# 61 resistor gives vo 6i 262 substituting eq 262

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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...
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## This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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