Tài liệu Bài giải mạch P2 pptx

Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Solution 2 p = v 2 /R → R = v 2 /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 +- Chapter 2, Solution 8 d c b a 9 A i 3 i 2 12 A 12 A i 1 8 A At node a, 8 = 12 + i 1 i 1 = - 4A At node c, 9 = 8 + i 2 i 2 = 1A At node d, 9 = 12 + i 3 i 3 = -3A Chapter 2, Solution 9 Applying KCL, i 1 + 1 = 10 + 2 i 1 = 11A 1 + i 2 = 2 + 3 i 2 = 4A i 2 = i 3 + 3 i 3 = 1A Chapter 2, Solution 10 At node 1, 4 + 3 = i 1 i 1 = 7A At node 3, 3 + i 2 = -2 i 2 = -5A 3 2 -2A 3A 1 4A i 2 i 1 Chapter 2, Solution 11 Applying KVL to each loop gives -8 + v 1 + 12 = 0 v 1 = 4v -12 - v 2 + 6 = 0 v 2 = -6v 10 - 6 - v 3 = 0 v 3 = 4v -v 4 + 8 - 10 = 0 v 4 = -2v Chapter 2, Solution 12 For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35v For loop 2, -10 +15 -v 2 = 0 v 2 = 5v For loop 3, -v 1 +v 2 +v 3 = 0 v 3 = 30v + 15v - loop 3 loop 2 loop 1 + 20v - + 10v - – 25v + + v 2 - + v 1 - + v 3 - Chapter 2, Solution 13 2A I 2 7A I 4 1 2 3 4 4A I 1 3A I 3 At node 2, 37 0 10 22 ++ = →=− IIA 12 2A 25A At node 1, II I I A 12 1 2 22 += →=−= At node 4, 24 24 44 =+ →=−=− II At node 3, 77 43 3 += →=−= II I Hence, IAI AIAI 12 34 12 10 5 2 ==−== ,,, A − V 11 8 Chapter 2, Solution 14 + + - 3V V 1 I 4 V 2 - I 3 - + 2V - + - + V 3 - + + 4V I 2 - I 1 V 4 + - 5V For mesh 1, −++= →= VV 44 250 7 For mesh 2, ++ + = →=−−=− 40 47 34 3 VV V V For mesh 3, −+ − = →=+=− 30 3 13 1 3 VV V V V For mesh 4, −− −= →=−−= VV V V V 12 2 1 20 26 Thus, VVVVV VV 123 4 86 11 =− = =− V7 = ,, , Chapter 2, Solution 15 + + + 12V 1 v 2 - - 8V + - v 1 - 3 + 2 - v 3 10V - + For loop 1, 812 0 4 22 −+= →= vvV V V For loop 2, −−− = →=− vv 33 810 0 18 For loop 3, −+ + = →=− vv v 13 1 12 0 6 Thus, vVvVv 123 64 =− = =− ,, V18 Chapter 2, Solution 16 + v 1 - + - + - 6V - + loop 1 loop 2 12V 10V + v 1 - + v 2 - Applying KVL around loop 1, –6 + v 1 + v 1 – 10 – 12 = 0 v 1 = 14V Applying KVL around loop 2, 12 + 10 – v 2 = 0 v 2 = 22V Chapter 2, Solution 17 + v 1 - + - + - + 10V 12V 24V loop 2 + v 3 - v 2 - loop 1 - + It is evident that v 3 = 10V Applying KVL to loop 2, v 2 + v 3 + 12 = 0 v 2 = -22V Applying KVL to loop 1, -24 + v 1 - v 2 = 0 v 1 = 2V Thus, v 1 = 2V, v 2 = -22V, v 3 = 10V Chapter 2, Solution 18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A -V ab + 5I + 8 = 0 V ab = 28V 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 = i 2 R = 4(3) = 12W Power supplied by the sources: p 12V = 12 (- -2) = 24W p 10V = 10 (-2) = -20W p 8V = (- -2) = -16W Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i 0 + 5i 0 = 0 i 0 = 4A Chapter 2, Solution 21 10 Ω + - 45V - + + v 0 - Apply KVL to obtain -45 + 10i - 3V 0 + 5i = 0 But v 0 = 10i, 3v 0 -45 + 15i - 30i = 0 i = -3A P 3 = i 2 R = 9 x 5 = 45W 5 Ω Chapter 2, Solution 22 4 Ω + v 0 - 10A 2v 0 6 Ω At the node, KCL requires that 0 0 v210 4 v ++ = 0 v 0 = –4.444V The current through the controlled source is i = 2V 0 = -8.888A and the voltage across it is v = (6 + 4) i 0 = 10 111.11 4 v 0 −= Hence, p 2 v i = (-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. i x 1 Ω + v x - 6A 2 Ω 3 Ω Applying current division, iAAvi xx = ++ == 2 213 62 12 () , V x = The current through the 1.2- resistor is 0.5i Ω x = 1A. The voltage across the 12- resistor is 1 x 4.8 = 4.8 V. Hence the power is Ω p v R W == = 22 48 12 192 . . Chapter 2, Solution 24 (a) I 0 = 21 RR V s + α −= 0 V I 0 ( ) 43 RR = 43 43 2 1 0 RR RR RR V + ⋅ + − α () () 4321 430 RRRR RR Vs ++ V − = α (b) If R 1 = R 2 = R 3 = R 4 = R, 10 42 R R2V V S 0 = α =⋅ α = α = 40 Chapter 2, Solution 25 V 0 = 5 x 10 -3 x 10 x 10 3 = 50V Using current division, I 20 = + )5001.0( 205 x = 5 0.1 A V 20 = 20 x 0.1 kV = 2 kV p 20 = I 20 V 20 = 0.2 kW Chapter 2, Solution 26 V 0 = 5 x 10 -3 x 10 x 10 3 = 50V Using current division, I 20 = + )5001.0( 205 x = 5 0.1 A V 20 = 20 x 0.1 kV = 2 kV p 20 = I 20 V 20 = 0.2 kW Chapter 2, Solution 27 Using current division, i 1 = = + )20( 64 4 8 A i 2 = = + )20( 64 6 12 A Chapter 2, Solution 28 We first combine the two resistors in parallel =1015 6 Ω We now apply voltage division, v 1 = = + )40( 614 14 20 V v 2 = v 3 = = + )40( 614 6 12 V Hence, v 1 = 28 V, v 2 = 12 V, v s = 12 V [...]... 6Ω 1A 1A 28V + 12V - + - 8Ω i1 = 2A 6Ω + 6V - 12 Ω 6Ω 4Ω 1A 0.6A 1A 28V Thus, v2 = + 12V - + - 12 Ω + 6V - + 15 Ω 3.6V v 13 (3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24 13 15 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 50V + - a + V1 i1 - 30 Ω I0 + 20 Ω i2 b V0 5 Ω - - 6Ω Combining the versions in parallel, 70 30 = i= 70x30 = 21Ω , 100... 3 V Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor: V + V R1 - Case 1 Hence p = V 2 p2 R1 , = R p1 R 2 + R2 - Case 2 p2 = R1 10 p1 = (12) = 30 W 4 R2 Chapter 2, Solution 81 Let R1 and R2 be in kΩ R eq = R 1 + R 2 5 (1) 5 R2 V0 = VS 5 R 2 + R 1 (2) From (1) and (2), 0.05 = 5 R1 2 = 5 R2 = 40 From (1),