2.53EXAMPLE 2-6Given us=5V, 02-3 V, and 04-10 V in the circuit shown in Figure 2-13, find ayand ps.SOLUTION:Inserting the given numerical values into Eq. (2-10) yields the following KVL equa-tion for loop 1:-D: +12-23=(-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 yields= -8 V. Using this value in the loop 2 KVL constraint in Eq. (2-10) produces-U3+14-05=-(+8) + (+10)+1=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)=02.532-53 Use circuit reduction to find vx, ix, and px in Figure P2-53.24 V+5 ΚΩW1.5 ΚΩ2 ks ix 1ko 1k0FIGURE P2-531+

Solution (2.53) - Given, The circuit diagram is shown below,

Answered: 2–53 Use circuit reduction to find vx,…

2–53 Use circuit reduction to find vx, ix, and px in Figure P2–53. 1.5 ΚΩ 5 ΚΩ Μ Μ Px 24 V + 2xΩΣΑ 1ΩΣ ΙΩ - FIGURE P2-53 +

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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Concept explainers

Compensation Techniques

The compensation techniques are the methods that are employed in the electrical control systems to enhance the performance of the system. The performance are generally affected by the factors like lag in the compensating devices, slow response of the system of the components like sensors used in the control system. The basic compensation technique involves lag compensation, lead compensation or the combination of the both.

Question

100%

2.53
EXAMPLE 2-6
Given us=5V, 02-3 V, and 04-10 V in the circuit shown in Figure 2-13, find ay
and ps.
SOLUTION:
Inserting the given numerical values into Eq. (2-10) yields the following KVL equa-
tion for loop 1:
-D: +12-23=(-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
= -8 V. Using this value in the loop 2 KVL constraint in Eq. (2-10) produces
-U3+14-05=-(+8) + (+10)+1=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
2.53
2-53 Use circuit reduction to find vx, ix, and px in Figure P2-53.
24 V
+
5 ΚΩ
W
1.5 ΚΩ
2 ks ix 1ko 1k0
FIGURE P2-53
1+

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