7.2----ALL INFORMATION IS PROVIDED TO FIND PR1 and PR2----R_L = Load ResistorV_s = 40VR_1 = [(k-alpha)R_o] / (alpha*k) R_1 = 2.76 kOhmsR_2 = [(k-alpha)R_o] / [alpha(1*k)]R_2 = 8.29 kOhms A voltage divider like that in the figure is to be designed sothat vo= kv, at no load (R₁ = ∞) and vo av, atfull load (RL = Ro). Note that by definition a

In this question we need to find the power dissipated in R1 and R2 when the load resistor is…

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Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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7.2----ALL INFORMATION IS PROVIDED TO FIND PR1 and PR2----

R_L = Load Resistor

V_s = 40V

R_1 = [(k-alpha)R_o] / (alpha*k)

R_1 = 2.76 kOhms

R_2 = [(k-alpha)R_o] / [alpha(1*k)]

R_2 = 8.29 kOhms

A voltage divider like that in the figure is to be designed so
that vo= kv, at no load (R₁ = ∞) and vo av, at
full load (RL = Ro). Note that by definition a <k < 1. (
Figure 1)
Figure
R₁
R₂
+
Vo
1 of 1
RL
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Part E
Assume the load resistor is accidentally short circuited. How much power is dissipated in R₁ and R₂?
Enter your answers numerically separated by a comma.
ΤΙ ΑΣΦΑΛΗ
PR₁. PR₂ =
Submit
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vec 6
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mW

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A voltage divider like that in the figure is to be designed so that vo= kv, at no load (R₁ = ∞) and vo av, at full load (RL = Ro). Note that by definition a