Answer ra 140 rb 70 rc 35 r1 r2 10 20 a b 40

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: in Fig. 2.50(b). PRACTICE PROBLEM 2.14 Transform the wye network in Fig. 2.51 to a delta network. Answer: Ra = 140 , Rb = 70 , Rc = 35 R1 R2 10 Ω 20 Ω a . b 40 Ω R3 c Figure 2.51 For Practice Prob. 2.14. EXAMPLE 2.15 i Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it to find current i . Solution: In this circuit, there are two Y networks and one network. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5- , 10- , and 20- resistors, we may select R1 = 10 , R2 = 20 , a a 10 Ω 12.5 Ω 5Ω 120 V + − c n 20 Ω 15 Ω R3 = 5 Thus from Eqs. (2.53) to (2.55) we have b Figure 2.52 R1 R2 + R2 R3 + R3 R1 10 × 20 + 20 × 5 + 5 × 10 = R1 10 350 = = 35 10 R1 R2 + R2 R3 + R3 R1 350 Rb = = = 17.5 R2 20 R1 R2 + R2 R3 + R3 R1 350 Rc = = 70 = R3 5 Ra = b For Example 2.15. | v v With the Y converted to , the equivalent circuit (with the voltage source removed for now) is shown in Fig. 2.53(a). Combining the three pairs of resistors in parallel, we obtain | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 30 Ω 54 PART 1 DC Circuits 70 30 = 17.5 = 12.5 15 70 × 30 = 21 70 + 30 12.5 × 17.5 = 7.2917 12.5 + 17.5 35 = 15 × 35 = 10.5 15 + 35 so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find Rab = (7.292 + 10.5) 21 = 17.792 × 21 = 9.632 17.792 + 21 Then i= 120 vs = = 12.458 A Rab 9.632 a 12.5 Ω 17.5 Ω a 70 Ω 30 Ω 7.292 Ω 21 Ω 35 Ω 15 Ω 10.5 Ω b b (b) (a) Figure 2.53 Equivalent circuits to Fig. 2.52, with the voltage removed. PRACTICE PROBLEM 2.15 i For the bridge network in Fig. 2.54, find Rab and i . 13 Ω a Answer: 40 24 Ω 100 V + − 20 Ω 30 Ω , 2.5 A. 10 Ω 50 Ω b Figure 2.54 For Practice Prob. 2.15. † 2.8 APPLICATIONS | v v Resistors are often used to model devices that convert electrical energy into heat or other forms of energy. Such devices include conducting wire, lightbulbs, electric heaters, stoves, ovens, and loudspeakers. In this | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 2 Basic Laws 55 section, we will consider two real-life problems that...
View Full Document

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.

Ask a homework question - tutors are online