Power Transmission with Transformers — Enrichment

Enrichment — beyond the explicitly named H2 Physics 9749 transformer outcome. This branch develops a useful application from the teacher anchor. The core syllabus requirement remains the simple iron-core transformer and ideal ratio.

Main topic: Transformers

Overview

Transformers make it convenient to transmit a fixed electrical power at high voltage and low current, then reduce the voltage before consumer use. The smaller line current greatly reduces resistive cable heating.

Core Ideas

• A step-up transformer is placed before the long transmission line.
• For fixed useful power, increasing voltage decreases current.
• Cable heating depends on $I_{\mathrm{rms}}^2{R}_{\mathrm{cable}}$, so the loss falls with the square of the current reduction.
• A step-down transformer reduces voltage after transmission; neither transformer creates extra power.

1. Where the transformers are used

A simplified transmission chain is:

1. an AC generator supplies electrical power;
2. a step-up transformer raises the transmission voltage;
3. cables carry power at high voltage and lower current;
4. step-down transformers reduce the voltage before use by consumers.

Figure: The step-up transformer is placed before the long transmission line, where reducing current matters most. A step-down transformer is placed after transmission so the consumer does not receive the very high line voltage. The transformers change voltage and current; they do not create the transmitted power.

2. Keep useful power and cable loss separate

In the simplified syllabus-style model, treat the useful transmitted power as fixed and use rms quantities:

$$P_{\mathrm{useful}}=V_{\mathrm{rms}}{I}_{\mathrm{rms}}.$$

This form assumes the model used in the question, such as a resistive or unity-power-factor load. Detailed AC power factor is beyond this branch.

For the resistive transmission cables,

$$P_{\mathrm{cable}\mathrm{loss}}=I_{\mathrm{rms}}^2{R}_{\mathrm{cable}}.$$

These are different powers:

• $P_{\mathrm{useful}}$ is the intended rate of energy transfer along the system;
• $P_{\mathrm{cable}\mathrm{loss}}$ is unwanted heating in the cable resistance.

3. Why raising voltage reduces loss

For fixed useful power,

$$I_{\mathrm{rms}}=\frac{P_{\mathrm{useful}}}{V_{\mathrm{rms}}}.$$

If transmission voltage is multiplied by a factor $k$, then current is divided by $k$:

$$V^{\prime }=kV⟹I^{\prime }=\frac{I}{k}.$$

Therefore,

$$P_{\mathrm{cable}\mathrm{loss}}^{\prime }={\left(\frac{I}{k}\right)}^{2}R=\frac{P_{\mathrm{cable}\mathrm{loss}}}{k^2}.$$

So multiplying the voltage by $10$ divides cable heating loss by $100$, provided useful transmitted power and cable resistance are unchanged.

Figure: Follow the causal chain, not merely the slogan “high voltage is efficient”. The comparison holds for the same useful power and the same cable resistance: higher voltage gives lower current, and the square in $I^{2}R$ makes the cable loss fall much more strongly.

4. Worked comparison

A system transmits $90\text{kW}$ through cables of total resistance $0.80\Omega$. Compare transmission at $3.0\text{kV}$ and $30\text{kV}$.

At $3.0\text{kV}$

$$I_1=\frac{90\times 10^3}{3.0\times 10^3}=30\text{A}.$$ $$P_{\mathrm{loss},1}=30^2(0.80)=720\text{W}.$$

At $30\text{kV}$

$$I_2=\frac{90\times 10^3}{30\times 10^3}=3.0\text{A}.$$ $$P_{\mathrm{loss},2}=3.0^2(0.80)=7.2\text{W}.$$

The voltage is ten times larger, the current is ten times smaller, and the cable loss is one hundred times smaller.

5. Common reasoning errors

• “Higher voltage reduces loss directly.” The causal link is: for fixed power, higher $V$ gives lower $I$; lower $I$ gives lower $I^{2}R$ loss.
• “The step-up transformer adds power.” An ideal transformer conserves power. It trades higher voltage for lower current.
• “Use the consumer current in the transmission cable.” Use the current in the section of cable whose loss is being calculated.
• “The cable loss is $VI$.” $VI$ describes the useful transmitted power in this simple model; cable heating is $I^2{R}_{\mathrm{cable}}$.
• “Voltage can be increased without limit.” Real systems face insulation, safety, clearance and equipment constraints; detailed engineering optimisation is beyond this branch.

Exam Relevance

This application is enrichment under the explicit 9749 transformer wording. If used in a school question or supplied context, keep useful transmitted power separate from cable loss and state the fixed-power assumption before using the $I=P/V$ and $I^{2}R$ chain.

Exam-style reasoning template

1. State that the useful transmitted power is fixed.
2. Use $I=P/V$ to show that a higher transmission voltage gives a lower rms current.
3. Use $P_{\mathrm{loss}}=I^{2}R$ to show that lower current reduces cable heating.
4. Conclude using the square dependence, not a vague statement about “less resistance”. The cable resistance has not been reduced in this comparison.

Links

• Transformers
• RMS and AC Power
• Alternating Current Generators