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

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – The Transformer

4.5.6 The Transformer

Learning Objective

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

Key Concepts

  • A transformer changes alternating current (AC) voltage levels while keeping the frequency constant.

  • Power transmitted, \$P{\text{out}}\$, is approximately the same as the power supplied, \$P{\text{in}}\$, neglecting small losses.

  • For a given transmitted power, increasing the voltage reduces the current: \$P = VI \;\Rightarrow\; I = \dfrac{P}{V}\$.

  • Resistive (I²R) losses in the transmission cables depend on the square of the current.

Why High \cdot oltage Reduces Cable Losses

Consider a transmission system that must deliver a power \$P_{\text{load}}\$ to a distant load.

  1. Using the power relation \$P = VI\$, the current required for a given power is inversely proportional to the voltage:

    \$I = \frac{P{\text{load}}}{V{\text{trans}}}\$

  2. The resistive loss in a cable of resistance \$R\$ is:

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

  3. From the expression above, if the transmission voltage \$V_{\text{trans}}\$ is doubled, the current is halved and the loss becomes one‑quarter of its original value:

    \$\frac{P{\text{loss,new}}}{P{\text{loss,old}}}= \left(\frac{V{\text{old}}}{V{\text{new}}}\right)^{2} = \left(\frac{1}{2}\right)^{2}= \frac{1}{4}\$

  4. Thus, higher transmission voltage → lower current → dramatically smaller \$I^{2}R\$ losses.

Typical Transformer Arrangement for Power Transmission

Suggested diagram: A step‑up transformer at the power station, high‑voltage transmission lines, and a step‑down transformer near the consumer.

Numerical Example

Assume a load requires \$P_{\text{load}} = 10\ \text{kW}\$ and the transmission line has a resistance \$R = 0.5\ \Omega\$.

Transmission \cdot oltage (V)Current (A)Power Loss \$I^{2}R\$ (W)
10 kV\$\displaystyle I = \frac{10\,000}{10\,000}=1.0\$\$\displaystyle P_{\text{loss}} = (1.0)^{2}\times0.5 = 0.5\$
100 kV\$\displaystyle I = \frac{10\,000}{100\,000}=0.10\$\$\displaystyle P_{\text{loss}} = (0.10)^{2}\times0.5 = 0.005\$

Increasing the voltage from 10 k \cdot to 100 k \cdot reduces the cable loss from 0.5 W to 0.005 W – a factor of 100, illustrating the \$V^{2}\$ advantage.

Summary Checklist

  • Power transmitted is \$P = VI\$.

  • For constant \$P\$, \$I = P/V\$ – current falls as voltage rises.

  • Resistive loss \$P_{\text{loss}} = I^{2}R\$ depends on the square of the current.

  • Doubling transmission voltage reduces \$I^{2}R\$ loss to one‑quarter.

  • Transformers enable high‑voltage, low‑current transmission, minimizing energy waste.