Capacitors, Magnetic Circuits, and Transformers is a free introductory textbook on the physics of capacitors, coils, and transformers. See the editorial for more information....

# Impedance Ratio

In many problems that involve transformers it is convenient to refer the impedance in one side of the transformer to the other side of the transformer.

 Figure 6-5. Equivalent circuits of (a) ideal transformer and connected load impedance; (b) load impedance referred to primary side of transformer.

For example, Fig. 6-5(a) shows an ideal transformer with a load impedance of ZL ohms connected across its secondary terminals. The load impedance referred to the primary is the value of impedance that, if connected directly across the source of voltage V1, would draw the same value of current I1 as the transformer with its connected load impedance ZL. This is shown in Fig. 6-5(b).

When Eq. 6-14 is divided by Eq. 6-17 the result is

or

 [6-18]

But the load impedance is

 [6-19]

and an impedance that would draw a current of I1 amp when connected directly across the source of voltage of V1 v must have a value of

 [6-20]

A comparison of Eqs. 6-18, 6-19, and 6-20 shows that

 [6-21]

where Z1 is the value of the load impedance referred to the primary of the transformer. The impedance ratio is, therefore

 [6-22]

or the square of the turns ratio of the transformer.

The turns ratio of a transformer is sometimes represented by the letter a, which means that

 [6-23]

and the impedance ratio is

 [6-24]

 Example 6-1: An electronic amplifier supplies the voice coil of a loudspeaker through a 20-to-1-ratio output transformer. The impedance of the voice coil is 9.6 ohms. Assume the transformer to be ideal and determine the value of the impedance that the transformer and loud-speaker place across the output of the amplifier. Solution: Turns ratio a = N1/N2, = 20. Impedance ratio = a2 = 400 Z1 = a2ZL = 400 x 9.6 = 3840 ohms.

Last Update: 2011-01-15