Explain the principle of operation of a simple iron‑cored transformer and describe its construction, performance relationships, real‑world uses and safety considerations (Cambridge IGCSE Physics 0625).
It is built from thin, insulated laminations; the insulation breaks up circulating eddy currents, thereby reducing core (iron) losses and keeping the transformer cooler.

When an alternating current flows in the primary winding it creates a time‑varying magnetic flux Φ in the core. By Faraday’s law a changing flux that links a coil induces an electromotive force (emf) in that coil.
For a coil of N turns:
\[
\mathcal{E} = -N\frac{d\Phi}{dt}
\]
Because the same core flux links both windings, the induced emfs are
\[
\mathcal{E}p = -Np\frac{d\Phi}{dt}, \qquad
\mathcal{E}s = -Ns\frac{d\Phi}{dt}
\]
Dividing the two equations gives the turns‑ratio relationship:
\[
\frac{Vs}{Vp}= \frac{Ns}{Np}
\]
From this we can define the two practical cases:
Given:
\[
Vs = Vp\frac{Ns}{Np}=240\ \text{V}\times\frac{200}{500}=96\ \text{V (rms)}
\]
\[
Ip = Is\frac{Ns}{Np}=2\ \text{A}\times\frac{200}{500}=0.8\ \text{A}
\]
\[
P{\text{in}} = Vp I_p = 240\ \text{V}\times0.8\ \text{A}=192\ \text{W}
\]
\[
P{\text{out}} = Vs I_s = 96\ \text{V}\times2\ \text{A}=192\ \text{W}
\]
(Both are equal because the transformer is assumed ideal.)
| Loss Type | Cause | Typical Effect on Performance |
|---|---|---|
| Core (iron) losses | Hysteresis + eddy currents (suppressed by laminated core) | Consumes a few % of input power; produces heat in the core. |
| Copper (I²R) losses | Resistance of the windings | Proportional to the square of the current; appears as heat in the windings. |
| Leakage flux | Flux that links only one winding | Reduces voltage regulation; limits the maximum transferable power. |
Quantitative illustration: A 100 W transformer with a total loss of 3 % delivers
\[
P_{\text{out}} = 100\ \text{W}\times(1-0.03)=97\ \text{W}
\]
When a load is connected, the secondary voltage falls slightly because of winding resistance and leakage flux. Regulation is defined as
\[
\text{Regulation (\%)} = \frac{V{\text{no‑load}}-V{\text{full‑load}}}{V_{\text{full‑load}}}\times 100\%
\]
Worked example: A transformer has \(V{\text{no‑load}} = 240\ \text{V}\) and \(V{\text{full‑load}} = 230\ \text{V}\).
\[
\text{Regulation} = \frac{240-230}{230}\times100\% \approx 4.3\%
\]
Thus the voltage drops by about 4 % under full load.
| Application | Purpose (why a transformer is needed) |
|---|---|
| Step‑up transformers in high‑voltage power transmission | Increase voltage to reduce I²R losses in long cables, allowing efficient bulk power transport. |
| Step‑down transformers in domestic lighting and appliances (e.g., 240 V → 12 V) | Provide safe low‑voltage supply for low‑power devices and LED lamps. |
| Power supplies for electronic devices (phone chargers, TV sets) | Convert mains voltage to a lower, regulated voltage suitable for sensitive electronics. |
| Isolation transformers in medical equipment | Separate the user‑accessible circuit from the mains for safety, preventing electric shock. |
| Run | Primary voltage (V rms) | Primary turns (\(N_p\)) | Secondary turns (\(N_s\)) | Coil separation (cm) | Secondary voltage (V rms) |
|---|---|---|---|---|---|
| 1 | |||||
| 2 | |||||
| 3 |
| Parameter | Ideal Transformer | Real Transformer |
|---|---|---|
| Voltage ratio | \(\displaystyle \frac{Vs}{Vp}= \frac{Ns}{Np}\) | Same, but small deviation due to regulation and losses. |
| Current ratio | \(\displaystyle \frac{Is}{Ip}= \frac{Np}{Ns}\) | Same, plus extra current to supply copper losses. |
| Power | \(Vp Ip = Vs Is\) (100 % efficient) | Input power > output power; difference appears as heat (core + copper losses). |
| Efficiency | 100 % | Typically 95 %–99 % (depends on size, frequency, core material). |
| Frequency | Unchanged (same as supply) | Unchanged – the transformer does not alter frequency. |
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