{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw8a - Chapter 4 Problem 40 Find the Thevenin equivalent at...

This preview shows pages 1–4. Sign up to view the full content.

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 = 1 + _ 70 V + V o 4 V o + 10 k 20 k b a b 20 4 V o 10 k + _ + V o I 2 a I 1 i i +

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

View Full Document
Chapter 4, Problem 43. Find the Thevenin equivalent looking into terminals a - b of the circuit in Fig. 4.110 and solve for i x . The 6 ohm resistor is a load resistance. Figure 4.110 Solution. To find R Th , consider the circuit in Fig. (a). R Th = 10||10 + 5 = 10 ohms To find V Th , consider the circuit in Fig. (b). v b = 2x5 = 10 V, v a = 20/2 = 10 V -v a + V Th + v b = 0, or V Th = v a – v b = 0 volts and i x = 0 A 2 (a) 10 R Th a b 10 5 (b) 10 20V + - 2 A 10 V Th + + v a - + v b - 5 b a
Chapter 4, Problem 48.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}