电路分析习题02解答.pdf

2-1 2-1 I = 16.5mAR = 2kΩR = 40 kΩR = 10 kΩR = 25 kΩ s s 1 2 3 I I I 1 2 3 Rs + I 1 I2 I3 Is U R 1 R2 R3 – 2-1 U I s (R1 // R2 // R3 ) 16.5 ×(40 // 10// 25) 100V I U R 100 40k 2.5mA 1 1 I U R 100 10k 10mA 2 2 I U R 100 25k 4mA 3 3 2-2 2-2 u = 100VR = 2 kΩR = 8 kΩ s 1 2 1R3 = 8 kΩ 2R = ∞R 3 3 3R = 0 R 3 3 3 u i i 2 2 3 + i2 i3 + us – R2 u2 R3 R 1 – 2-2 1R3 = 8 kΩR2 //R3 = 4 kΩ u 100 s 200 u × R R × 2 R +R // R ( 2 // 3 ) 2 +4 4 3 V 1 2 3 200 3 i u R 25 3mA 2 2 2 8k 200 3 i u R 25 3mA 3 2 3 8k 2R = ∞R 3 3 u 100 u s R ×8 80V 2 2 R +R 2 +8