Sinosoidal Steady State Power Calculations-Electric Circuits 8th Edition Nilsson-Electrical Circuital Analysis-Solution Manual, Exercises of Electronic Circuits Analysis

Download Sinosoidal Steady State Power Calculations-Electric Circuits 8th Edition Nilsson-Electrical Circuital Analysis-Solution Manual and more Exercises Electronic Circuits Analysis in PDF only on Docsity! —10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/—45°V, T= 20/15°A Therefore P= 5 (100)(20) eos|—45 —(15))=500W, AB Q = 1000 sin —60° = —866.03 VAR, BoA [b] V = 100/— 45°, I= 20/165° P = 1000cos(—210°) = —-866.03W, BoA Q = 1000sin(—210°) = 500 VAR, A—>B [e] V=100/—45°, 1 = 20/— 105° P = 1000cos(60°) =500W, A—B Q = 1000sin(60°) = 866.03 VAR, A—>B [d]} V=100/0°, 1=20/120° P = 1000cos(—120°)=—500W, BoA Q = 1000sin(—120°) = —866.03 VAR, BoA AP 10.2 pf = cos(, — 6;) = cos[15 — (75)] = cos(—60°) = 0.5 leading rf =sin(0, — 6;) = sin(—60°) = —0.866 8 10-1 docsity.com 10-2 CHAPTER 10. Sinusoidal Steady State Power Calculations AP 10.3 iy. 018 From Ex. 9.4 Jeg = Wao a4 P=2,R= ca) (5000) = 54 Ww AP 10.4 [a] Z = (39 + 726)||(—j52) = 48 — j20 = 52/— 22.62°2 250/0° fi = 22 _ a gg age Therefore I, B—720+1494 4.85/18.08° A (rms) Vi, = Zz = (52/ = 22.62°) (4.85/18.08°) = 252.20/— 4.54° V(rms) ns Raine. L= 394 726 > 5.38/— 38.23° A(rms) [b]_ Sy = Vit, = (252.20/— 4.54°)(5.38/-+ 38.23°) = 1357/33.69° = (1129.09 + 9752.73) VA P, = 1129.09 W; Qt = 752.73 VAR. {c] Pe = |Ic|?1 = (4.85)?-1=23.52W; Qe = [I-24 = 94.09 VAR [d] S,(delivering) = 2501; = (1152.62 — 7376.36) VA Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR. _ [MiP _ (252.20)? le] Qeap = 5 = “Ey = — 1223.18 VAR Therefore the capacitor is delivering 1223.18 magnetizing VAR. Check: 94.09 + 752.73 + 376.36 = 1223.18 VAR and 1129.09 + 23.52 = 1152.62 W AP 10.5 Series circuit derivation: = 2501" = (40,000 — 730,000) Therefore I* = 160 — j120 = 200/— 36.87° A(rms) I = 200/36.87° A(rms) Vv 250 z . =— = —_—__ =]. — 36.87° = (1 — 70. Qa 2= 7 = appear = 1 25L= 36.87 = (1 — j0.75) Therefore R=1Q, Xo = -0.750 3 docsity.com Problems 10-5 (b] LL. -l, =0.4—70.32A P75 = lh — I,|?(375) = 49.20 W [ce] P, = 5 (248)(08) = 99.20W > Pats = 50+ 49.2 = 99.20W (checks) AP 10.10 [a] Vin = 210V; V2 = $V; I=1h Short circuit equations: 840 = 801, — 20In + Vi 0 = 200, — ly) — Ve 210 \ b=MA; Rn = 5 = 150 [b] Pax = (0)"15 = 735 = (2) 5= AP 10.11 [a] Vrn = —4(146/0°) = -584/0° V(rms) V2 =4Vi; I, = 4h Short circuit equations: 146/0° = 801, — 2012 + Vi 0 = 20(p —)) — Ve . Ip =-146/365=-0.40A; Rm = = = 14602 a 2 [b] P= (Sm) 1460 = 58.40 W 3 docsity.com 10-6 CHAPTER 10. Sinusoidal Steady State Power Calculations Problems P10.1 p=P+ Pcos2ut— Qsin Qt; op = —2wP sin 2wt — 2wQ cos Qwt dt #9 when — 2wPsin2wt = 2wQcos2wt or tan 2ut = —2 P 2 2 -2 P+Q P . Q 2ut = —————; in Qwt = ———=— ONS TPE SNS pe Let @ = tan-!(—Q/P), then p is maximum when 2ut = 0 and p is minimum when 2ut = (+7). . P Q(-Q) / 21 = <——————_—_—_—_— —- —— P 2. 2. Therefore Pmax = P +P PTO VPPTOE + (P?+Q P and Pmin =P ~ P- ppg ~Q: pty = P— (P+ P102 [a] P= 5 (340)(20) cos(60 — 15) = 3400cos 45° = 2404.16 W (abs) Q = 3400sin 45° = 2404.16 VAR (abs) [b] P= 5(16)(75) cos(—15 — 60) = 600 cos(~75°) = 155.29W (abs) Q = 600sin(—75°) = —579.56 VAR (del) [c] P= 5 (625)(4) cos(40 — 150) = 1250 cos(—110°) = —427.53W (del) Q = 1250sin(—110°) = —1174.62 VAR (del) [d] P= 5 (180)(10) cos(130 — 20) = 900cos(110°) = —307.82W (del) Q = 900sin(110°) = 845.72 VAR. (abs) 3 docsity.com Problems 10-7 P 10.3 [a] coffee maker = 1200W radio = 71 W television = 145 W portable heater = 1322 W XP = 2738 W 2738 Therefore Ing = 70 = 22.82 A Yes, the breaker will trip. [b] 5° P = 2738 — 1200 = 1538 W; Loe = = =12.82A ‘Yes, the breaker will not trip if the current is reduced to 12.82 A. P104 I, =30/0°mA; 1 ne tof =-j10 “= * jwE ~ 725 x 105)(40) jwL = j(25 x 10%)(40) x 10-6 = 710 22 52 AW AW 30/0°m + -310 f 4310 Z, = —jil|(5+ 91) =0.2- 712 Zeq = 2+ Z, =2.2- 712 2 . ) Py = |Inms|?Re{ Zeq} = (3 x 10) (2.2) = 990 nW P 10.5 + __ 1 _ _os090 “ wO ~ (5000)(80) — __ —32500(7500) = = 750 — j22500 «= F500 — 7500 ~ (0 ~ 92250 Z, = 15000 Z 750 — 52250 | SS = ——__ = 0.5—- 91.5 Z 1500 OS Vo= #y,, V, =4/0°V qj 3 docsity.com 10-10 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.8 P 10.9 _1(90) _ fl P= 97350 ~ 3 _ 1 (0)? Q= 2 (1012.5) AVAR Prax = P + \/P?+Q? =3+ \/(3)? + (4? = 8 W(del) [b] Pmin = 3 — 5 = —2 W(abs) [ec] P=4W_ from (a) [d] Q=4VAR_ from (a) {e] absorb, because Q > 0 [f] pf = cos(6, — 6;) poy 1350 71012.5 *. p£ =cos(0+ 53.13°) = 0.6 lagging = 0.0667 — 0.08889 = 111.11/—53.13° mA [g] rf = sin(53.13°) = 0.8 [a] From the solution to Problem 9.56 we have: T, Tbe x 2¢£450 z 1 F-j102 A2.s%, vzsQ (H15/0°a V_ = 72 + 596 = 120/53.13° V Sy = -5V, = —5 (72 + §96)(15) = —540 — j720VA Therefore, the independent current source is delivering 540 W and 720 magnetizing vars. I= ¥ = 15/53.13° A Pao = 5(15)°(8) = 900 W Therefore, the 8 resistor is absorbing 900 W. In = ~ = -9.6 + 97.2 = 12/143.13° A docsity.com Problems 10-11 D Qea= 3 (12)"(-10) = —720VAR Therefore, the —j10Q. capacitor is delivering 720 magnetizing vars. 2.51, = —24+4 j18V Vo-2-5In _ 72+ 596 + 24 — j18 I= 7D i = 15.6 — 719.2 A = 24.72/—50.91° A 1 Qi5 = g/L?) = 1530 VAR Therefore, the j5Q inductor is absorbing 1530 magnetizing vars. Sost, = = $(2.51,) 5 = 3(—24 + j18)(15.6 + 719.2) = —360 — j90 VA Thus the dependent source is delivering 360 W and 90 magnetizing vars. [b] }> Peen = 360 + 540 = 900 W = de Pav [c] 5° Qgen = 720 + 90 + 720 = 1530 VAR = S Qabs P 10.10 [a] From the solution to Problem 9.57 we have 20Q -j20Q AWN 60/0°vG » 45a j90V t, t, I, = 2.25 — 2.25A; I, = —-6.75+j0.75A; I,=9-—Jj3A Seov = — 5(60)K: = —30(2.25 + 72.25) = 67.5 — j67.5VA Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars. Soov = —3(j90)If, = —j45(—6.75 — 50.75) = —33.75 + 7303.75 VA Thus, the 90 V source is delivering 33.75 W and absorbing 303.75 magnetizing vars. 1 Pog = gltl?(20) = 101.25 W docsity.com 10-12 CHAPTER 10. Sinusoidal Steady State Power Calculations Thus the 20 resistor is absorbing 101.25 W. Q_j200 = 5lisi°(-20) = —461.25 VAR, Thus the —j20 capacitor is developing 461.25 magnetizing vars. Qisa = lel) = 225 VAR Thus the j5Q inductor is absorbing 225 magnetizing vars. [b] > Paev = 67.5 + 33.75 = 101.25 W = D> Pars fe] 3° Qeev = 67.5 + 461.25 = 528.75 VAR DY Qube = 225 + 303.75 = 528.75 VAR = > Qaev v2 totT 4/2 10.11 Wa, = £T; = = dt E Ws RR” We t. Rk v2. totT 2 Mdep = [ vs at R te R 1 tot+T Ve= Th, wv? dt 1 stetT Vac = \[ = U2 dt = Vis = Vag T Sty P 10.12 [a] Tor = 60/110 ¥0.545A; — [b] Jur = (60 + 80)/110 & 1.273. A P 10.13 [a] Area under one cycle of v2: A = (400)(4)(20 x 10-*) + 10,000(2)(20 x 10-5) = 21,600(20 x 10-*) Mean value of v3: A_____ 21,600(20 x 10-5) MV. = 359x103 = — 190 x 10-8 = 3600 “. Vims = V3600 = 60 V(rms) 2 {b] P= Yen — 560 _ sow docsity.com Problems 10-15 [ Is =—2- = 5(82-+24)(1 — jl) =2.8- jO.4A Qis0 = sits?) = 20 VAR(abs) 0-08 = 5ltal*(—10) = —80 VAR(dev) Y Qavs = 20 + 60 = 80 VAR = S> Qaev P1018 (al 502 800 WA 2 ANA 5 _ 340/0@) -31000 J " f 4609 V(rems) aL ° I, _ ~*~ Vo V. — 340 Vo + —ji00 + 50 + 80+ 5007 ° Vo = 238 — j34V 340 — 238 + 734 l= ee = 2.04+ j0.68A Sy = Vol, = (340)(2.04 — 70.68) = 693.6 — j231.2VA [b] Source is delivering 693.6 W. [c] Source is absorbing 231.2 magnetizing VAR. [d] 1, = = 0.344 j2.38A Vo =7100 Sy = Voli = (238 — 734) (0.34 — 52.38) =0-j578VA Vo _ 238 — j34 I =1L7-fl. 2= 304760 80+ 760 ~17~I17A Sp = Vol} = (238 — j34)(1.7 + j1.7) = 462.4 + j346.8VA Sson = |1y|?(50) + 70 = (2.15)?(50) = 231.2 W 3 docsity.com 10-16 CHAPTER 10. Sinusoidal Steady State Power Calculations fe] > Pia = 693.6 W DY Paiss = 462.4 + 231.2 = 693.6 W 3 Pia Y Paiss = 693.6 W [f] 37 Qaus = 231.2 + 346.8 = 578 VAR YD Qaev = 578 VAR “. 5 mag VAR dev = > mag VAR abs = 578 P 10.19 [a] Let Vz = Vp,/0°: 1 48Q 240/e° " i V (rms) v,,L0 5, St = 250(0.6 + 70.8) = 150 + j200VA = 10, 200, 5 _ 150 200 =, Vat “Wa, Ve 240/0 = Vin + (= - i”) (1 +38) 240Vin/0 = V2 + (150 — j200)(1 + 78) = V2 +1750 + 71000 240Vn cos = V2+1750; 240Vn sin = 1000 (240)?V2 = (V2 + 1750)? + 1000? 57,600V;2 = V,4 + 3500V2 + (3.0625 + 1) x 10° t or VA — 54,100V,2 + 4,062,500 = 0 Solving, V2 = 27,050 + 26,974.8; Vin = 232.43 V and V,, = 8.67 V Tf Vin = 232.43 V: 1000 in Qe 0170: “. 0 =1.03° sind (232.43) (240) 0.0179; 1.03 If Vin = 8.67 V: 1000 si = ——__——_ = 0. * 2. 0 = 28.72° sin 8 (8.67)(240) 0.4805; 8.72' docsity.com Problems [b] ° 240/1.03°v 232.43/0°v ° T,~ 1.08/-53.13 A 240/28.72 Vv 8.67/0 Vv 28.84/-53.13°V .52,800 P 10.20 Sp = 52,800 — 08 (0.6) = 52,800 — 739,600 VA Si = 40,000(0.96 + 70.28) = 38,400 + 711,200 VA So = Sp — S, = 14,400 — 750,800 = 52,801.52/— 74.17? VA rf = sin(—74.17°) = —0.9621 pf = cos(—74.17°) = 0.2727 leading P 10.21 [a] Z = 12+ j(2m)(60)(15 x 107%) = 13.27/25.23° pf = cos(25.23°) = 0.9 lagging rf = sin(25.23°) = 0.43 =s0-—__/ — 64,24° 2a = 80 — sencia xis 7 184-08/ = 64.04".0 pf = cos(—64.24°) = 0.43 leading rf = sin(—64.24°) = —0.9 23 = 400+ Z, Z= jwL(1/jwC) _ jwk ? jwL+1/jw0 1—wLC 10-17 docsity.com 10-20. CHAPTER 10. Sinusoidal Steady State Power Calculations [c] Vi lags V, by 2.21° or 102.31 ps Vv g 2.21° L P 10.26 [a] 9,05 I, 45 pso/o ay (rms) 500052000. L= 5 40 — j16 A (rms) 1,2 =i = = 30-12 (rms) [y= S020 _ 59-4 50:8 (eins) 250 *. Ig, = 72 — 316A (rms) T, =l, —In = 10-74 A (rms) Tyo = 62 — j12A Voi = 0.051g: + 125 + j0 + 0.141, = 130 — 71.36 V(rms) Vo2 = —0.141, + 125 + j0 + 0.05142 = 126.7 — 70.04 V(rms) Soi = [(130 — 71.36)(72 + 716)] = [9381.76 + 71982.08] VA Sy2 = [(126.7 — j0.04) (62 + j12)] = [7855.88 + 1517.92] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. {b] Po.os = |Ip1|?(0.05) = 272 W Pos = |In|?(0.14) = 16.24 W Po.os = |Ig2|?(0.05) = 199.4 W >? Pais = 272 + 16.24 + 199.4 + 5000 + 3750 + 8000 = 17,237.64 W 3 docsity.com P 10.27 [al Problems > Piey = 9381.76 + 7855.88 = 17,237.64 W = > Pais YX Qaus = 2000 + 1500 = 3500 VAR DY Qaer = 1982.08 + 1517.92 = 3500 VAR = 7 Qats , 120/0°Q) Ly V(rms) I, 2120 f 480 120/0°C L, Vv (rms) I, 1201; = 1800+ j600; 9. Ty = 15 — 75 A(rms) 12015 = 1200-7900; *. Ip = 10 +. 7.5 A(rms) 240 240 = 47 =20-j 3 = F5 + 548 0 — j5 A(rms) In =, +15 =35—j10A Sor = 120(35 + j10) = 4200 + 71200 VA 10-21 Thus the V,i source is delivering 4200 W and 1200 magnetizing vars. Igo = In + Iz = 30 + 72.5 A(rms) S2 = 120(30 — 2.5) = 3600 — 7300 VA Thus the V4 source is delivering 3600 W and absorbing 300 magnetizing vars. [b] 37 Pren = 4200 + 3600 = 7800 W 2. SS Pabs = 1800 + 1200 + oer = 7800 W = > Peen Y Qua = 1200 + 900 = 2100 VAR 240)? YZ Qube = 300 + 600 + oer = 2100 VAR = 3° Qaa docsity.com 10-22 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.28 P 10.29 S1 = 1200 + 1196 + 516 + j0 = 2912+ jOVA . 2912, , 5 T= sop +90 = 24.274 j0A So = 600 + 279 + 88 + 512 + j0 = 1479+ jOVA 1479 = +70=12. j I, 120 +j0 33+ j0A Sg = 4474 + 12,200 + j0 = 16,674 + j0VA 16,674 . Ts = Jyp +10 = 69.484 JOA In. =) + Ig = 93.75 + j0A Tyo = 1p + Ig = 81.814 j0A Breakers will not trip since both feeder currents are less than 100 A. 1Q 48Q as > I, + ae , 2s00/o°v | 250kva 0.96 lag -i%, ess 240,000 — 770,000 _ L= 2500 = 96 — 728 A(rms) 2500 _ 2500 _ =9X sw Ke = JAG c= Ty = 96 — 928 + jIc = 96 + j(Ic — 28) V., = 2500 + (1 + 78)[96 + j(Ic — 28)] = (2820 — 81c) + §(740 + Ic) |V.|? = (2820 — 8c)? + (740 + Ic)? = (2500)? *. 6514 — 43,640[¢ + 2,250,000 = 0 docsity.com Problems [b] V..(before) = 4800 + (375 — 7500)(0.02 + 70.16) = 4887.5 + j50 = 4887.5/0.59° V(rms) |V.(before)| = 4887.76 V(rms) V, (after) = 4800 + (500 — 7145.83)(0.02 + j0.16) = 4833.33 + 77.08 = 4833.95/0.91° V(rms) |V.,(after)| = 4833.95 V(rms) 125/0° 125 P 10. Uy re Oe ah 032 al ti 20+ 734+5+ 716 254750 1724 jwM j50 Ls 241) = 2" _1 4 2 = Fe = 3004 j150 "1 7) = 0.44 — 70.08 = 0.45/— 10.30° A Vi = (150 — 7100)(0.44 — 0.08) = 58 — j56 = 80.62/— 43.99° V |Vi| = 80.62 V [b] P,(ideal) = 125(1) = 125 W P, (practical) = 125 — |I,|?(5) = 125 — 25 = 100 W P, = [IL |?(150) = 30 W 0 100 P 10.35 [al 19 % delivered = (100) = 30% V (rms) 7 3 20 = J2(I — Ip) + jl (la — Is) — f1( — Is) 0 = 1p + jl (Ip — Is) + 910 — Tn) + 2(Ip — Lh) — f(a — I) 10-25 docsity.com 10-26 [b] CHAPTER 10. Sinusoidal Steady State Power Calculations 0=—J1(Iz —1,) + j1(I3 — Lb) — 1, -— I,) + 1 Solving, T, = 20—j20A(rms); I, = 20+ j0A(rms);_ I; = 0 A(rms) I, =], = 20— 720A i, =, -l=—j20A I, =I, =20A ly =I3 —Ip = -20A I, =I, —I3 = 20— 720A T;=1;=0A + ve = 19 7 et¥, - ih ¢ Ye-¢ 320 j19Q x, J + J Ny : afc vy-j19 ey, Vins) €, - <, Tt, - V. = 204 j0V Vy = j2ly — fla = 40 + j20V V. = II, = 20+ j0V Va = jlla — jl, = —20 — j20V V. = fll, = —20-j20V Vp = 1; =0V S, = —201% = —400 — j400 VA Sp = Vile = —400 + 7800 VA So = VeIt = 400 + j0VA Sa = Vali = 400 + 7400 VA Se = Vel = 0 — j800VA Sp = Vet =0+4j0VA [ce] > Paey = 400 W YS Pavs = —400 + 400 + 400 = 400 W Note that the total power absorbed by the coupled coils is zero: —400 + 400 =0=FA,+ Py docsity.com Problems 10-27 [d] > Qeev = 400 + 800 = 1200 VAR Both the source and the capacitor are developing magnetizing vars. YS Qavs = 400 + 800 = 1200 VAR > Q absorbed by the coupled coils is Q, + Qa P 10.36 i 4300 102 ~=—- 450 4209 Wi SY. SON. e e e 340/0° ¢ j100nt 4409 #400 V (rms) x) 4700 L, 340/0° = 101, + 7501, + j70(Iy — In) — 73015 +3701, — j40I2 + j100(I, — Ip) 0 = j100(Iz — I,) — j701, + 7401 + 52015 +j40(Ip — I) — 7301, + 40Iy Solving, I, = 5 — j1 A(rms); T, = 6/0° A(rms) Pron = (6)?(40) = 1440 W [b] P, (developed) = (340)(5) = 1700 W 4. [e] Zn = x -10= a — 10 = 55.38 + 713.08 = 56.91/13.28° Q : = [d] Pro = [Z_|?(10) = 260 W DY Pais = 1440 + 260 = 1700 W = S> Piey 3kQ P 10.37 [al v 30/0" @) -L Vv (rms) tr 1k 30 = 30001, + Vi + 1000(I; — Ip) 3 docsity.com 10-30 CHAPTER 10. Sinusoidal Steady State Power Calculations 90/0°(60) . Von = = = 45(1 — fl) = 45V2/— 45° [b] Vin 60+ 760 45(1 — j1) = 45V2/— 45°V 30kQ -j10kQ 30kQ WA 1}-———"v- — I 45fa5c V(rems) ‘Y j10kQ e 45y/2/— 45° - = So = 0: [{— 45° Tm So tor = 0-75V2/—45°mA [ms] = 0.75 mA Prada = (0.75)? x 10-°(30 x 10%) = 16.875mW 240 — j80— 480 | 240-780 _ P 10.42 [a] tae _ —100(240 + 780) . Zen = 240 — 780) = 80+ 602 Zi, = 80 — 7602 480/0° = — = CAT 3 [bh] I=755 oe 3/0° A(rms) P = (9)(80) = 720W 50 -j109 vy. 109 P 10.43 [al] 2S ek ae * 250 fo 0.2v, v, £150 Vom v (rms) ae _ —e Vi — 250 Vi Men OV, ==0 s=yi0 +0495 _ —J5Vi _ “Vi 710495 2471 _ §0.2V1 —0.2V, = 255r 1, 902, 1 7) _ 250 "15-710 2+71 ' 10+ 95] 5-710 3 docsity.com Problems 10-31 Thus, V, = 10(10 + 5) V1 = 750 = 50/90° V(rms) pd Vn, = Tm 10+ 95 Short circuit current: 52 -j100 109 Wie e+ a | —— H 8 Wa 250 /0C 450 V(rms) | 4 250/02 _ 50 A(rms) Ie = 15 — 710 ~ 3= 72 = Ig ih eds e m= 7 = 5g 2n 330 22 5 + i 50 /90@) vy, +430 V(rms) 50/90° L= a = 12.5/90° A(rms) P = (12.5)?(2) = 312.50W [b] Vi = (2 — j3)(j12.5) = 37.5 + 525 V(rms) 52 -j100 100 20 WA als AM L Wie —I ° + = ee ‘ 250 /0 C) 0.2y v. Ye '50v, 4-330 Vi(rms) + I eu Vi _ 37.5 + 925 . a SE TBA 75 75 j7.5 A(rms) Ih =1 +I =5— 97.5 + 712.5 =5 + 75 A(rms) 3 docsity.com iL 10-32 P 10.44 [al CHAPTER 10. Sinusoidal Steady State Power Calculations Ves = Vi + 10Iy = 37.5 + 25 + 50 + j50 = 87.5 + 775 V(rms) Vo = -Vi, = —37.5 — 525 0.2V, = —-7.5 — 35 Ses = —Vealt, = —(87.5 + §75)(—7.5 + 75) = 1031.25 + j125VA Therefore, the dependent source is absorbing 1031.25 W and 125 magnetizing vars. Only the independent voltage source is developing power. I, = —0.2V, +1, =7.5+j5+5+ 95 =1254+ 510A Sq = —2501, = —3125 + 72500 VA Paey = 3125 W F 312.5 % delivered = 3195 ape (100) = 10% Thus, 10% of the developed power is delivered to the load. Checks: Pron = (5V2)?10 = 500 W Poo = 312.5W Psq = (V256.25)°5 = 1281.25 W DY Piev = > Paws = 500 + 312.5 + 1281.25 + 1031.25 = 3125 W VAR Check: The 250 V source is absorbing 2500 vars; the dependent current source is absorbing 125 vars; the j5Q inductor is absorbing [37.5 + j25|?/5 = 406.25 vars. Thus, YS Qabs = 2625 + 406.25 = 3031.25 VAR DX Qaev = (12.5)?(3) + 256.25(10) = 3031.25 VAR = 37 Qavs 100 340n 360 /o" 3402 2700 We ae sa0H a) Fa rms) i, 360/0° = 101, + 3401, + j30(I2 — 11) — 7301, + 740(1, — I.) docsity.com Problems 10-35 [b] Poss = [11 |?(15) + |Ie[?(30) = 780 + 187.5 = 967.5 W 967.50 ea 1530 P 10.47 [a] Zm=8+j15+ ~~ (100) = (=724)(18 + 6) 18 —j18 63.24% = 24+ 97 = 25/16.26°Q = Yeni = = 252 b =i [b] Ven = 18+ — 420, PB 4 V(rms) _ 4202/0? ~ 49457’ pe (2000) P 10.48 [al B50)@) 05) 94 (630/0°) = 420 — 5420 = 420V/2/— 45° V(rms) 24Q jJ7Q i 25Q in) = 00V2 ~ 50 3600 W = 3.6kW -O5Y, o 600/0 -j20Q fe $ 5100 Yon Vs~600 | Vs 10” ji0 —0.05Vy =0 *. Vy = 240 + 7480 V(rms) Von = V5 + 0.05V4(—j20) = Vo(1 — j1) = 720 + j240 V(rms) Short circuit current: docsity.com 10-36 CHAPTER 10. Sinusoidal Steady State Power Calculations ZG. Ae 1h 102 + 4202 cv0/0"d y V (rms) & ®§ 31002 Tye Va / Tue = 0.05Vg + —55 = (0.05 + 30.05) Ve Ver M8 4 Me Ao . V6 = 480 + 7240 V(rms) Tec = (0.05 + 30.05) (480 + j240) = 12 + 736 A(rms) Vim _ 720-+ 3240 Zon = Th = OE = 12 — 516 = 20/— 53.13° t= 12+ 736 12 — 716 = 20/— 53.13° Ry = 202 [b] r 12Q -416Q 72043240 3) 202 Virms) 720 + 5240 ; == = PAI 32-716 15 + j15 = 15/2/45° A(rms) P = (15/2)?(20) = 9000 W = 9kW Ie] r Wy Li o AWN 12Q -316Q 12Q 720+42404 : Virms) 160 e t= met A = 30+ 710 A(rms) P = (V1000)?(12) = 12kW 3 docsity.com  Problems 10-37 [d] ZUM ad V5 - 1092 ~3200 — Seat iioa (rns) WV 1 2 e T, + I, +12 600/o0 4 § solo", Ye j10 «2004600 F ste I V (rms) 3 - ry - 2 Vo = 600 | Ve | Vo ~ 200— 7600 _ 10 j10 —j20 7 V5 = 200 + 7200V 0.05Vg = 10+ 710A 10+ j10+I1c = 30+ 10; Ip =204+j0A V6 ; = —* = 20-2 I. Fo 20 — 720A Tk =Ic +I, = 40-j20A I, = Ix + 0.05Vy = 50 — 710 A(rms) Sq = —600I; = —30,000 — 76000 VA 600 = V5 + 200+ 7600; V., = 400 — j600V Sea = (400 — 3600)(10 — 710) = —2000 — 710,000 VA SY Paev = 30,000 + 2000 = 32,000 W = 32kW ‘ 12 % delivered to Z, = 3g (100) = 37.50% Check: D. Pabs = 12,000 + 1 (10) = 32kW = > Pay SY Qeev = 6000 + 10,000 + [Ic ]?(20) = 24kVAR De Qave = [HL ?(10) + T|?(16) = 24KVAR = > Qaov P 10.49 [a] First find the Thévenin equivalent: 1 10° . joe ae —j1002 docsity.com 10-40 CHAPTER 10. Sinusoidal Steady State Power Calculations [b] P = |1,[?(140) = 560W [ce] P, = (240)(6.4) = 1536 W 560 % delivered = i530 (100) = 36.46% 240/09 240/0° P 10.53 [a] Von = 20+ 74079) + 30 + 740-7) = 480 + j240 V(rms) 20Q j1602 Wi e e a0 °c “ay sa00§ ges yy ‘sc From the solution to Problem 10.49 we can write 240 = (20+ j40)I, — j100I,. 0 = —j1001, + 7320I,. Solving, I,, = 3.15 — j1.377 _ Van _ 480+ 5240 = = = = j120 = 156.20/50.19° Q Zn 1. 315 — 1.377 100 + 7120 = 156.20/50.19 Ry, = 156.202 bl 1002 j120Q AM a Sr 480+43240C 156.202 V (rms) -e 536.66/26.57° 5 a 282.92/25.10° ore P = |I|?(156.20) = 562.05 W 3 docsity.com Problems P 10.54 [a] 202 j160Q NW e A ¢ ° oohe $1402 ‘5 5 240 ‘ J V (rms) “a jana x wy 240 = 201, + j40(I; — Ip) + j80KIn 0 = j40(Iz — Ih) — j80KI2 + 71601 + j80K(I; — Ip) + 1401, or 12 = (1+ j2)I, + j(4k — 2)I, 0 = 9(4k — 2)T, + [7 + j(10 — 8k)]Ip No = —j(4k — 2)(12); I, = 0 when Np =0 V. = 0 when Ip = 0 k=0.5 [b] When Ip = 0 12 = isp 2.4 — j4.8 A(rms) P, = (240)(2.4) = 576 W Check: Prous = |Ii|2(20) = 576. W P 10.55 [a] Vin = (950) = = 380/16.26°V Zap, = 31 + 100+ ay (28 — 796) = 38 +776 Z, = 38 — 7769 1, = 2800626" = 4.9 4 51.4 = 5/16.26° A(rms) 76 = [I,|?(38) = 950 W 10-41 docsity.com 10-42 CHAPTER 10. Sinusoidal Steady State Power Calculations 450 8Q 3562 202 YY 312 380 Wy AMY 7 . AW e AMY nen os —>t, —>(4.84+41.4)a Vv (rms) j40Q j1009 7-j76Q —o— 760/0° = 1,(28 + 596) — 750(4.8 + 71.4) _ 690 + 7240 * 100/73.74° 5q(delivered) = 760(4.24 + j5.95) = 3219.36 + j4523.52 VA = 7.31/— 54.56° = 4.24 — j5.95 A Pros = |Ii|?(8) = 426.96 W Py (transformer) = 3219.36 — 426.96 = 2792.40 W . 950 % delivered to Z, = 3793.4 (10) = 34.02% P 10.56 [a] jwL1 = j(5000)(2 x 10-*) = 7102 jwL2 = j(5000)(8 x 10-*) = 7402 3109 10Q 70 = (10+ j10)I, + 7101, 0 = j101, + (30+ 740)I, Solving, I, =4—-j3A; IL=-1A Thus, tg = 5.cos(5000t — 36.87°) A zz = 1cos(5000¢ — 180°) A M 2 = =05 Til, V16 [b] k= docsity.com Problems 10-45 Ip=2+j1A Solving, 1, =4/A Z, = 80/4 = 20+ j02 P 10.58 10Q_.r, rm r, ANN 122.8. 1:4 ° e ° + + 100v A mp owe || Ew EIT Eo, Fae ideal | @— ideal] Va V.=4V:,; 4I, =I,; therefore Lo 2502 a -V, Vv 2. MeeVee 2.514; therefore V8 — 2 _ sng I, 6.25 Therefore I}, = [100/(10 + 40)] = 2.A (rms); since the ideal transformers are lossless, Pixq = Pion, and the power delivered to the 4kQ resistor is 2?(40) or 160 W. P 10.59 [al 10Q_1:2.5 La 4% ae e ie... 10010 >) y, } y, 3 Vi Veems)O 1, : vy, Eo ideal ideal 10+Vi=100; I,=-25h; Vi =—-V2/2.5 10(—2.5I,) — V2/2.5 = 100 Ip=al.=0; V2=Vec/a; —10[—2.5(0)] — Voc/2.5a = 100 Voc = —250a 3 docsity.com 10-46 CHAPTER 10. Sinusoidal Steady State Power Calculations 1002 1:2.5 lia +* + Ly J*s 100/0° - vem Mi | 4, | a - ae - Ig ideal ideal 101, + Vi = 100; I, = —2.51,; Vy = —V2/2.5 10(—2.51,) — V2/2.5 = 100 V2 = V3/a = 0; I, = al,,; 10[—2.5(aI,¢)] — 0 = 100 I... = 100/(—2.5a) = —4/a Voe _ —250a ~ T —4/a For maximum power to the 4 kQ load, 4000 = Zyy, = 62.5a?; so a=8 [b] The circuit, with everything to the left of the 4 kQ load resistor replaced by its Thevenin eanivalent: = 62.50" 4kQ 2000/0° . Viems) We4kQ _ Vi _ (-1000)? _ = 3909 = 4000 = 250 . 20)? F . © P 10.60 [a] Zr = 32+ 71244 (=) (3 — j4) = 80 + j60 = 100/36.87°Q Ze» = 1002 A. 2 = TEN NP (1+ i /N2)? = 3600/100 = 36 ~ M/No=5 or Np =N,/5 . No = 300 turns 3 docsity.com Problems 240/0° [b] Vin = 0/0" (520) = 960/36.87° V 3+ 94 80Q j602 | St, 960 46.874 af. C 1002 960/36.87° ° I= 191700 = 1.6V10/18.43° A(rms) || = 1.6V/10 A(rms) P =(I/?(100) = 2560 W [e] 3Q 420Q 32Q 4.84+51.6)A IAS 31.6) 1002 240/0° = (3 + 74)Ih — j20(4.8 + 71.6) *. I, = 40.32 — j21.76 A(rms) Peon = (240)(40.32) = 9676.80 W Paiss = 9676.80 — 2560 = 7116.80 W a 7116.80. % dissipated = 9676.80 (190) = 73.54% P 10.61 [al 5 20k 240° [ Vv (rms) Zan ? For maximum power transfer, 7, = 20kQ Ni? Zn = (1-72) Ly Ny)? _ 20,000 _ (-%) = = = 400 10-47 docsity.com 10-50 CHAPTER 10. Sinusoidal Steady State Power Calculations 19 10A 20¢ 10v 192 V (rms) P = (10)?(1) = 100W {b] 10 a 10a + lov - 10 4Q . 0.25Q WV AWN: ah AW ,| dow, +e e+ 45 25 fi ¢ vy, vy 210 Vv (ems) a 25 = (10+1,)(1) +4, + Vi A = 49, (0.25) + (41, + 10)(1) Solving, L=-1A *. Prource = (25)(10 — 1) = 225 W s 100 % delivered = 395 (100) = 44.44% [e] Paey = 25(10 — 1) = 225W Pio = (9)°(1) =81W; Pin = (-1)°(4) = 4W Pio = (10)°(1) = 100W; Phase = (—4)°(0.25) = 4W Pro = (10 — 4)?(1) = 36 W DY Pats = 81 +4 + 100-44 +36 = 225 W = D> Paey | 3 docsity.com Problems 10-51 P 10.63 [a] Open circuit voltage: 20kQ T3 5kQ 1kQ re 10kQ We AMY . ANN <9 fo (Z,42,) I, +e et i + 00°C %, ov, rn V (rms) = = ° 100/0° = 5000(1; + Is) + 20,0001; + Vn, I, = —5I3 100 = 5000(—5I3 + Is) + 20,00013 + Vin Solving, Vin = 100/0° V Short circuit current: M 20kQ a §kQ 1kQ 10kQ AWA AN ss Wy , | ety) I, +e e+ S 100 A : v, S¥, Isc Vv (rms) 100/0° = 50001, + 50001; + 10001, + V; 5V, = 25,000(1,/5); <. Vi = 10001, 100/0° = 70001, + 500013 Also, 100/0° = 5000(I, + I3) + 20,0001; Solving, I, = 13.33 mA; I; = 1.33 mA; I,, =1)/5+1; =4mA docsity.com 10-52 CHAPTER 10. Sinusoidal Steady State Power Calculations eo cet ee Ron L. 0.004 5kO 25kQ AMY 0 > roof 2 25kQ Vv (rms) 100/0° 5 I= 50,000 = 2/0° mA(rms) P = (0.002)?(25,000) = 100 mW [b] i 20kQ Ta 5kQ uwQ 10kQ ANA AM . Wy I, + @+ I,7+ (I,+Z,) 1 a wf * ° ¥, sy, > SOVE25kQ V (rms) = 100 = 5000(I; + Iz) + 20,0001; + 50 5V, = 10,000 (3) +50 100 = 5000(I, + Is) + 10001, + V; I, = 14.82 mA; I; = —0.963 mA; I, +I = 13.857/0° mA Pioov (developed) = 100(13.857 m) = 1386 mW 100 1386 le] Pe, =100mW; Pigg = (2.96 m)?(10k) = 87.9mW ko = (0.963m)°(20k) = 18.6mW; Py. = (13.857m)?(5000) = 960.1 mW % delivered = (100) = 7.22% Pool Pike = (14.82 m)?(1000) = 219.6 mW Dd Pave = 100 + 87.9 + 18.6 + 960.1 + 219.6 = 1386 mW = YS Pav 3 docsity.com [c] Note that the HIGH setting has R; and Ry in parallel: Vv? 120? Pics = TR, ~ 28.8] 288 = 1000 W 10-55 If the HIGH setting has required power other than 1000 W, this problem could not have been solved. In other words, the HIGH power setting was chosen in such a way that it would be satisfied once the two resistor values were calculated to satisfy the LOW and MEDIUM power settings. v? v2 10. PF, = >—;5y =— P 10.67 [a] P, R, +R, Ry + Re B v2 Ve Pu = Ry R= Pa y= V?(Ri + Re) a= Ri v? v2 ov? Rit R= pi Mapp, B V?V?/PL PuPiPu H™ 7 _v2)\)()~ P(PR,—P) (mH) (Re) Pen.) Ph ee (750)2 =— =] [b] Pa (750 — 250) 125W P 10.68 First solve the expression derived in P10.67 for Py as a function of PR, and Py. Thus Pz P2 Pu Pi= pe or Beth =0 Pi — PuPa + PuPa = 0 2 A= * ae (=) — PLP _ Pa i zt) = HPaylZ cS For the specified values of P, and Py Pu = 500 + 1000/0.25 — 0.24 = 500 + 100 docsity com 10-56 CHAPTER 10. Sinusoidal Steady State Power Calculations P 10.69 *. Py = 600 W; Pua = 400 W Note in this case we design for two medium power ratings If Pui = 600 W (120)? a= 00 = 240 (120)2 240 R, = 60 — 24 = 362 (120)2(60) (36) (24) Ry+ R= = 602 CHECK: Py = = 1000 W If Pu2 = 400 W Ri + Re = 60 (as before) Ry = 242 CHECK: Py = 1000 W 120)? Ri + Ro+ Rs = ce = 240 (120)? 900 Ry = 24-16 = 80 120)? Rs + Rl|Ro = teu =129 8Re 8+ Ro Ro+ R3= = 162 16—Ro+ =12 8Ry Ro - =4 B+ Re 8R2 + RZ -8Ry = 32+4R, F3—4R, -32=0 Ry = 24 V1F32 =246 Ry =89; Ry =82 docsity.com P 10.70 Rp = (220)2 = 96.82 5007 = 968 Ry + Ry = 22)" _ 93.60 pene" 50 Ry, = 96.82 CHECK: Rj||R2 = 48.40 (220)2 P= BA = 1000 W Problems 10-57 docsit com