# 16.2.2 Transformer

The transformer (despite its exciting sounding name) consists of just two coils of wires wrapped around a soft iron core.

The primary coil is to be connected to the power supply; the secondary coil is where we tap the output voltage or power. The circuit symbol for a transformer looks like this

To explain how the transformer works, we will be using Vp and Vs to denote the voltage across the primary and secondary coils respectively. Likewise, Np and Ns denote the number of turns in the primary and secondary coils respectively.

Open Circuit Operation

To keep things simple, let’s leave the secondary coil open for the time being. When a sinusoidal voltage Vp is applied, the primary coil is magnetized like an electromagnet. This results in an alternating magnetic flux f in the soft iron core.

Thanks to this alternating magnetic flux f, the primary coil now experiences a changing magnetic flux linkage Fp. Faraday’s Law of EMI dictates that an emf Ep is induced in the primary coil[1]:

$\displaystyle {{E}_{p}}=\frac{{d{{\Phi }_{p}}}}{{dt}}={{N}_{p}}\frac{{d\phi }}{{dt}}\text{ }\cdots \cdots \cdots (1)$

Since the secondary coil is wrapped around the same core, it is also experiencing the same alternating magnetic flux f. A similar emf Es is thus induced in the secondary coil:

$\displaystyle {{E}_{s}}=\frac{{d{{\Phi }_{s}}}}{{dt}}={{N}_{s}}\frac{{d\phi }}{{dt}}\text{ }\cdots \cdots \cdots (2)$

Dividing the equation (2) by equation (1), we obtain

$\displaystyle \frac{{{{E}_{s}}}}{{{{E}_{p}}}}=\frac{{{{N}_{s}}}}{{{{N}_{p}}}}$

If the primary coil has zero resistance, the self-induced emf Ep must match Vp exactly[2]. Similarly, if the secondary coil has zero resistance, Vs is exactly equal to Es.

Replacing Es and Ep with Vs and Vp respectively, we obtain

$\displaystyle \frac{{{{V}_{s}}}}{{{{V}_{p}}}}=\frac{{{{N}_{s}}}}{{{{N}_{p}}}}$

So there we have it: the voltage ratios $\displaystyle \frac{{{{V}_{s}}}}{{{{V}_{p}}}}$ follows the turns ratio $\displaystyle \frac{{{{N}_{s}}}}{{{{N}_{p}}}}$. A transformer with $\displaystyle \frac{{{{N}_{s}}}}{{{{N}_{p}}}}>1$ is a step-up transformer, since it transforms a lower voltage into a higher one. Conversely, a transformer with $\displaystyle \frac{{{{N}_{s}}}}{{{{N}_{p}}}}<1$ is a step-down transformer.

Let’s now connect a load R (i.e. a fixed resistor) to the secondary coil.

Closed Circuit Operation

With the circuit now closed, Vs is going to cause a current Is to flow in the secondary coil. Ohm’s Law applies. So

$\displaystyle {{I}_{s}}=\frac{{{{V}_{s}}}}{R}$

Once Is is flowing, a current Ip must also flow in the primary coil. Why?

Well, fundamentally it is because of something called mutual induction. But mutual induction is complicated and beyond the H2 syllabus[3]. Instead, you’re only expected to use the principle of conservation of energy to obtain the relationship between Ip and Is.

Firstly, we note that power is delivered to the load R when Is flows. This power obviously must ultimately come from the power supply connected to the primary coil. If the transformer is lossless, the power drawn from the supply must be equal to the power delivered to the load. So

${{V}_{p}}{{I}_{p}}={{V}_{s}}{{I}_{s}}$

From here, we can show that the current ratio is equal to the inverse of the turn ratio.

$\displaystyle \frac{{{{V}_{p}}}}{{{{V}_{s}}}}=\frac{{{{I}_{s}}}}{{{{I}_{p}}}}\text{ }\Rightarrow \text{ }\frac{{{{I}_{s}}}}{{{{I}_{p}}}}=\frac{{{{N}_{p}}}}{{{{N}_{s}}}}$

It is worth highlighting that a step-up transformer steps down the current (by the same factor it steps up the voltage), and vice versa.

Why Iron Core?

Before we end this section, let’s acknowledge the crucial role played by the soft iron core.

• Because soft iron has a very high permeability, the magnetic flux f (and thus $\displaystyle \frac{{d\phi }}{{dt}}$) experienced by the coils is increased dramatically. An air-core transformer has no meaningful induction because the f (and thus $\displaystyle \frac{{d\phi }}{{dt}}$) is too small.
• Because soft iron has so much higher permeability than air, all the magnetic flux f is completely kept within the core. (Just like electric current flows within a copper wire in an electric circuit, the magnetic flux “flows” within the soft-iron core in this magnetic circuit) This means that the magnetic flux by either coil is completely linked to the other coil. Zero flux leakage is crucial for the efficiency of the transformer.

Demonstrations

Turn Ratio (using the Lenz’s Jumping Ring demo set)

Lenz’s Hovering Ring

Concept Test

3223

Beyond Syllabus

Non-Conservative Field (using the Lenz’s Jumping Ring demo set)

[1] Do you realize that there is something about this induction that is different from the ones you have encountered in the EMI chapter? This coil is having an emf induced by its own changing magnetic flux. This is called self-induction, the detail of which is beyond the H2 syllabus.

[2] Do you realize that this self-induced emf Ep is preventing the primary coil from overheating? If not for Ep, the current through the primary coil would have been infinitely large since it has zero resistance and it is shorted across Vp. But because the induced Ep cancels out the supply voltage Vp (almost completely), the (open circuit) current in the primary coil is a negligibly small current. Often called the magnetizing current Im, this current is responsible for maintaining the alternating flux in the core, and should be differentiated from the additional current Ip that flows in the primary coil when a load is connected across the secondary coil.

[3] Because the two coils are magnetically connected through the core, the change in flux ΔΦ produced by Is induces Ip in the primary coil, which produces a corresponding change in flux to cancel out ΔΦ. There is zero delay between the onset of Is and the response of Ip, thus preserving the magnetising flux in the core, which explains why Vp and Vs are totally unaffected by Ip and Is. Never mind. As I said, this is not in the H2 syllabus.