Nilsson SM Ch01

# Electric Circuits (8th Edition)

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Unformatted text preview: Circuit Variables Assessment Problems AP 1.1 To solve this problem we use a product of ratios to change units from dollars / year to dollars / millisecond. We begin by expressing \$10 billion in scientiﬁc notation: \$100 billion = \$100 x 109 Now we determine the number of milliseconds in one year, again using a product of ratios: 1 year 1 day I 1 hour 1 min 1 sec _ 1 year 365.25 days 24 hours 60 mins 60 secs 1000 ms — 31.5576 X 109 ms Now we can convert from dollars / year to dollars / millisecond, again with a product of ratios: \$100 x 109 1 year _ 100 1 year 31.5576 x 109 ms ‘ 31.5576 = \$3.17/ms AP 1.2 First, we recognize that 1 ns = 10‘9 s. The question then asks how far a signal will travel in 10“9 s if it is traveling at 80% of the speed of light. Remember that the speed of light 0 = 3 x 108 m/s. Therefore, 80% of c is (0.8)(3 X 108) = 2.4 X 108 m/s. Now, we use a product of ratios to convert from meters / second to inches/ nanosecond: .._._..._...._._._.._____._......____.—, __.____.....__._.._..__.— ls 109 ns 1 m 2.54 cm “ (109)(2.54) 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45” in a nanosecond. 1—2 AP 1.3 AP 1.4 CHAPTER 1. Circuit Variables Remember from Eq. (1.2), current is the time rate of change of charge, or 2' :- Z’lﬁ In this problem, we are given the current and asked to ﬁnd the total charge. To do this, we must integrate Eq. (1.2) to ﬁnd an expression for charge in terms of current: q(t) = [image We are given the expression for current, i, which can be substituted into the above expression. To ﬁnd the total charge, we let t ——> oo in the integral. Thus we have 0° 20 °° 20 = —5000.’r d = ~5000\$ : *00 M ’0 qtotal [0 206 m “50008 0 —5000(6 ‘9 ) 20 20 Recall from Eq. (1.2) that current is the time rate of change of charge, or i 2 %§. In this problem we are given an expression for the charge, and asked to ﬁnd the maximum current. First we will ﬁnd an expression for the current using Eq. (1.2): ,is_i[i_(3+3_)e-at] *dt—dt a2 a 0:2 Now that we have an expression for the current, we can ﬁnd the maximum value of the current by setting the ﬁrst derivative of the current to zero and solving for t: dz' d a; = aﬂte‘mt) = 6“” + t(—04)e"“t 2 (1 —— awe—‘3“t = 0 Since 6““ never equals 0 for a. ﬁnite value of t, the expression equals 0 only when (1 —~ at) = 0. Thus, t = 1/a will cause the current to be maximum. For this value of t, the current is 1 2-: ___e—a/a ___ O! a Problems 1*?» Remember in the problem statement, oz = 0.03679. Using this value for 04, 1 0.036796 10 ’L: AP 1.5 Start by drawing a picture of the circuit described in the problem statement: - 1 2013' + 2 +41% Also sketch the four ﬁgures from Fig. 1.6: 4* 2' 4— 2‘ + l + l V v (31' (131' (d) [a] Now we have to match the voltage and current shown in the ﬁrst ﬁgure with the polarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2 is the same as 4A of current leaving Terminal 1. We get (a)v=—20V, i=~4A; (b)v=—20V, i=4A (c)v=20V, i=—4A; (d)'u=20V, i=4A [b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p z m’ = (“20)(—4) : 80 W. Since the power is greater than 0, the box is absorbing power. [0] From the calculation in part (b), the box is absorbing 80 W. AP 1.6 Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig. 1.5, p :2 ml. From Eq. (1.3), we know that power is the time rate of change of energy, or p = %‘f«. If we know the pOWer, we can ﬁnd the energy by integrating Eq. (1.3). To begin, ﬁnd the expression for power: 1) = m; 2: (10,OOOe_5000t)(206‘5000t) = 200,0006—10’000t = 2 x 105.e*1‘“")‘)t w 1—4 CHAPTER 1. Circuit Variables Now ﬁnd the expression for energy by integrating Eq. (1.3): w(t) = /:p(a:) dzr Substitute the expression for power, p, above. Note that to ﬁnd the total energy, we let t —+ 00 in the integral. Thus we have oo 2 x 105 °° = 5 -10,000x d = —10,000:c w /02X106 m _10,0008 0 2x105 0 2x105 2><105 = -°° a z 0 — 1 = = 20 \$10,000“) 6 ) —10,000( ) 10,000 J AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative. Thus, using the passive Sign convention, p 2 *vi. Substituting the values of voltage and current given in the ﬁgure, 1) z «(800 x 103)(1.8 x 103) = *1440 x 106 = #1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line. Chapter Problems (250 x 106x440) P 1.1 109 = 110 glga-Watt hours . 5280 ft 2526 lb 1 kg P12 4 d,- 4 .#.W.__: _ 6 ( 0°” ) (8 5 m1) 1 mi 1000 ft 2.21b 20 5 x 10 kg 1000 songs :15 songs P 1.3 M z [a] (32)(24) (2.1) mm3 1 mm3 (1000)(1) 33.. — r: 0.62 3—minute songs, or about 111.6 seconds of music P 1.4 P 1.5 P 1.6 P 1.7 Problems 1—5 h ._ 4x 106 MB [ ] (32)(24)(2.1) mm3 "” (0.1)3 1111113 ” (4 x 109)(0.001) 3” “ W (320)(240) pixels 2 bytes 30 frames 1 pixel = 2480 bytes 2 4.608 x 106 bytes/sec 1 frame 1 sec (4.608 X 106 bytes/sec)(a: secs) : 30 X 109 bytes 30 x 109 2: m = 6510 880 = 108.5 min of Video 2'} [a] We can set up a ratio to determine how long it takes the bamboo to grow 10 pm First, recall that 1 mm 2 103mm. Let’s also express the rate of growth of bamboo using the units mm/s instead of mm / day. Use a product of ratios to perform this conversion: 250 mm 1 day 1 hour 1 min __ 250 ~ 10 mm/S 1 day 24 hours 60 min 60 sec _ (24)(60)(60) —P 3456 Use a ratio to determine the time it takes for the bamboo to grow 10 ,um: 10/3456 X 10—3 m 10 x 10-6 m 10 x 10—6 1 s ‘ a: s SO "3 " 10/3456 >< 10~~3 _ 3456 E” 1 cell 3600 s (24)(7) hr _ ' S ' 1 hr ' W — cells/week Volume 2 area >< thickness Convert values to millimeters, noting that 10 m2 = 106 mm2 106 = (10 x 106)(thickness) 106 :> thickness = 1.6022 x 10“19 C X 1 electron 1029 electrons 1 m3 C/m3 = = 1.6022 x 101°C/m3 Cross-sectional area of wire 2 71"!“2 = 7r(1.5 x 10‘“3 m)2 = 7.07 X 10*6 m2 C/m = (1.6022 x 101°C/m3)(7.07 x 10451112) = 113.253 x 103 0/111 Therefore, 2‘ = (113.253 x 103) (C) x avg vel (311-) SEC III S 2' 1200 m = m = 0-0106m/s Thus, average velocity = 145 P 1.8 P 1.9 P 1.10 P 1.11 P 1.12 CHAPTER 1. Circuit Variables 35 x 10-6 C/s = W =2. 1 14 , . 1.6022 x 10—19 C/elec 18 X 0 elec/s TI, First we use Eq. (1.2) to relate current and charge: . dq 2 —— a —- 24 cos 4000t Therefore, dq 2: 24 cos 4000t dt To ﬁnd the charge, we can integrate both sides of the last equation. Note that we substitute .r for q on the left side of the integral, and y for t on the right side of the integral: q(t) t / dm = 24 / cos4000y dy £100 0 We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: t 24 24 24 sin 4000y . . ~—————— :2 0t -— 2 0 4000 sm400 400031n4000(0) 4000 4000 q(t) — q(0) = 24 sin 400015 But (1(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 6 X 10“3 sin 4000tC = 6sin 400015 mC w = qV = (1.6022 x 10“19)(6) = 9.61 x 10“19 = 0.961 aJ 3600 s 5hr 1hr p = (9)(100 x 10-3) = 0.9 W; = 18,000 8 18,000 i 10(75):]0 pdt 10(18,()00)=/0 0.9dt=0.9(18,000)= 16.2 kJ Assume we are standing at box A looking toward box B. Then, using the passive Sign convention p = vi, since the current i is ﬂowing into the + terminal of the voltage 11. Now we just substitute the values for v and i into the equation for power. Remember that if the power is positive, B is absorbing power,,so the power must be ﬂowing from A to B. If the power is negative, B is generating power so the power must be ﬂowing from B to A. [ p = (120)(5) = 600 w [b] p = (250)(—8) = 42000 w [c] p = (—150)(16) = ~2400 w [d] p = (—480)(—~10) :2 4800 w 600 W from A to B 2000 W from B to A 2400 W from B to A 4800 W from A to B a] Problems 1*? P 1.13 [a] p = m’ :2 (40)(—10) = —400 W Power is being delivered by the box. b] Leaving P 1.14 a] p = m' = (~60)(—10) = 600 W, so power is being absorbed by the box. [ [c] Gaining [ [b] Entering [c] Losing 410A P 1.15 [a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i ﬂows into the + terminal of the battery of Car A). Therefore using the passive sign convention, p 2 vi : (30)(12) == 360 W. Since the power is positive, the battery in Car A is absorbing power, so Car A must have the ” dead” battery. -t [b] w(t) =/ deL'; 1 min: 60 s 0 60 112(60): O 360d11: w = 360(60 — 0) = 360(60) = 21,600 J = 21.6 kJ t P116 p=vi; wm/pdx .0 Since the energy is the area under the power vs. time plot, let us plot 1) vs. 75. 10 180x103 120x10'3 I 238 ks f Note that in constructing the plot above, we used the fact that 80 hr 2: 288,000 s = 288 ks 19(0) : (9)(20 x 10*) = 180 x 10—3 w 1~8 CHAPTER 1. Circuit Variables p(288 ks) = (6)(20 x 10‘3) = 120 x 10—3 W w = (120 x 10’3)(288 x 103) + \$7180 x 10-3 — 120 x 10’3)(288 x 103) = 43.2 kJ P 1.17 [a] p = m: = 3065‘” — 30am“ — 40610“ + 50e~2000t —— log-300W p(1 ms) 2 3.1 mW [b] “’(t) a [of (306—500‘” -- 30645003” ._ 408—10009;+ 508-20001: _ 106-3000a3)d\$ = 21.67 —~ 608'500t + 20645001: + 406—1000t* 256—200“ + 3.3353000th w(1 ms) : 1.24%} [c] wtotal = 21.67/11 P 1.18 [a] 0(10 ms) : 4006‘1 81112 2 133.8 V i(10 ms) = 5e“1 \$1112 2 1.67 A p(10 1118) = m’ = 223.79 W b _— vi = 20006 20‘” \$1112 20075 1 1 P 1 1 = 200012—20“ [—2- — 5 (1054004 =: mama—200‘ —~ 10006-20“ cos400t 00 00 w = / 1000.240015 dt —/ 10006-20“ cos 400tdt 0 0 e—200t 00 = 1 000 \$200 0 640% 40m 400 ‘ 400t 00 _1000 W[—200ms ,+ sm ] O 200 = _ M = _ 1 5 100014 x 104 + 16 x 104] 5 w=4J Problems 1—9 P119 [a] Os§t<4sz v=2.5t V; i=1,u.A; p=2.5t,LLW 4s<t§85: vzlﬂV; i=0A; p=0W Ss_<_t<16s: v=-2.5t+30 V; i=—-1/1,A; p=2.5t—30,U.W 168<t§20sz v=—10V; i=0A; p=0W 2Os§t<3652 'uzt—BO V; i=0.4p.A; 19:0.4t—«12/JJW 36s<t£4ﬁsz v=6V; i=0A; p=0W 4ﬁsgt<505: v=—1.5t+75 V; i:—0.6,uA; p=0.9tw45/LW t>5031 11:0V; i=0A; p=OW [b] Calculate the area under the curve from zero up to the desired time: 10(4) 2 %(4)(10) = 20m] M12) = M4) ~ 5(4)(10) = 0 J w(36) 2-: w(12) + §(4)(10) —- §(10)(4) + %(6)(2.4) = 7.2 #J w(50) = w(36) — —;—(4)(3.6) : 0 J 1‘10 CHAPTER 1. Circuit Variables P 1.20 [a] p = m‘ = (0.05e*10°°t)(75 — 751240000 2 (3.755100% ‘ 3756—20001) W g? = —375Oe‘10°°t + 7500e~2°°°t = 0 so 2e~2°°°t = e-1000t 2 = 61000it so 1112 = 100075 thus p is maximum at t 2: 693.15 [1.8 :1 pmax : MS) 2 IIIW [szﬁ 3.75 3.75 P 1.21 [a] p = 02' z 900 sin(2007rt) cos(2007rt) 2 450 sin(4007rt) W Therefore, pm.X = 450 W [b] pmax(extracting) = 450 W 00 3-75 (10001“ 3‘75 63-20001 — 1000 —2000 [3.755100% — 3.756200%] dt = [ 5x10—3 [c] pavg = 200 f0 45OSin(4007rt)dt _ 5x10-3 = 9x104 [Cffogww 22273—5[1—00327r]=0 0 [d] I)an = 1:2[1 — cos 2.571'] = \$29- = 57.3 W P 1.22 [a] (1 area under 2' vs. 73 plot mew+umm+aem+ewrewmﬂxm3 [10 + 40 + 16 + 48 + 9]103 = 123,000 C [b] w = /pdt=/m'dt II II‘ [I 11 = 0.2x 10”3t+9 Ogtg 15ks 0 g t 3 40003 2' = 15 m 1.25 x 10”3t p = 135 — 8.25 x 10‘315 — 0.25 x 10-6t2 4000 . wl = A (135 a 8.25 x 10*31 ~ 0.25 x 10-612) dt = (540 — 66 ~— 5.3333)103 :- 468.667 kJ 4000 S t 5 12,000 2 = 12 — 0.5 X 10‘3t p x 108 — 2.1 x 10—31 ~— 0.1 x 104512 12,000 202 = f (108 —— 2.1 x 10~3t — 0.1 x 10-6712) dt 4000 == (864 — 134.4 —~ 55.467)103 = 674.133 kJ Problems 1~11 12,000 S t 3 15,000 2‘ = 30—2x 10—31 p = 270 — 12 x 10‘3t —— 0.4 x 10‘6t2 15,000 103 = f (270 ~ 12 x 10~3t — 0.4 x 10612) dt 12,000 2 (810 - 486 — 219.6)103 = 104.4 kJ wT = 1111 + 1112 + 103 = 468.667 + 674.133 + 104.4 = 1247.2 kJ P 1.23 [a] p = m‘ = [16,000t + 20)e“800t][(128t + 0.16)e-800t] = 2048 x 103t26"160°t + 512Ot8—1600t + 326—160“: 2 3219—16001:[640,000t2 + 16006 + 1] E3? = 3.2{e‘1600t[1280 x 10315 + 1600] ~ 16008—1600t[640,000t2 + 160075 + 1]} 3 r: ~3.28‘1600t[128 X 104(800t2 + t)] = ~409.6 x 1046—1600ti(800t + 1) ‘ Therefore, % = 0 when t = 0 so pmX occurs at t 4—: 0. pmax = 3-28—0[0 + 0 + = 3.2 W t [c] w = / pda: 0 t t t 3w—2 = / 640,0005026‘16009” d\$+ / 1600378—1600x dx+ / 6—1600“ dx . 0 0 0 640 00063-16003“ * = W 25 1 4 2 320 2 _4096X106[ 6x 039+ 0513+]0+ 16006—16002: t 8—160032 ’3 W -1 _. 256 x 104 ( 6005” 1) 0 + —1600 0 When t —-> 00 all the upper limits evaluate to zero, hence w _ (640,000)(2) 1600 1 3.7 _ 4096 x 106 + 256 x 104 + 1600 w=10‘3+2>< 10*3+2>< 10—3=5mJ. P 1.24 [a] We can ﬁnd the time at which the power is a maximum by writing an expression for p(t) = v(t)z’(t), taking the ﬁrst derivative of p(t) and setting it to zero, then solving for t. The calculations are shown below: 1*12 CHAPTER 1. Circuit Variables p z: 0 t<0, p20 t>3s p = vi=t(3—t)(6~—4t) = 18t—18t2—I—4t3 mw 0933s 1’; = 18 — 36t + 12252 = 120:2 — 3t + 1.5) Egg : 0 whent2~3t+1.5=0 t z 3 :I: gm 2 3 2/3 251 = 3/2 - \/3/2 = 0.634 8; t2 = 3/2 + «3/2 = 2.366 s p(t1) = 18(0.634) — 13(0634)2 + 4(0.634)3 = 5.196 mW peg) = 18(2.366) — 13(2366)2 + 4(2366)3 = —5.196 mW Therefore, maximum power is being delivered at t = 0.634 s. [b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 5.196 mW (delivered) [.0] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t r: 2.366 s. [d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmax 2 5.196 mW (extracted) t t [e] w = /0pdm 2 f0 (1895 —— 18m2 + 4:133)d.r 2 9t2 —— 6163 + t4 26(0) = OmJ 10(2) = 4mJ w(1) = 4 mJ 111(3) = 0 mJ To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below: 2.5 2.0 1.5 E 1.0 0.5 0.0 1.00 2.00 3.00 -0.5 i (ma) 1.00 .00 3.00 mLﬁJDMnm 1(9) Problems 1—13 5?: E, 6 5 a. 4 E. 3 g 2 1 GB 1.00 2.00 .00 KS) P 1.25 [a] p = m' =.- 400 x 103t26“8°0t + mow—80‘” + 0.256340% = «3-800‘t[400,000t2 + 70015 + 0.25] % 2 {e-80W[800 x 1032: + 700] « 8006‘300‘[400,000t2 + 700t + 0.251} = [~3,200,000t2 + 2400t + auntie-80°t Therefore, d3? = 0 when 3,200,000t2 — 240015 — 5 = 0 so pmax occurs at t = 1.68 ms. [b] pm 2 [400,000(.00168)2+700(.00168)+0.25]e—800(.00168) = 666 mw t / pdm 0 t t t w 2: / 400,0003326—80093 dw+ / 70050680” (195+ / 0256—8003” d3: 0 0 t 0 400,0008-‘8001 “512 X 106 7006—8005” ' e 64 x 104 0 + 0'25 —800 0 When t z 00 all the upper limits evaluate to zero, hence (400,000) (2) 700 0.25 =—————+-——-—-—+——~==2.97mJ. w 512 x 106 64 x 104 800 P 1.26 We use the passive Sign convention to determine whether the power equation is p z m' or p 2 —m' and substitute into the power equation the values for v and i, as shown below: II [C] w it [64 x 10%2 + 16002: + 2] f + 0 45009: t (~800x — 1) 1-14 P 1.27 CHAPTER 1. Circuit Variables Pa Pb 1% Pet Pa Pf mW. II N vaia = (0.150)(0.6) = 90 mW _’Ucic —(0.100)(~—0.8) = 80 mW raid 2 (0.250)(—0.8) = —200 mW "’Ue’ie ——(0.300)(~2) = 600 mW em = (~0.300)(1.2) : —360 mW Remember that if the power is positive, the circuit element is absorbing power, whereas is the power is negative, the circuit element is developing power. We can add the positive powers together and the negative powers together —~ if the power balances, these power sums should be equal: ZPdev = 210 + 200 + 360 = 770 mW; ZIP”, = 90 + 80 + 600 = 770 mW Thus, the power balances and the total power developed in the circuit is 770 [a] From the diagram and the table we have Pa Pb, Po Pd Pe Pf Pg Ph 2 __""‘ I1 H II Pdel abs Therefore, ZPdel % ZPabS and the subordinate engineer is correct. [b] The difference between the power delivered to the circuit and the power absorbed by the circuit is ~4700 + 3500 = 1200 W One-half of this difference is 600 W, so it is likely that 196 is in error. Either the voltage or the current probably has the wrong Sign. (In Chapter 2, we will discover that using KCL at the top node, the current 0,, should be ——3.0 kV, not 3.0 kV!) If the sign of pc is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: was, —(5000)(—0.150) = 750 w vbib = (2000) (0.250) = 500 W = —(3000)(0.200) = —600 w vdz'd a (—5000)(0.400) = ~2000 w were —(1000)(—0.050) = 50 w um = (4000)(0.350) :2 1400 w —vgz'g .—_- ~(—2000)(0.400) = 800 W aha = -(—6000)(—0.350) = —2100 w 600 + 2000 + 2100 = 4700 w 750 + 500 + 50 + 1400 + 800 = 3500 w "Doric P 1.28 P 1.29 Pa Pb Pa Pd Pe Pf 1’8 Ph P1 ZPdel = Zpabs = 2000 + 2100 2 4100 W balances for the circuit. H II 45343, = ~(36)(250 x 10—6) = —9 mW 5.341, x (44)(—250 x 10-6) = «11 mW vcic = (28)(—250 X 10‘6) = ~7 mW odid : (—108)(100 x 10-6) = —10.8 mW me = (432x150 x 1076) = —4.8 mw 4W} = —(60)(—350 x 10-6) 2 21 mW vgz‘g = (~48)(—200 x 1076) = 9.6 mW vhih = (80)(—150 x 1076) = —12 mW wig- = —(80)(—~300 >< 10*6) = 24 mW Therefore, ZPabs = 21 + 9.6 + 24 = 54.6 mW :Pdel:9+11+7+10.8+4.8+12=54.6W ZPabs : Zpdel Thus, the interconnection satisﬁes the power check Pa Pb PC Pd Pe Pf Pg P5 191' H H H II —vaia = ——(1.6)(0.080) = ~128 mW «6134b : —(2.6)(0.060) = 4-156 mW me = (-4.2)(—0.050) = 210 mW wvdid = —(1.2)(0.020) = —24 mW me = (1.8)(0.030) = 54 mW ——'ufz'f = ——(——1.8)(~0.040) 2 ~72 mW rugig = (—3.6)(—0.030) = 108 mW vhz'h = (3.2)(-0.020) = ~64 mW 42141- : -(—2.4)(0.030) = 72 mW 213161 = 128 + 156 + 24 + 72 + 64 = 444 mW Zpabs = 210 + 54 + 108 + 72 = 444 mW Problems 750 + 500 + 600 + 50 + 1400 —I— 800 1: 4100 W Now the power delivered equals the power absorbed and the power 1—15 1—16» CHAPTER 1. Circuit Variables Therefore, ZPdel = 2P3},S = 444 mW Thus, the interconnection satisﬁes the power check P 1.30 [a] From an examination of reference polarities, elements a, b, e, and f absorb power, While elements (3, d, g, and h supply power. [b] p, = var, = (0.300)(25 x 10-6) = 7.5 MW Pb = “"vb'ib = ~(—-0.100)(10 X 10'6) z: 1 MW pa = me = (—0.200)(15 x 10‘6) r: —3 MW Pd = *vdid = —(~0.200)(—35 x 104‘) = *7 MW pe = ——ve'ie = —(0.350)(~25 >< 10-6) = 8.75 ,uW pf = 'Uf'if = (0.200)(10 X 10—6) == ZMW pg 2 ugz‘g : (—0.250)(35 x 10—6) 4 —8.75 MW ph = Uh’ih = X 10‘6) =1 MW gab, = 7.5 + 1 + 8.75 + 2 = 19.25 MW ZPdel = 3 + 7 + 8.75 + 0.5 = 1925 MW Thus, 19.25 MW of power is delivered and 19.25 MW of power is absorbed, and the power balances ...
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