61 resistor gives vo 6i 262 substituting eq 262

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Figure 2.23 6Ω (b) For Example 2.6. Solution: We apply KVL around the loop as shown in Fig. 2.23(b). The result is −12 + 4i + 2vo − 4 + 6i = 0 Applying Ohm’s law to the 6- (2.6.1) resistor gives vo = −6i (2.6.2) Substituting Eq. (2.6.2) into Eq. (2.6.1) yields −16 + 10i − 12i = 0 ⇒ i = −8 A and vo = 48 V. PRACTICE PROBLEM 2.6 10 Ω Find vx and vo in the circuit of Fig. 2.24. Answer: 10 V, −5 V. + vx − 35 V + − 5Ω + − 2vx + vo − | v v Figure 2.24 | e-Text Main Menu | Textbook Table of Contents | For Practice Prob. 2.6. Problem Solving Workbook Contents 40 PART 1 DC Circuits EXAMPLE 2.7 Find current io and voltage vo in the circuit shown in Fig. 2.25. a + vo − 0.5io Solution: Applying KCL to node a , we obtain io 4Ω 3A 3 + 0.5io = io For the 4- Figure 2.25 ⇒ io = 6 A resistor, Ohm’s law gives vo = 4io = 24 V For Example 2.7. PRACTICE PROBLEM 2.7 Find vo and io in the circuit of Fig. 2.26. io io 4 2Ω 6A Figure 2.26 + vo − 8Ω Answer: 8 V, 4 A. For Practice Prob. 2.7. EXAMPLE 2.8 Find the currents and voltages in the circuit shown in Fig. 2.27(a). 8Ω i1 30 V + − + v2 − i3 8Ω i2 + v1 − a + v1 − 3Ω + v3 − 6Ω 30 V + − Loop 1 (a) Figure 2.27 i1 i3 a i2 + v2 − 3Ω Loop 2 + v3 − 6Ω (b) For Example 2.8. Solution: We apply Ohm’s law and Kirchhoff ’s laws. By Ohm’s law, v1 = 8i1 , v2 = 3i2 , v3 = 6i3 (2.8.1) Since the voltage and current of each resistor are related by Ohm’s law as shown, we are really looking for three things: (v1 , v2 , v3 ) or (i1 , i2 , i3 ). At node a , KCL gives i1 − i2 − i3 = 0 (2.8.2) Applying KVL to loop 1 as in Fig. 2.27(b), | v v −30 + v1 + v2 = 0 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 2 Basic Laws 41 We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain −30 + 8i1 + 3i2 = 0 or i1 = (30 − 3i2 ) 8 (2.8.3) Applying KVL to loop 2, −v2 + v3 = 0 ⇒ v3 = v2 (2.8.4) as expected since the two resistors are in parallel...
View Full Document

Ask a homework question - tutors are online