I need steps R = 30 N Xc, = 12 N23-25 In Fig. 23–31, solve for= 120 Nа. Z.b. 4.= 100 VXc3äheoXcc. VR,-d. Ve, Ve, and Ve:120 N80 x 12048%3D200 Ω.80+120= 20 Ne. V, and V,40f. 0z.mttXc =80SECTION 23-7 SERIES-PARALLELREACTANCE AND RESISTANCE25. а. Z, 3 50b. = 2 Ac. V = 60 Vd. V. = 24 V, V = 240 Vand V. = 96 Ve. V = 144 V andV = 40 Vf. 0, = -53.13°C3

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23-25 In Fig. 23-31, solve XL, = 120 N Z. Xca = b. 4. = 100 V 80 x 120 8. +120 c. VR, d. Vc, Vc, 120 N = 48 and Vc; = 200 N Xoz X, = 20 N e. V, and V. f. 07. 30+40 X =80 SECTION 23-7 SERIES-PARALLEL REACTANCE AND RESISTANCE 25. a. Z,= 50 N b. , = 2 A c. VR = 60 V d. V. = 24 V V. = 240 V

23-25 In Fig. 23-31, solve XL, = 120 N Z. Xca = b. 4. = 100 V 80 x 120 8. +120 c. VR, d. Vc, Vc, 120 N = 48 and Vc; = 200 N Xoz X, = 20 N e. V, and V. f. 07. 30+40 X =80 SECTION 23-7 SERIES-PARALLEL REACTANCE AND RESISTANCE 25. a. Z,= 50 N b. , = 2 A c. VR = 60 V d. V. = 24 V V. = 240 V

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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Coulomb’s Law

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I need steps

R = 30 N Xc, = 12 N
23-25 In Fig. 23–31, solve for
= 120 N
а. Z.
b. 4.
= 100 V
Xc3
äheo
Xc
c. VR,-
d. Ve, Ve, and Ve:
120 N
80 x 120
48
%3D
200 Ω.
80+120
= 20 N
e. V, and V,
40
f. 0z.
mtt
Xc =80
SECTION 23-7 SERIES-PARALLEL
REACTANCE AND RESISTANCE
25. а. Z, 3 50
b. = 2 A
c. V = 60 V
d. V. = 24 V, V = 240 V
and V. = 96 V
e. V = 144 V and
V = 40 V
f. 0, = -53.13°
C3

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