# Hence eqs 220 and 221 remain the same when voltage

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Unformatted text preview: would have been +v1 , −v5 , +v4 , −v3 , and −v2 , which is the same as before except that the signs are reversed. Hence, Eqs. (2.20) and (2.21) remain the same. When voltage sources are connected in series, KVL can be applied to obtain the total voltage. The combined voltage is the algebraic sum of the voltages of the individual sources. For example, for the voltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL. A single-loop circuit illustrating KVL. | v v −Vab + V1 + V2 − V3 = 0 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 38 PART 1 DC Circuits or Vab = V1 + V2 − V3 (2.23) To avoid violating KVL, a circuit cannot contain two different voltages V1 and V2 in parallel unless V1 = V2 . a + + − + − b V2 − + Vab V1 V3 a + + V =V +V −V 1 2 3 −S Vab − b − (a) (b) Figure 2.20 Voltage sources in series: (a) original circuit, (b) equivalent circuit. EXAMPLE 2.5 For the circuit in Fig. 2.21(a), ﬁnd voltages v1 and v2 . 2Ω + v1 − + v1 − v2 3Ω 20 V + − + (a) Figure 2.21 i v2 3Ω + − + − − 20 V 2Ω (b) For Example 2.5. Solution: To ﬁnd v1 and v2 , we apply Ohm’s law and Kirchhoff ’s voltage law. Assume that current i ﬂows through the loop as shown in Fig. 2.21(b). From Ohm’s law, v1 = 2i, v2 = −3i (2.5.1) Applying KVL around the loop gives −20 + v1 − v2 = 0 (2.5.2) Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain −20 + 2i + 3i = 0 5i = 20 or ⇒ i =4A Substituting i in Eq. (2.5.1) ﬁnally gives | v v v1 = 8 V, | e-Text Main Menu | Textbook Table of Contents | v2 = −12 V Problem Solving Workbook Contents CHAPTER 2 Basic Laws 39 PRACTICE PROBLEM 2.5 4Ω Find v1 and v2 in the circuit of Fig. 2.22. Answer: 12 V, −6 V. + v1 − + − v2 8V − 10 V + − + 2Ω Figure 2.22 For Practice Prob. 2.5 EXAMPLE 2.6 Determine vo and i in the circuit shown in Fig. 2.23(a). i 12 V 4Ω 4Ω 2vo +− + − 4V − + 2vo +− i 12 V + − − 4V + 6Ω + vo − + vo − (a)...
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