# 2 23 this 12 resistor is in series with the 10

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Unformatted text preview: resistance to give a combined resistance of 1 + 2 = 3 . Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2- and 3- resistors in parallel to get 2×3 2 3= = 1.2 2+3 This 1.2- resistor is in series with the 10- resistor, so that Rab = 10 + 1.2 = 11.2 PRACTICE PROBLEM 2.10 20 Ω Find Rab for the circuit in Fig. 2.39. Answer: 11 5Ω 8Ω . a 18 Ω Rab 20 Ω 9Ω 1Ω 2Ω b | v v Figure 2.39 For Practice Prob. 2.10. | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 2 Basic Laws 47 EXAMPLE 2.11 5S Find the equivalent conductance Geq for the circuit in Fig. 2.40(a). Solution: The 8-S and 12-S resistors are in parallel, so their conductance is Geq 12 S 8S 6S 8 S + 12 S = 20 S (a) This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is 20 × 5 =4S 20 + 5 5S Geq 6S 20 S This is in parallel with the 6-S resistor. Hence Geq = 6 + 4 = 10 S (b) We should note that the circuit in Fig. 2.40(a) is the same as that in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, they are expressed in ohms in Fig. 2.40(c). To show that the circuits are the same, we ﬁnd Req for the circuit in Fig. 2.40(c). Req = 1 6 = 1 6 1 6 11 + 58 × + 1 4 1 4 = 1 12 = 1 6 1 1 + 5 20 = 1 6 1 5 Req 1 6 Ω 1 8 Ω 1 4 Ω 1 12 Ω (c) 1 10 Figure 2.40 Geq = For Example 2.11: (a) original circuit, (b) its equivalent circuit, (c) same circuit as in (a) but resistors are expressed in ohms. 1 = 10 S Req This is the same as we obtained previously. PRACTICE PROBLEM 2.11 Calculate Geq in the circuit of Fig. 2.41. Answer: 4 S. 8S 4S Geq 2S Figure 2.41 12 Ω 6S For Practice Prob. 2.11. EXAMPLE 2.12 Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3- resistor. Solution: The 6- and 3- resistors are in parallel, so their combined resistance is | v v 6 | 3 e-Text Main Menu = 6×3 =2 6+3 | Textbook Table of Contents | Problem Solvi...
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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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