Electric Circuits, 9th Edition P80 pdf

766 Answers to Selected Problems 2.34 a) i = 385 mA,so a warning sign should be posted and precautions taken. b) Use the following resistors: 390 ft, 47 (1, and 220 ft. 2.35 2.36 a) P arm = 59.17 W;/> leg = 29.59 W; P tmnk = 7.40 W b) ^m = 1414.23 s; r leg = 7071.13 s; 'trunk = 70.677.37 S c) All values are much greater than a few minutes. 2.37 a) 40 V b) No, 12 V/800 ft = 15 mA will cause a shock. 2.38 3000 V Chapter 3 3.1 a) 6 kfl and 12 kft, 9 kfl and 7 kft; simplified circuit is 10 V b) 3 kft, 5 kft, and 7 kft; simplified circuit is 3 mA c) 300 ft, 400 ft and 500 ft; simplified circuit is 1200 Q 200 mV 3.2 a) 10 ft and 40 ft, 10011 and 25 ft; simplified circuit is 60 V b) 9 kfl and 18 kft, and 6 kfl; simplified circuit is 50 mA c) 600 ft, 200 ft, and 300 ft; simplified circuit is 250 fl 0.2 A 3.3 3.4 3.15 3.16 3.18 3.23 3.24 a) b) c) a) b) c) a) b) c) a) b) 21.2 ft 10 kft 1600 ft 30 ft 5 kft 80 fl 66 V 1.88 W, 1.32 W 17,672 ft, 12,408 ft 1200 fl, 300 ft 1 W 1875 ft, 3750ft, 7.5 kf a) b) c) d) e) a) b ) c) d) R 1 ) 36 V 2A 0.96 A 24 V 6.4 V 25 mA 250 V 50 V 25 mA 4.167 mA Answers to Selected Problems 767 3.31 a) 49,98012 b) 498012 c) 23012 d) 5ft 3.35 a) R,„ = 50fl; (25/12) 3.53 3.56 3.58 3.72 3.73 50 + (25/12) (k 25 b) 2500 mcas c) Yes 3.51 a) 1500 (1 b) 28.8 mA c) 750 a, 276.48 mW d) 100012,92.16 mW 23.2 V, 21 V a) A-connected R2—R3—R4 becomes Y-connected 512—2012—4ft; equivalent resistance is 3312. b) Y-connected R 2 —R4—R>, becomes A-connected 10012 —8012—20 ft; equivalent resistance is 3312. c) Convert the delta connection R 4 —R s —Rf, to its equivalent wye. Convert the wye connection Ri—Rj — Rh to its equivalent delta. 9012 R { = 1.0372ft, R 2 = 1.1435 ft,/? 3 = 1.212, R 4 = 1.1435 ft, R 5 = 1.0372 ft, R a = 0.0259 ft, R b = 0.006812, R c = 0.006812 R L{ = 0.025912 P diss = 624W = P dd 3.74 a) R ] = 0.426912, R 2 = 0.4617 ft, R 3 = 0.48 ft, R 4 = 0.4617ft, R 5 = 0.4269 ft, R a = 0.0085 ft, R h = 0.0022 ft, rt c = 0.002212, R d = 0.008512 b) i, = 26.51 A, /?/?, = 300 W or 200 W/m; i 2 = 25.49 A, i\R 2 = 300 W or 200 W/m; i b = 52 A, ilR h = 6 W or 200 W/m; P dd = 1548 W =P {Sss Chapter 4.1 a) b) c) d) e) f) g) 4 11 10 9 s 6 4 6 4.2 a) 8 b) 3 c) 4 d) Avoid the topmost mesh and the leftmost mesh, which both contain current sources. 4.3 a) 2 b) 5 c) 7 d) 1,4,7 4.4 4.8 4.9 4.13 4.17 4.19 4.24 4.26 4.27 4.33 4.34 4.38 4.39 4.42 4.44 4.48 a) 5 b) 3 c) -i s + ij + U = 0; —/1 + /4 + /3 = 0; /5 — / 2 — /3 = 0 d) 2 e) /?,/! + tf 3 / 3 - R 2 i 2 = 0; /?3* 3 + R 5 i 5 - R4I4 = 0 120 V, 96 V 4V a) -6.8 A, 2.7 A, -9.5 A, 2.5 A, -12 A b) 3840W a) 8800W b) 8800W 750 W 3.2 V a) -37.5 V, 75 W b) -37.5 V, 75 W c) Part (b), because there are fewer equations to write and solve. -20 V a) 5.6 A, 3.2 A,-2.4 A b) -8.8 A,-1.6 A, 7.2 A a) 17,940 W b) 17,940 W 259.2 W 2700 W a) 162.92 W b) 518.52 W c) Power delivered equals power absorbed. a) 2 raA b) 304 mW c) 0.9 mW 740 W 768 Answers to Selected Problems 4.51 a) 5.7 A, 4.6 A, 0.97 A,-1.1 A, 3.63 A b) 5>dev= £/^ = 1319.685W 4.52 a) The constraint equations are easier to formu- late in the node voltage method, making it the preferred method. b) 480 mW 4.53 a) Minimize the number of equations to write and solve by using the mesh current method. b) 4mW c) No, since the mesh current method still minimizes the number of equations to write and solve. d) 200 mW 4.59 a) -1mA b) -1 mA 4.60 a) -0.85 A b) -0.85 A 4.63 60 V source, positive at the top, in series with a 1011 resistor 4.64 1 mA current source, with current flowing from top to bottom, in parallel with a 3.75 O resistor 4.71 a) 51.3 V b) -5% 4.74 160 V source, positive at the bottom, in series with a 56.4 kft resistor 4.77 8 ft (The voltage source is zero because there are no independent sources in the circuit.) 4.83 2.5 H and 22.5 O 4.87 a) 6ft b) 24 W 4.91 a) 50 V b) 250W 4.96 30 V 4.105 v { = 39.583 V,v 2 = 102.5 V 4.106 t»i = 37.5 V,v 2 = 105 V 4.107 V] = 52.0833 V,i> 2 = 117.5 V Chapter 5 5.1 a) inverting input non-inverting input positive power supply output negative power supply b) The input resistance; i n = 0 c) The open-loop voltage gain; (v p - v n ) = 0 d) ^ = 9V 5.2 -1 mA 5.3 a) -15 V (saturates) b) -10 V c) -4V d) 7V e) 15 V (saturates) f) -1.08 V < v a < 4.92 V 5.8 a) Many possible designs; one uses a single 3.3 kft input resistor and three series-connected 3.3 kft resistors in the feedback path. b) ±15 V 5.9 a) 0 < or < 0.40 b) 556.25 At A 5.11 0<i? f <60kft 5.12 a) Inverting summing amplifier b) -6V c) -0.5V<v„<2V 5.14 a) 14 V b) 3.818 V < v a < 9.273 V 5.17 a) Non-inverting amplifier b) 2v s c) -6V < v s < 4 V 5.18 a) 10.54 V 5.25 5.26 5.28 5.33 5.34 5.43 5.45 b) -4.55 V < v g < 4.55 V c) 181.76 kft a) -15.1V b) 34.3 kft c) 250 kft 20 kft a) 16 V b) -4.2 V < v b <3.8V 19.93 kft < R x < 20.07 kft a) 24.98 b) -0.04 c) 624.5 a) -19.9844 b) 736.1 jaV c) 5003.68 ft d) -20,0 V, 5000 ft a) 13.49 b) 999.446 mV, 999.834 mV c) 387.78/AV d) 692.47 pA e) 13.5,1V,0V,0A Answers to Selected Problems 769 5.49 a) 2kfl b) 12 mfl Chapter 6 6.1 » f (mV) e) v(V) 6.4 4 3 2 1 0 ( -1 -2 -3 -4 — ) 1 2 l /*""~3 l 4 1.4 1.2 1 F 0.8 0.6 0.4 - 0.2 - 0 0 1 a) / = 0 i = 50/ A / = 0.5 - 50/A / = 0 b) v = 0 v = lV v = -\V v = 0 p = 0 p = 50/ W /? = 50/ - 0.5 W /7 = 0 w = 0 w = 25/ 2 J /(s) /(s) 4 / < 0 0 < / < 5 ms 5 < / < 10 ms 10 ms < / / < 0 0 < / < 5 ms 5 < / < 10 ms 10 ms < / / < 0 0 < / < 5 ms 5 < / < 10 ms 10 ms < / / < 0 0 < / < 5 ms u> = 25/ 2 - 0.5/ + 0.0025 J 5 < / < 10 ms to = 0 10 ms < / 6.16 a) -50 X 10 4 / + 15 V b) 10 6 /V c) 1.6 X 10 6 / - 12 V d) 52 V 6.17 /(jxs) t (ms) 6.21 8 H 6.25 a) -5e~ Al A b) -4e~ 4? - 6 A c) -e _4i + 6 A d) 40 J e) 400 J f) 360 J 9) 1(4)(-6) 2 + j(16)(6) 2 = 360 J (checks) 6.26 2 ^cF, initial voltage is 25 V 6.31 a) -20e" 25 'V b) -16<T 25 ' + 21 V c) -4<T 25/ - 21 V d) 320/AJ e) 2525/AJ f) 2205/xJ g) \(2 X 10" 6 )(21) 2 + |(8 X 1()" 6 )(-21) 2 = 2205 /xJ 6.39 a) 16 ^ + 32/ 2 = 2-± at at b) -16<T f + 32e _2f + 32e - ' - 32e~ 2 ' = 16e" c) 34<T' - 4e~ 2t V d) 30 V, which is consistent with the circuit's behavior 6.44 a) 2721.6 mJ b) 2721.6 mJ c) 518.4 mJ d) 518.4 mJ 770 Answers to Selected Problems 6.45 a) -4.5 A b) No 6.48 a) 50mH,2.4 b) 0.2 X 10~ 6 Wb/A,0.2 X 10 _6 Wb/A 6.49 0.8 nWb/A, 1.2 nWb/A 6.51 v(t) = 0333v s (t) 6.53 There is no difference in the output voltage for these two circuits. Chapter 7.4 7.5 7.7 7.23 7.26 7.35 7.36 a) b) c) d) e) a) b) c) d) e) f) 9) h) i) i) k) I) a) b) c) d) a) b) c) a) b) c) a) b) a) b) c) d) 7 5 mA, 15 mA 5 mA, -5 mA 5,,-20.000/ mA -5e- 20 - ,)00 'mA The current in a resistor can change instantaneously. 0A 160mA 65 mA 160 mA 225 mA 0A 160e" 2n "' mA OV -3.2 V OV -32e~ 2m V 225 - 160e" 200 ' mA 2A 20 ms 2e~ 5i)t A, -160e~ 5lM V, -144<T 50 ' V 69.92% 1.6e~ 5()f mA,32<r 5() ' + 8V, -8<T 50 ' + 8V 800/d 160/xJ,640/J 24«-«" raA - Se -5om + 80V 2880/d -2-3e- 500,, 'A,48-48e- 5,K,(, 'V 60V,0V -5 mA 0.333 mA 5/AS 0.333 -5.333<r 2(,a(,0,) 'mA 7.37 7.51 7.53 7.55 7.56 7.68 7.69 7.71 7.78 7.85 7.87 7.95 7.96 7.103 7.104 7.105 a) b) a) b) c) d) e) 5 + 15e- 1000 'A 50-450<T 1(loa, V 60 - 60e~ im V 1 - 0.6e- m)l mA 1 + 2Ae~ m)t mA 4 - 2Ae~ m ' mA 3.4 mA -60 + 90e~ 20,)0 ' V a) b) c) d) e) f) a) b) c) d) a) b) c) d) e) a) b) c) d) e) 50 V -24 V 0.1 fis -18.5 A -24 + 74<T lt,7 'V -18.5e- 1()7 'A 90 V -60 V 1000/AS 916.3/AS 4-4<T 20 'A 80e" 20f V 2.4 - 2.4e~ 20 ' A 1.6 - 1.6<T 20 'mA Yes 40 - 40e" 500( " mA 10e -50oo, v 16 - \6e' 5m)t mA 24 - 24e" 5000 ' mA Yes -5.013 V -5V, 0 < t < 5 s; -5£>- (U( '" 5) V, 5 s < t < 83.09 ms a) b) 2.25 272.1 /AS -1600/ + 8V,-15 + lle- 200 'V, 23 80 a) b) a) b) a) b) c) - 1600* - lle" 200 'V ms 1.091 MO 0.29 s 8.55 flashes/min 559.3 kH 24.32 flashes/min 99.06 mA $43.39 per year CO Answers to Selected Problems 771 Chapter 8 8.1 a) -10,000 rad/s,-40,000 rad/s b) overdamped c) 3125 ft d) -16,000 +/12,000 rad/s, -16,000 - /12,000 rad/s e) 2500ft 8.7 a) 25 nF b) 2500ft c) 75 V d) 30 mA 8.8 8.9 8.18 8.19 8.20 8.29 8.30 8.31 8.50 e) -8000, (30 cos 6000? + 71.25 sin 6000?) mA a) 8 kft, 40 H, 625 rad/s, 500 rad/s b) -U- 25()l + 4e~ imt mA, O.&T 250 * -0.8<T 10() " f mA, 02e- 25{)[ - 32e- xmt mA a) lkft,l/xF,6000V/s,8V b) (-3000? + 2)^ 5(lll, mA 8.11 a) 500 rad/s, 400 rad/s, 1.5625 H,4/xF, -15 mA, 60 mA b) 18.75e" 200 ' - 18.75<r 8()0 'V c) 75<T 200 ' - 75^ 800 ' mA 800/ d) -60e- 200 ' + 15(T ftUUf mA 5e~ 2im + \Qe-* m) <V \5e~ 25m cos 3122.5? + 721e~ 2mi sin 3122.5? V 15e- 4m] 'V 60 - 120c-™ + 15e- 2iumi mA 60 - 105e" W)00 ' cos 6000? - 90e" 800<)r sin 6000? mA 60 - 750,000?e" 104 ' - 105eT 104 ' mA 60-80ff- 8n0r + 2Off- 32Wf V 8.51 60 - 120,000?e -2000( - 60e 2000( V 8.52 60 - 60e>- 200 (" cos 1500? - 80<r 20(K)/ sin 1500? V 8.63 a) 0 < ? < 0.5" s: v n (t) = l0t 2 V,v lA (t) = -1.6? V; 0.5 + < ? < ? sal : v 0 (t) = -5? 2 + 15? - 3.75 V, v ol (t) = 0.8? - 1.2 V b) 3.5 s 8.64 0 < ? < 0.5 ~ s: v a {t) = 10 - 20e"' + Hfe~*V, v oi (t) = -0.8 + 0.8e~ a V, 0.5 + < ? < ? sal : Vo {t) = -5 + 19.42ff" (t-a5) - 12.87e _2(f - a5) V, v ol (0 = 0.4 - 0.91e- 2( ' _05) V 8.68 a) 55.23fis b) 262.42 V c) ? max = 53.63 /xs, v(t max ) = 262.15 V 8.69 a) 40 mJ b) -27,808.04 V c) 568.15 V Chapter 9 9.1 a) 80 V b) 500 Hz c) 3141.59 rad/s d) -0.5236 rad e) -30° f) 2 ms g) 166.67 ^s h) -80sinl0007r?V i) 333.33/AS j) 166.67jiis 9.4 a) 600 Hz b) 1.67 ms c) 10V d) 6 V e) -53.13°,-0.9273 rad f) 662.64/AS g) 245.97^s 2 a) -195.72<r l066 - 67f + 200cos(800? - 11.87°) mA b) -195.72^- 1066 - 67/ mA, 200 cos(800? - 11.87°) mA c) 28.39 mA d) 200 mA, 800 rad/s, -11.87° e) 36.87° 9.11 a) 111.8 cos(500? - 3.43°) b) 102.99 cos(377? + 40.29°) c) 161.59 cos(100? - 29.96°) d) 0 9.13 a) 502,654.82 rad/s b) 90° c) -39.79 ft d) 0.05/AF e) -/39.79 ft 9.14 a) 400 Hz b) -90° 9.8 9.9 772 Answers to Selected Problems 9.85 9.15 9.24 9.28 9.29 9.32 9.37 9.45 9.46 9.49 9.55 9.59 9.60 9.64 9.76 9.77 9.83 9.84 c) 5fi d) 1.99 mH e) /5 ft a) 40 a /40 a AAA /~W~lT\ j VVv 600/20° vf J 7- b) 8.32 /76.31° A ^ -/100 ft c) 8.32cos(8000f + 76.31°) A a) 200 /36.87° mS b) 160 mS c) 120 mS d) 10 A 500 rad/s 42.43 cos(50,000f + 45°) V 42.43 cos(2000? + 45°) V 2/3 ft 227.68 / -18.43° V, (3.6 + ./10.8)(1 2 /-36.87° A, (100 - /50) ft 10/-45° A, (1.6 + /3.2) ft 188.43/-42.88° V /80 = 80 /90° V 36 cos 2000 fV 56.57 cos(10,000f -45°)V a) 0.3536 b) 2 A a) 5 cos(5000f - 36.87°) A, 1 cos(5000f - 180°) A b) 0.5 c) 9mJ,12mJ 512,000/60° ft I V =(V /2)-1 R o v in / m v a) 247.11/1.68° V b) -32 ft, 241.13/1.90° V c) -26.90 ft 9.88 a) l { = 24 /0° A, 1 2 = 2.04 /0° A, I 3 = 21.96/0° A, I 4 = 19.40/0° A, I 5 = 4.6/0° A, I 6 = 2.55/0° A b) 0.42 /0° A 9.89 a) 0A b) 0.436/0° A c) Yes; when the loads are equal, no power is lost to the neutral line, so the cost of power is lower. Chapter 10 10.1 a) 409.58 W (abs), 286.79 VAR (abs) b) 103.53 W (abs), -386.37 VAR (del) c) -1000W (del),-1732.05 VAR (del) d) -250 W (del), 433.01 VAR (abs) 10.2 a) Yes b) Yes 10.15 a) 15.81 V(rms) b) 62.5 W 10.18 a) 6.4 W, 4.8 VAR, 8 VA b) 6.4 W c) 4.8 VAR 10.26 a) 0.96 lagging, 0.28; 0.8 leading,-0.6; 0.6 leading, -0.8 b) 0.74 leading,-0.67 10.27 a) 1.875 + /0.625 ft b) 0.9487 lagging 10.44 a) 20 + /20 ft b) 20 W c) With 22 ft and 1 mH, the load impedance is 22 + /5 ft and the load power is 17.7 W 10.47 a) 360 mW b) 4000ft,0.1/AF c) 443.1 mW; yes d) 450 mW e) 4000 ft, 66.67 nF f) Yes 10.48 a) 4123.1 ft, 0.1/xF, 443.18 mW b) Yes c) Yes 10.64 90 W 10.65 a) 10 b) 250W 10.66 a) 28.8 ft b) 28.8 ft c) Yes Answers to Selected Problems 773 10.67 a) P L = PH = V 1 R\ + R 2 v\Ri + R 2 ) R1R2 PM = Yl Ri V 2 V 2 Pi. (Yl _ vL\(¥l\ (p L p ¥ )yp,) PM - PL b) 1125W 10.68 36ft, 24 ft Chapter 11 11.2 a) acb b) abc 11.3 a) Balanced, negative phase sequence b) Balanced, positive phase sequence c) Balanced, negative phase sequence d) Balanced, positive phase sequence e) Unbalanced, due to unequal amplitudes f) Unbalanced, due to unequal phase angle separation 11.7 v AB = 13,198.23 cos a>t V, v BC = 13,198.23 cos(wr + 120°) V, v CA = 13,198.23 cos(w/ - 120°) V 11.9 a) 15.24 A(rms) b) 6583.94 V(rms) 11.11 a) I aA = 5/-36.87° A, I b B = 5/83.13° A, I cC - 5/-156.87° A b) V ab = 216.51 /-30°V,V bc = 216.51 /90° V, V ca = 216.51/-150° V c) V AN = 122.23/-1.36° V, V BN = 122.23/118.64° V, V CN = 122.23/-121.36° V d) V AB = 211.72/-31.36° V, V BC = 211.72/88.64° V, V CA = 211.72/-151.36° V 11.12 a) 1+/3H 20/If V( 150° /+\ rms) \-J aA 39 - /33 n b) 0.4 /-173.13° A(rms) c) 35.39/176.63° V(rms) 11.13 21.64/121.34° V(rms) 11.16 159.5 /29.34° V(rms) 11.22 6120/36.61° VA 11.24 a) 1833.46 /22° VA b) 519.62 V(rms) 11.43 a) W 2 ~ W x = V L / L [cos (d - 30°) - cos (6 + 30°)] = 2V L I L $in0sm30 o = V L I L sm6 Thus, V3(W 2 - W x ) = Vw L I L sm 0 = Q T b) 2592 VAR, -2592 VAR, 3741.23 VAR, -4172.80 VAR 11.44 197.26 W, 476.64 W 11.52 a) 1.70 MVA 1.2MVAR 1.2 MW b) 1.2 MW 11.53 a) 16.71 AIF b) 50.14/aF 11.56 |V ab | = 12,548.8 V, so the voltage is below the acceptable level of 13 kV. Thus, when the load at the substation drops to zero, the capacitor bank must be switched off. 11.57 P L(be fore) = 81.66 kW, P L(after) = 40.83 kW Chapter 12 12.2 12.3 12.7 12.9 12.10 12.14 a) (t + \0)u(t + 10) - 2tu(t) + (t - 10)«(f - 10) b) -8(t + 3)u(t + 3) + 8(7 + 2)u(t + 2) + 8(r + l)«(f + 1) - 8(f - l)u(t -1)- 8(/ - 2)u{t -2) + 8(/ - 3}M(/ - 3) a) 5t[u{t) - u{t - 2)] + 10[«(f -2)- u(t - 6)] + (-5? + A0)[u(t - 6) - u(t - 8)] b) 10 sin irt[u{t) - u(t - 2)] c) 4t[u(t) - u{t - 5)] a) 1.0 b) 0 c) oo a) 26 b) 2.25 2/9 at SC ° 3) f + c, 2 M ^ 2 b) * + J c) 2 d) check 774 Answers to Selected Problems 12.17 a) b) c) d) e) 12.22 a) b) 12.40 a) (.v + a) 2 CO 2 i ~> S + CO co cos 0 + s 2 + 1 -> sr sinh 0 + (r - 1 v(.v + a) 1 .v sin 6 -> of y[cosh 0] - 1) 13.9 a) 12.41 12.42 12.50 12.55 12.56 s(s + a) [«?"' + 5e' 2 ' + 2e- 4 ']u(t) b) [6 + 4e~ 2t + 2e~ 4t + e~ ( "]u(t) c) [4e' 1 + 20cT'cos(2f + 36.87°)]M(?) d) [490 + 250e" 7 'cos(r - 163.74°)]«(r) a) [20/ - 4 + 4e-*]u(t) b) [250 - lOOte' 1 - 250e~ f ]u(t) c) [30r - 8 + K)e 3l cos(t + 36.87°) ]u(t) d) [20 - 2.5fV - I5te~' - 20e"']«(0 e) [16 + S9A4te~ 2t cos (t + 26.57°) + I13.14e _2r cos0 + 98.13°) ]u(t) c) 5'(t) - 105(f) + [3(k' _5/ + 20e~ 10f ]«(f) a) /(()') = 8,.f(oo) = 0 b) /(0^) = 13,/(00) = 6 c) /(0 + ) = 20, /(oo) = 0 d) /(0 + ) = 250, /(oo) = 490 0.947 588 Chapter 13 13.4 a) — 8 X 10 7 s 13.6 a) .v 2 + 40,000* + 256 X 10 6 b) Zero at 0; poles at -8000 rad/s and -32,000 rad/s s 2 + 8000* + 25 x 10 6 b) Zeros at -4000 + /3000 rad/s and -4000 - /30()() rad/s; pole at 0 16 x Uf -^vw— 5000 (2 SI 150 V-s b) '© -150s + V„ 2.5 s a (s + 400)(.v + 1600) c) (SO*-*** - 200e- l60()f )«(r)V 13.10 a) A/W- + 10012 b) c) 13.12 a) 5 x 10 s n s ' V. o.ob' n 75.v 2 + 812,500.v + 6875 X sis 1 + 10 4 v + 5 X 10 7 ) / + \ 137.5 / 1.25 mV -Hv ^_7 10 6 [137.5 + 8().04^ 50()() 'cos(5000/ + 14134 0 ) ]K(0 ii, V-s R sC re© 4 £0 v. b) c) •48(s + 8000) .v 2 + 8000* + 25 X 10 6 2.4(.v + 4875) .v 2 + 8000s + 25 X 10 6 d) [80e"-" )00f cos(3000f + 126.87°) ]u(t)V e) [2.5e- 4mt cos(3(mt - 16.26°)]H(0A 13.21 a) [35 + 5.73e~' cos(7r + 167.91 °)]H(I)V b) Compare solution at t = 0 and t = oo to circuit at t = 0 and t = oo 13.22 a) [10 - Kk>"" a5 'cos0.5/]u(f)A b) 7.07e _OA cos(0.5r - 45°)w(r) V c) Compare solutions at t = 0 and t = oo to circuit at t = 0 and t = oo Answers to Selected Problems 775 13.29 a) 65 2 + 6s - 18 -9r - 30.v 18 s(s + 2)(5 + 3)' s(s + 2)(5 + 3) b) Initial values: 6 A, -9 A; final values: -3 A, -3 A c) [-3 + 3e~ 2 ' + 6e- 3, ]u(t)A, [-3 - 3e~ 2 ' - 3e" 3 ']u(t)A 13.34 63.25*" ia *cos(50> + 71.57°)w(0 mA 240(5 + 40) 13.39 a) — ' s(s + 20)(5 + 80) b) Initial value is 0. final value is 6 A c) (6 - 4e~ m - 2e~ m )u(t) A 13.40 a) (-2e~ m + 2e^ ) ')«(0mA b) (2< -2()( 2e -hl >(0 mA 13.42 a) 480(5 + 2.5) 5(5 + 4)(5 + 6) b) [50 + 90e~ 4 ' - \4Qe~ ( "]u(t)V 13.50 a) b) c) d) e) 250 5 + 250 5 5 + 250 V 5 + 8000 8000 5 + 8000 , no zeros, pole at -250 racl/s , zero at 0, pole at -250 rad/s , zero at 0, pole at -8000 rad/s , no zeros, pole at -8000 rad/s 100 , no zeros, pole at -500 rad/s 5 + 500 13.62 (e - l)e" ? V 13.63 (1 - e)e-'V 13.77 16.97 cos (3/ + 8.13°) V 5(5 + 30,000) 13.79 a) 1 (5 + 5000)(5 + 8000) b) (5e- 5l)m - 4Ae-* m ')u(t)V 13.80 a) =1*2 ; (5 + 400)(5 + 1000) b) 13.13 cos(400f - 156.8°) V 13.87 a) 0.8 A b) 0.6 A c) 0.2 A d) -0.6 A e) Q.6e~ zxm '\t(t)A f) -0.6e- 2xl %(f)A g) -1.6 x 1(T 3 8(0 - 72QOe' 2xlift u(t)V 13.88 a) 80 V b) 20 V c) 0 V d) 325(f)juA e) 16 V f) 4V g) 20 V 13.92 a) i 2 (0~) = * 2 <0 + ) = 0A; *'L(0~) = idQ + ) = 35.36 A 1440TT( 122.92 V25 - 3000TT\/2 b) V Q = = =— + J (5 + 1475TT)(5 2 + 14,400TT 2 ) 300 V2 5 + 1475 w v a = 252.89<T l475jr/ + 172.62 cos(1207it + 6.85°) V y„(0 + ) = 424.26 V c) V„ = 122.06/6.85° V(rms) d ) »„ (V) t (ms) 13.93 a) -20.58^ -1475 ^ + 172.62 cos( 12()77-/ - 83.15°) V b) v„ (V) t (ms) c) Voltage spikes in Problem 13.92 but not here.