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OBIECTIVE 2-6 CIRCUIT REDUCTION (SECT. 2-6)

OBIECTIVE 2-6 CIRCUIT REDUCTION (SECT. 2-6)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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2.51 OBJECTIVE 2-6 CIRCUIT REDUCTION (SECT. 2-6) Given a linear resistive circuit, find selected signal variables using successive application of series and parallel equivalence, source transformations, and voltage and current division. See Examples 2-25 to 2-28 and Exercises 2-27 to 2-31. 2-51 Use circuit reduction to find ux and ix in Figure P2-51. 100 mA 220 Ω + 100 2 x FIGURE P2-51 100 Ω Σ EXAMPLE 2-6 Given u=5V, 02-3 V, and 04-10V in the circuit shown in Figure 2-13, find o and vs. SOLUTION: Inserting the given numerical values into Eq. (2-10) yields the following KVL equa- tion for loop 1: -02+12+13=-(-5)+(-3)+(03)-0 The sign outside the parentheses comes from the loop 1 KVL constraint in Eq. (2-10). The sign inside comes from the actual polarity of the voltage. This equation yields vy=+8 V. Using this value in the loop 2 KVL constraint in Eq. (2-10) produces -03 +94-05=-(+8) + (-10) +5=0 The result is us=-2 V. The minus sign here means that the actual polarity of us is the opposite of the assigned reference polarity indicated in Figure 2-13. The results can be checked by substituting all the aforementioned values into the loop 3 KVI. con- straint in Eq. (2-10). These substitutions yield -(+5)+(-3)+(-10)-(-2)-0
Transcribed Image Text:2.51 OBJECTIVE 2-6 CIRCUIT REDUCTION (SECT. 2-6) Given a linear resistive circuit, find selected signal variables using successive application of series and parallel equivalence, source transformations, and voltage and current division. See Examples 2-25 to 2-28 and Exercises 2-27 to 2-31. 2-51 Use circuit reduction to find ux and ix in Figure P2-51. 100 mA 220 Ω + 100 2 x FIGURE P2-51 100 Ω Σ EXAMPLE 2-6 Given u=5V, 02-3 V, and 04-10V in the circuit shown in Figure 2-13, find o and vs. SOLUTION: Inserting the given numerical values into Eq. (2-10) yields the following KVL equa- tion for loop 1: -02+12+13=-(-5)+(-3)+(03)-0 The sign outside the parentheses comes from the loop 1 KVL constraint in Eq. (2-10). The sign inside comes from the actual polarity of the voltage. This equation yields vy=+8 V. Using this value in the loop 2 KVL constraint in Eq. (2-10) produces -03 +94-05=-(+8) + (-10) +5=0 The result is us=-2 V. The minus sign here means that the actual polarity of us is the opposite of the assigned reference polarity indicated in Figure 2-13. The results can be checked by substituting all the aforementioned values into the loop 3 KVI. con- straint in Eq. (2-10). These substitutions yield -(+5)+(-3)+(-10)-(-2)-0
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