Bài giải mạch P2

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Bài giải mạch P2

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  1. Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Solution 2 p = v2/R → R = v2/p = 14400/60 = 240 ohms Chapter 2, Solution 3 R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4 (a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6 n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7 7 elements or 7 branches and 4 nodes, as indicated. 30 V 1 20 Ω 2 3 +++- + - 2A 30 Ω 60 Ω 40 Ω 10 Ω 4
  2. Chapter 2, Solution 8 12 A a i1 8A b i3 i2 12 A c 9A d At node a, 8 = 12 + i1 i1 = - 4A At node c, 9 = 8 + i2 i2 = 1A At node d, 9 = 12 + i3 i3 = -3A Chapter 2, Solution 9 Applying KCL, i1 + 1 = 10 + 2 i1 = 11A 1 + i2 = 2 + 3 i2 = 4A i2 = i3 + 3 i3 = 1A Chapter 2, Solution 10 2 4A -2A i2 1 i1 3 3A At node 1, 4 + 3 = i1 i1 = 7A At node 3, 3 + i2 = -2 i2 = -5A
  3. Chapter 2, Solution 11 Applying KVL to each loop gives -8 + v1 + 12 = 0 v1 = 4v -12 - v2 + 6 = 0 v2 = -6v 10 - 6 - v3 = 0 v3 = 4v -v4 + 8 - 10 = 0 v4 = -2v Chapter 2, Solution 12 + 15v - loop 2 – 25v + + 10v - + v2 - + + + 20v loop 1 v1 loop 3 v3 - - - For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v For loop 2, -10 +15 -v2 = 0 v2 = 5v For loop 3, -v1 +v2 +v3 = 0 v3 = 30v Chapter 2, Solution 13 2A I2 7A I4 1 2 3 4 4A I1 3A I3
  4. At node 2, 3 + 7 + I2 = 0 →  I 2 = −10 A At node 1, I1 + I 2 = 2 →  I 1 = 2 − I 2 = 12 A At node 4, 2 = I4 + 4 →  I 4 = 2 − 4 = −2 A At node 3, 7 + I4 = I3 →  I3 = 7 − 2 = 5 A Hence, I 1 = 12 A, I 2 = −10 A, I 3 = 5 A, I 4 = −2 A Chapter 2, Solution 14 + + - 3V V1 I4 V2 - I3 - + 2V - + - + V3 - + + 4V I2 - V4 I1 5V + - For mesh 1, −V4 + 2 + 5 = 0 →  V4 = 7V For mesh 2, +4 + V3 + V4 = 0 →  V3 = −4 − 7 = −11V For mesh 3, −3 + V1 − V3 = 0 →  V1 = V3 + 3 = −8V For mesh 4, −V1 − V2 − 2 = 0 →  V2 = −V1 − 2 = 6V Thus, V1 = −8V , V2 = 6V , V3 = −11V , V4 = 7V
  5. Chapter 2, Solution 15 + + + 12V 1 v2 - - 8V + - v1 - 3 + 2 - v3 10V - + For loop 1, 8 − 12 + v2 = 0 →  v2 = 4V For loop 2, − v3 − 8 − 10 = 0 →  v3 = −18V For loop 3, − v1 + 12 + v3 = 0 →  v1 = −6V Thus, v1 = −6V , v2 = 4V , v3 = −18V Chapter 2, Solution 16 + v1 - + + loop 1 6V - 10V v1 - +- +- 12V loop 2 + v2 - Applying KVL around loop 1, –6 + v1 + v1 – 10 – 12 = 0 v1 = 14V Applying KVL around loop 2, 12 + 10 – v2 = 0 v2 = 22V
  6. Chapter 2, Solution 17 + v1 - loop 1 + - + 24V - v2 v3 + - 10V + - loop 2 -+ 12V It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0 v2 = -22V Applying KVL to loop 1, -24 + v1 - v2 = 0 v1 = 2V Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A -Vab + 5I + 8 = 0 Vab = 28V
  7. Chapter 2, Solution 19 Applying KVL around the loop, we obtain -12 + 10 - (-8) + 3i = 0 i = -2A Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W p8V = (- -2) = -16W Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Solution 21 Apply KVL to obtain 10 Ω -45 + 10i - 3V0 + 5i = 0 + v0 - But v0 = 10i, + - 45V - + 3v0 -45 + 15i - 30i = 0 i = -3A P3 = i2R = 9 x 5 = 45W 5Ω
  8. Chapter 2, Solution 22 4Ω + v0 - 6Ω 10A 2v0 At the node, KCL requires that v0 + 10 + 2 v 0 = 0 v0 = –4.444V 4 The current through the controlled source is i = 2V0 = -8.888A and the voltage across it is v0 v = (6 + 4) i0 = 10 = −11.111 4 Hence, p2 vi = (-8.888)(-11.111) = 98.75 W Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. ix 1Ω + vx - 6A 2Ω 3Ω
  9. Applying current division, 2 ix = (6 A) = 2 A, v x = 1i x = 2V 2 + 1+ 3 The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω resistor is 1 x 4.8 = 4.8 V. Hence the power is v 2 4.8 2 p= = = 1.92W R 12 Chapter 2, Solution 24 Vs (a) I0 = R1 + R2 αV0 R3 R4 V0 = −α I0 (R3 R4 ) = − ⋅ R1 + R 2 R3 + R4 V0 − αR3 R4 = Vs (R1 + R2 )(R3 + R4 ) (b) If R1 = R2 = R3 = R4 = R, V0 α R α = ⋅ = = 10 α = 40 VS 2R 2 4 Chapter 2, Solution 25 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, 5 I20 = (0.01x50) = 0.1 A 5 + 20 V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW
  10. Chapter 2, Solution 26 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, 5 I20 = (0.01x50) = 0.1 A 5 + 20 V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27 Using current division, 4 i1 = (20) = 8 A 4+6 6 i2 = (20) = 12 A 4+6 Chapter 2, Solution 28 We first combine the two resistors in parallel 15 10 = 6 Ω We now apply voltage division, 14 v1 = (40) = 20 V 14 + 6 6 v2 = v3 = (40) = 12 V 14 + 6 Hence, v1 = 28 V, v2 = 12 V, vs = 12 V
  11. Chapter 2, Solution 29 The series combination of 6 Ω and 3 Ω resistors is shorted. Hence i2 = 0 = v2 12 v1 = 12, i1 = = 3A 4 Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30 8Ω i1 i + 9A 6Ω v 4Ω - 12 By current division, i = (9) = 6 A 6 + 12 i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31 The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω . Hence by the voltage division principle, 5 v= (20V) = 10 V 5+5 by ohm's law, v 10 i= = = 1A 4 + 6 4+ 6 pp = i2R = (1)2(4) = 4 W
  12. Chapter 2, Solution 32 We first combine resistors in parallel. 20 x30 20 30 = = 12 Ω 50 10x 40 10 40 = = 8Ω 50 Using current division principle, 8 12 i1 + i 2 = (20) = 8A, i 3 + i 4 = (20) = 12A 8 + 12 20 20 i1 = (8) = 3.2 A 50 30 i2 = (8) = 4.8 A 50 10 i3 = (12) = 2.4A 50 40 i4 = (12) = 9.6 A 50 Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i 4S i + + 9A v 1S 9A v 1S 4S 2S - - 6x3 6 S 3S = = 25 and 25 + 25 = 4 S 9 Using current division, 1 i= (9) = 6 A, v = 3(1) = 3 V 1 1+ 2
  13. Chapter 2, Solution 34 By parallel and series combinations, the circuit is reduced to the one below: i1 8Ω 10 x15 10 ( 2 + 13 ) = = 6Ω 25 15 x15 + 15 (4 + 6) = = 6Ω 28V + v1 25 - 6Ω - 12 (6 + 6) = 6Ω 28 Thus i1 = = 2 A and v1 = 6i1 = 12 V 8+6 We now work backward to get i2 and v2. i1 = 2A 8Ω 6Ω 1A 1A + + + 6V 28V - 12V 12 Ω 6Ω - - i1 = 2A 8Ω 6Ω 1A 4Ω 0.6A 1A + + + + 6V 3.6V 28V - 12V 12 Ω 15 Ω 6Ω - - - 13 v Thus, v2 = (3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24 15 13 p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W Chapter 2, Solution 35 i + 70 Ω V1 30 Ω + i1 - I0 50V a b - + 20 Ω V0 5 Ω i2 -
  14. Combining the versions in parallel, 70x30 20x 5 70 30 = = 21Ω , 20 15 = =4 Ω 100 25 50 i= =2 A 21 + 4 vi = 21i = 42 V, v0 = 4i = 8 V v v i1 = 1 = 0.6 A, i2 = 2 = 0.4 A 70 20 At node a, KCL must be satisfied i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A Hence v0 = 8 V and I0 = 0.2A Chapter 2, Solution 36 The 8-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0 is the voltage across the 6Ω resistor. 4 4 I0 = = =1 A 2 + 3 16 4 v0 = I0 (3 6 ) = 2I 0 = 2 V
  15. Chapter 2, Solution 37 Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the 6 R combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor. 16 Hence, I = = 1 A. 16 20 6R But I = =1 4= 6R= R = 12 Ω 16 + 6 R 6+R Chapter 2, Solution 38 Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel. 6I0 = 2 x 3 R0 = 1 A The total current through the 4Ω resistor = 1 + 2 = 3 A. Hence vS = (2 + 4 + 2 3 ) (3 A) = 24 V vS I= = 2.4 A 10 Chapter 2, Solution 39 (a) Req = R 0 = 0 R R (b) Req = R R + R R = + = R 2 2 (c) Req = (R + R ) (R + R ) = 2R 2R = R 1 (d) Req = 3R (R + R R ) = 3R (R + R ) 2 3 3Rx R = 2 =R 3 3R + R 2  R ⋅ 2R  (e) Req = R 2R 3R = 3R    3R  2 3Rx R = 3R 2 R= 3 = 6R 3 2 11 3R + R 3
  16. Chapter 2, Solution 40 Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω 10 10 I= = = 2A Re q 5 Chapter 2, Solution 41 Let R0 = combination of three 12Ω resistors in parallel 1 1 1 1 = + + Ro = 4 R o 12 12 12 R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R ) 60(14 + R ) 50 = 30 + 74 + R = 42 + 3R 74 + R or R = 16 Ω Chapter 2, Solution 42 5x 20 (a) Rab = 5 (8 + 20 30) = 5 (8 + 12) = = 4Ω 25 (b) Rab = 2 + 4 (5 + 3) 8 + 5 10 (6 + 4) = 2 + 4 4 + 5 5 = 2 + 2 + 2.5 = 6.5 Ω Chapter 2, Solution 43 5x 20 400 (a) Rab = 5 20 + 10 40 = + = 4 + 8 = 12 Ω 25 50 −1  1 1 1  60 (b) 60 20 30 =  + +  = = 10Ω  60 20 30  6 80 + 20 Rab = 80 (10 + 10) = = 16 Ω 100
  17. Chapter 2, Solution 44 (a) Convert T to Y and obtain 20 x 20 + 20 x10 + 10 x 20 800 R1 = = = 80 Ω 10 10 800 R2 = = 40 Ω = R3 20 The circuit becomes that shown below. R1 a R3 R2 5Ω b R1//0 = 0, R3//5 = 40//5 = 4.444 Ω Rab = R2 / /(0 + 4.444) = 40 / /4.444 = 4Ω (b) 30//(20+50) = 30//70 = 21 Ω Convert the T to Y and obtain 20 x10 + 10 x 40 + 40 x 20 1400 R1 = = = 35Ω 40 40 1400 1400 R2 = = 70 Ω , R3 = = 140 Ω 20 10 The circuit is reduced to Ω shown below. 15 that 11 Ω R1 R2 R3 30 Ω 21 Ω 21 Ω Combining the resistors in parallel
  18. R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 11 Ω 10.5 Ω 21 Ω 140 Ω 21 Ω 21 Ω Coverting the T to Y leads to the circuit below. 11 Ω 10.5 Ω R4 R5 R6 21 Ω 21x140 + 140 x 21 + 21x 21 6321 R4 = = = 301Ω = R6 21 21 6321 R5 = = 45.15 140 10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94 Rab = 11 + 17 .94 = 28.94Ω Chapter 2, Solution 45 (a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8 Rab = 5 + 50 + 4.8 = 59.8 Ω (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = 5 + 12.8 + 15 = 32.5Ω
  19. Chapter 2, Solution 46 30x 70 60 + 20 (a) Rab = 30 70 + 40 + 60 20 = + 40 + 100 80 = 21 + 40 + 15 = 76 Ω (b) The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted. 20x30 20 30 = = 12Ω 50 40x 60 40 60 = = 24 100 Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω Chapter 2, Solution 47 5x 20 5 20 = = 4Ω 25 6x3 6 3= = 2Ω 9 10 Ω 8Ω a b 2Ω 4Ω Rab = 10 + 4 + 2 + 8 = 24 Ω
  20. Chapter 2, Solution 48 R 1 R 2 + R 2 R 3 + R 3 R 1 100 + 100 + 100 (a) Ra = = = 30 R3 10 Ra = Rb = Rc = 30 Ω 30x 20 + 30x50 + 20x 50 3100 (b) Ra = = = 103.3Ω 30 30 3100 3100 Rb = = 155Ω, R c = = 62Ω 20 50 Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω Chapter 2, Solution 49 RaRc 12 + 12 (a) R1 = = = 4Ω Ra + Rb + Rc 36 R1 = R2 = R3 = 4 Ω 60x30 (b) R1 = = 18Ω 60 + 30 + 10 60 x10 R2 = = 6Ω 100 30x10 R3 = = 3Ω 100 R1 = 18Ω, R2 = 6Ω, R3 = 3Ω Chapter 2, Solution 50 Using R ∆ = 3RY = 3R, we obtain the equivalent circuit shown below: R 30mA 3R 3R 30mA 3R 3R/2 3R R
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