tailieunhanh - Bài giải phần giải mạch P4

Chapter 4, Solution 1. 1Ω i 5Ω io 1V + − 8Ω 3Ω 8 (5 + 3) = 4Ω , i = 1 1 = 1+ 4 5 io = 1 1 i= = 2 10 Chapter 4, Solution 2. 6 (4 + 2) = 3Ω, i1 = i 2 = 1 A 2 5Ω 4Ω io = 1 1 i1 = , v o = 2i o = 2 4 i1 io i2 1A 8Ω 6Ω 2Ω If is = 1µA, then vo = Chapter 4, Solution 3. R 3R io 3R Vs 3R R + vo − (a) (b) 1V + − + − 3R .(a) We transform the Y sub-circuit to the equivalent ∆ . R 3R. | Chapter 4 Solution 1. 3 Q 8 5 3 4Q 1 i ----------- 1 4 1 5 i- 2i 10 Chapter 4 Solution 2. 6 4 2 3 ii i2 2 A 2 Q If is 1 pA then vo V Chapter 4 Solution 3. 1 V 3R b a We transform the Y sub-circuit to the equivalent A. 3R2 3 3 3 3_ R 3R ----- R R R R 4R 4 4 4 2 V vo independent of R io Vo R When vs 1V Vo io b When Vs 10V Vo 5V io 5A c When Vs 10V and R 10Q Vo 5V io 10 10 500mA Chapter 4 Solution 4. If Io 1 the Voltage across the 6Q resistor is 6V so that the current through the 3Q resistor is 2A. 2A 2 Q 2 Q b a 3 6 2Q Vo 3 4 12V i1 3A. Hence Is 3 3 6A If Is 6A--------- Io 1 Is 9A----- Io 6 9 Chapter 4 Solution 5. If Vo If Vs 6 Q 1V V1 1 2 V J 2 10 s 3 V 3 10 3 Vo 1 Then vs 15 Vo x15 V 10 ----- Chapter 4 Solution 6 Let Rt R2 R3 R2 R3 then V Rt Vs T 23 r2 r3 o rt r1 s R2 R3 k Vo RT R2 R3 R2 R3_ Vs Rt R1 R2 R3 r R1R2 R2 R3 R3 R1 R2 R3