86 mA158 k \u03a9 203 V E Th 165 V 203 V 368 V 19 a R Th R Th 51 k \u03a9 10 k \u03a9 836 k \u03a9

# 86 ma158 k ω 203 v e th 165 v 203 v 368 v 19 a r th

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= (12.86 mA)(1.58 k Ω ) = 20.3 V E Th = 16.5 V + 20.3 V = 36.8 V 19. a. R Th : ¨ R Th = 51 k Ω || 10 k Ω = 8.36 k Ω E Th : E Th = 10 k (20 V) 10 k + 51 k = 3.28 V R Th 5 mA 5 mA 1.58 k Ω E Th + + 1.2 k Ω 3.3 k Ω + I' 7.86 mA 7.86 mA
102 CHAPTER 9 b. I E R E + V CE + I C R C = 20 V but I C = I E and I E ( R C + R E ) + V CE = 20 V or I E = 20 V - V CE R C + R E - 20 V - 8 V 2.2 k + 0.5 k = 12 V 2.7 k = 4.44 mA c. E Th - I B R Th - V BE - V E = 0 and I B = E Th - V BE - V E R Th = 3.28 V - 0.7 V - (4.44 mA)(0.5 k ) 8.36 k = = 43.06 µ A d. V C = 20 V - I C R C = 20 V - (4.44 mA)(2.2 k Ω ) = 20 V - 9.77 V = 10.23 V 20. a. E Th = 20 V I = 1.6 mA = = 12.5 k Ω b. E Th = 60 mV , R Th = 2.72 k Ω c. E Th = 16 V , R Th = 2.2 k Ω 21. R Th = 4 Ω || (2 Ω + 2 Ω ) = 4 2 = 2 Ω V 4 Ω = 2 (6 V) 2 + 4 + 2 = 12 V 8 = 1.5 V V 2 Ω = V 4 Ω = 1.5 V E Th = V 4 Ω + V 2 Ω = 1.5 V + 1.5 V = 3 V
CHAPTER 9 103 22. a. From Problem 9, R N = R Th = 6 Ω R T = 6 Ω + 3 Ω || 4 Ω = 6 Ω + 1.714 Ω = 7.714 Ω I s = E R T = 18 V 7.714 = 2.333 A I N = 3 (2.333 A) 3 + 4 = 1 A b. R Th = 6 Ω , E Th = I N R N = (1 A)(6 Ω ) = 6 V c. same results 23. a. From Problem 10, R N = R Th = 4.1 k Ω I ¢ = 2.4 k (120 mA) 2.4 k + (1.2 k || 3.3 k ) = 87.80 mA I N = 1.2 k (87.80 mA) 1.2 k + 3.3 k = 23.41 mA b. R Th = 4.1 k Ω , E Th = I N R N = (23.41 mA)(4.1 k Ω ) = 96 V c. same results. 24. From Problem 12, R N = R Th = 2.18 Ω I N = 6 A 1.5 A = 4.5 A 25. From Problem 13, R N = R Th = 1.58 k Ω I N = 8 mA 7.27 mA = 0.73 mA 26. From Problem 14, R N = R Th = 7.56 Ω 3 Ω 10 Ω 8 Ω 12 V 18 V I N + + 6 A 8 Ω I N 3 Ω 1.5 A
104 CHAPTER 9 I ¢ = 4 (2 A) 4 + 3 + 6 || 2 = 4 (2 A) 7 + 1.5 = 0.941 A I ¢¢ = 2 ¢ I 2 + 6 = 12 (0.941 A) 8 = 0.235 A I N = 2 A I ¢¢ = 2 A 0.235 A = 1.765 A 27. From Problem 16, R N = R Th = 10 Ω R T = 20 Ω + 5 Ω (12 Ω + 1.778 Ω ) = 23.67 Ω I s = E T R T = 20 V 23.67 = 844.95 mA I 12 Ω = 5 (844.95 mA) 5 + (12 + 1.778 ) =224. 98 mA I N = 16 (224.98 mA) 16 + 2 = 200 mA 28. From Problem 18, R N = R Th = 6.08 k Ω I N : Starting with figure from problem 18: and coverting sources: I ¢ = 12.42 V - 6 V 1.58 k + 3.3 k + 1.2 k = 1.06 mA 3 Ω 2 Ω 4 Ω I '' 6 Ω 2 A I ' 2 A 1.58 k Ω 1.2 k Ω 3.3 k Ω 7.86 mA I N 5 mA I N I' 1.58 k Ω 3.3 k Ω 12.42 V + 16 V + 1.2 k Ω I'