# 6.3 The Ideal Transformer

The transformer shown in Fig. 6-5(a) consists of two coils wound around a common magnetic core. The primary coil has N 1 turns and is connected to an ac voltage source V 1(t). The secondary coil has N 2 turns and is connected to a load resistor R L. In an ideal transformer the core has infinite permeability (μ=∞), and the magnetic flux is confined within the core.

# Figure 6-5 In a transformer, the directions of I 1 and I 2 are such that the flux Φ generated by one of them is opposite to that generated by the other. The direction of the secondary winding in (b) is opposite to that in (a), and so are the direction of I 2 and the polarity of V 2.

Figure 6 5 Full Alternative Text

On the primary side of the transformer, the voltage source V 1 generates current I 1 in the primary coil, which establishes a flux Φ in the magnetic core. The flux Φ and voltage V 1 are related by Faraday’s law:

V 1=−N 1 d Φ d t.(6.14)

A similar relation holds true on the secondary side:

V 2=−N 2 d Φ d t.(6.15)

The combination of Eqs. (6.14) and (6.15) gives

In an ideal lossless transformer, all the instantaneous power supplied by the source connected to the primary coil is delivered to the load on the secondary side. Thus, no power is lost in the core, and

P 1=P 2.(6.17)

Since P 1=I 1 V 1 and P 2=I 2 V 2, and in view of Eq. (6.16), we have

Thus, whereas the ratio of the voltages given by Eq. (6.16) is proportional to the corresponding turns ratio, the ratio of the currents is equal to the inverse of the turns ratio. If N 1/N 2=0.1, V 2 of the secondary circuit would be 10 times V 1 of the primary circuit, but I 2 would be only I 1/10.

The transformer shown in Fig. 6-5(b) is identical to that in Fig. 6-5(a) except for the direction of the windings of the secondary coil. Because of this change, the direction of I 2 and the polarity of V 2 in Fig. 6-5(b) are the reverse of those in Fig. 6-5(a).

The voltage and current in the secondary circuit in Fig. 6-5(a) are related by V 2=I 2 R L. To the input circuit, the transformer may be represented by an equivalent input resistance R in, as shown in Fig. 6-6, defined as

R i n=V 1 I 1.(6.19)

Use of Eqs. (6.16) and (6.18) gives

R in=V 2 I 2 N 1 N 2 2=N 1 N 2 2 R L.(6.20)

# Figure 6-6 Equivalent circuit for the primary side of the transformer.

Figure 6 6 Full Alternative Text

When the load is an impedance Z L and V 1 is a sinusoidal source, the phasor-domain equivalent of Eq. (6.20) is