It would be difficult to think about an electrified world — one in which we generate lots of power in a few locations and send it over wires to homes and businesses in cities and towns — without this amazing, passive device.

**Transformers** convert potential (voltage) and current to higher/lower levels while maintaining the total electric power $(P = IV)$. They are indispensable for moving large amounts of electric power from place to place with minimal loss to the (albeit small) resistance of wires.

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.

Here's a look at the basic anatomy of a simple transformer.

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.

Electric current in commonly used in two ways, **direct current (DC)** and **alternating current (AC)** Here is a graph of a direct current over time, this one at a potential of 50 V. The potential and current remain constant over time — boring, but useful in many situations.

When it comes to induction, DC is of limited use, because any magnetic field it produces is a static field, and therefore can't induce a current in a conductor.

Compare that to ...

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.

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

The key thing to notice about transformers is

There is no direct path for current to flow from the primary to the secondary side.

While current could flow through the core material, that's not how transformers work. The winding wires on either side are insulated and not meant to be in electrical contact with the core. When that happens, we say that the transformer has a **short**. It could even be dangerous in that situation.

Consider this transformer in which a direct current (DC) source is hooked up the primary wires. Faraday's laws tell us that the current in the coils will induce a magnetic field (B-field) in the core. Now look at the secondary side. Those coils of wire are subject to a *static* magnetic field, thus a current is *not* induced in them. We might just as well call this an electromagnet with some other wires dangling from it; it's a dead end for current.

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.

The potential difference between the input and output sides of a transformer is a function of the number of windings on each side. The transformer equation is

$$\frac{V_p}{V_s} = \frac{n_p}{n_s},$$

where the $V$ are the voltages on the primary and secondary sides, and the $n$'s are the numbers of windings on the corresponding sides.

The principal of conservation of energy tells us that we can't lose or create energy during the transformation process, and we express this as a conservation of electrical **power** (energy divided by time).

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.

The input (primary, 1˚) and output (secondary, 2˚) potentials (voltages) on either side of a transformer are related to the number of wire windings on each side by the transformer equation:

$$\frac{V_p}{V_s} = \frac{n_p}{n_s},$$

where $V_p$ and $V_s$ are the primary and secondary voltages and $n_p$ & $n_s$ are the numbers of wire windings on the primary and secondary sides.

A certain transformer has 1,000 windings on the primary side and 440 on the secondary. If 12,000 Watts of power at a voltage of 1,200 V is applied to the primary side, calculate the maximum voltage and current on the secondary side.

**Solution**

$$ \begin{align} \frac{V_p}{V_s} &= \frac{n_p}{n_s} \\[5pt] V_s &= \frac{n_s V_p}{n_p} \\[5pt] &= \frac{440 \cdot 1200}{1000} \\[5pt] &= 528 \, \text{V} \end{align}$$

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.

Let's say we want to design a transformer to convert electricity at 50,000 V (50 KV) and 1.0 A to 5050 V. If the transformer is to have 980 windings on the primary side, how many do we need on the secondary side? What will the current be on the secondary side?

**Solution**

$$ \begin{align} \frac{V_p}{V_s} &= \frac{n_p}{n_s} \\[5pt] n_s &= \frac{n_p V_s}{V_p} \\[5pt] &= \frac{980 \cdot 5050}{50,000} \\[5pt] &= 99 \, \text{windings} \end{align}$$

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:

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

X
### qualitative / quantitative

A **qualitative** description of a phenomenon uses words and ideas, like "The rock was heavy." A **quantitative** description involves numbers, usually of a precision appropriate to the thing being described: "The mass of the rock was 22.4 Kg."

X
*ferromagnetic*

A **ferromagnetic** material is one that can be magnetized by another magnetic field. Ferromagnetic materials are those to which permanent magnets will stick. Contrast these to magnetic or nonmagnetic materials.

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