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# 242b is divided between the 4 and 2 resistors hence 2

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Unformatted text preview: ng Workbook Contents 48 PART 1 i 4Ω io a + vo − 6Ω 12 V + − 3Ω 12 =2A 4+2 and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4- and 2- resistors. Hence, 2 vo = (12 V) = 4 V 2+4 Similarly, io can be obtained in two ways. One approach is to apply Ohm’s law to the 3- resistor in Fig. 2.42(a) now that we know vo ; thus, (a) 4Ω a + vo − 12 V + − Thus our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is not affected by the combination of the resistors because the resistors are in parallel and therefore have the same voltage vo . From Fig. 2.42(b), we can obtain vo in two ways. One way is to apply Ohm’s law to get i= b i DC Circuits 2Ω b 4 A 3 Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know i , by writing vo = 3io = 4 (b) Figure 2.42 For Example 2.12: (a) original circuit, (b) its equivalent circuit. ⇒ io = 6 2 4 i = (2 A) = A 6+3 3 3 The power dissipated in the 3- resistor is io = po = vo io = 4 4 3 = 5.333 W PRACTICE PROBLEM 2.12 i1 Find v1 and v2 in the circuit shown in Fig. 2.43. Also calculate i1 and i2 and the power dissipated in the 12- and 40- resistors. 12 Ω + v1 − Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V, i2 = 250 mA, p2 = 2.5 W. 6Ω i2 + 15 V 10 Ω + − Figure 2.43 v2 − 40 Ω For Practice Prob. 2.12. EXAMPLE 2.13 | v v For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vo , (b) the power supplied by the current source, (c) the power absorbed by each resistor. Solution: (a) The 6-k and 12-k resistors are in series so that their combined value is 6 + 12 = 18 k . Thus the circuit in Fig. 2.44(a) reduces to that | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 2 Basic Laws 49 shown in Fig. 2.44(b). We now apply the current division technique to ﬁnd i1 and i2 . 18,000 (30 mA) = 20 mA i1 = 9000 + 18,000 i2 = 6 kΩ 30 mA...
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