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Engineering Circuit Analysis, Eighfh
Edition
Practice Problem Solutions
=
(60 w)(24 UdaD(60 minlh)(60 s/min)
;
(60 WXI kW1000 V't)(24Uday)(7 daylwk) (12.5 centslkWh):
2.4
(pl3)
2.s
(pls)
VI:-V2
2.6
:
(pl7)
Pub,
2.7
SO V2:-Vl
- (0.220)(4) -
(p 17)
We have a current - 1.75 A flowing out ofour positive voltage reference terminal, or +
A flowing into the positive reference terminal.
Thus, applyng the passive sign convention we find
Pobs: 1.75 x
2.8
-
3.8
: - 6.65 W or a generated power of - Po*: -
(pl7)
We see thata current of + 3.2 A flows out of our positive voltage reference terminal, so a
current - 3.2 A flows into that positive reference terminal.
Applying the passive sign convention,
Pobr:8e-lm x P*" (t = 5 ms) =
=
3.2:
-25.6 e-loot
w
(15 .6)e-r0o(sxr0-3 )
- 15.53 W
Engineering Circuit Analpis, 8ft Edition
Copynght @2012 McGraw-Hill, Inc.
All Rights
Resen/ed
1.75
2.e
Engineering Circuit Analysis, Eighth Edition Practice Problem Solutions
Chapters One through Six
(p20)
Moving from left to right and applying the passive sign convention,
Pobr:
Pob, :
Pob, =
Pobr:
Pob, :
(Check
2.r0
-7x8:
2x8:
-5x12:
8x20=
0.25v*x20 :[0.25 x -
-
56
16
- 60 + 160 - 60:0
)
@24)
v - Ri, so R
2.tt
+
l2fx20:
- vli -
-441-2xl 0-6:
@24)
Pub,
2.t2
ll2x1o3
:
@24)
Pub,
2.13
-'l/R -
- i2R - (3xlo-e)'(4.7xlou) _
@27)
28
Thus,
wire is 25.7
C)
per 1000 ft, so 500 ft is 32.65
C).
v- Ri - (32.65X0.1): 3.265V
Engineering Circuit Analysis, 8ft Edition
Copyright @2012 McGraw-Hill, Inc.
All
Rights Ressrved