Transformers are also used for safety. They allow us to separate users from direct current. The transformers we mostly use are induction-based devices that rely on an alternating current source (AC). On this page we'll see how they work using Faraday's laws.

Step-up transformers convert an incoming voltage to a higher voltage at the cost of decreasing the current to maintain the total power. Step-down transformers do the opposite: They reduce voltage and increase current.

A transformer is a passive device. It just sits there with no moving parts. It consists of a core of a ferromagnetic metal that may or may not be shaped like the one shown here. Often these are made of stacked metal plates. This stacking eliminates some of the loss of power that all transformers undergo.

Any transformer has two sides that we call primary (often denoted 1˚, the input side) and secondary (2˚, the output side).

On each side, insulated (that's important!) wires are wound around the core a known number of times. On the 1˚ side we call the number of windings $n_1$ and on the 2˚ side it's $n_2$. We place a potential (voltage) across the wires of the primary side and see how it changes on the secondary side. That's where the magic of electromagnetic induction kicks in, so let's look at that.

alternating current, in which the direction of the current changes periodically, as shown in this graph. The voltage is continually switching from positive to negative, thus driving the current in alternate directions. As we'll see below, the changing magnetic fields (B-fields) produced by AC circuits are perfect for inducing secondary currents in a nearby conductor.

Alternating current: V vs. time

In this graph, the voltage is ±120 V. In the U.S., the switching frequency of AC is 60 Hz or so.

Now let's replace the input current with an alternating current (AC) source. The current changes direction with some frequency (in the U.S. it's usually 60 Hz). That means that the direction of the B-field in the core is flipping back and forth at the same frequency. Now think about the 2˚ coils: They are experiencing a changing B-field, and Faraday's laws tell us that a current will be induced in those coils.

So as long as we use AC, the transformer essentially passes current through, but it is not flowing through the core. But that's not all, because the primary and secondary voltages are related to the number of windings around each side. Thus we can increase the voltage from 1˚ to 2˚ or decrease it, with decreases/increases in current, respectively. The law of conservation of energy tells us that the power is conserved from side to side. Recall that power is just energy divided by time. So as voltage drops, current must rise, and vice versa, to keep the power the same.

Because

$$P = IV,$$

if the potential increases (a "step-up" transformer), then the current has to decrease in order to preserve the power: $V \uparrow \cdot I \downarrow = P$. If the potential decreases from primary to secondary windings, then the current must increase: $V \downarrow \cdot I \uparrow = P$.

We can rearrange the transformer equation to determine the action of a transformer on the input, or to design a transformer to do what we want.

We get the current from conservation of the power across the transformer:

$$P = IV \; \rightarrow \; I = \frac{P}{V} = \frac{12000 \, W}{528 \, V} = 22.7 \, \text{A}$$

For comparison, the current on the primary side was

$$I = \frac{P}{V} = \frac{12000 \, W}{1200 \, V} = 10 \, \text{A}$$

12,000 W is the maximum power we'd expect on the output side of such a transformer. There are always some small losses due to a number of factors, including just the loss of some energy to heat.

The total power is $P = IV = (1 \, A)(50,000 \, V) = 50,000 \; W$.

So the current on the output side will be quite large:

$$I = \frac{P}{V} = \frac{50,000 \, W}{5050 \, V} = 9.9 \, \text{A}$$