电路分析基础(英文版)课后答案第三章.pdf

3 Techniques of Circuit Analysis Drill Exercises DE 3 1 a 11 8 resistors 2 independent sources 1 dependent source b 9 c 9 R4 R5forms an essential branch as does R8 10 V The remaining seven branches contain a single element d 7 e 6 f 4 g 6 DE 3 2 Solution given in text DE 3 3 Solution given in text DE 3 4 Solution given in text DE 3 5 a The two node voltage equations are 15 v1 60 v1 15 v1 v2 5 0 5 v2 2 v2 v1 5 0 Solving v1 60 V and v2 10 V Therefore i1 v1 v2 5 10 A b p15A del 15 60 900 W c p5A 5 10 50 W DE 3 6 Use the lower node as the reference node Let v1 node voltage across 1 resistor and v2 node voltage across 12 resistor Then v1 1 v1 v2 8 4 5 53 54CHAPTER 3 Techniques of Circuit Analysis v2 12 v2 v1 8 v2 30 4 0 Solving v1 6 Vv2 18 V Thus i v1 v2 8 1 5 A v v2 2i 15 V DE 3 7 Use the lower node as the reference node Let v1 node voltage across the 8 resistor let v2 node voltage across the 4 resistor Then v1 50 6 v1 8 v1 v2 2 3i1 0 5 v2 4 v2 v1 2 3i1 0 i1 50 v1 6 Solving v1 32 V v2 16 V i1 3 A p50V 50i1 150 W delivering p5A 5 v2 80 W delivering p3i1 3i1 v2 v1 144 W delivering DE 3 8 Use the lower node as the reference node Let v1 node voltage across the 7 5 resistor and v2 node voltage across the 2 5 resistor Place the dependent voltage source inside a supernode between the node voltages v and v2 The node voltage equations are node 1 v1 7 5 v1 v 2 5 4 8 supernode v v1 2 5 v 10 v2 2 5 v2 12 1 0 We also have v ix v2and ix v1 7 5 Solving this set of equations for v gives v 8 V DE 3 9 v1 60 2 v1 24 v1 60 6i 3 0 i 60 6i v1 3 Thereforev1 48 V DE 3 10 vo 40 vo 10 10 vo 20i 20 0 i 10 vo 10 10 20i 30 Thereforevo 24 V Problems55 DE 3 11 De ne three clockwise mesh currents i1 i2 and i3in the lower left upper and lower right windows The three mesh current equations are 80 31i1 5i2 26i3 0 5i1 125i2 90i3 0 26i1 90i2 124i3 a Solving i1 5 A therefore the 80 V source is delivering 400 W to the circuit b Solving i3 2 5 A therefore p8 6 25 8 50 W DE 3 12 a b 8 n 6 b n 1 3 b De ne three clockwise mesh currents i1 i2 and i3in the upper lower left and lower right windows The three mesh current equations are 3v 19i1 2i2 3i3 0 25 10 2i1 7i2 5i3 10 3i1 5i2 9i3 We also havev 3 i3 i1 Solving for i1and i3givesi1 1 A i3 3 A Thereforev 12 Vandp3v 3v i1 36 W DE 3 13 Let ia lower left mesh current cw let ib upper mesh current cw and ic lower right mesh current cw Then 25 14ia 6ib 8ic 0 6ia 16ib 8ic 0 8ia 8ib 16ic 5i i ia ia 4 A ic 2A vo 8 ia ic 16 V DE 3 14 Mesh 1 30 11i1 2i2 Mesh 2 0 2i1 19i2 5i3 56CHAPTER 3 Techniques of Circuit Analysis Current source i3 16 A Solution givesi1 2 A i2 4 A i3 16 A The current in the 2 resistor isi1 i2 6 A P2 6 2 2 72 W DE 3 15 Mesh a 7ia 2ib 5ic 75 Current sources ib 10 A ic 2v 5 Dependent variable v 5 ia ic Solution ia 15 A ib 10 A ic 10 A v 25 V DE 3 16 Supermesh a b 2ia 4ib 4ic 10 Mesh c 2ia 2ib 5ic 6 Current source ia ib 2 A Solution ia 7 A ib 5 A ic 6 A p1 i2 c 1 6 2 1 36 W Problems57 DE 3 17 Let v1denote the voltage across the 2 A source Let v1be a voltage rise in the direction of the 2 A current v1 20 15 2 v1 25 10 0 v1 35 V p2A 35 2 70 Wp2A del 70 W DE 3 18 Mesh 1 12i1 6i2 4i3 128 Mesh 2 6i1 14i2 3i3 30ix 0 Current source i3 4 A Dependent variable ix i1 i3 Solution i1 9 A i2 6 A i3 4 A ix 5 A v4A 3 i3 i2 4ix 10 V The power delivered by the 4A source isp4A 10 4 40 W DE 3 19 To nd the Th evenin resistance deactivate the independent voltage source and note that RTh 5k20 8 k12 6 With the terminals a b open the current delivered by the 72 V source is 72 24 or 3 A The current left to right in the 5 resistor is 20 25 3 2 4 A and the current left to right in the 12 resistor is 5 25 3 or 0 6 A The Th evenin voltage vTh vabis the drop across the 8 resistor plus the drop across the 20 resistor Thus vTh 8 0 6 20 3 64 8 V 58CHAPTER 3 Techniques of Circuit Analysis DE 3 20 After one source transformation the circuit becomes ThereforeIN 6 A RN 20k12 7 5 DE 3 21 Find the Th evenin equivalent with respect to A B VTh 36 12 000 VTh 60 000 0 018 0 VTh 150 V RTh 15 000 60 000 12 000 72 000 25 k Therefore vmeas 150 100 000 125 000 120 V DE 3 22 Summing the currents away from node a where vTh vabWe have vTh 8 4 3ix vTh 24 2 0 ix vTh 8 Solving for vThyields vTh 8 V iT 4ix vT 2 ix vT 8 ThereforeiT vTandRTh vT iT 1 Problems59 DE 3 23 Use the bottom node as the reference Let v1be the node voltage across the 60 resistor Then v1 60 v1 vTh 160i 20 4 0 vTh 40 vTh 80 vTh 160i v1 20 0 i vTh 40 thereforevTh 30 V Let iTbe the test current into terminal a iT vT 80 vT 40 vT 160 vT 40 80 iT vT 1 10 Therefore RTh 10 DE 3 24 First nd the Th evenin equivalent circuit To nd vTh use the bottom node as the reference Let vTh vaband v1 node voltage across the 20 V 4 branch The two node Voltage equations are vTh 100 v 4 vTh v1 4 0 v v1 20 v1 100 4 v1 20 4 v1 vTh 4 0 Solving for vThgives vTh 120 V To nd RTh deactivate the two independent sources and apply a test voltage source across a b Let vTbe positive at a and iTdirected into a Then the two node Voltage equations are vT v 4 vTh v 4 iT v 4 v 4 v vT 4 0 Thereforev vT 3and12iT 4vT SoRTh vT iT 3 a For maximum power transfer RL RTh 3 b pmax 120 6 2 3 1200 W 60CHAPTER 3 Techniques of Circuit Analysis DE 3 25 When RL 3 the voltage across RLis 60 V As before let v1be the node voltage across the 20 V 4 branch then v v1 20 and 60 3 60 v1 4 60 100 v 4 0 Therefore v1 60 V and v 40 V The current out of the plus terminal of the 100 V source is i1 100 60 4 100 40 60 4 10 20 30 A a Therefore 100 V is delivering 3000 W to the circuit b The current out of the plus terminal of the dependent source is 20 A Therefore the dependent source is delivering 800 W to the circuit c The load power is 1200 3800 100 or 31 58 of this generated power Problems P 3 1 a Five b Three c Sum the currents at any three of the four essential nodes a b c and d Using nodes a b and c we get ig i1 i2 0 i1 i4 i3 0 i5 i2 i3 0 Problems61 d Two e Sum the voltages around two independent closed paths avoiding a path that contains the independent current source since the voltage across the current source is not known Using the upper and lower meshes formed by the ve resistors gives R1i1 R3i3 R2i2 0 R3i3 R5i5 R4i4 0 P 3 2 2 vo 4 vo 55 5 0 vo 20 V p2A 20 2 40 W absorbing P 3 3 Let v2be the node voltage across the 80 resistor positive at the upper terminal Then 4 v1 20 v2 80 v2 40 0 Note we have created a super node in writing this expression v1 60 v2 v1 20 V v2 80 V pdel 60igwhere igis the current out of the positive terminal 4 ig v1 20 ig 3 A pdel 60 3 180 W 62CHAPTER 3 Techniques of Circuit Analysis P 3 4 a v1 48 v1 128 8 v1 v2 18 0 v2 20 v2 v1 18 v2 70 10 0 Solving v1 96 V v2 60 V ia 128 96 8 4 A ib 96 48 2 A ic 96 60 18 2 A id 60 20 3 A ie 60 70 10 1 A b pdev 128 4 70 1 582 W P 3 5 Use the lower terminal of the 5 resistor as the reference node vo 60 10 vo 5 3 0 Solving vo 10 V P 3 6 a Problems63 v1 110 2 v1 v2 8 v1 v3 16 0 v2 v1 8 v2 3 v2 v3 24 0 v3 110 2 v3 v2 24 v3 v1 16 0 Solving v1 74 64 V v2 11 79 V v3 82 5 V Thus i1 110 v1 2 17 68 Ai4 v1 v2 8 7 86 A i2 v2 3 3 93 Ai5 v2 v3 24 3 93 A i3 v3 110 2 13 75 Ai6 v1 v3 16 9 82 A b XP dev 110i1 110i3 3457 14 W XP dis i21 2 i22 3 i23 2 i24 8 i25 24 i26 16 3457 14 W P 3 7 2 4 v1 125 v1 v2 25 0 v2 v1 25 v2 250 v2 375 3 2 0 Solving v1 25 V v2 90 V CHECK p125 25 2 125 5 W p25 90 25 2 25 169 W p250 90 2 250 32 4 W p375 90 2 375 21 6 W p2 4A 25 2 4 60 W Xp abs 5 169 32 4 21 6 60 288 W Xp dev 90 3 2 288 W CHECKS 64CHAPTER 3 Techniques of Circuit Analysis P 3 8 a v1 50 v1 640 5 v1 v2 2 5 0 v2 v1 2 5 v2 v3 5 12 8 0 v3 2 5 v3 v2 5 12 8 0 Solving v1 380 V v2 269 V v3 111 V b ig 640 380 5 52 A pg del 640 52 33 280 W P 3 9 v1 144 4 v1 10 v1 v2 80 0so29v1 v2 2880 3 v2 v1 80 v2 5 0so v1 17v2 240 Solving v1 100 V v2 20 V P 3 10 v1 75 20 v1 50 v1 v2 40 0 Problems65 v2 v1 40 v2 75 800 6 v2 200 0 Solving v1 115 V v2 287 V vo 115 75 40 V P 3 11 va 4 V 7 vb 3 vb vc 1 0 2vx vc vb 1 vc va 2 0 vx vc va vc 4 Solving vo vb 1 5 V P 3 12 v1 v2 30 15 v1 v2 31 25 v1 25 4 0 v1 v2 30 15 v2 v3 50 v2 v1 31 25 0 66CHAPTER 3 Techniques of Circuit Analysis v3 v2 50 v3 50 1 0 Solving v1 76 V v2 46 V v3 2 V i30V 0 A p4A 4v1 4 76 304 W del p1A 1 2 2 W del p30V 30 0 0 W p15 0 2 15 0 W p25 v2 1 25 762 25 231 04 W p31 25 v1 v2 2 31 25 302 31 25 28 8 W p50 lower v2 v3 2 50 482 50 46 08 W p50 right v2 3 50 4 50 0 08 W Xp diss 0 231 04 28 8 46 8 0 08 306 W Xp dev 304 2 306 W CHECKS P 3 13 v1 70 v1 v2 10 v1 80 5 0 v2 12 v2 v1 10 v2 80 5 0 Solving v1 70 V v2 60 V Thus io v1 v2 10 1 A Problems67 P 3 14 a v0 60 10 vo 5 3 0 vo 10 V b Let vx voltage drop across 3 A source vx vo 100 3 290 V p3A developed 3 290 870 W c Let ig current into positive terminal of 60 V source ig 10 60 10 5 A p60V developed 5 60 300 W d Xp dis 5 2 10 3 2 100 10 2 5 1170 W Xp dis 300 870 1170 W e vois independent of any nite resistance connected in series with the 3 A current source P 3 15 a From the solution to Problem 3 5 we know vo 10 V therefore p3A 3vo 30 W p3A developed 30 W b The current into the negative terminal of the 60 V source is ig 60 10 10 5 A p60V 60 5 300 W p60V developed 300 W c p10 5 2 10 250 W p5 10 2 5 20 W Xp dev 300 W Xp dis 250 20 30 300 W P 3 16 a v1 20 25 103 v1 0 25 103 11 10 3 v1 10 0 5 103 0 68CHAPTER 3 Techniques of Circuit Analysis v1 5 V i1 20 5 25 000 1 mA i2 v1 250 5 250 20 mA i5 10 5 500 10 mA i4 10 1000 10 mA i4 i3 11 i5 0 i3 11 i4 i5 11 10 10 31 mA b p20V 20i1 20 1 10 3 20 mW p10V 10i3 10 31 10 3 310 mW v11mA v1 10 v11mA 10 5 5 V p11mA 11v11mA 55 mW del Xp dev 20 310 330 mW p25k 25 103i2 1 25 mW p0 25k 0 25 103i2 2 100 mW p0 5k 0 5 103i2 5 50 mW p1k 1 103i2 4 100 mW Xp diss 25 100 50 100 55 330 mW Xp diss Xp dev 330 mW P 3 17 a Problems69 v2 500 4 v2 v4 2 v2 v3 3 0 v3 v2 3 v3 6 v3 v5 2 0 v4 v2 2 v4 500 11 v4 v5 4 0 v5 v3 2 v5 4 v5 v4 4 0 Solving v2 300 V v3 180 V v4 280 V v5 160 V i5 500 v4 11 500 280 11 20 A p5 20 2 5 2000 W b i500V v1 v2 4 v1 v4 11 500 300 4 500 280 11 50 20 70 A p500V 35 000 W Check XP dis 50 2 4 40 2 3 30 2 6 20 2 11 10 2 2 30 2 4 10 2 2 40 2 4 35 000 W P 3 18 a vo v1 R vo v2 R vo v3 R vo vn R 0 nvo v1 v2 v3 vn vo 1 n v1 v2 v3 vn 1 n Xn k 1vk b vo 1 3 150 200 50 100 V 70CHAPTER 3 Techniques of Circuit Analysis P 3 19 Place v 5 inside a supernode and use the lower node as a reference Then v1 50 10 v1 30 v1 v 5 39 v1 v 5 78 0 134v1 6v 3900 v 50 v1 Solving v1 30 V v 20 V vo 30 v 5 30 4 26 V P 3 20 vo 80 5 vo 50 vo 75i 25 0 i vo 50 Solving vo 50 V i 1 A io 50 75 1 25 5 A p75i 75i io 375 W The dependent voltage source delivers 375 W to the circuit P 3 21 3 vo 200 vo 5i 10 vo 80 20 0 i vo 80 20 a Solving vo 50 V b ids vo 5i 10 i 50 80 20 1 5 A ids 4 25 A 5i 7 5 V pds 5i ids 31 875 W c p3A 3vo 3 50 150 W del p80V 80i 80 1 5 120 W del Xp del 150 120 270 W CHECK p200 2500 200 12 5 W Problems71 p20 80 50 2 20 900 20 45 W p10 4 25 2 10 180 625 W Xp diss 31 875 180 625 12 5 45 270 W P 3 22 a io v2 v3 50 2io v1 100 v1 v2 25 0 v2 v1 25 v2 200 v2 v3 50 v3 v2 50 v3 5io 5 v3 38 5 20 0 Solving v1 50 V v2 30 V v3 2 5 V b io v2 v3 50 30 2 5 50 0 65 A i3 v3 5io 5 2 5 5 0 65 5 1 15 A ig 38 5 2 5 20 1 8 A Xp dis Xp dev Calculate Xp devbecause we don t know if the dependent sources are developing or absorbing power Likewise for the independent source p2io 2iov1 2 0 65 50 65 W dev p5io 5ioi3 5 0 65 1 15 3 7375 W dev pg 38 5 1 8 69 30 W dev Xp dev 69 3 65 3 7375 138 0375 W 72CHAPTER 3 Techniques of Circuit Analysis CHECK Xp dis 2500 100 900 200 400 25 0 65 2 50 1 15 25 1 8 2 20 138 0375 W Xp dev Xp dis 138 0375 W P 3 23 a 5 v1 15 v1 v2 5 0 v2 v1 5 v2 30 v2 10 v2 5i 30 0 i v1 v2 5 Solving v1 30 V v2 15 V i 3 A io 15 15 30 1 A p5i 15 1 15 W del p5A 5 30 150 W del pdev 165 W b Xp abs 30 2 15 15 2 30 15 2 10 3 2 5 1 2 30 165 W Xp dev Xp abs 165 W P 3 24 i v3 v4 4 235 222 4 3 25 A 30i 30 3 25 97 5 V v1 30i v4 v1 v4 30i 222 97 5 124 5 V v3 v 250 v 250 235 15 V 3 2v 3 2 15 48 A ig 250 124 5 2 250 235 1 77 75 A Problems73 p250V 250ig 250 77 75 19 437 5 W del i30i i v4 40 48 0 i30i i 222 40 48 3 25 5 55 48 50 3 A p30i 30i i30i 97 5 50 3 4904 25 W dev p3 2v 3 2v v4 48 22 10 656 W abs Xp dev 19 437 5 4904 25 24 341 75 W p10 v2 1 10 124 5 2 10 1550 025 W p2 250 124 5 2 2 7875 125 W p1 250 235 2 1 225 W p20 235 2 20 2761 25 W p4 3 25 2 4 42 25 W p40 222 2 40 1232 10 W Xp diss 10 656 1550 025 7875 125 225 2761 250 42 25 1232 1 24 341 75 W Thus Xp dev Xp diss Agree with analyst P 3 25 74CHAPTER 3 Techniques of Circuit Analysis Node equations v1 20 v1 20 2 v3 v2 4 v3 80 3 125v 0 v2 40 v2 v3 4 v2 20 1 0 Constraint equations v 20 v2 v1 35i v3 i v2 40 Solving v1 20 25 V v2 10 V v3 29 V Let igbe the current delivered by the 20 V source then ig 20 20 25 2 20 10 1 30 125 A pg delivered 20 30 125 602 5 W P 3 26 18 3i1 9i2 15 6i2 2i1 0 i2 i1 3 Solving i1 0 6 A i2 2 4 A p18V 18i1 10 8 W diss p3 0 6 2 3 1 08 W p2 0 6 2 2 0 72 W p9 2 4 2 9 51 84 W Problems75 p6 2 4 2 6 34 56 W Xp diss 99 W vo 15i2 15 36 15 21 V p3A 3vo 63 W dev p15V 15i2 36 W dev Xp dev 99 W Xp diss P 3 27 100 5i1 15i2 15 0 5i1 15i2 115 i2 i1 3 i2 i1 3 15i2 15i1 45 20i1 70 i1 3 5 A i2 6 5 A vo 15i2 15 97 5 15 82 5 V p100V 100i1 350 W dev p3A 3vo 247 5 W dev p15V 15i2 97 5 W dev Xp dev Xp dis 695 W Xp dis 3 5 2 5 6 5 2 15 695 W 76CHAPTER 3 Techniques of Circuit Analysis P 3 28 a Summing around the supermesh used in the solution to Problem 3 27 gives 10 5i1 15i2 15 0 i2 i1 3 i1 2 A i2 1 A p10V 10 2 20 W del vo 15i2 15 0 V p3A 3vo 0 W p15V 15i2 15 W del Xp diss 2 2 5 1 2 15 35 W Xp dev 35 W Xp diss b With 3 A current source replaced with a short circuit i1 2 A i2 1 A XP diss 2 2 5 1 2 15 35 W c A 3 A source with zero terminal voltage is equivalent to a short circuit carrying 3 A P 3 29 a 40 50i1 45i2 64 45i1 50 5i2 Solving i1 9 8 A i2 10 A ia i1 9 8 A ib i1 i2 0 2 A ic i2 10 A b If the polarity of the 64 V source is reversed we have 40 50i1 45i2 64 45i1 50 5i2 i1 1 72 Aandi2 2 8 A ia i1 1 72 A ib i1 i2 1 08 A ic i2 2 8 A Problems77 P 3 30 600 25 6i1 16i2 5 6i3 424 16i1 20i2 0 8i3 30 i3 Solving i1 35 A i2 8 A i3 30 A a v30A 0 8 i2 i3 5 6 i1 i3 0 8 8 30 5 6 35 30 10 4 V p30A 30v30A 30 10 4 312 W abs Therefore the 30 A source delivers 312 W b p600V 600 35 21 000 W del p424V 424 8 3392 W abs Therefore the total power delivered is 21 000 W c p4 35 2 4 4900 W p3 2 8 2 3 2 204 8 W p16 35 8 2 16 11 664 W p5 6 35 30 2 5 6 140 W p0 8 30 8 2 0 8 387 2 W Xp resistors 17 296 W Xp abs 17 296 312 3392 21 000 W CHECKS 78CHAPTER 3 Techniques of Circuit Analysis P 3 31 a 110 12 17i1 10i2 3i3 0 10i1 28i2 12i3 12 70 3i1 12i2 17i3 Solving i1 8 A i2 2 A i3 2 A p110 110i1 880 W del p12 12 i1 i3 120 W del p70 70i3 140 W del Xp dev 1140 W b p4 8 2 4 256 W p10 6 2 10 360 W p12 4 2 12 192 W p2 2 2 2 8 W p6 2 2 6 24 W p3 10 2 3 300 W Xp abs 1140 W P 3 32 a 2400 i1 0 0025 1500i1 150 i1 0 0025 0 Problems79 i1 1 5 mA i i1 2 5 1 0 mA b vo 0 0025 400 0 001 2400 3 4 V p2 5mA 3 4 2 5 8 5 mW p2 5mA deliver 8 5 mW c 150i 150 1 0 10 3 0 15 V pdep source 150i i1 0 15 0 0015 0 225 mW pdep source absorbed 0 225 mW P 3 33 Mesh equations 25i1 5i2 2 5i3 0 75 5i1 12 5i2 7 5i3 Constraint equations i3 0 2v v 5 i2 i1 Solving i1 3 6 A i2 13 2 A i3 9 6 A v 48 V vcs 125 v 2 5 i3 i1 125 48 2 5 9 6 3 6 62 V pvc 62 9 6 595 2 W abs p50V 50 i2 i3 50 13 2 9 6 180 W abs 80CHAPTER 3 Techniques of Circuit Analysis p125V 125i2 125 13 2 1650 W del Thus the total power developed is 1650 W CHECK p17 5 3 6 2 17 5 226 8 W p5 13 2 3 6 2 5 460 8 W p2 5 9 6 3 6 2 2 5 90 W p7 5 13 2 9 6 2 7 5 97 2 W Xp abs 226 8 460 8 90 97 2 180 595 2 1650 W P 3 34 80 31i1 16i2 7i3 0 16i1 27i2 4i3 0 7i1 4i2 31i3 24i2 0 Solving i1 3 5 A p8 3 5 2 8 98 W Problems81 P 3 35 a 200 85i1 25i2 50i3 0 75i1 35i2 150i3 super mesh i3 i2 4 3 i1 i2 Solving i1 4 6 A i2 5 7 A i3 0 97 A ia i2 5 7 A ib i1 4 6 A ic i3 0 97 A id i1 i2 1 1 A ie i1 i3 3 63 A b 10i2 vo 25 i2 i1 0 vo 57 27 5 84 5 V p4 3id vo 4 3id 84 5 4 3 1 1 399 685 W dev p200V 200 4 6 920 W dev XP dev 1319 685 W XP dis 5 7 210 1 1 2 25 0 97 2100 4 6 2 10 3 63 2 50 1319 685 W XP dev XP dis 1319 685 W P 3 36 660 30i1 10i2 15i3 82CHAPTER 3 Techniques of Circuit Analysis 20i 10i1 60i2 50i3 0 15i1 50i2 90i3 i i2 i3 Solving i1 42 A i2 27 A i3 22 A i 5 A 20i 100 V p20i 100i2 100 27 2700 W p20i developed 2700 W CHECK p660V 660 42 27 720 W dev XP dev 27 720 2700 30 420 W XP dis 42 2 5 22 2 25 20 2 15 5 2 50 15 2 10 30 420 W P 3 37 Mesh equations 50i1 20i2 25ig 0 20i1 120i2 30i 100ig 0 Constraint equations ig 4 i i1 Problems83 Solving i1 4 A i2 5 A i25 4 i1 0 A i20 i2 i1 1 A i100 4 i2 1 A i5 i1 4 A v4A 100 4 i2 100 V p4A v4Aig 100 4 400 W abs v30i 30i 30i1 120 V p30i 30i i2 120 5 600 W Therefore the dependent source is developing 600 W all other elements are absorbing power and the total power developed is thus 600 W CHECK p5 16 5 80 W p25 0 W p20 1 20 20 W p100 1 100 100 W p4A 400 W Xp abs 80 0 20 100 400 600 W CHECKS P 3 38 a 10 18i1 16i2 0 16i1 28i2 4i 4 8i Solving i1 1 A i2 0 5 A i 0 5 A v0 16 i1 i2 16 0 5 8 V 84CHAPTER 3 Techniques of Circuit Analysis b p4i 4i i2 4 0 5 0 5 1 W abs p4i deliver 1 W P 3 39 Mesh equations 2 65v 40i1 15i2 25i3 0 15i1 150i2 100i3 125 25i1 100i2 210i3 125 Constraint equations v 100 i2 i3 Solving i1 7 A i2 1 2 A i3 2 A v 100 i2 i3 100 1 2 2 80 V p2 65v 2 65v i1 1484 W Therefore the dependent source is developing 1484 W CHECK p125V 125i2 150 W left source p125V 125i3 250 W right source Xp dev 1484 250 1734 W p35 1 2 2 35 50 4 W Problems85 p85 2 2 85 340 W p15 7 1 2 2 15 504 6 W p25 7 2 2 25 625 W p100 1 2 2 2 100 64 W Xp diss 50 4 340 504 6 625 64 150 1734 W P 3 40 a Mesh equations 50 6i1 4i2 9i 0 9i 4i1 29i2 20i3 0 Constraint equations i i2 i3 1 7v v 2i1 Solving i1 5 A i2 16 A i3 17 A v 10 V 9i 9 16 144 V ia i2 i1 21 A ib i2 i3 1 A vb 20ib 20 V p50V 50i1 250 W absorbing p9i ia 9i 21 144 3024 W delivering p1 7V 1 7v vb i3vb 17 20 340 W delivering b XP dev 3024 340 3364 W XP dis 250 5 2 2 21 2 4 16 2 5 1 2 20 3364 W 86CHAPTER 3 Techniques of Circuit Analysis P 3 41 a Mesh equations 15 30i1 25i2 2i3 10 25i1 30i2 i3 Constraint equations i3 1 2v v 25 i1 i2 Solving i1 10 A i2