376 Sinusoidal Steady-State Power Calculations The average power lost in the line results from the current flowing through the line resistance: *U = |IJ 2 * = (89.44) 2 (0.05) = 400 W Note that the power supplied totals 20,000 + 400 = 20,400 W, even though the loads require a total of only 20,000 W. c) As we can see from the power triangle in Fig. 10.15(c), we can correct the power factor to 1 if we place a capacitor in parallel with the existing loads such that the capacitor supplies 10 kVAR of magnetizing reactive power. The value of the capacitor is calculated as follows. First, find the capacitive reactance from Eq. 10.37: X = ivy 2 Q (250) 2 -10,000 the power factor is 1, the apparent power and the average power are the same, as seen from the power triangle in Fig. 10.16(c). Therefore, the apparent power once the power factor has been corrected is \S\ = P = 20 kVA. The magnitude of the current that supplies this apparent power is 20,000 250 = 80 A. The average power lost in the line is thus reduced to Pimc = \h\ 2 R = (80) 2 (0.05) = 320 W. Now, the power supplied totals 20,000 + 320 = 20,320 W. Note that the addition of the capaci- tor has reduced the line loss from 400 W to 320 W. = -6.25 a. Recall that the reactive impedance of a capacitor is -l/o)C, and w = 2TT(60) = 376.99 rad/s, if the source frequency is 60 Hz. Thus, C = -1 -1 o)X (376.99)(-6.25) 424.4 ^F. The addition of the capacitor as the third load is represented in geometric form as the sum of the two power triangles shown in Fig. 10.16. When 22.36 kVA^~- ^^ t 20 kW (a) lOkVAR + 20 kW (c) -lOkVAR (b) Figure 10.16 A (a) The sum of the power triangles for loads 1 and 2. (b) The power triangle for a 424.4 ^tF capacitor at 60 Hz. (c) The sum of the power triangles in (a) and (b). Example 10.7 Balancing Power Delivered with Power Absorbed in an ac Circuit a) Calculate the total average and reactive power delivered to each impedance in the circuit shown in Fig. 10.17. b) Calculate the average and reactive powers asso- ciated with each source in the circuit. c) Verify that the average power delivered equals the average power absorbed, and that the magnet- izing reactive power delivered equals the magnet- izing reactive power absorbed. 'VW orv-v-> i a /20 + 1 nn -/16 n; + v •—wv in |., /3 0 39 I t V s = 150/0°V V, = (78 - /104) V Ij = (-26 - /52) A V 2 = (72 + /104) V l v = (-2+ /6) A V 3 = (150 - /130) V I 2 = (-24 - /58) A Figure 10.17 A The circuit, with solution, for Example 10.7. 10.5 Power Calculations 377 Solution a) The complex power delivered to the (1 + /2) ft impedance is Si = |v,II = /», + /Qi = ^(78 - /104)(-26 + /52) = -(3380 + /6760) = 1690 +/3380 V A. Thus this impedance is absorbing an average power of 1690 W and a reactive power of 3380 VAR. The complex power delivered to the (12 - /16) ft impedance is S 2 = 2 V 2^ = Pi + jQi = -(72 + /104)(-2 - /6) = 240 - /320 VA. Therefore the impedance in the vertical branch is absorbing 240 W and delivering 320 VAR. The complex power delivered to the (I + /3) ft impedance is S 3 = ~V 3 IS = P 3 + jQ 3 = -(150 - /130)(-24 + /58) = 1970 + /5910 V A. This impedance is absorbing 1970 W and 5910 VAR. b) The complex power associated with the inde- pendent voltage source is s s = -|v,ii = p s + /a = -^(150)(-26 + /52) = 1950 - /3900 VA. Note that the independent voltage source is absorbing an average power of 1950 W and delivering 3900 VAR. The complex power asso- ciated with the current-controlled voltage source is S x = |(39I r )(I|) = P x + jQ x = |(-78 + /234)(-24 + /58) = -5850 - /5070 VA. Both average power and magnetizing reactive power are being delivered by the dependent source. c) The total power absorbed by the passive imped- ances and the independent voltage source is Absorbed = P\ + Pi + i>3 + A = 5850 W. The dependent voltage source is the only circuit element delivering average power. Thus delivered = 5850 W. Magnetizing reactive power is being absorbed by the two horizontal branches. Thus ^absorbed = Q\ + Qi = 9290 VAR. Magnetizing reactive power is being delivered by the independent voltage source, the capacitor in the vertical impedance branch, and the dependent voltage source. Therefore Qdelivered = 9290 VAR. 378 Sinusoidal Steady-State Power Calculations I/ASSESSMENT PROBLEMS Objective 1—Understand ac power concepts, their relationships to one another, and how to calculate them in a circuit 10.4 The load impedance in the circuit shown is shunted by a capacitor having a capacitive reac- tance of -52 0,. Calculate: a) the rms phasors V L and I L , b) the average power and magnetizing reactive power absorbed by the (39 + /26) £1 load impedance, c) the average power and magnetizing reactive power absorbed by the (1 + /4) O line impedance, d) the average power and magnetizing reactive power delivered by the source, and e) the magnetizing reactive power delivered by the shunting capacitor. l a •AW- /4 0 -TTTA. O 250/0! V(rms) + V,. 39 n ./26 a Source Line— —***- Load (c) 23.52 W, 94.09 VAR; (d) 1152.62 W, -376.36 VAR; (e) 1223.18 VAR. 10.5 The rms voltage at the terminals of a load is 250 V The load is absorbing an average power of 40 kW and delivering a magnetizing reactive power of 30 kVAR. Derive two equivalent impedance models of the load. Answer: 1 H in series with 0.75 Q, of capacitive reactance; 1.5625 H in parallel with 2.083 H of capacitive reactance. 10.6 Find the phasor voltage Y s (rms) in the circuit shown if loads L x and L 2 are absorbing 15 kVA at 0.6 pf lagging and 6 kVA at 0.8 pf leading, respectively. Express V v in polar form. Answer: (a) 252.20 /-4.54° V (rms), 5.38 /-38.23° A (rms); (b) 1129.09 W, 752.73 VAR; NOTE: Also try Chapter Problems 10.18,10.26, and 10.27. /10 + (_") 20070° V (rms) —n— I <2 v s Answer: 251.64 /15.91° V. 10.6 Maximum Power Transfer a»- Generalized linear network operating in the sinusoidal steady state b« Figure 10.18 • A circuit describing maximum power transfer. Recall from Chapter 4 that certain systems—for example, those that trans- mit information via electric signals—depend on being able to transfer a maximum amount of power from the source to the load. We now reexam- ine maximum power transfer in the context of a sinusoidal steady-state network, beginning with Fig. 10.18. We must determine the load imped- ance Z L that results in the delivery of maximum average power to termi- nals a and b. Any linear network may be viewed from the terminals of the load in terms of a Thevenin equivalent circuit. Thus the task reduces to finding the value of Z L that results in maximum average power delivered to Z L in the circuit shown in Fig. 10.19. For maximum average power transfer, Z L must equal the conjugate of the Thevenin impedance; that is, Condition for maximum average power transfer • Z T = Z Th' (10.38) 10.6 Maximum Power Transfer 379 We derive Eq. 10.38 by a straightforward application of elementary calcu- lus. We begin by expressing Z Th and Z L in rectangular form: Z Th - i? Th + jX Th , (10.39) Z L = R L + jX L , (10.40) In both Eqs. 10.39 and 10.40, the reactance term carries its own algebraic sign—positive for inductance and negative for capacitance. Because we are making an average-power calculation, we assume that the amplitude of the Thevenin voltage is expressed in terms of its rms value. We also use the Thevenin voltage as the reference phasor. Then, from Fig. 10.19, the rms value of the load current I is I = *"rh (R-i- h + R L ) + j(x Th + x L y (10.41) Figure 10.19 • The circuit shown in Fig. 10.18, with the network replaced by its Thevenin equivalent. The average power delivered to the load is P = HI 2 /*,, (10.42) Substituting Eq. 10.41 into Eq. 10.42 yields P = |v Th |X (R Th + R L ) 2 + (X Th + X L ) 2 (10.43) When working with Eq. 10.43, always remember that Vr h , i? Th , and X Th are fixed quantities, whereas R L and X L are independent variables. Therefore, to maximize P, we must find the values of /? L and X L where dP/3R L and dP/BX h are both zero. From Eq. 10.43, dP -\\ Th \ 2 2R L (X L + X Th ) dX L [(R L + R Th ) 2 + (X L + X Th ) 2 } 2 ' (10.44) *P_ = IVlhPK^L + ^Th) 2 + (XL + *Th) 2 - 2R L (RL + *Th)] dR L [(R L + R Th ) 2 + (X L + X Th ) 2 ] 2 From Eq. 10.44, dP/dX L is zero when (10.45) X, = -X riv (10.46) From Eq. 10.45,3P/dR L is zero when R L = VWh + (*L + x Th y (10.47) Note that when we combine Eq. 10.46 with Eq. 10.47, both derivatives are zero when Z L = Zj h . 380 Sinusoidal Steady-State Power Calculations The Maximum Average Power Absorbed The maximum average power that can be delivered to Z L when it is set equal to the conjugate of Z Th is calculated directly from the circuit in Fig. 10.19. When Z L = Zj h , the rms load current is V Th /2JR L , and the max- imum average power delivered to the load is P = 1 max 4Rr 1 ghj 4 R, (10.48) If the Thevenin voltage is expressed in terms of its maximum amplitude rather than its rms amplitude, Eq. 10.48 becomes 1VJ, 8 R L (10.49) Maximum Power Transfer When Z is Restricted Maximum average power can be delivered to Z L only if Z L can be set equal to the conjugate of Z Th . There are situations in which this is not pos- sible. First, R L and X L may be restricted to a limited range of values. In this situation, the optimum condition for R L and X L is to adjust X L as near to -X Th as possible and then adjust R L as close to VRjh + (X L + X Th ) 2 as possible (see Example 10.9). A second type of restriction occurs when the magnitude of Z L can be varied but its phase angle cannot. Under this restriction, the greatest amount of power is transferred to the load when the magnitude of Z L is set equal to the magnitude of Z Th ; that is, when \Z L \ = |ZnJ. (10.50) The proof of Eq. 10.50 is left to you as Problem 10.40. For purely resistive networks, maximum power transfer occurs when the load resistance equals the Thevenin resistance. Note that we first derived this result in the introduction to maximum power transfer in Chapter 4. Examples 10.8-10.11 illustrate the problem of obtaining maximum power transfer in the situations just discussed. Example 10.8 Determining Maximum Power Transfer without Load Restrictions a) For the circuit shown in Fig. 10.20, determine the impedance Z L that results in maximum average power transferred to Z L . b) What is the maximum average power transferred to the load impedance determined in (a)? Solution a) We begin by determining the Thevenin equiva- lent with respect to the load terminals a, b. After two source transformations involving the 20 V source, the 5 fl resistor, and the 20 il resistor, we simplify the circuit shown in Fig. 10.20 to the one shown in Fig. 10.21. Then, V Th = 16 /0° "Th (-/6) 4 + /3 - /6 = 19.2 /-53.13° = 11.52 - /15.36 V. Figure 10.20 • The circuit for Example 10.8. 10.6 Maximum Power Transfer 381 Figure 10.21 • A simplification of Fig. 10.20 by source transformations. which we replaced the original network with its Tlievenin equivalent. From Fig. 10.22, the rms magni- tude of the load current I is 19.2/V2 '•« = im = U785 A ' The average power delivered to the load is We find the Thevenin impedance by deactivat- ing the independent source and calculating the impedance seen looking into the terminals a and b. Thus, z Th = ( 7 y6)( ! + J ? = 5 - 76 - /i- 68 a - Th 4 + /3 - /6 J m ~ n -/1.6811 5.76 ft ,/ a 19.2/-53.13Y + V It 5.76 0 4-/1.68 0- For maximum average power transfer, the load impedance must be the conjugate of Z T h, so Z L = 5.76 + /1.68 ft. b) We calculate the maximum average power deliv- ered to Z L from the circuit shown in Fig. 10.22, in Figure 10.22 A The circuit shown in Fig. 10.20, with the original network replaced by its Thevenin equivalent. P = I 2 cU (5.76) = 8W. Example 10.9 Determining Maximum Power Transfer with Load Impedance Restriction a) For the circuit shown in Fig. 10.23, what value of Z L results in maximum average power transfer to Z L ? What is the maximum power in milliwatts? b) Assume that the load resistance can be varied between 0 and 4000 ft and that the capacitive reactance of the load can be varied between 0 and -2000 ft. What settings of R L and X L transfer the most average power to the load? What is the maximum average power that can be transferred under these restrictions? Solution a) If there are no restrictions on R L and X L , the load impedance is set equal to the conjugate of the output or the Thevenin impedance. Therefore we set R L = 3000 ft and X L = -4000 ft, or 3000 a /4000 a -vw 10 V (rms Figure 10.23 • The circuit for Examples 10.9 and 10.10. Because the source voltage is given in terms of its rms value, the average power delivered to Z L is 1 10 2 25 P = 4 3000 = T mW = 833mW - b) Because R L and X L are restricted, we first set X L as close to -4000 ft as possible; thus Xj = -2000 ft. Next, we set R L as close to vRjh + (X L + X Th ) 2 as possible.Thus Z L = 3000 - /4000 ft. R L = V3000 2 + (-2000 + 4000) 2 = 3605.55 ft. 382 Sinusoidal Steady-State Power Calculations Now, because R L can be varied from 0 to 4000 ft, we can set R L to 3605.55 ft. Therefore, the load impedance is adjusted to a value of Z L = 3605.55 - /2000 ft. With Z L set at this value, the value of the load current is 10 /0° leff 6605.55 + /2000 1.4489 /-16.85° mA. The average power delivered to the load is P = (1.4489 X Kr 3 ) 2 (3605.55) = 7.57 mW. This quantity is the maximum power that we can deliver to a load, given the restrictions on R L and X L . Note that this is less than the power that can be delivered if there are no restrictions; in (a) we found that we can deliver 8.33 mW. Example 10.10 Finding Maximum Power Transfer with Impedance Angle Restrictions A load impedance having a constant phase angle of -36.87° is connected across the load terminals a and b in the circuit shown in Fig. 10.23. The magnitude of Z L is varied until the average power delivered is the most possible under the given restriction. a) Specify Z L in rectangular form. b) Calculate the average power delivered to Z L . Solution a) From Eq. 10.50, we know that the magnitude of Z L must equal the magnitude of Z Th . Therefore \ZL\ = |Z T hl = 13000 + /4000| = 5000 ft. Now, as we know that the phase angle of Z L is -36.87°, we have Z L = 5000 /-36.87° = 4000 - /3000 ft. b) With Z L set equal to 4000 - /3000 ft, the load current is 10 !eff = ^^ ^7^ = 1-4142 /-8.13° mA, 611 7000 + /1000 L and the average power delivered to the load is P = (1.4142 X 10" 3 ) 2 (4000) = 8mW. This quantity is the maximum power that can be delivered by this circuit to a load impedance whose angle is constant at —36.87°. Again, this quantity is less than the maximum power that can be delivered if there are no restrictions on Z L . ^ASSESSMENT PROBLEM Objective 2—Understand the condition for maximum real power delivered to a load in an ac circuit 10.7 The source current in the circuit shown is 3.6 mH 3cos5000r A. a) What impedance should be connected across terminals a,b for maximum average power transfer? b) What is the average power transferred to the impedance in (a)? c) Assume that the load is restricted to pure resistance. What size resistor connected across a,b will result in the maximum aver- age power transferred? d) What is the average power transferred to the resistor in (c)? Answer: (a) 20 - /10 ft; (b)18W; (c) 22.36 ft; (d) 17.00 W. 4H NOTE: Also try Chapter Problems 10.44,10.47, and 10.48. 10.6 Maximum Power Transfer 383 Example 10.11 Finding Maximum Power Transfer in a Circuit with an Ideal Transformer The variable resistor in the circuit in Fig. 10.24 is adjusted until maximum average power is delivered to R L . a) What is the value of R L in ohms? b) What is the maximum average power (in watts) delivered to R { 1 60O ideal a ^^Tl 4:1 O 840/Q! V (rms) r*L 20 n Figure 10.24 • The circuit for Example 10.11. Solution a) We first find the Thevenin equivalent with respect to the terminals of R L . The circuit for determining the open circuit voltage in shown in Fig. 10.25. The variables V], V 2 , l\, and I 2 have been added to expedite the discussion. 60 n -VA— <• a I, + «~Ii 840/0!^+ V (rms) 4:1 Ideal -•a V 2 rh 20 n »b Figure 10.25 A The circuit used to find the Thevenin voltage. First we note the ideal transformer imposes the following constraints on the variables Vj, V 2 , Ii, and I 2 : V 2 = Jv,. I, = -\h. The open circuit value of I? is zero, hence I] is zero. It follows that Vi = 840 /QP_ v, V 2 = 210 /0°V. From Fig. 10.25 we note that \ rh is the negative of V 2 , hence V Th = -210 /0° V. The circuit shown in Fig. 10.26 is used to deter- mine the short circuit current. Viewing l { and I 2 as mesh currents, the two mesh equations are 840 /0° = goij - 20I 2 + \ h 0 = 20I 2 - 201! + V 2 . 60(1 I, + k V ' I 840Z0°/ + \ r (rms)W 4 -I' 1 c v 2 Ideal ]• + ( :20O i II Figure 10.26 • The circuit used to calculate the short circuit current. When these two mesh current equations are com- bined with the constraint equations we get 840 /(F = -40I 2 + Vj, Vi 0 = 25I 2 + -i. 4 Solving for the short circuit value of I 2 yields I 2 = -6 A. Therefore the Thevenin resistance is -210 R Th - -6 = 35 n. 384 Sinusoidal Steady-State Power Calculations Maximum power will be delivered to R L when R L equals 35 ft. b) The maximum power delivered to R L is most easily determined using theThevenin equivalent. From the circuit shown in Fig. 10.27 we have 210/0°/- V(rms")U 35 O P = 1 max -210 70 (35) = 315 W. Figure 10.27 • The Thevenin equivalent loaded for maximum power transfer. ^ASSESSMENT PROBLEMS Objective 3—Be able to calculate all forms of ac power in ac circuits with linear transformers and ideal transformers 10.8 Find the average power delivered to the 100 Q resistor in the circuit shown if v g = 660 cos 5000r V. 10 mH 100 ft Answer: 612.5 W. 10.9 a) Find the average power delivered to the 400 fl resistor in the circuit shown if v t = 248 cos 10,000f V. b) Find the average power delivered to the 375 H resistor. c) Find the power developed by the ideal volt- age source. Check your result by showing the power absorbed equals the power developed. NOTE: Also try Chapter Problems 10.64 and 10.65. 40 mH 50 mH lOOmH —l'YW^— :37512 400 O Answer: (a) 50 W; (b)49.2W; (c) 99.2 W, 50 + 49.2 = 99.2 W. 10.10 Solve Example 10.11 if the polarity dot on the coil connected to terminal a is at the top. Answer: (a) 15 Q,; (b) 735 W. 10.11 Solve Example 10.11 if the voltage source is reduced to 146 /0° V rms and the turns ratio is reversed to 1:4. Answer: (a) 1460 H; (b)58.4W. Practical Perspective Heating Appliances A handheld hair dryer contains a heating element, which is just a resistor heated by the sinusoidal current passing through it, and a fan that blows the warm air surrounding the resistor out the front of the unit. This is shown schematically in Fig. 10.28. The heater tube in this figure is a resis- tor made of coiled nichrome wire. Nichrome is an alloy of iron, chromium, and nickel. Two properties make it ideal for use in heaters. First, it is more resistive than most other metals, so less material is required to achieve the needed resistance. This allows the heater to be very compact. Second, unlike many other metals, nichrome does not oxidize when heated red hot in air. Thus the heater element lasts a long time. A circuit diagram for the hair dryer controls is shown in Fig. 10.29. This is the only part of the hair dryer circuit that gives you control over the heat Heater tube Hot air 4 Controls Fan and motor Power cord Figure 10.28 • Schematic representation of a handheld hair dryer. Practical Perspective 385 • 1 Thermal fuse ; xfc R^ < R-> f t T t OFF L M H Figure 10.29 A A circuit diagram for the hair dryer controls. \ Thermal fuse " * is? "1 5 ft 2 t t t t OFF L M H (a) R\ > R 2 Figure 10.30 • (a) The circuit in Fig. 10.29 redrawn for the LOW switch setting, (b) A simplified equiva- lent circuit for (a). setting. The rest of the circuit provides power to the fan motor and is not of interest here. The coiled wire that comprises the heater tube has a connec- tion partway along the coil, dividing the coil into two pieces. We model this in Fig. 10.29 with two series resistors, R^ and R 2 . The controls to turn the dryer on and select the heat setting use a four-position switch in which two pairs of terminals in the circuit will be shorted together by a pair of sliding metal bars. The position of the switch determines which pairs of terminals are shorted together. The metal bars are connected by an insulator, so there is no conduction path between the pairs of shorted terminals. The circuit in Fig. 10.29 contains a thermal fuse. This is a protective device that normally acts like a short circuit. But if the temperature near the heater becomes dangerously high, the thermal fuse becomes an open circuit, discontinuing the flow of current and reducing the risk of fire or injury. The thermal fuse provides protection in case the motor fails or the airflow becomes blocked. While the design of the protection system is not part of this example, it is important to point out that safety analysis is an essential part of an electrical engineer's work. Now that we have modeled the controls for the hair dryer, let's deter- mine the circuit components that are present for the three switch settings. To begin, the circuit in Fig. 10.29 is redrawn in Fig. 10.30(a) for the LOW switch setting. The open-circuited wires have been removed for clarity. A simplified equivalent circuit is shown in Fig. 10.30(b). A similar pair of fig- ures is shown for the MEDIUM setting (Fig. 10.31) and the HIGH setting (Fig. 10.32). Note from these figures that at the LOW setting, the voltage source sees the resistors R± and R 2 in series; at the MEDIUM setting, the volt- age source sees only the R 2 resistor; and at the HIGH setting, the voltage source sees the resistors in parallel. Thermal fuse • • •• • /¾ t t t t OFF L M H (a) ft, (b) Figure 10.31 • (a) The circuit in Fig. 10.29 redrawn for the MEDIUM switch setting, (b) A simplified equiv- alent circuit for (a). Thermal fuse v • • • 11 I * * * IJ_ ft, R 2 t t t t OFFL M H (a) ft, ft? (b) Figure 10.32 A (a) The circuit in Fig. 10.29 redrawn for the HIGH switch setting, (b) A simplified equiva- lent circuit for (a). NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 10.66-10.68. . Recall from Chapter 4 that certain systems—for example, those that trans- mit information via electric signals—depend on being able to transfer a maximum amount of power from the source to. of this example, it is important to point out that safety analysis is an essential part of an electrical engineer's work. Now that we have modeled the controls for the hair dryer, let's