What are transformers?

You have probably built an electromagnet before by attaching a battery to a wire wrapped around a metal core.  Transformers operate on the same basic principle that when current passes through a wire, it creates a magnetic field around the wire (see Figure 1).  The basic structure for a transformer is shown in Figure 2.  Two wires are wound around an iron core.  The winding attached to the source is called the primary winding while the winding attached to the load is called the secondary winding.


Transformers are used to step up or down the voltage (or current) from one level to another.  Unlike amplifiers, power is conserved from the primary to the secondary.  This means that a transformer that increases the voltage from the primary to the secondary must compensate by lowering the current from the primary to the secondary (and vice versa).  You can find transformers in those “wall-wart” power supplies that eat up space on surge protectors.

You might think that because the two windings are not electrically connected, they do not interact.  However, transformers work because of the connection between magnetic and electric fields.  Transformers use changing magnetic fields to transfer energy from one circuit to another.  Changing current in the primary circuit produces a change in the magnetic field.  The changing magnetic field then induces a voltage in the secondary circuit.  Because a changing current is necessary to produce a changing magnetic field, transformers require an AC signal in order to work.  DC signals are not transferred through a transformer.

Notice how the two voltages in Figure 2 have the same polarity.  For the primary and secondary’s voltages to match up, their windings must be wrapped in opposite directions.  As you move around the core, you can see that the primary winding is wrapped counterclockwise while the secondary winding is wrapped clockwise.

How to analyze transformers

Transformers are fairly easy to analyze.  Recall that the power in the secondary is the same as the power in the primary and that power is equal to the product of voltage and current.  If the power is conserved, the voltage can be altered without wasting power.

P_{1} = P_{2} \rightarrow I_{1} V_{1} = I_{2} V_{2}

Using the equation above, we can step up the voltage from the primary to the secondary.  Doing so requires the current on the secondary to be proportionally smaller to ensure that the power is the same on both sides.  Likewise, a step down in voltage from the primary to the secondary will create a proportional step up in current from the primary to the secondary.

Voltage transformation depends on the number of windings on each side of the transformer.  The ratio of the number of turns on the primary to the number of turns on the secondary is known as the turns ratio (N).  The turns ratio is directly proportional to the ratio of the voltage on the primary to the voltage on the secondary.

N = \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}

Remember that power must be preserved from the primary to the secondary.  A voltage increase from the primary to the secondary must be balanced with a proportional drop in current.  This is reflected in the relationship between the turns ratio and the current ratio.  Whereas the voltage ratio was directly proportional to the turns ratio, the current ratio is inversely proportional to the turns ratio.

N = \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}}


Example: Find the secondary voltage and both the primary and secondary currents for the circuit in Figure 3.

Solution: Identify all of the given variables and plug them into the equation above.  In this instance, the given variables are V1 = 120 VRMS and N = 12:1.

N = \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}} \rightarrow N = \frac{12}{1} = \frac{120}{V_{2}} = \frac{I_{2}}{I_{1}}

Using this equation, solve for V2.

N = \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} \rightarrow \frac{12}{1} = \frac{120}{V_{2}} \rightarrow \frac{I_{2}}{I_{1}}

Once V2 is known, use Ohm’s law to find I2.

V_{2} = I_{2} R_{2} \rightarrow I_{2} = \frac{I_{2}}{R_{2}} = \frac{10V}{1k \Omega} = 10mA

Plug in these new values into the transformer equations and solve for I1.

N = \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} = \frac{I_{2}}{I_{1}} \rightarrow \frac{12}{1} = \frac{120V}{10V} = \frac{10mA}{0.83mA}

If all calculations are correct, power should be preserved from the primary to the secondary.  That is, the power delivered by the source should be equal to the power dissipated by the load.

P_{1} = I_{1} V_{1} = 0.83 mA \cdot 120V = 100mW (delivered)

P_{2} = I_{2} V_{2} = 10 mA \cdot 10V = 100mW (dissipated)

Transformers are way more exciting in this form than in the movie, wouldn’t you all say?  I think so.  Thanks to Ryan Eatinger ( for contribution of the post.

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