{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EE302_spring_2008_hw8a

# EE302_spring_2008_hw8a - EE 302 HW 8 solutions Chapter 4...

This preview shows pages 1–5. Sign up to view the full content.

EE 302 HW 8 solutions 1 Chapter 4, Problem 40. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.107. Figure 4.107 For Prob. 4.40. Solution. To obtain V Th , we apply KVL to the loop. 70 (10 20) 4 0 o kI V + + + = But 10 o V kI = 70 70 1 kI I mA = → = 70 10 0 60 V Th Th kI V V + + = → = To find R Th , one approach is to remove the 70-V source and apply a 1-V source at terminals a-b, as shown in the circuit below. 1 V We notice that V o = -1 V. 1 1 1 20 4 0 0.25 mA o kI V I − + + = → = 2 1 1 0.35 mA 10 V I I k = + = 2 1 1 2.857 k 0.35 Th V R k I = = Ω = Ω + _ 70 V + V o 4 V o + 10 k Ω 20 k Ω moonfull moonfull b a b 20 Ω 4 V o 10 k Ω + _ + V o I 2 a I 1 trianglecentrt trianglecentup + I

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EE 302 HW 8 solutions 2 Chapter 4, Problem 42. For the circuit in Fig. 4.109, find Thevenin equivalent between terminals a and b . Figure 4.109
EE 302 HW 8 solutions 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EE 302 HW 8 solutions 4 Chapter 4, Problem 44. For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals (a) a - b (b) b - c Figure 4.111 Chapter 4, Solution 44.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

EE302_spring_2008_hw8a - EE 302 HW 8 solutions Chapter 4...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online