2.51OBJECTIVE 2-6 CIRCUIT REDUCTION (SECT. 2-6)Given a linear resistive circuit, find selected signal variablesusing 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 mA220 Ω+100 2 xFIGURE P2-51100 Ω ΣEXAMPLE 2-6Given u=5V, 02-3 V, and 04-10V in the circuit shown in Figure 2-13, find oand 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)-0The 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 yieldsvy=+8 V. Using this value in the loop 2 KVL constraint in Eq. (2-10) produces-03 +94-05=-(+8) + (-10) +5=0The result is us=-2 V. The minus sign here means that the actual polarity of us is theopposite of the assigned reference polarity indicated in Figure 2-13. The results canbe 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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Engineering

Electrical Engineering

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

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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