12.9 PSpice for Three-Phase Circuits

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12.9 PSpice for Three-Phase Circuits

PSpice can be used to analyze three-phase balanced or unbalanced cir cuits in the same way it is used to analyze single-phase ac circuits. However, a delta-connected source presents two major problems to PSpice. First, a delta-connected source is a loop of voltage sources—which PSpice does not like. To avoid this problem, we insert a resistor of neg ligible resistance (say, 1 μΩ per phase) into each phase of the deltaconnected source. Second, the delta-connected source does not provide a convenient node for the ground node, which is necessary to run PSpice. This problem can be eliminated by inserting balanced wye-connected large resistors (say, 1 MΩ per phase) in the delta-connected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this.

For the balanced Y-∆ circuit in Fig. 12.27, use PSpice to find the line current IaA, the phase voltage VAB, and the phase current IAC. Assume that the source frequency is 60 Hz.

Solution:

The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT are inserted in the appropriate lines to obtain IaA and IAC, while VPRINT2 is inserted between nodes A and B to print differential voltage VAB. We set the attributes of IPRINT and VPRINT2 each to AC = yes, MAG = yes, PHASE = yes, to print only the magnitude and phase of the currents and voltages. As a single-frequency analysis, we select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 60, and Final Freq = 60. Once the circuit is saved, it is simulated by selecting Analysis/Simulate. The output file includes the following:

FREQ	V(A,B)	VP(A,B)
6.000E+01	1.699E+02	3.081E+01
FREQ	IM(V_PRINT2)	IP(V_PRINT2)
6.000E+01	2.350E+00	-3.620E+01
FREQ	IM(V_PRINT3)	IP(V_PRINT3)
6.000E+01	1.357E+00	-6.620E+01

Example 12.11

Schematic for the circuit in Fig. 12.27.

From this, we obtain

$$I_{aA} = 2.35 \underline{/-36.2^{\circ}} A$$ $$V_{AB} = 169.9 \underline{/30.81^{\circ}} V, \quad I_{AC} = 1.357 \underline{/-66.2^{\circ}} A$$

Practice Problem 12.11

Refer to the balanced Y-Y circuit of Fig. 12.29. Use PSpice to find the line current IbB and the phase voltage VAN. Take f = 100 Hz.

Answer: 100.9⧸ 60.87° V, 8.547⧸−91.27° A.

Consider the unbalanced ∆-∆ circuit in Fig. 12.30. Use PSpice t o find the generator current Iab, the line current IbB, and the phase current IBC.

Example 12.12

Solution:

• 1. Define. The problem and solution process are clearly defined.
• 2. Present. We are to find the generator current flowing from a to b, the line current flowing from b to B, and the phase current flowing from B to C.
• 3. Alternative. Although there are different approaches to solving this problem, the use of PSpice is mandated. Therefore, we will not use another approach.
• 4. Attempt. As mentioned above, we avoid the loop of voltage sources by inserting a 1-μΩ series resistor in the delta-connected source. To provide a ground node 0, we insert balanced wyeconnected resistors (1 MΩ per phase) in the delta-connected source, as shown in the schematic in Fig. 12.31. Three IPRINT pseudocomponents with their attributes are inserted to be able

Figure 12.31 Schematic for the circuit in Fig. 12.30.

to get the required currents Iab, IbB, and IBC. Since the operating frequency is not given and the inductances and capacitances should be specified instead of impedances, we assume ω = 1 rad/s so that f = 1∕2π = 0.159155 Hz. Thus,

$$L = \frac{X_L}{\omega} \quad \text{and} \quad C = \frac{1}{\omega X_C}$$

We select Analysis/Setup/AC Sweep and enter Total Pts = 1, Start Freq = 0.159155, and Final Freq = 0.159155. Once the schematic is saved, we select Analysis/Simulate to simulate the circuit. The output file includes:

FREQ	IM(V_PRINT1)	IP(V_PRINT1)
1.592E-01	9.106E+00	1.685E+02
FREQ	IM(V_PRINT2)	IP(V_PRINT2)
1.592E-01	5.959E+00	-1.772E+02
FREQ	IM(V_PRINT3)	IP(V_PRINT3)
1.592E-01	5.500E+00	1.725E+02

which yields

I_{ab} = 5.595 \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \

5. Evaluate. We can check our results by using mesh analysis. Let the loop aABb be loop 1, the loop bBCc be loop 2, and the loop ACB be loop 3, with the three loop currents all flowing in the clockwise direction. We then end up with the following loop equations:

Loop 1

(54 + j10)I¹ − (2 + j5)I² − (50)I3 = 208⧸ 10° = 204.8 + j36.12

Loop 2

$$-(2+j5)I1 + (4+j40)I2 - (j30)I3 = 208/ -110o$$

= -71.14 - j195.46

Loop 3

$$-(50)I_1 - (j30)I_2 + (50 - j10)I_3 = 0$$

Using MATLAB to solve this we get,

>>Z = [(54+10i),(-2-5i),-50;(-2-5i),(4+40i),
-30i;-50,-30i,(50-10i)]

Z =

54.0000+10.0000i-2.0000-5.0000i-50.0000
-2.0000-5.0000i 4.0000 + 40.0000i 0-30.0000i
-50.0000 0-30.0000i 50.0000-10.0000i

>>V = [(204.8+36.12i);(-71.14-195.46i);0]
V =
1.0e+002*
2.0480+0.3612i
-0.7114-1.9546i
 0
>>I = inv(Z)*V
I =
8.9317+2.6983i
0.0096+4.5175i
5.4619+3.7964i
   IbB = −I1 + I2 = −(8.932 + j2.698) + (0.0096 + j4.518)
     = −8.922 + j1.82 = 9.106⧸168.47° A Answer checks
    IBC = I2 − I3 = (0.0096 + j4.518) − (5.462 + j3.796)
       = −5.452 + j0.722 = 5.5⧸172.46° A Answer checks

Now to solve for Iab. If we assume a small internal impedance for each source, we can obtain a reasonably good estimate for Iab. Adding in internal resistors of 0.01 Ω, and adding a fourth loop around the source circuit, we now get

Loop 1

$$(54.01 + j10)I1 - (2 + j5)I2 - (50)I3 - 0.01I4 = 208/10o$$

= 204.8 + j36.12

Loop 2

$$-(2+j5)I1 + (4.01+j40)I2 - (j30)I3 - 0.01I4$$

= 208/ $-110^{\circ}$ = -71.14 - j195.46

Loop 3

$$-(50)I1 - (j30)I2 + (50 - j10)I3 = 0$$

Loop 4

$$-(0.01)I_1 - (0.01)I_2 + (0.03)I_4 = 0$$

Z = [(54.01+10i),(-2-5i),-50,-0.01;(-2-5i), (4.01+40i),-30i,-0.01;-50,-30i,(50-10i), 0;-0.01,-0.01,0,0.03] Z = 54.0100 + 10.0000i -2.0000-5.0000i, -50.0000 -0.0100 -2.0000-5.0000i 4.0100-40.0000i 0-30.0000i 0.0100 -50.0000 0-30.0000i 50.0000-10.0000i 0 -0.0100 -0.0100 0 0.0300 >>V = [(204.8 + 36.12i);(-71.14-195.46i);0;0]

V =
1.0e+002*
2.0480+0.3612i
-0.7114-1.9546i
 0
 0
>>I = inv(Z)*V
I =
8.9309+2.6973i
0.0093+4.5159i
5.4623+3.7954i
2.9801+2.4044i
   Iab = −I1 + I4 = −(8.931 + j2.697) + (2.98 + j2.404)
= −5.951 − j0.293 = 5.958⧸−177.18° A. Answer checks.

6. Satisfactory? We have a satisfactory solution and an adequate check for the solution. We can now present the results as a solution to the problem.

Practice Problem 12.12 For the unbalanced circuit in Fig. 12.32, use PSpice to find the generator current Ica, the line current IcC, and the phase current IAB.

Answer: 24.68⧸−90° A, 37.25⧸ 83.79° A, 15.55⧸−75.01° A.

12.10 Applications

Both wye and delta source connections have important practical applications. The wye source connection is used for long distance transmission of electric power, where resistive losses ( I 2 R) should be minimal. This

is due to the fact that the wye connection gives a line voltage that is √ __ 3 greater than the delta connection; hence, for the same power, the line current is √ __ 3smaller. In addition, delta connected are also undesirable due to the potential of having disastrous circulating currents. Sometimes, using transformers, we create the equivalent of delta connect source. This conversion from three-phase to single-phase is required in residential wiring, because household lighting and appliances use single-phase power. Three-phase power is used in industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta-connected. For example, both connections are satisfac tory with induction motors. In fact, some manufacturers connect a motor in delta for 220 V and in wye for 440 V so that one line of motors can be readily adapted to two different voltages.

Here we consider two practical applications of those concepts co vered in this chapter: power measurement in three-phase circuits and residential wiring.

12.10.1 Three-Phase Power Measurement

Section 11.9 presented the wattmeter as the instrument for measuring the average (or real) power in single-phase circuits. A single wattmeter can also measure the average power in a three-phase system that is bal anced, so that P1 = P2 = P3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are necessary to measure power if the system is unbalanced. The threewattmeter method of power measurement, shown in Fig. 12.33, will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected. The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is con stantly changing. The total average power is the algebraic sum of the three wattmeter readings,

$$P_T = P_1 + P_2 + P_3 \tag{12.61}$$

where P1, P2, and P3 correspond to the readings of wattmeters W1, W2, and W3, respectively. Notice that the common or reference point o in Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point o can be connected to the neutral point n. For a delta-connected load, point o can be connected to any point. If point o is connected to point b, for example, the voltage coil in wattmeter W2 reads zero and P2 = 0, indicating that wattmeter W2 is not necessary. Thus, two wattmeters are sufficient to measure the total power.

The two-wattmeter method is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to an y two phases, as sho wn typically in Fig. 12.34. Notice that the current coil of each w attmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line v oltage. Also notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected. Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readings equals the total a verage power absorbed by the load, regardless of whether it is wye- or delta-connected, balanced or

unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings,

$$P_T = P_1 + P_2 \tag{12.62}$$

We will show here that the method works for a balanced three-phase system.

Consider the balanced, wye-connected load in Fig. 12.35. Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the abc sequence and the load impedance ZY = ZY^⧸θ. Due to the load impedance, each voltage coil leads its current coil by θ, so that the po wer factor is cos θ. We recall that each line voltage leads the corresponding phase voltage by 30°. Thus, the total phase dif ference between the phase current Ia and line v oltage Vab is θ + 30°, and the a verage po wer read by wattmeter W1 is

$$P_1 = \text{Re}[\mathbf{V}_{ab}\mathbf{I}_a^*] = V_{ab}I_a \cos(\theta + 30^\circ) = V_L I_L \cos(\theta + 30^\circ) \qquad (12.63)$$

Figure 12.35 Two-wattmeter method applied to a balanced wye load.

Similarly, we can show that the average power read by wattmeter 2 is

$$P_2 = \text{Re}[\mathbf{V}_{cb}\mathbf{I}_c^*] = V_{cb}I_c \cos(\theta - 30^\circ) = V_L I_L \cos(\theta - 30^\circ) \qquad (12.64)$$

We now use the trigonometric identities

$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$

\n

$$\cos(A - B) = \cos A \cos B + \sin A \sin B$$

(12.65)

to find the sum and the difference of the two wattmeter readings in Eqs. (12.63) and (12.64):

$$P_1 + P_2 = V_L I_L [\cos(\theta + 30^\circ) + \cos(\theta - 30^\circ)]$$

= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$

• $\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)$
  = $V_L I_L 2 \cos 30^\circ \cos \theta = \sqrt{3} V_L I_L \cos \theta$ (12.66)

since 2 cos 30° = √ __ 3 . Comparing Eq. (12.66) with Eq. (12.50) shows that the sum of the wattmeter readings gives the total average power,

$$P_T = P_1 + P_2 \t\t(12.67)$$

Similarly,

$$P_1 - P_2 = V_L I_L [\cos(\theta + 30^\circ) - \cos(\theta - 30^\circ)]$$

= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$
$- \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ)$ (12.68)
= $-V_L I_L 2 \sin 30^\circ \sin \theta$
$P_2 - P_1 = V_L I_L \sin \theta$

since 2 sin 30° = 1. Comparing Eq. (12.68) with Eq. (12.51) shows that the difference of the wattmeter readings is proportional to the total reactive power, or

$$Q_T = \sqrt{3}(P_2 - P_1)$$

(12.69)

From Eqs. (12.67) and (12.69), the total apparent power can be obtained as

$$S_T = \sqrt{P_T^2 + Q_T^2}$$

(12.70)

Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power fac tor angle as

$$\tan \theta = \frac{Q_T}{P_T} = \sqrt{3} \frac{P_2 - P_1}{P_2 + P_1}$$

\n(12.71)

from which we can obtain the power factor as pf = cos θ. Thus, the two-wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67), (12.69), and (12.71), we conclude that:

• 1. If P2 = P1, the load is resistive.
• 2. If P2 > P1, the load is inductive.
• 3. If P2 < P1, the load is capacitive.

Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real power in a three-phase four-wire system.

Three wattmeters W1, W2, and W3 are connected, respectively, to phases a, b, and c to measure the total power absorbed by the unbalanced wye-connected load in Example 12.9 (see Fig. 12.23). (a) Predict the wattmeter readings. (b) Find the total power absorbed.

Solution:

Example 12.13

Part of the problem is already solved in Example 12.9. Assume that the wattmeters are properly connected as in Fig. 12.36.

Figure 12.36 For Example 12.13.

(a) From Example 12.9,

$$\mathbf{V}_{AN} = 100 \underline{\text{/}0^{\circ}}, \qquad \mathbf{V}_{BN} = 100 \underline{\text{/}120^{\circ}}, \qquad \mathbf{V}_{CN} = 100 \underline{\text{/} -120^{\circ}} \text{ V}$$

while

$$\mathbf{I}_a = 6.67 \underline{\bigcirc}^{\circ}, \qquad \mathbf{I}_b = 8.94 \underline{\bigcirc} 3.44^{\circ}, \qquad \mathbf{I}_c = 10 \underline{\bigcirc} -66.87^{\circ} \text{ A}$$

We calculate the wattmeter readings as follows:

$$P_1 = \text{Re}(\mathbf{V}_{AN}\mathbf{I}_{a}^{*}) = V_{AN}I_a \cos(\theta_{\mathbf{V}_{AN}} - \theta_{\mathbf{I}_a})$$

= 100 × 6.67 × cos(0° – 0°) = 667 W

$$P_2 = \text{Re}(\mathbf{V}_{BN}\mathbf{I}_{b}^{*}) = V_{BN}I_b \cos(\theta_{\mathbf{V}_{BN}} - \theta_{\mathbf{I}_b})$$

= 100 × 8.94 × cos(120° – 93.44°) = 800 W

$$P_3 = \text{Re}(\mathbf{V}_{CN}\mathbf{I}_{c}^{*}) = V_{CN}I_c \cos(\theta_{\mathbf{V}_{CN}} - \theta_{\mathbf{I}_c})$$

= 100 × 10 × cos(-120° + 66.87°) = 600 W

(b) The total power absorbed is

$$P_T = P_1 + P_2 + P_3 = 667 + 800 + 600 = 2067$$

W

We can find the power absorbed by the resistors in Fig. 12.36 and use that to check or confirm this result

$$P_T = |I_a|^2(15) + |I_b|^2(10) + |I_c|^2(6)$$

= 6.67²(15) + 8.94²(10) + 10²(6)
= 667 + 800 + 600 = 2067 W

which is exactly the same thing.

Practice Problem 12.13	Repeat Example 12.13 for the network in Fig. 12.24 (see Practice Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B. Answer: (a) 13.175 kW, 0 W, 29.91 kW, (b) 43.08 kW.
Example 12.14	The two-wattmeter method produces w attmeter readings P1 = 1560 W and P2 = 2100 W when connected to a delta-connected load. If the line voltage is 220 V, calculate: (a) the per-phase average power, (b) the per phase reactive power, (c) the power factor, and (d) the phase impedance.

Solution:

We can apply the given results to the delta-connected load. (a) The total real or average power is

$$P_T = P_1 + P_2 = 1560 + 2100 = 3660$$

W

The per-phase average power is then

$$P_p = \frac{1}{3}P_T = 1220 \text{ W}$$

(b) The total reactive power is

$$Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(2100 - 1560) = 935.3 \text{ VAR}$$

so that the per-phase reactive power is

$$Q_p = \frac{1}{3}Q_T = 311.77 \text{ VAR}$$

(c) The power angle is

$$\theta = \tan^{-1} \frac{Q_T}{P_T} = \tan^{-1} \frac{935.3}{3660} = 14.33^{\circ}$$

Hence, the power factor is

$$\cos \theta = 0.9689(\text{lagging})$$

It is a lagging pf because QT is positive or P2 > P1. (d) The phase impedance is Zp = Zp⧸θ. We know that θ is the same as the pf angle; that is, θ = 14.33°.

$$Z_p = \frac{V_p}{I_p}$$

We recall that for a delta-connected load, Vp = VL = 220 V. From Eq. (12.46),

2.46),
\n

$$P_p = V_p I_p \cos \theta \implies I_p = \frac{1220}{220 \times 0.9689} = 5.723 \text{ A}$$

Hence,

$$Z_p = \frac{V_p}{I_p} = \frac{220}{5.723} = 38.44 \ \Omega$$

and

$$Z_p = 38.44 / 14.33^{\circ} \,\Omega$$

Let the line voltage VL = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be P1 = −560 W and P2 = 800 W. Determine:

(a) the total average power

• (b) the total reactive power
• (c) the power factor

(d) the phase impedance

Is the impedance inductive or capacitive?

Answer: (a) 240 W, (b) 2.356 kV AR, (c) 0.1014, (d) 18.25 ^⧸ 84.18° Ω, inductive.

Practice Problem 12.14

Example 12.15

The three-phase balanced load in Fig. 12.35 has impedance per phase of ZY = 8 + j6 Ω. If the load is connected to 208-V lines, predict the read ings of the wattmeters W1 and W2. Find PT and QT.

Solution:

The impedance per phase is

$$\mathbf{Z}_{Y} = 8 + j6 = 10/36.87^{\circ} \,\Omega$$

so that the pf angle is 36.87°. Since the line voltage VL = 208 V, the line current is __

$$I_L = \frac{V_p}{|\mathbf{Z}_Y|} = \frac{208/\sqrt{3}}{10} = 12 \text{ A}$$

Then

$$P_1 = V_L I_L \cos(\theta + 30^\circ) = 208 \times 12 \times \cos(36.87^\circ + 30^\circ)$$

= 980.48 W

$$P_2 = V_L I_L \cos(\theta - 30^\circ) = 208 \times 12 \times \cos(36.87^\circ - 30^\circ)$$

= 2478.1 W

Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since P2 > P1, the load is inductive. This is evident from the load Z*^Y* itself. Next,

$$P_T = P_1 + P_2 = 3.459 \text{ kW}$$

and

$$Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(1497.6)$$

VAR = 2.594 kVAR

If the load in Fig. 12.35 is delta-connected with impedance per phase of Zp = 30 + j40 Ω and VL = 220 V, predict the readings of the wattmeters W1 and W2. Calculate PT and QT. Practice Problem 12.15

Answer: 200.6 W, 1.5418 kW, 1.7424 kW, 2.323 kVAR.

12.10.2 Residential Wiring

In the United States, most household lighting and appliances operate on 120-V, 60-Hz, single-phase alternating current. (The electricity may also be supplied at 110, 115, or 117 V, depending on the location.) The local power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down to 120/240 V with a transformer (more details on transformers in the next chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up to zero.

That is,

$$\mathbf{V}_W = 0/\underline{0}^\circ$$

, $\mathbf{V}_B = 120/\underline{0}^\circ$ , $\mathbf{V}_R = 120/\underline{180}^\circ = -\mathbf{V}_B$ .
\n

$$\mathbf{V}_{BR} = \mathbf{V}_B - \mathbf{V}_R = \mathbf{V}_B - (-\mathbf{V}_B) = 2\mathbf{V}_B = 240/\underline{0}^\circ \qquad (12.72)$$

Figure 12.37

A 120/240 household power system. Source: A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd edition, © 1975, p. 324. Pearson Education, Inc., Upper Saddle River, NJ.

Figure 12.38

Single-phase three-wire residential wiring.

Since most appliances are designed to operate with 120 V, the light ing and appliances are connected to the 120-V lines, as illustrated in Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are connected in parallel. Hea vy appliances that consume lar ge currents, such as air conditioners, dishwashers, ovens, and laundry machines, are con nected to the 240-V power line.

Because of the dangers of electricity , house wiring is carefully regulated by a code dra wn by local ordinances and by the National Electrical Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit breakers are used. Modern wiring codes require a third wire for a separate ground. The ground wire does not carry po wer like the neutral wire but enables appliances to have a separate ground connection. Figure 12.40 shows the connection of the receptacle to a 120-V rms line and to the ground. As shown in the figure, the neutral line is connected to the ground (the earth) at man y critical locations. Although the ground line seems redundant, grounding is important for many reasons. First, it is required by NEC. Second, grounding provides

Figure 12.39

A typical wiring diagram of a room. Source: A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd edition, © 1975, p. 325. Pearson Education, Inc., Upper Saddle River, NJ.

a convenient path to ground for lightning that strik es the po wer line. Third, grounds minimize the risk of electric shock. What causes shock is the passage of current from one part of the body to another . The human body is lik e a big resistor R. If V is the potential dif ference between the body and the ground, the current through the body is determined by Ohm’s law as

$$I = \frac{V}{R} \tag{12.73}$$

The value of R varies from person to person and depends on whether the body is wet or dry. How great or how deadly the shock is depends on the amount of current, the pathway of the current through the body, and the length of time the body is e xposed to the current. Currents less than 1 mA may not be harmful to the body , but currents greater than 10 mA can cause se vere shock. A modern safety de vice is the ground-fault circuit interrupter (GFCI), used in outdoor circuits and in bathrooms, where the risk of electric shock is greatest. It is essentially a circuit break er that opens when the sum of the currents iR, iW, and iB through the red, white, and the black lines is not equal to zero, or iR + iW + iB ≠ 0.

The best w ay to a void electric shock is to follo w safety guide lines concerning electrical systems and appliances. Here are some of them:

• Never assume that an electrical circuit is dead. Always check to be sure.
• Use safety de vices when necessary , and wear suitable clothing (insulated shoes, gloves, etc.).
• Never use tw o hands when testing high-v oltage circuits, since the current through one hand to the other hand has a direct path through your chest and heart.
• Do not touch an electrical appliance when you are wet. Remember that water conducts electricity.
• Be extremely careful when working with electronic appliances such as radio and TV because these appliances ha ve lar ge capacitors in them. The capacitors tak e time to dischar ge after the po wer is disconnected.
• Always ha ve another person present when w orking on a wiring system, just in case of an accident.

12.11 Summary

• 1. The phase sequence is the order in which the phase v oltages of a three-phase generator occur with respect to time. In an abc sequence of balanced source v oltages, Van leads Vbn by 120°, which in turn leads Vcn by 120°. In an acb sequence of balanced v oltages, Van leads Vcn by 120°, which in turn leads Vbn by 120°.
• 2. A balanced wye- or delta-connected load is one in which the threephase impedances are equal.
• 3. The easiest way to analyze a balanced three-phase circuit is to transform both the source and the load to a Y-Y system and then analyze the single-phase equivalent circuit. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four possible configurations.
• 4. The line current IL is the current flowing from the generator to the load in each transmission line in a three-phase system. The line voltage VL is the v oltage between each pair of lines, e xcluding the neutral line if it e xists. The phase current Ip is the current flowing through each phase in a three-phase load. The phase voltage Vp is the voltage of each phase. For a wye-connected load,

$$V_L = \sqrt{3} V_p \qquad \text{and} \qquad I_L = I_p$$

For a delta-connected load,

$$V_L = V_p \qquad \text{and} \qquad I_L = \sqrt{3}I_p$$

• 5. The total instantaneous po wer in a balanced three-phase system is constant and equal to the average power.
• 6. The total comple x po wer absorbed by a balanced three-phase Y-connected or ∆-connected load is

$$\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta}$$

where θ is the angle of the load impedances.

• 7. An unbalanced three-phase system can be analyzed using nodal or mesh analysis.
• 8. PSpice is used to analyze three-phase circuits in the same w ay as it is used for analyzing single-phase circuits.
• 9. The total real power is measured in three-phase systems using either the three-wattmeter method or the two-wattmeter method.
• 10. Residential wiring uses a 120/240-V, single-phase, three-wire system.

Review Questions

12.1 What is the phase sequence of a three-phase motor for which VAN = 220⧸−100° V and VBN = 220⧸ 140° V?

(a) abc (b) acb

12.2 If in an acb phase sequence, Van = 100⧸−20°, then Vcn is:

(a)

$$100 \div 140^{\circ}$$

(b) $100 \div 100^{\circ}$

(c)

$$100\div 50^{\circ}$$

(d) $100\div 10^{\circ}$

12.3 Which of these is not a required condition for a balanced system:

$$(a) |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}|$$

(b)

$$\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0$$

(c)

$$V_{an} + V_{bn} + V_{cn} = 0$$

• (d) Source voltages are 120° out of phase with each other.
• (e) Load impedances for the three phases are equal.

12.4 In a Y-connected load, the line current and phase current are equal.

(a) True (b) False

12.5 In a ∆-connected load, the line current and phase current are equal.

(a) True (b) False

12.6 In a Y-Y system, a line voltage of 220 V produces a phase voltage of:

(a) 381 V (b) 311 V (c) 220 V (d) 156 V (e) 127 V

12.7 In a ∆-∆ system, a phase voltage of 100 V produces a line voltage of:

(a) 58 V (b) 71 V (c) 100 V (d) 173 V (e) 141 V

Problems1

Section 12.2 Balanced Three-Phase Voltages

12.1 If Vab = 400 V in a balanced Y-connected threephase generator, find the phase voltages, assuming the phase sequence is:

(a) abc (b) acb

• 12.2 What is the phase sequence of a balanced threephase circuit for which Van = 120⧸30° V and Vcn = 120⧸−90° V? Find Vbn.
• 12.3 Given a balanced Y-connected three-phase generator with a line-to-line voltage of Vab = 100⧸45° V and Vbc = 100⧸165° V, determine the phase sequence and the value of Vca.
• 12.4 A three-phase system with abc sequence and VL = 440 V feeds a Y-connected load with ZL = 40⧸30° Ω. Find the line currents.
• 12.5 For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are:

vAN = 120 cos(ωt + 32°) V vBN = 120 cos(ωt – 88°) V vCN = 120 cos(ωt + 152°) V

Write the time-domain expressions for the line-toline voltages vAB, vBC, and vCA.

Section 12.3 Balanced Wye-Wye Connection

12.6 Using Fig. 12.41, design a problem to help other students better understand balanced wye-wye connected circuits.

12.8 When a Y-connected load is supplied by voltages in abc phase sequence, the line voltages lag the corresponding phase voltages by 30°.

(a) True (b) False

12.9 In a balanced three-phase circuit, the total instantaneous power is equal to the average power.

(a) True (b) False

12.10 The total power supplied to a balanced ∆-load is found in the same way as for a balanced Y-load.

(a) True (b) False

Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b, 12.9a, 12.10a.

Figure 12.41

For Prob. 12.6.

• 12.7 Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page.
• 12.8 In a balanced three-phase Y-Y system, the source is an acb sequence of voltages and Vcn = 120⧸ 35° V rms. The line impedance per phase is (1 + j2) Ω, while the per-phase impedance of the load is (11 + j14) Ω. Calculate the line currents and the load voltages.
• 12.9 A balanced Y-Y four-wire system has phase voltages

$$\mathbf{V}_{an} = 120 \underline{\text{O}^{\circ}}, \qquad \mathbf{V}_{bn} = 120 \underline{\text{O} - 120^{\circ}}$$ $$\mathbf{V}_{cn} = 120 \underline{\text{O} \cdot 120^{\circ}} \text{ V}$$

The load impedance per phase is 19 + j13 Ω, and the line impedance per phase is 1 + j2 Ω. Solve for the line currents and neutral current.

¹ Remember that unless stated otherwise, all given voltages and currents are rms values.

Problems 543

12.10 For the circuit in Fig. 12.43, determine the current in the neutral line.

For Prob. 12.10.

Section 12.4 Balanced Wye-Delta Connection

12.11 In the Y-∆ system shown in Fig. 12.44, the source is a positive sequence with Van = 440⧸ 0° V and phase impedance Zp = 2 – j3 Ω. Calculate the line voltage VL and the line current IL.

Figure 12.44 For Prob. 12.11.

Figure 12.46 For Prob. 12.13.

12.14 Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page.

For Prob. 12.14.

12.15 The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line voltage of 210 V. If Zl = 1 + j1 Ω, Z∆ = 24 − j30 Ω, and ZY = 12 + j5 Ω, determine the magnitude of the line current of the combined loads.

For Prob. 12.15.

• 12.16 A balanced delta-connected load has a phase current IAC = 5⧸−30° A.
  • (a) Determine the three line currents assuming that the circuit operates in the positive phase sequence.
  • (b) Calculate the load impedance if the line voltage is VAB = 440⧸ 0° V.
• 12.17 A positive sequence wye-connected source where Van = 120⧸ 90° V, is connected to a delta-connected load where Z*^L* = (60 + j45) Ω. Determine the line currents.
• 12.18 If Van = 220⧸ 60° V in the network of Fig. 12.49, find the load phase currents IAB, IBC, and ICA.

Figure 12.49 For Prob. 12.18.

Section 12.5 Balanced Delta-Delta Connection

12.19 For the ∆-∆ circuit of Fig. 12.50, calculate the phase and line currents.

Figure 12.50

For Prob. 12.19.

12.20 Using Fig. 12.51, design a problem to help other students better understand balanced delta-delta connected circuits.

For Prob. 12.20.

12.21 Three 440-V generators form a delta-connected source that is connected to a balanced deltaconnected load of ZL = (8.66 + j5) Ω per phase as shown in Fig. 12.52. Determine the value of IBC and IaA. What is the pf of the load?

b

• ‒

• 12.22 Find the line currents IaA, IbB, and IcC in the three-phase network of Fig. 12.53. Take ZL = (114 + j87) Ω and Z*^l* = (2 + j) Ω.
• 12.23 A balanced delta connected source is connected to a balanced delta connected load where ZL = (80 + j60) Ω and Zl = (2 + j) Ω. Given that the load voltages are VAB = 100⧸ 0° V, VBC = 100⧸ 120° V, and VCA = 100⧸−120° V. Calculate the source voltages Vab, Vbc, and Vca.
• 12.24 A balanced delta-connected source has phase voltage Vab = 880⧸ 30° V and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as 60⧸ 30° Ω and line impedance per phase as 1 + j1 Ω.
• 12.26 Using Fig. 12.55, design a problem to help other students better understand balanced delta connected sources delivering power to balanced wye connected loads.

Figure 12.55 For Prob. 12.26.

10 ‒ j8 Ω

10 ‒ j8 Ω

Section 12.6 Balanced Delta-Wye Connection

12.25 In the circuit of Fig. 12.54, if

$$\mathbf{V}_{ab} = 440/10^{\circ}$$

,
\n $V_{bc} = 440/-110^{\circ}$ , $V_{ca} = 440/130^{\circ}$ V, find the line currents.

3 + j2 Ω 10 ‒ j8 W

Ib

Ia

^{a 3+ j}2 W

Vab

• 12.27 A ∆-connected source supplies power to a Yconnected load in a three-phase balanced system. Given that the line impedance is 2 + j1 Ω per phase while the load impedance is 6 + j4 Ω per phase, find the magnitude of the line voltage at the load. Assume the source phase voltage Vab = 208⧸ 0° V rms.
• 12.28 The line-to-line voltages in a Y-load have a magnitude of 880 V and are in the positive sequence at 60 Hz. If the loads are balanced with Z1 = Z2 = Z3 = 25⧸30°, find all line currents and phase voltages.

Section 12.7 Power in a Balanced System

• 12.29 A balanced three-phase Y-∆ system has Van = 240⧸ 0° V rms and Z∆ = 51 + j45 Ω. If the line impedance per phase is 0.4 + j1.2 Ω, find the total complex power delivered to the load.
• 12.30 In Fig. 12.56, the rms value of the line voltage is 208 V. Find the average power delivered to the load.

Figure 12.56

For Prob. 12.30.

• 12.31 A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line voltage of 480 V. Each load phase draws 24 kW at a lagging power factor of 0.8. Find:

  • (a) the load impedance per phase
  • (b) the line current
  • (c) the value of capacitance needed to be connected in parallel with each load phase to minimize the current from the source

• 12.32 Design a problem to help other students better understand power in a balanced three-phase system.

• 12.33 A three-phase source delivers 4.8 kVA to a wyeconnected load with a phase voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage.

• 12.34 A balanced wye-connected load with a phase impedance of 10 – j16 Ω is connected to a balanced three-phase generator with a line voltage of 220 V. Determine the line current and the complex power absorbed by the load.

• 12.35 Three equal impedances, 60 + j30 Ω each, are delta-connected to a 230-V rms, three-phase circuit. Another three equal impedances, 40 + j10 Ω each, are wye-connected across the same circuit at the same points. Determine:

  • (a) the line current
  • (b) the total complex power supplied to the two loads
  • (c) the power factor of the two loads combined

• 12.36 A 4200-V, three-phase transmission line has an impedance of 4 + j Ω per phase. If it supplies a load of 1 MVA at 0.75 power factor (lagging), find:

  • (a) the complex power
  • (b) the power loss in the line
  • (c) the voltage at the sending end

• 12.37 The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading. If the line voltage is 440 V, calculate the line current IL and the load impedance ZY.

• 12.38 Given the circuit in Fig. 12.57 below, find the total complex power absorbed by the load.

Figure 12.57 For Prob. 12.38.

For Prob. 12.39.

12.40 For the three-phase circuit in Fig. 12.59, find the average power absorbed by the delta-connected load with Z∆ = 21 + j24 Ω.

Figure 12.59 For Prob. 12.40.

• 12.41 A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the three-phase system has an effective line voltage of 400 V, find the line current.
• 12.42 A balanced three-phase generator delivers 7.2 kW to a wye-connected load with impedance 30 – j40 Ω per phase. Find the line current IL and the line voltage VL.
• 12.43 Refer to Fig. 12.48. Obtain the complex power absorbed by the combined loads.
• 12.44 A three-phase line has an impedance of 1 + j3 Ω per phase. The line feeds a balanced delta-connected load, which absorbs a total complex power of 12 + j5 kVA. If the line voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor.
• 12.45 A balanced wye-connected load is connected to the generator by a balanced transmission line with an impedance of 0.5 + j2 Ω per phase. If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator.
• 12.46 A three-phase load consists of three 100-Ω resistors that can be wye- or delta-connected. Determine which connection will absorb the most average

power from a three-phase source with a line voltage of 110 V. Assume zero line impedance.

12.47 The following three parallel-connected three-phase loads are fed by a balanced three-phase source:

Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf

If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero.

• 12.48 A balanced, positive-sequence wye-connected source has Van = 240⧸ 0° V rms and supplies an unbalanced delta-connected load via a transmission line with impedance 2 + j3 Ω per phase.
  • (a) Calculate the line currents if ZAB = 40 + j15 Ω, ZBC = 60 Ω, ZCA = 18 – j12 Ω.
  • (b) Find the complex power supplied by the source.
• 12.49 Each phase load consists of a 20-Ω resistor and a 10-Ω inductive reactance. With a line voltage of 480 V rms, calculate the average power taken by the load if:

(a) the three-phase loads are delta-connected (b) the loads are wye-connected

12.50 A balanced three-phase source with VL = 240 V rms is supplying 8 kVA at 0.6 power factor lagging to two wye-connected parallel loads. If one load draws 3 kW at unity power factor, calculate the impedance per phase of the second load.

Section 12.8 Unbalanced Three-Phase Systems

12.51 Consider the wye-delta system shown in Fig. 12.60. Let Z¹ = 100 Ω, Z² = j100 Ω, and Z³ = –j100 Ω. Determine the phase currents, IAB, IBC, and ICA, and the line currents, IaA, IbB , and IcC.

Figure 12.60 For Prob. 12.51.

$$\mathbf{V}_{an} = 220/120^{\circ}, \qquad \mathbf{V}_{bn} = 220/0^{\circ}$$ $$\mathbf{V}_{cn} = 220/-120^{\circ} \text{ V}$$

If the impedances are

$$\mathbf{Z}_{AN} = 20/60^\circ, \qquad \mathbf{Z}_{BN} = 30/0^\circ$$ $$\mathbf{Z}_{cn} = 40/30^\circ \ \Omega$$

find the current in the neutral line.

12.53 Using Fig. 12.61, design a problem that will help other students better understand unbalanced threephase systems.

• For Prob. 12.53.
• 12.54 A balanced three-phase Y-source with VP = 880 V rms drives a Y-connected three-phase load with phase impedance ZA = 80 Ω, ZB = 60 + j90 Ω, and ZC = j80 Ω. Calculate the line currents and total complex power delivered to the load. Assume that the neutrals are connected.
• 12.55 A three-phase supply, with the line-to-line voltage of 240 V rms, has the unbalanced load as shown in Fig. 12.62. Find the line currents and the total complex power delivered to the load.

Figure 12.62 For Prob. 12.55.

12.56 Using Fig. 12.63, design a problem to help other students to better understand unbalanced three-phase systems.

Figure 12.63

For Prob. 12.56.

12.57 Determine the line currents for the three-phase circuit of Fig. 12.64. Let Va = 220⧸ 0°, Vb = 220⧸−120°, Vc = 220⧸ 120° V.

For Prob. 12.57.

Section 12.9 PSpice for Three-Phase Circuits

• 12.58 Solve Prob. 12.10 using PSpice or MultiSim.
• 12.59 The source in Fig. 12.65 is balanced and exhibits a positive phase sequence. If f = 60 Hz, use PSpice or MultiSim to find VAN,VBN, and VCN.

For Prob. 12.59.

12.60 Use PSpice or MultiSim to determine Io in the single-phase, three-wire circuit of Fig. 12.66. Let Z1 = 15 – j10 Ω, Z2 = 30 + j20 Ω, and Z3 = 12 + j5 Ω.

12.61 Given the circuit in Fig. 12.67, use PSpice or MultiSim to determine currents IaA and voltage VBN.

Figure 12.67

For Prob. 12.61.

12.62 Using Fig. 12.68, design a problem to help other students better understand how to use PSpice or MultiSim to analyze three-phase circuits.

12.63 Use PSpice or MultiSim to find currents IaA and IAC in the unbal anced three-phase system shown in Fig. 12.69. Let

$$Zl = 2 + j, \t Z1 = 40 + j20 Ω,\nZ2 = 50 - j30 Ω, \t Z3 = 25 Ω$$

Figure 12.69

For Prob. 12.63.

• 12.64 For the circuit in Fig. 12.58, use PSpice or MultiSim to find the line currents and the phase currents.
• 12.65 A balanced three-phase circuit is shown in Fig. 12.70 on the next page. Use PSpice or MultiSim to find the line currents IaA, IbB, and IcC.