Analyse the operation of ideal and real transformers, including the turns ratios V_s/V_p = N_s/N_p and I_p/I_s = N_s/N_p, energy losses, and the role of step-up and step-down transformers in AC power transmission

What this dot point is asking

NESA wants you to derive and apply the ideal-transformer voltage and current ratios from Faraday's law, recognise that an ideal transformer conserves power, identify the four main loss mechanisms in real transformers and how the design mitigates them, and explain why AC transmission relies on stepping voltage up for transmission and down for distribution.

The answer

How a transformer works

A transformer is two coils (the primary and secondary) wound on the same ferromagnetic core. An alternating current in the primary produces a changing flux in the core. The same changing flux links the secondary, inducing an EMF in it (Faraday's law).

Because both coils share the same flux $\Phi$ (in the ideal case):

$V_p = N_p \, d\Phi / dt$ and $V_s = N_s \, d\Phi / dt$.

Dividing:

The voltage ratio equals the turns ratio.

Power conservation and the current ratio

An ideal transformer has no losses. Power in equals power out:

Combining with the voltage ratio:

Currents are in the inverse ratio to voltages. A step-up transformer ($N_s > N_p$) raises the voltage and lowers the current; a step-down transformer ($N_s < N_p$) does the reverse.

Why transformers need AC

Faraday's law requires $d\Phi / dt \neq 0$ to induce a secondary EMF. A DC primary current produces a constant flux, so the secondary EMF is zero (except during the brief switch-on transient). AC, by changing direction many times per second, produces the continuous flux change required.

The four losses in a real transformer

Typical efficiencies: 95 percent for small transformers, above 99 percent for large grid transformers.

Step-up and step-down in AC transmission

Transmitting electrical power $P = VI$ over a long line of resistance $R_{\text{line}}$ wastes power as $P_{\text{loss}} = I^2 R_{\text{line}}$. The loss depends on the current squared, not the voltage. So for the same transmitted power $P$, a higher transmission voltage means a smaller current and dramatically smaller line losses.

Typical Australian grid:

1. Generation at a power station: about 11 to 25 kV from the generator.
2. Step-up transformer at the station raises this to 132, 220, 330, 500 kV or higher for long-distance transmission.
3. Transmission lines carry power at high voltage (low current, low $I^2 R$ loss).
4. Step-down transformer at a sub-transmission substation drops to 33 or 66 kV.
5. Distribution transformer at street level drops to 11 kV, then a final transformer drops to 415 V (three-phase) / 240 V (single-phase) for delivery to homes and businesses.

Without transformers, this voltage manipulation would not be possible, and long-distance AC transmission would lose most of the generated power as heat in the wires. This is why Tesla's AC system, with its easy transformer-based voltage conversion, won out over Edison's DC system in the 1890s. (Modern HVDC links exist today, but they require expensive electronic converters at each end.)

Worked example: transmission line saving

A small power station delivers $1.0$ MW to a town through a transmission line of total resistance $5.0$ ohms. Compare the line losses at $1000$ V transmission versus $100$ kV transmission.

At $V = 1000$ V:

$I = P / V = 10^6 / 10^3 = 1000$ A.

$P_{\text{loss}} = I^2 R = (1000)^2 \times 5 = 5.0 \times 10^6$ W = 5 MW.

The line cannot even deliver 1 MW: the losses exceed the power.

At $V = 100$ kV:

$I = 10^6 / 10^5 = 10$ A.

$P_{\text{loss}} = (10)^2 \times 5 = 500$ W.

The losses drop from 5 MW to 500 W, a factor of 10,000. This factor is exactly the square of the voltage ratio ($100^2 = 10\,000$), illustrating the $V^2$ dependence of transmission efficiency.

Worked example: turns ratio

A neon sign requires $5.0$ kV from a $240$ V mains supply. Find the turns ratio, the primary current when the sign draws $50$ mA, and the input power.

Turns ratio:

$N_s / N_p = V_s / V_p = 5000 / 240 \approx 20.8$.

Primary current:

$I_p = (V_s / V_p) I_s = 20.8 \times 0.050 = 1.04$ A.

Input power:

$P = V_p I_p = 240 \times 1.04 = 250$ W (equal to $V_s I_s = 5000 \times 0.050 = 250$ W, as expected for an ideal transformer).

Examples in context

Example 1. TransGrid Liddell step-up transformer. Snowy Hydro's generator output of $V_p = 20 \text{ kV}$ feeds a step-up transformer with $N_p = 200$ primary turns. To reach the TransGrid backbone at $V_s = 330 \text{ kV}$, the secondary needs $N_s = N_p \times V_s/V_p = 200 \times 330/20 = 3300$ turns. Power throughput is $P = 600 \text{ MW}$, so primary current is $I_p = P/V_p = 6.0 \times 10^8 / 2.0 \times 10^4 = 30{,}000 \text{ A}$, while secondary current is only $I_s = I_p N_p/N_s = 30{,}000 \times 200/3300 = 1818 \text{ A}$. The high-voltage low-current secondary feeds the transmission line.

Example 2. $I^2 R$ loss saving by stepping up to 330 kV. Suppose the Liddell-to-Sydney transmission line is $300 \text{ km}$ of aluminium conductor with total resistance $R = 6.0 \Omega$. At $20 \text{ kV}$, delivering $600 \text{ MW}$ requires $I = 30{,}000 \text{ A}$ and the line dissipates $I^2 R = (3.0 \times 10^4)^2 \times 6.0 = 5.4 \times 10^9 \text{ W} = 5400 \text{ MW}$ - more than the power being sent. At $330 \text{ kV}$, $I = 1818 \text{ A}$, so $I^2 R = 1818^2 \times 6.0 = 1.98 \times 10^7 \text{ W} = 19.8 \text{ MW}$, or just $3.3\%$ loss. Stepping up the voltage by a factor of $16.5$ cuts the loss by $16.5^2 = 272$.

Try this

Q1. Write the ideal-transformer voltage ratio and explain why transformers do not work on DC. [2 marks]

• Cue. $V_s/V_p = N_s/N_p$. DC produces no $d\Phi/dt$ in the iron core, so no induced EMF in the secondary.

Q2. A transformer steps $240 \text{ V}$ down to $12 \text{ V}$ with $400$ primary turns. (a) Find the number of secondary turns. (b) If the secondary delivers $5.0 \text{ A}$ to a load, find the primary current (ideal). [2+2 marks]

• Cue. (a) $N_s = 400 \times 12/240 = 20$. (b) $I_p = I_s N_s/N_p = 5.0 \times 20/400 = 0.25 \text{ A}$.

Q3. A $33 \text{ kV}$ rural feeder delivers $5.0 \text{ MW}$ over a line of resistance $4.0 \Omega$ (total return path). (a) Calculate the line current. (b) Calculate the $I^2 R$ loss as a percentage of the transmitted power. (c) State two real (non-ideal) energy losses in a transformer and how each is mitigated. [2+2+2 marks]

• Cue. (a) $I = P/V = 5.0 \times 10^6 / 3.3 \times 10^4 = 152 \text{ A}$. (b) $I^2 R = 152^2 \times 4 = 9.24 \times 10^4 \text{ W}$; $9.24 \times 10^4 / 5.0 \times 10^6 = 1.85\%$. (c) Eddy (laminated core), hysteresis (soft iron), copper $I^2 R$ (thick winding), flux leakage (closed core).