iron core – thin sheets of silicon‑steel insulated from each other; lamination reduces eddy‑current loss.
  • Solid iron or ferrite core – used in low‑power devices where losses are less critical.

• Primary winding (N_p) – connected to the supply voltage.
• Secondary winding (N_s) – delivers the transformed voltage to the load.
• Step‑up transformer – N_s > N_p, therefore V_s > V_p. Used at power stations to raise the generated voltage.
• Step‑down transformer – N_p > N_s, therefore V_s < V_p. Used near homes and factories to obtain a safe domestic voltage.

2. Ideal transformer relations

For an ideal (loss‑free) transformer the following equations hold:

\[

\frac{V{p}}{V{s}}=\frac{N{p}}{N{s}}, \qquad

\frac{I{s}}{I{p}}=\frac{N{p}}{N{s}}, \qquad

V{p}I{p}=V{s}I{s}

\]

The last relation shows that the input power equals the output power; consequently, increasing the turns on the secondary raises the voltage while reducing the current in proportion.

3. Principle of operation (mutual induction)

1. An alternating current in the primary creates a changing magnetic flux Φ in the core.
2. The changing flux links the secondary winding and induces an emf given by Faraday’s law:

   \[

   \mathcal{E}{s}= -N{s}\,\frac{d\Phi}{dt}

   \]

   The magnitude of the induced emf is directly proportional to the rate of change of flux (|dΦ/dt|).

3. The frequency of the induced emf is identical to the supply frequency; a transformer does not change frequency.

4. Loss mechanisms & efficiency

• I²R (winding) losses – resistive heating in the copper windings:

  \[

  P_{\text{winding}} = I^{2}R

  \]

• Core (iron) losses

  • Hysteresis loss – proportional to the frequency and to the maximum flux density raised to a material‑dependent exponent:

    \[

    P{\text{hyst}} \propto f\,B{\max }^{\,n}\quad (n\approx 1.6\text{–}2.0)

    \]

  • Eddy‑current loss – caused by circulating currents in the core material; reduced by lamination:

    \[

    P{\text{eddy}} \propto f^{2}\,B{\max }^{2}\,t^{2}

    \]

    where *t* is the lamination thickness.

• Leakage‑flux loss – a small fraction of the magnetic flux does not link both windings; it appears as a resistive‑like loss and is usually only a few per cent of the total loss.
• Protective devices – real transformers are fitted with safety equipment such as:

  • Buchholz relay (detects gas from oil decomposition)
  • Oil temperature and pressure sensors
  • Thermal overload relays

Overall efficiency is therefore

\[

\eta = \frac{P{\text{out}}}{P{\text{in}}}\times100\%

\]

with \( \eta < 100\% \) because of the losses listed above.

5. Why high transmission voltage reduces cable losses

Power that must be delivered to a load is

\[

P{\text{load}} = V{\text{trans}} I_{\text{trans}}

\]

Hence the current required for a given load power is

\[

I{\text{trans}} = \frac{P{\text{load}}}{V_{\text{trans}}}

\]

The resistive loss in a transmission cable of resistance \(R\) is

\[

P{\text{loss}} = I{\text{trans}}^{2} R = \left(\frac{P{\text{load}}}{V{\text{trans}}}\right)^{2} R

\]

Thus the loss varies with the square of the inverse of the transmission voltage. Doubling the voltage reduces the loss to one‑quarter.

Numerical example – realistic transmission values

Assume a load of 100 MW (typical for a small city) and a line resistance of 0.2 Ω per kilometre over a 200 km line (total \(R = 40\;Ω\)).

The table shows that raising the voltage from 10 kV to 400 kV reduces the cable loss from an impossible 4 000 MW to only 2.5 MW – a reduction of more than three orders of magnitude.

6. Practical application & safety

• Power‑station side – a large step‑up transformer raises the generator voltage (typically 10–30 kV) to transmission levels of 110 kV, 220 kV, 400 kV or higher.
• Transmission lines – high‑voltage, low‑current cables minimise \(I^{2}R\) losses and allow economical delivery over hundreds of kilometres.
• Consumer side – a step‑down transformer near the premises reduces the voltage to the standard domestic level (230 V in the UK, 240 V in many countries).
• Safety measures

  • Insulation, adequate clearances and reliable earthing are essential for high‑voltage equipment.
  • Transformers are protected by devices such as Buchholz relays, oil temperature/pressure sensors, and thermal overload relays.
  • Never touch exposed conductors; always follow lock‑out/tag‑out procedures before working on any equipment.

7. Summary checklist

• Power transmitted: P = VI.
• For a constant load power, current varies as I = P/V – higher V → lower I.
• Resistive loss in cables: P_loss = I²R (depends on the square of the current).
• Doubling the transmission voltage reduces I²R loss to one‑quarter; increasing voltage from 10 kV to 400 kV can cut losses by a factor of > 1 000.
• Ideal transformer relations: \(\displaystyle \frac{V{p}}{V{s}}=\frac{N{p}}{N{s}},\; \frac{I{s}}{I{p}}=\frac{N{p}}{N{s}},\; V{p}I{p}=V{s}I{s}\).
• Real transformers have:

  • I²R (winding) losses
  • Core losses – hysteresis \(\propto fB{\max }^{n}\) and eddy‑current \(\propto f^{2}B{\max }^{2}\)
  • Leakage‑flux losses (small percentage)

• Protective devices (e.g., Buchholz relay) are fitted to detect faults and prevent damage.
• High‑voltage transmission → low current → minimal energy waste; step‑down transformers make the voltage safe for domestic use.