NilssonSM_Ch04-01

NilssonSM_Ch04-01 - 4—22 CHAPTER 4. Techniques of Circuit...

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Unformatted text preview: 4—22 CHAPTER 4. Techniques of Circuit Analysis [d] When a conductor joins the lower nodes of the two separate parts, there is now only a single part in the circuit. There would now be 4 nodes, because the two lower nodes are now joined as a. single node. The number of branches remains at 7, where each branch contains one of the seven individual circuit. components. P 4.4 [a] From Problem 4.2(d) there are 10 essential branches were the current is unknown, so we need 10 simultaneous equations to describe the [b] From Problem 4.2(f), there are 5 essential nodes, so we can apply KCL at (5 — l) = 4 of these essential nodes. There would also be two dependent source constraint equations. [c] The remaining 4 equations needed to describe the circuit will be derived from KVL equations. [(1] We must avoid using the meshes containing current sources, as we have no way of determining the voltage drop across a current source. P 4.5 [a] At node 1: —t'g+’£1+’i2=0 Atnoch: —i2+é3+t4=0 Atnode3: E‘Q‘Fé1—1‘l3—i‘1ZO [b] There are many possible solutions. For example, solve the equation at node 1 for i9: i9 = a + a Substitute this expression for 2'9 into the equation at node 3: (awn—s ~é3~é4=0 so tg—ts—azo Multiply this last equation by -1 to get the equation at node 2: —(é2—-t3—i4)=—O so —i2+t3+i4=0 P 4.6 Use the lower terminal of the 5 Q resistor as the reference node. 129—60 co m 3: 10 +5+ 0 Solving, co = 10 V Problem 4—23 P 4.7 [a] From the solution to Problem 4.5 we know 1)., z 10 V, therefore p3A # 300 = 30 W '. pm (developed) 2 #30 W [b] The current into the negative terminal of the 60 V source is Psov = —60(5) = —300 w pew (developed) = 300 W [6] pm = (5)2(10) : 250 W 175:: = (ling/5 = 20 W Dam, = 300 w Zeus = 250 + 20 + 30 == 300 W ‘i‘Fo — log P48 [3.] 10 +5 [b] Let v3 2 voltage drop across 3 A source 1.13 = 120 » (10)(3) = w20 v pm (developed) = (3)(20) = 60 W +3=0; wo=10V [B] Let i5. = be the current into the positive terminal of the 60 V source is = (10 _ 60)/10 = '5 A paw (developed) = = 300 W [d1 Does = (5)2(10) + (3)2(10) + (10)2/5 = 360 W Elms = 300+60=360 w [e] no is independent of any finite resistance connected in series with the 3 A current emu-(3e =0 1J1 U1+v2 P4.9 . — 24+125+ 25 02 — 'Ul 112 ’02 _ 25 +25o+375_3'2*0 Solving, v1 = 25 V; ’02 = 90 V 4—24 CHAPTER 4. Techniques of Circuit Analysis CHECK: P1259 = = 5 P2511 = QU—E—fi): = 169 W 10259;; = = 32.4 W 1037511 = (90: = 21.6 W 39311 = (25)(2.4) = 60 W 2pm =5-+169+32.4+21.6+60=288 W 2888, = (90)(3.2) = 288 w (CHECKS) P 4.10 [a] '01 — '01 U1 - ’02 8 +48 18 =0 152*?)1 ’02 112—70 _ 18 +20+ 10 _ D In standard form, (1+1+1)+1(1) _ 128 U18 48 18 ’12 18 _ 8 118 218 20 10 _10 Solving, 01:96V; U2=60V 3_fi1284§6=4A 8 96 "=—-—=2A ‘b 48 ic296—6022A 18 4—26 CHAPTER 4. Techniques of Circuit Analysis P 4.12 111*144 1J1 Ul—v2__ _ 4 U nga +v 80 — 0 so 2901 1);; H 2880 —3+ 2801+€=0 so —v1+171)2=240 Solving, 111 = 100 V; 112 = 20 V P 4.13 ’00 = 20 V pm = (20](2) = 40 W (absorbing) P 4.14 [a] — 640 — “I “I + “I “2 : 0 so 311:1 — 2002 + 0123 = 6400 50 5 2 5 “2 1"“ + “2 “)3 + 12.8 = 0 so «2U1+3t'2 ~vs = ~64 2.9 5 U3 U3 — 112 53+ 5 —12.8=D so Owl—v2+3v3==64 Solving,v1 = 380 V; w = 269 V; 1.13 = 111 V, , 640 — 380 [b] {"9 = = 52 A Pg(del) : (6%)(52) = 33,280 W Problems 4—27 P 4.15 1J1 — (‘02 + 'U] "' ’02 1}} fl _ _ 4 = 0 15 + 31.25 + 25 =0 'ul — ('02 + 30) 1:2 - 1:3 1:2 — v1 [ 15 + 50 + 31.25 U3 -* U2 “3 50 50+]:0 Solving, v1 = T6 V; 02 = 46 V; U3 = —2 V; éaw = 0 A Pm = —4'u1 I H406) = —304 w ((191) 101A = (IX—2) = ‘2 w ((191) 103W = (30)(0) = 0 W P159 2=(0)2(15]= 0 W 0% 762 =—=—=2 1.04W F250 25 25 3 (111 — ‘02)2 3132 1331.259 = = 31.25 = 28.8 W _ (1:2 —113)2 _ 482 _ pmflower] — 50 — 50 — 416.08 W U2 4 't:i=—=u 1950001th 50 50 03 w 2de = 0 + 231.04 + 28.8 + 46.08 + 0.08 2 306 w 23de = 304 + 2 = 306 w (CHECKS) 4+28 CHAPTER 4. Techniques of Circrfif; Analysis 0" o'" 0—” vet—“n P4.16[a]”R”1+”R“2+”R3+...+ R =0 ‘. nvo=v1+vz+v3+---+vfl 1 1 ‘. yo x Eful + ‘Ug + 1J3 + ' ' ' + Ural : EEZ=Ivk [b] 00 x $150 + 200 — 50) = 100 v uo+5aA 0040* . «00—80 '00 . ‘H —' 0‘ = P417 3+ 200+ 10 + 20 , an 20 [a] Solving, '00 =2 50 V . ‘Ua + 5%.A 1b] “3 _ 10 éA = (50 — 80)/20 = —1.5 A ids = 4-25 A; 50; = —7.5 V: pas = (—50A)(éd5) :— 31.875 w [a] pm = —3va = —3(50) : “150 w ((19.1) pm = 80in = 80(—1.5) = —120 w (del) 2%. = 150 + 120 = 270 w CHECK: P2000. = 2500/200 = 12.5 W page = (80 — 50)2/20 = 900/20 = 45 W pm = (425)200) = 180.625 w 2pm 2 31.875 + 180.025 + 12.5 + 45 = 270 W P 4.18 [a] P 4.19 Problems 4—29 _2éo+1m+ 25 =8 so 5v1H8o2+4o3=0 U2 — ’01 ’02 U2 — ’Ug _ 25 + SO 8'01 + 13’02 41);), -—— 0 v3 -— 1:2 'U'3 -— 55}, 1:3 -— 38.5 = 0 -— 4 = 1 .5 50 5 20 80 0'01 ‘02 + 29113 92 Solvmg,v1 "' ’50 U), — “:50 V, 03 = V 112—113 _ “50—515 __ [b] to +- 50 — 50 .- 055 A _ ' _ r _ £3 _ v3 510 = 2 5 o( 0.65) _ 1-15 A 5 5 38.5 —— 2 5 59 w 20 — 1.8 A Spas = Z:de Calculate Zpdev because we don’t know if the dependent sources are developing or absorbing power. Likewise for the independent source. 1025,, = #5025 = —2(—U.65)(—50) = —55 W(dev) png 2 55,53 : 5(~0.65)(1.15) = “3.7375 W(dev) pg = #38503) = —69.30 W(dev) )3de = 59.3 + 65 + 3.7375 = 138.0375 w CHECK 2% 250” 900 + 3’9 + (055)2(50) + (1.15% + (1.8)2(2{}) 100 +§fifi 25 = 138.0375W 1i Dew = 2pm = 138-0375 w 4-30 CHAPTER 4. Techniques of Circuit Analysis . _ 50—(—75)(1) _ so — 25 *5 A pm, 2 75m; = “375 w The dependent voltage source delivers 375 W to the circuit. P420 ['3] #5+:J—;+”1;02:0 ‘50 401*302+0'5A=75 “02—111 '02 U2 v2+5i3 . 5 +30+10+ 30 50 1+ “2+5” 0 iA=v1r—v2 SO Ul—Ug-‘51'ASU I.) , Solving,v1=30V; “2:15v; éA=3A; 10:15:);1031A Pam = (“450(1) : *15 “(0191) Pa = ~5(30) = —150 W(dei) pdev = 165 W [b] £10m; = (1‘22 + (13532 + (1332 + (3)2(5) + (1)2(30) = 165 W P 4.21 The two node voltage equations are: 1J1 — U1 ‘01 — 1J2 80 + % 40 = 0 H2 —— ‘01 U3 ’U2 *- - 0.75 —— : 40 + 200 + 800 0 Place theee equations in standard form: 1; + 1.!(H1) — 50 1 30 50 40 2 40 ‘ 80 ll 1 1 1 1 1:1 + W (— + — + —-—) 0.75 + 5% Problems 4w35 P 4.26 This ciI'Cuit has a. supernode includes the nodes 111, v2 and the 40 V source. The supernode equation is ‘01 ‘02 ‘02 0 3000 + 20,000 + 10,000 : —0.05 + The supemode constraint equation is o; + —?)1 = 40 Place these two equations in standard form: 1 1 1 01(8000) + “2(20,000+10,000) = 0'05 v1(—1) + 1&(1) = 40 Solving, v1 = 160 V and 112 = 200 V, so no = v2 = 200 V. ’01 mmzmmA em=0mu P40v = —(40)z'.m = ~(40)(0.03) = —1.2 w The 40 V source delivers 1.2 W. P 4.27 Place 155/5 inside a supernode and use the lower node as a reference. Then U] — '01 ‘01 — 33/5 1!] - TIA/5 _ 10 30+ 39 + 78 ‘0 13401 -— 61),; = 3900; ‘03 = 50 -- ‘01 Solving, 01 =30 V; 1:13:20 V; wro=3O-vA/5=30—4=25V P428 i:¢=”3;”‘ =235—22-2—2=3.25A 303's = 30(3.25) = 97.5 V 4-38 CHAPTER 4. Techniques of Circuit Analysis P 4.31 [a] 40 = 50151 — 4512 54 = —45’i1 + 50.53} Solving, i1 = 9.8 A; 2'2 2 10 A ia = i1 29.8 A; ib =é1w'i2 = #02 A; 1}; = wig = —10 A- [b] If the polarity of the 64 V source is reversed, we have 40 = 503} w 4552 ——64 = —45é1 + 50.535 3'1 = —1-72 A and i2 2 ~18 A éa=é1 =—1.72 A; ib=é1 —1'2 =1.08 A; in: —i2=2.8A P 4.32 [a] 110 + 12 z 173'] -10%'2 — 313 0 = -10é1 + 28152 — 121'3 "-12 - 70 = -3i1— 1212 + 1713 Solving, i1 = 8 A; i2 = 2 A; 153 = —2 A Pun = —110i1 : H880 W(del) p12 = —12(z‘1 — 3‘3) x —120 W(del) p70 = 70% 2 #140 W(de1) 2de = 1140 w ...
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This note was uploaded on 04/20/2008 for the course ECE 202 taught by Professor Xou during the Spring '08 term at Clemson.

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NilssonSM_Ch04-01 - 4—22 CHAPTER 4. Techniques of Circuit...

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