# Transformers via Inductors…

Now that we’ve had a brief, serious walkthrough over current & inductors, transformers wouldn’t be a big trouble. Unfortunately, today’s demonstration can’t be done using DC. Why? Because DC has got only two amplitudes (waveform’s maximum height, which is the value of voltage), either positive or zero. In  order for the transformers to function, you’d need a continuously varying voltage, which is happily, AC. We’ll get into it soon. For now, recall our last discussion. Well, Brady has done a nice video on Sixty Symbols addressing inductors. It’s worth recalling via that, instead of digging into my boring post again…

## Faraday’s experiment (this time, two coils)…

So, our magnetic fella “Faraday” did another experiment to demonstrate electromagnetic induction. This time, he made use of two coils (let’s name it 1 & 2). I don’t think I need to sketch that (as it’s pretty much easy to imagine). He took the first coil and connected it to a battery. Then, the second coil to a galvanometer and placed it closer to the first one such that the coils have a common central axis. At first look, it seems like both the circuits have no influence on each other.

But, something terrible happens. When he closes the first circuit, the galvanometer connected to the second coil shows a sudden deflection. It’s just a momentary deflection. Beeeeeng to some extent towards one end, and the pointer comes back to mean position again. Then, he opens the circuit. Now, the galvanometer shows the same momentary deflection in the opposite direction (like a recoil). Today, we have a fancy name for this property. It’s called mutual induction and this kind of “coupling” of inductors is what’s gonna be helpful for us to look into a transformer.

Okay, now what was all that beeeng & beeeng? Last time, I told you that the back emf induced, depends on the amount of magnetic flux linked to the coil (and also, the number of turns) and that it opposes the current that feeds it. Keep in mind that a current-carrying coil is just an electromagnet. It produces its own field. The more the current, the more the field intensity. It’s also worth noting that the more the change in current, the more the induced back emf opposes it. When Faraday first closed the circuit, current flows through it and a magnetic field is produced along the middle of the coil. Also, there’s a change (initially, there was no current, but now, current flows). The other coil is still in this field, and it can smell the change. Based on its distance (i.e) how much it’s influenced by the field, an emf is induced and this drives the current through the other circuit.

For the purpose of our discussion, let’s write his law. It’s silly to shorten it. I’ll expand…

$V=-N$ $\frac{d\phi}{dt}$

$V=-NA$ $\frac{dB}{dt}$

Here, $B$ is the magnetic field intensity and $A$ is the area enclosed by the coil. Was that really necessary now? Yep. If we proceed with the first expression, you may not be aware that the flux is nothing but the magnetic lines of force crossing through a given area. Or the other way around, magnetic field intensity is flux per unit area. Okay, that’s enough.

## Just a couple of inductors…

A transformer is nothing but a couple of inductors wound on a common soft iron core (probably, laminated steel to minimize eddy currents – another fancy name for the current driven by the back-emf which forms closed loops on metal plates). The usage of a ferromagnetic cores in coils is more or less the same reason why dielectrics are used for capacitors. It multiplies the field intensity by a factor of a few thousand (or more, say “million”) times. How? For now, let’s stick to the idea that electrons inside atoms are more like spinning tops. Their spin is either upward or downward (based on some classical rule). I should also confuse you by telling that the spin of a quantum particle isn’t very well defined. The alignment of the electron spins is what gives rise to a magnetic field in the macroscopic scale. In ferromagnets like iron, a small magnetic field is far enough to align the spin of electrons. Once they’re aligned, the electrons produce their own magnetic field and thereby, multiply the intensity by a tremendous magnitude. So, that explains the usage of iron cores in electromagnets, inductors, transformers, etc.

Bring the picture… This thing here is the sketch of a transformer (I’m not kidding you, I’m explaining all these mysteries and along the way, I should show you the picture of it)

It’s quite easy to explain a core-type transformer (shown above) though the underlying principle involved in both shell & core types are essentially the same. What do you see here? Two coils are wound on a core, and names are given to the voltage, current and turns of both the coils (primary & secondary). Duh…

## Stepping up & down

Before we get deeper into the subject, it’s necessary to skim through power. Electrical power is crudely spoken out by a lot of people as $P=VI$. What does that mean? It requires our definition of voltage and current. $V$, the voltage is the amount of energy obtained by a free electron in the applied electric field. Current $I$ is the rate of flow of free electrons (i.e) the number of stuff that goes within a second. The product of both the terms simply gives the rate of flow of energy, which is power.

Here comes a common misconception. This question comes to me often, “Why doesn’t Ohm’s law $V=IR$ hold good for transformers?” It doesn’t. Because, the law is based on a number of assumptions like steady current should flow through the conductor, temperature should be constant, etc. So, it fails in cases like diodes, superconductors, etc. It’s just like Hooke’s law for elasticity which is also merely an approximation. While this has nothing to do with transformers, I said this because I wanna make it clear that it’s an approximation and not some kind of universal axiom or something…

What’s the issue with transformers then? Power is a constant here. Whenever people think of a current-carrying conductor and apply Ohm’s law, they’re crudely assuming that they have sufficient supply of power. If 12kW power comes in through the primary, the same 12kW should go out through the secondary. You can’t create energy out of nothing. What the transformer does by changing the number of windings? It’s just redistributing the energy. “A step-down transformer decreases voltage” means that the thing catches a large number of electrons of some energy over a period of time and transfers it to a relatively small number of electrons, and kicks them out within the same period of time (so that each have a larger quantity of energy than the former). A larger number of turns means that it covers a larger area and the magnetic field is generated by a large bulk of electrons over the primary. In the secondary, this magnetic flux is linked to a smaller area (i.e) less electrons. Now, you have to squeeze the hell lot of energy obtained from the bulk electrons into these thinner ones. It comes to the same point again & again. You get some-some from a lot and give a lot to the some-some. A step-up transformer does it the other way around.

What I said above can be written in this expression… (“P” stands for primary, “S” stands for secondary)

$\frac{V_P}{V_S}=\frac{N_P}{N_S}=\frac{I_S}{I_P}$

Note that voltage is directly proportional to the number of turns (if you recall Lenz’ correction to Faraday) while current varies as inverse (if you recall power $P=VI$, where voltage should increase as you decrease current in order for power to remain constant)

## Where’s the up-down useful?

Simple. Transmission of power over long distances. Electric power generated from the water or steam turbine of something (say, nuclear reactor, thermal plant, etc.) should get to the national power grid, then the local grid and finally distributed to our homes. From the reactor, it’s stepped up by a transformer to around a million volts. When it reaches our home, it’s barely 230V. What’s going on here, and why should this be done?

We have not yet succeeded in power transmission using superconductivity. I mean, there’s no “thing” which has zero resistivity. Even the wires offer resistance. When there’s resistance to the flow of electrons, the voltage (their energy) is dissipated as heat. A guy named “Joule” found this effect. He formulated a law based on his observations, that the heat produced is proportional to the power (i.e) invoking Ohm’s law to include resistance,

$Q\propto VI\implies Q\propto I^2 R$

The “proportional” term is used because we’ve to consider a lot of constraints such as distance, temperature, etc. But, they aren’t necessary now. I remember the wonderful example given in our high school (probably the only example for this concept). Say, you’re to transmit 120,000 watts of power at 1200 volts and 100 amperes, and the wires offer a resistance of about 5 ohms, the rate of heat energy dissipated is simply,

$Q\propto (100)^2 \times 5$

$\mathrm{Power\ loss\propto 50,000\ W}$ which isn’t good…

In such a case, about 40% of the power is wasted as heat. Moreover, you’d need thicker wires (thin wires would melt down) and stronger posts (to hold them) to transmit such a large quantity of current. Now, consider the situation with same power, but transmitted at 60,000 V and 2 amperes. The power lost is just 20 W, which is around 4% of initial power.

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You may well ask, “Hey..!!! I can substitute $V=IR$ and write the equation as $Q\propto V^2/R$. Now, the power loss for the second case would be greater…”

Hmm… Yeah, go on with your calculation and you’d get a power loss of 720 million watts. Not so lucky, It goes like HELL..!!! Now you ask me, “Where did I go wrong?” In your substitution of Ohm’s law. It’s often a misconception. The power $VI$ makes use of the voltage provided by the source, whereas the power loss (i.e) heat dissipated (Ohm’s law itself) makes use of the voltage drop across the wire (i.e) the voltage drops according to the resistance. Let’s see…

Say you’ve got a power source that outputs 101 W. It’s connected to a load resistor of 100 ohms through a wire of resistance 1 ohm. The power dissipated by the resistance is 100 W, whereas the power lost due to wire is just 1 watt. No matter how many resistances you connect in the circuit, the voltage drops accordingly. Well, if you measure the voltage drop across the load, it shows 100 volts. Apply the same treatment to your calculation. You’ll conclude that the voltage drop across the wires is just 10 V. Henceforth, the power loss is again, 20 W. Now you say, “Damn… Physics..!!!”