Ideal Transformer Principles

A transformer is an electrical device that transfers AC electrical energy from one circuit to another through electromagnetic induction. Transformers are critical in modern power grids because they allow power to be transmitted efficiently at very high voltages and then safely stepped down for consumer use.

An ideal transformer is a theoretical model that assumes no energy losses (no core losses, no copper losses), perfect magnetic coupling (no leakage flux), and infinite core permeability.

Turns Ratio and Voltage/Current Relationships

A transformer consists of two coils (windings) wrapped around a common magnetic core:

• The primary winding has $N_1$ turns and is connected to the source.
• The secondary winding has $N_2$ turns and is connected to the load.

The ratio of the turns on the secondary to the turns on the primary is defined as the turns ratio ($n$): $n = \frac{N_2}{N_1}$

For an ideal transformer, the voltage across the windings is directly proportional to the number of turns: $\frac{V_2}{V_1} = \frac{N_2}{N_1} = n$

Because an ideal transformer is 100% efficient, the power into the primary equals the power out of the secondary ($V_1 I_1 = V_2 I_2$). Therefore, the current is inversely proportional to the turns ratio: $\frac{I_1}{I_2} = \frac{N_2}{N_1} = n$

If $n > 1$, the transformer steps up the voltage. If $n < 1$, it steps down the voltage.

Impedance Reflection

A critical engineering application of transformers is impedance matching. A load impedance ($Z_L$) connected to the secondary winding will "appear" differently to the source connected to the primary winding.

The equivalent impedance seen at the primary winding ($Z_{in}$) is: $Z_{in} = \frac{V_1}{I_1}$

Substituting the voltage and current relationships: $Z_{in} = \frac{V_2 / n}{n I_2} = \frac{1}{n^2} \frac{V_2}{I_2}$

Since $V_2 / I_2 = Z_L$: $Z_{in} = \frac{Z_L}{n^2} = \left(\frac{N_1}{N_2}\right)^2 Z_L$

This means an ideal transformer changes the magnitude of an impedance by the square of the inverse turns ratio.

Worked Example

Problem: An ideal transformer with 500 primary turns and 50 secondary turns is connected to a 120 V AC source. A $4\Omega$ resistor is connected across the secondary. Calculate the secondary voltage, the secondary current, and the equivalent impedance seen by the source.

Solution:

1. Identify the parameters:
   • $N_1 = 500$, $N_2 = 50$, $V_1 = 120\text{ V}$, $Z_L = 4\Omega$
2. Calculate the turns ratio $n$: $n = \frac{N_2}{N_1} = \frac{50}{500} = 0.1$
3. Calculate the secondary voltage $V_2$: $V_2 = n \times V_1 = 0.1 \times 120 = 12\text{ V}$
4. Calculate the secondary current $I_2$: $I_2 = \frac{V_2}{Z_L} = \frac{12}{4} = 3\text{ A}$
5. Calculate the equivalent primary impedance $Z_{in}$: $Z_{in} = \frac{Z_L}{n^2} = \frac{4}{(0.1)^2} = \frac{4}{0.01} = 400\Omega$

Engineering Check

While the ideal model works excellently for basic circuit analysis, real transformers deviate from this behavior. Real transformers exhibit winding resistance (copper losses), core hysteresis and eddy currents (iron losses), and leakage inductance. These non-ideal factors cause voltage drops under load and generate heat, necessitating cooling systems in large power transformers.