Bài giải phần giải mạch P4

Chapter 4, Solution 1. 1Ω i 5Ω io1V+ −8Ω3Ω8 (5 + 3) = 4Ω , i =1 1 = 1+ 4 5io =1 1 i= = 0.1A 2 10Chapter 4, Solution 2.6 (4 + 2) = 3Ω, i1 = i 2 =1 A 25Ω 4Ωio =1 1 i1 = , v o = 2i o = 0.5V 2 4i1ioi2 1A 8Ω 6Ω 2ΩIf is = 1µA, then vo = 0.5µVChapter 4, Solution 3. R 3R io 3R Vs 3R R + vo − (a) (b) 1V+ −+ −3R1.5R.(a) We transform the Y sub-circuit to the equivalent ∆ . R 3R...
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Bài giải phần giải mạch P4Chapter 4, Solution 1. 1Ω i 5Ω io + 1V 8Ω 3Ω − 1 18 (5 + 3) = 4Ω , i = = 1+ 4 5 1 1io = i= = 0.1A 2 10Chapter 4, Solution 2. 16 (4 + 2) = 3Ω, i1 = i 2 = A 2 1 1io = i1 = , v o = 2i o = 0.5V 2 4 5Ω i1 4Ω io i2 1A 8Ω 6Ω 2ΩIf is = 1µA, then vo = 0.5µVChapter 4, Solution 3. R 3R io 3R 3R + + 3R 1.5R + R 1V Vs vo − − − (b) (a)(a) We transform the Y sub-circuit to the equivalent ∆ . 3R 2 3 3 3 3R 3R = = R, R + R = R 4R 4 4 4 2 vsvo = independent of R 2io = vo/(R)When vs = 1V, vo = 0.5V, io = 0.5A(b) When vs = 10V, vo = 5V, io = 5A(c) When vs = 10V and R = 10Ω, vo = 5V, io = 10/(10) = 500mAChapter 4, Solution 4.If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ωresistor is 2A. 2A 2Ω 2Ω 1A 3A 3A i1 + 3Ω 6Ω 4Ω Is 2Ω 4Ω Is v1 − (a) (b) vo3 6 = 2Ω , vo = 3(4) = 12V, i1 = = 3A. 4Hence Is = 3 + 3 = 6AIf Is = 6A Io = 1 Is = 9A Io = 6/(9) = 0.6667AChapter 4, Solution 5. 2Ω v1 3Ω vo + Vs 6Ω 6Ω 6Ω − 1If vo = 1V, V1 =   + 1 = 2V 3 2 10 Vs = 2  + v1 = 3 3 10If vs = vo = 1 3 3Then vs = 15 vo = x15 = 4.5V 10Chapter 4, Solution 6 R2 R3 RT Let RT = R2 // R3 = , then Vo = Vs R2 + R3 RT +R1 R2 R3 V RT R2 + R3 R2 R3 k= o = = = Vs RT + R1 R2 R3 R1 R2 + R2 R3 + R3 R1 + R1 R2 + R3Chapter 4, Solution 7We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider thecircuit below. 3Vx 5Ω 5Ω + + 4V 15 Ω VTh - 6Ω - + Vx -From the figure, 15 V x = 0, VTh = (4) = 3V 15 + 5To find RTh, consider the circuit below: 3Vx 5Ω 5Ω V1 V2 + 4V 15 Ω 1A - 6Ω + Vx -At node 1,4 − V1 V V − V2 = 3V x + 1 + 1 , V x = 6 x1 = 6  → 258 = 3V2 − 7V1 (1) 5 15 5At node 2, V1 − V21 + 3V x + =0 → V1 = V2 − 95 (2) 5Solving (1) and (2) leads to V2 = 101.75 V 2 V V 9RTh = 2 = 101.75Ω, p max = Th = = 22.11 mW 1 4 RTh 4 x101.75Chapter 4, Solution 8.Let i = i1 + i2,where i1 and iL are due to current and voltage sources respectively. 6Ω i2 i1 6Ω 4Ω 5A + 4Ω 20V − (a) ...