Consider Figure 4.8a which shows an ideal transformer with a load impedance ZL (of an apparatus or a circuit element) connected to its secondary terminals. Assume that all variables involved are given in phasors. Therefore, the impedance ZL is defined as the ratio of the phasor voltage across it to the phasor current flowing through it. Hence,
Z V
L = I2
2
(4.28)
*Polarities result from the relative directions in which the two windings are wound on the core.
† According to the ANSI, additive polarities are required in large (greater than 200 kVA) high-voltage (higher than 8660 V) power transformers. To reduce voltage stress between adjacent leads, small transformers have subtractive polarities.
For further information, see Gönen (2008).
(a)
(b)
(c)
I1
Z΄L= a2ZL V1
+
–
Z΄L N1: N2
I2 I1
V1 +
–
Z΄L N1: N2 V1
I1 I2
+
–
V2 +
–
ZL
FIGURE 4.8 Illustration of impedance transfer across an ideal transformer: (a) an ideal transformer with a load impedance, (b) after the transfer of the impedance to the source side, and (c) the resultant equivalent circuit.
Here, not only the voltages V1 and V2 are in phase, but also the currents I1 and I2. As shown in Figure 4.8b, the apparent impedance of the primary circuit of the transformer is
′ =
Z V
L I1
1 (4.29)
where the primary voltage and current, respectively, are
V1= aV2 (4.18)
I I
1= 2
a (4.24)
Substituting Equations 4.18 and 4.24 into Equation 4.29, the apparent impedance of the primary becomes
The resulting equivalent circuit is shown in Figure 4.8c. Note that the a2 is known as the impedance ratio of the transformer. Therefore, as far as the source is concerned, the three circuits shown in Figure 4.8 are the same.
The impedance Z′L is simply the result of impedance transformation of the load impedance ZL through the transformer. Transferring an impedance from one side of the transformer to the other in this manner is known as referring the impedance* to the other side. Thus, Z′L is known as the load impedance referred to the primary side. Using Equations 4.18 and 4.24, voltages and currents can also be referred to one side or the other.
Similarly, an impedance located at the primary side of a transformer can also be referred to the secondary side as
Impedance transfer is very beneficial in calculations since it helps to get rid of a coupled circuit in an electrical circuit and thus simplifies the circuit.
Furthermore, it can be used in impedance matching to determine the maximum power transfer from a source with an internal impedance Zs to a load impedance ZL. Here, it is necessary to select the turns ratio so that
*It s also known as reflecting, transferring, or scaling the impedance.
′ =
= =
ZL N L L s
N1 Z a Z Z
2 2
2 (4.33)
For maximum power transfer when ZL may be complex,
′ =
= =
ZL N ZL ZL Zs
N1 a
2 2
2 (4.34)
where Zs* is the conjugate of Zs.
4.4.3 relaTIonshIP BeTween InPuT and ouTPuT Powers of an Ideal TransforMer
The input power provided to a transformer by its primary circuit is
Pin=V I1 1cosθ1 (4.35)
where θ1 is the angle between the primary voltage and the primary current. The power output of a transformer through its secondary circuit to its load is
Pout =V I2 2cosθ2 (4.36)
where θ2 is the angle between the secondary voltage and the secondary current. In an ideal transformer,
θ1=θ2=0
Therefore, the same power factor is seen by both the primary and secondary windings. Also, since V2 = V1/a and I2 = aI1, substituting them into Equation 4.36,
P V I V
a aI
out = =
2 2 1
cosθ ( 1) cosθ
or
Pout =V I1 1cosθ=Pin (4.37)
Therefore, in an ideal transformer, the output power is equal to its input power. This makes sense because, by definition, an ideal transformer has no internal power losses. One can extend the same argument to reactive and apparent powers. Therefore,
Qin = V I1 1sinθ= V I2 2sinθ=Qout (4.38)
Sin=V I1 1=V I2 2=Sout (4.39)
Example 4.2
Assume that a 60 Hz, 250 kVA, 2400/240 V distribution transformer is an ideal transformer and determine the following:
(a) Its turns ratio.
(b) The value of load current (i.e., I2), if a load impedance connected to its secondary (i.e., low-voltage side) terminals makes the transformer fully loaded.
(c) The value of the primary-side (i.e., high-voltage side) current.
(d) The value of the load impedance referred to the primary side of the transformer.
Solution
(a) The turns ratio of the transformer is
a N N
V
= 1 =V =
2 1 2
2,400 V 240 V =10
(b) Since the transformer is an ideal transformer, it has no losses.
S V I= 1 1=V I2 2
from which
I S
2 V
2
= = 250,000 VA=
240 V 1,041.67 A
(c) The corresponding primary current
I I
1= a2 =1,041.67 A
10 =104.167 A
or
I S
1 V
1
= =250,000 VA
2,400 V =104.167 A
(d) The value of the load impedance at the secondary side is
Z V
L= I2 V =
1,0 A
2 = 240 Ω
41.67 0.2304
Thus, its value referred to the primary side is
′ =
= = × =
ZL N L L
N1 Z a Z
2 2
2 102 0 2304 23 04. . Ω
Example 4.3
A single-phase, 60 Hz transformer is supplying power to a load of 3 + j5 Ω through a short power line with an impedance of 0.2 + j0.6 Ω, as shown in Figure 4.9a. The voltage at the generator bus (i.e., bus 1) is 277∠0° V. Determine the following:
(a) If the current at bus 1 is equal to the current at bus 2 (i.e., Iline = Iload), find the voltage at the load bus and the power losses that take place in the power line.
(b) If two ideal transformers T1 and T2 are inserted at the beginning and end of the line, as shown in Figure 4.9b, find the voltage at the load bus and the power losses in the line.
Solution
(a) Figure 4.10 shows the line-to-neutral diagram of the one-line diagram of the given system. Using the generator terminal voltage as the reference phasor, the line current can be found as
I V
and the line losses (i.e., the copper losses) are
Pline loss=I Rline line2 =( .42 9471 A) ( .2 02Ω)=368 89 1 W. 0
(b) Figure 4.9b shows the one-line diagram of the system with the step-up transformer T1 (with a turns ratio* of a1 = 1/10) and the step-down transformer T2 (with a turns ratio of a2 = 10/1). The load impedance referred to the power-line side of transformer T2 is
*The given numbers in 10:1 and 1:10 simply represent the turns ratios, respectively, rather than representing the actual number of turns in each winding.
Load
FIGURE 4.9 One-line diagram of the power system given in Example 4.3: (a) without the up and step-down transformers and (b) with the transformers.
′ = =
The resulting equivalent impedance is
Zeq Zline Zload
Referring this Zeq to the generator side of transformer T1, the new equivalent impedance is found as
Thus, the generator current can be calculated from
I V
Therefore, working back through transformer T1,
N I1G= N2Iline
and the line current can be found as
Iline N IG
FIGURE 4.10 The line-to-neutral diagram of the system given in Figure 4.9a.
Similarly, working back through transformer T2, Nline lineI =N2Iload
and the load current can be found as
Iload N Iload
= N2 = ∠ − ° = ∠ − °
1
10
1 ( .4 7455 59 05. A) 47 455. 59 05. A
Hence, the voltage at the load bus is
Vload=IloadZload=( .47 455∠−59 05. °A)(3+j5Ω)=276 7093. ∠−0 0. 1 V° The line losses are
Pline loss=I Rline line2 =( .4 7455 A ()2 0 2. Ω)=4 5039. W
Notice that the percent reduction in the line losses, after adding the up and the step-down transformers is
Assume that the impedances of the transformers T1 and T2, given in Part (b) of Example 4.3, are not small enough to ignore, and that they are ZT1= + .15 Ω0 j0 and ZT2= 0+j .150 Ω, respectively. Also assume that they are referred to the high-voltage sides of each transformer, respectively. Solve Part (b) of Example 4.3, accordingly.
Solution
Figure 4.11a shows the one-line diagram of the given system that includes T1 the step-up and T2
the step-down transformer. Figure 4.11b shows the line-to-neutral diagram of the same system.
Figure 4.11c shows the load impedance referred to power-line side of transformer T2 as well as the impedance of the transformer (which is given as already referred to its high-voltage side).
Therefore, the resulting equivalent impedance can be found as
Z Z Z Z Z
Referring this Zeq to the generator side of transformer Tl, as shown in Figure 4.11d, the new equivalent impedance is found as
′ = =
Therefore, the generator current can be found from
Thus, working back through transformer T1,
Iline N IG
= N1 = ∠ − ° = ∠ − °
2
1
10( .47 4339 59 6.0 ) 4 74339. 59 6 A.0
Similarly, working back through transformer T2,
Iload N Iline
FIGURE 4.11 One-line diagram of the power system given in Example 4.4.
Hence, the voltage at the load bus is
Vload=IloadZload=( .47 4339∠ −59 6 3.0 °)( +j5)=276 5873. ∠ −0 02. °
The line losses are
Pline loss=I Rline line2 =( .4 74339) ( . )2 02 =4 4999 W.
The percent reduction in line losses, after adding the step-up and the step-down transformer, is
Reduction in W
A real transformer differs from an ideal transformer in many respects. For example, as illustrated in Figure 4.12a, (1) the primary and secondary winding resistances R1 and R2 are not negligible, (2) the leakage fluxes and Φ 1andΦ 2 exist, (3) the core losses are not negligible, (4) the permeability of the core material is not infinite and therefore a considerable mmf is required to establish mutual flux Φm in the core, and (5) the core material saturates. The resulting representation of this transformer is shown in Figure 4.12b. Here, X 1 and X 2 are the leakage fluxes, respectively. Therefore,
X L
L 2 is the leakage inductance of the primary winding
L N
L 1 is the leakage inductance of the secondary winding
L N
P 1 is the permeance of the leakage flux path of the primary winding P 2 is the permeance of the leakage flux path of the secondary winding