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전기 공학 문제 및 솔루션

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Chapter P학km,
t.tn/pKe1=96ox64opixels=Hrame
1.2
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쓔쏺
2.46×10 있세 x
length
i.
1.3
A)
10mm
8 GB
=
b) 100µm
400mm3
=
233Bt의
=
金刎迦
3wphotosl.ba
쏘혾0쌾
)
=
10-3 mm
"
8mm3-2
✗
F임g-4회
64 GB ✗
698 pounds
=
X2mm
× 20mm
length hand)
x
me
x
3
어mm3
/
214748 byl.es
=
=
쌎맖y幾
=
b)
4NE5 day
귔擧 갔쁪
✗
1750mm
=
1쓰
1750mm ×
1.7
t
i
V
a)
-
은 ( t)
"
=
125어 앉A ,
t.ie?zotECtt
쓽
상디
q뱌
b) f
)
If
=
1.14
A
5NE%
=
5시0초
.to
also
안
=
-
E대7kg
5시엔
a)
→
C)
→
Car
2>0
b)
V70
,
form is Hong from
A to B
은 70
,
→
has
He
하다 의 ) 씨어
2.44 ×
Air
is
Thing from B to A
de b, Hey
i.co
106CC )
v40
,
Power is How;ng from A
to B
Fw
V40
d)
之化 ,
→
pmeeeisflaingf.in/3toA
2
> o
200W
String from A to B
which
_
卜沙妙 t.no#=5xi1tE7
100 W
∴
1750000
에 length
5×10%-0"722
320W
1.15에 Power is
=
"
=
=
B
許李式
=
( ㄷ) di
=
C) GC 5시에
10Mm
is B
b)
P
=
12V
Energy Hans Hard
×
=
=
40A
=
480 W
480W × 1.5TX
다/ S )
43200J
FT똇
1.18
N
=
E
=
A)
3이하 ( V )
5이하 (
Power
"
E가역 t.tn
0.0136 W
5시어 et
1.5NE2
=
1
=
A )
m
-
=
.
.
3
.
T.l.to?ettdt=_t5NotitTO
1.5시에 J
=
1.25
+
호
V
-
max.powerdelivered.to He element
a )
Target
1170W
=
mail.powerextractedfrmthee/ementiF9SIn8toit1A
v.ie//7osinnooatC)AVeraGeP=ojfpctidtt
N
=
b)
260대 850파 ( V)
=
)
per
" 70W
a as
=
=
PCt )
d)
dt
/
averagepeaoo.pe
1.26
16
8
i
24
i
16
-
-
-
-
:
i
i
-
에
0.556 W
=
NA)
UV)
24
0.016525
/
142.474 W
,
i
,
I
4
8
12
1 6 20
,
,
t 1k$
,
ii.ir
4
8
12
16
t (k s )
20
t.tn/chargetmnsfer edfldt=yAreawnderthegraph=48t
192t 32 = 272KC
=
b)
t.tn/EnergytmnskrredVoHage=IJ/
c]
/
VH ) .int )
과
'
,
빠기
'
charge
:
240.000 C
Wang Voyage
:
22V
t : 0N130005
태와매
t i 130005
272,000 C
→
240,000C
~
✗
22 V
=
5289000J
16.0005 → charge i 32,000 C
average.io/tagei23VEnergyi32,o00CX23V=
736,000 J
九右 / Energy
=
6.0/6×10
6J
Chapter Problem s
a) Wid
2.1
b) ahsorb.mg : 30V
de.ve/oping:20V,8AtwC)30V3OV.8A=24OW(absorbing
de.ve/oping)8ALl0V)C8A)=-8owCdevehping)240W240W=0ebsorbi
-
)
20V
(20V ) 8A
.
-
=
100W (
-
g) Chef )
d)
a) V에d.
t.
tt
de.ve/oping:30V
20V전암
고
ov
-7
30V
b) Power
+
8A
±
Power
50V
C)
-
2.4
8A 전류원양단에 50V의 전압이 걸린다
8A 전류원
being :
160W
1에서
5A +2A
Ti
a
240W +
node
핥,
.
8A
=
50V = 400W
.
K다
.
7A 二之
=
K다
node 2에서
7A -5A +3A = 5A # O
•
Node 2
2.6
in void
→
K4만록X
a) at node 1
,
K다 t 400mA
V,
•
b)
20V
400mA
.
20V
=
=
=
V50
20V
8W
( about.mg
)
•
node /
2.7
A
.3인이
) What
→
left
loop .NL at node
E
gi.io ✓
=
1
-20mA
-20mA -0=-10 V
500
4 =
b)
(10V ) ( 20mA )
0.2W ( absorb.mg )
=
C) absorb.mg
2.9
kW
10t
loop
What
t
loop
30V-40V HOT0
→
30V
at
loop@
_
→
10V
기아 +40V
1800회
-
-180in
+
=
-
60V t.co
30
in 二一
F 50mA
a)
2.11
V
=
2 R
20T
=
b)
t匯
C)
리어
Power
U
=
Power
20mA
.
2K2
40 V
absorbe_dbytheresistor-V.ie
-
40V
absorb_edbyttere.sk
40V20mA
=
0.8W
=
=
=
원­
20mA
=
in void ( ? )
40V
.
20mA
0.8W
i.RS/0V=Ez).2kR.'i=-5mAb)Pmersup liedbytheroHage
a)
2.12
U =
source
昨 i)
=
C)
with
蚩二 5mA
i.
Letter
0.05W
=
snppliedbyte.ro/tagesource=l0V.5mA=
Power
0.05W
2.14
N =
12000 i.
→
12KM저항
→
0hm Y kw
int
口适우EE
v
2.16
V ( V)
V =
25
50×103 i.
50 KR reactor
50
5
7
10 O
125
150
嫩
2.17
A) NL → Model
in -1.5
100M
저항 :
150st 250R
,
4002
b)
No
=
i, .ws
i,
=
1.2A .ws
=
/ 전류비
No 로같은
1 2A
.
,
ii.IE 4:1
:
立二 0.3A
120 V
=
C)
전압
:
t.tn/parerdeveloped=l.5A:l.5Atl20V)=-l8OW=l8owtotnl
pwerdissipnted-iloomti.EC/5osL)tir4250r)=l8ow
1.44.100
218
12혜
A)
I
0.09.400
t
i
=
=
회 : ib
=
b)
C)
20
=
1.61.757
.
.in
75 : 300
pwerdelivered.by Hero V
20V
.
2A
:
=
4
1.6A
(2540R
source
4의
가마이마
noorre.si#or
40R resist.ir
→
=
0.4A
ib
Power dissipate.to
d)
二些
e)
=
2A
=
1
'
180W
40R +6052=1007
200%con
전류 비
=
=
160-t
→
¢432.30082
=
48W
→
(1.72.751
-
二凸
2.22
A) Find in
nA
"
lynfojhy
迦陶 .net
E25
yni.t2.ir
③
.
다
"
-
.li 4
node 3
-
I,
-
2
-
I4
=
E3
=
b) iii, ti .
=
ii.
=
=
522-70회
=
一悲
70회
-
25호 ,
0/180-120-35
-3524=0
4=0
(
采健 二緋
70 i,
-25L f)
70회
+
I5-2
7회
二娜
pwF-diipated.in
f
5R
70이
-
-35T
60
=
∴
o
=
0
회遐
f -6=0
+
A
芻尸 5.ir
257 :
(f) 놔 .TW for
健 ) 2.70뿖와
35r :
(判仔二 Tw
:
2.25
:
260대에
node 3
M47t츏.n!岱
Leslie洪山一
I
→ 125V
"
전치히로에 대해 KVL
→
15M저항에 걸리는 전압 : 45V
node I 에 대해 KCL
→
i, +3A
오른쪽아재m매에 대해
node 2에 대해
K나
5A
=
:
3023
i, t i,
NL )
_
E,
∴
10V
-
i.
=
=
454152=3 A
→
2A
is
80V = 0
-
3A
5A
in
0
=
소비하는 전력
(5A가 전
7R )
b)
=
롱아 ⇒
13A가 301
16이
(5A가 16R
ㅋ
iqtn.de 3에서
=
제2.15R의35N
175W
152
t
(
270W
52
→
(제2-5n
=
20W
=
0
400W
K다
:
125V 전압원이 공급하는 전력
C)
0
許選
⇐
is
-
loop ①②③④
522-120+2523=0
O
.
爬一 2-i.no
180
0
=
tis-i.io
②
는④
KU for
for node 1.33.4
Ka
각저항들이 소비하는 전력의 합
→
5A +3A
125 VX8A
:
=
-
E4
IE8A
1000W
175Wt 135m270Wt
20W1-400W = 1000W
2.29
Met
R호골이
"
node / 에서 K다
한
i.
※를
식 ①
¥+7
→
it
0
=
.
.
-
①
뼐이므로 저항R양단에 걸리는전압
i
:
Ut
븜
=
ft ix. 0.ir ※
다0일때 하다 이므로
,
Ut
,
i. tix.it
→
n
=
회로를 구성해보면
로
t
6.tl?ILx=-l0A-l-.
※ 一舵 .is Ut
b)
후also
이A
80일때 하
=
t
6
=
Device 가
曲爬二
예하에 전달할전력
20R
ti-1
适山
Ut
i.in
一魅
20R
i.in
1nd
Power
=
8A
II R
:
a)
2.30
10A
.
i,
∴
4:1
=
b)
i.cmA)
int2 A
,
C2A )
=
Es
'
17.5
'
'
'
.
2052
=
80W
o 스Vs스가V
=
-
°
20.0
'
25m쨔A
→
+20
vs
i
=
_
#
Disney
0
20mA67m
15.0
12.5
힚
a.
.
iii,
vs
Ellen
4.0
0.0
I
I
I
1
255075100751¢
Toy
20mA
紙牟
ois
d)
25KM
Current del Need
→
I-U
Open
=
전류비
:
'
(V )
is t o
2.5 : 10대4
Int
a
V5
20=0
1401V )
→
실제 회로에서는
도선의 저항 전류원의 저항을 무시학수 없기 때문
,
2.32
"
粃
K4 for node 1
Ext를 -50A
→
KVLforrightme.int
식
②
t
E50
No
=
8in
4 ix
전력을 공급능난로자는 50A 독립전류원 → 양단에 계는 전압
∴
_
3 ii,
=
一竹
V0
to A
0
vx
150MA
25V
IN
(
_
=
E)
0
Ex
-
떅
:
t
-
-
-
②
YA
TV
50A
.TV
=
ET W
.
15V
( 전업분배
.HR = -45
5.3 )
→
=
No
Power
absorbedinhecirai.tt
40m사300n
50mA가
(
.
=
50mA
2.33
.
W200世
4
=
二愍
atualopen.cirai.tw/tge
4mA , 16mA ( 재에의
f)
Grand )
Circuit What
5001
t
(150mA5.302
=
8.75 W
.
vs +20
Cm A )
에
2.34
50V
v,
20mA
=
.
5M
하 50%n
100V
12.5mA
b) Power
⇒
.
100V
12.5mA
=
rate d
in The Circuit
100V t
12.5mA
gene
20mA
=
2.625W_pwerabsorbed.int
.
50V
=
(on A ) 2.5KM + C2 메린1KM3
⇒
.
=
2.38
iE
→
KR )
2쁘
EBti.ci
=
ㅌㅌ
i z
he Circuit
=
M3
E f) 는 B
=
=
E
회
-
EB
tiERE-I iDREO-i.Rtl/c -C2iDRz=o-IiCRt DtVc tEBRz=0i=
V0
眺毋
+
成叩冊
'
tiERE.CM#+iR2jiDRz=0VotCltf)iBRE-(Rkcj-iBR'T)Rz=0iBIREl
V0
tp
i.in
=
+
申門幽)
+
V0
-
旽姿
낪삾밊早cp쯾
) RE
=
0
Chapter 尻化
m
7.1
is
A)
0.0.00
<
1뜮
RA
=
diri.im
current
by
KCL
③
Junction
at
ANT
en
紙幼
.tw
r
⇒
Junction
紅
"
4-12
-
-
x
,
I2
=
→
1의TV 3 k¢
y
8A
-
→
at
-12A +4A +8A
=
i,
o
_
iii.
x
→
-12A +12A
y
→
-12A +12A E 0
4.18=0
K V2 for loop ②
4.18
8.3
-
-
8.6
=
0
KW for loop ③
120
-
4-12
-
8.3 -8.6=0
32
dissipated.4rnedstrt.IR
Power
a)
•
o
3.3
=
( 125.4
(8) ? 3
IR
18이 reiator.IR
=
(4) 2.18
resistor.IR
=
(8가 6
3R
.br
by 120V source
editor
=
r
=
b)
Porter del ; wie d
C)
T.la/Pmerdissipated=576Wtl92Wt288Wt384W=l440WEqual !
A)
o
a
(a )
→
( b)
→
3 KR
4 KR
=
120V
:
( 1000R , 1500R )
,
,
b)
=
=
=
=
12A = 1440W
Series
,
1200R 1007 )
.CC ) > ( 25M 107 , 15R )
0
-
( d) t ( 607.5m )
,
=
,
,
( 18Th 22R )
,
( 65740R)
=
Series
=
Series
Series
(a)
6kt
a
MT97
가이속 紅r
te.lt
6001
다에
(b)
Radio
und
2500
3002
(C)
mE에
1-951
迹臧
.se
쉾얘댁얂延
w
✓
istiiti.io
12A -4A -8A
b) KU for loop ①
120
i, -4A
→
4r
x
쎺
=
.sn
576W
192W
288W
384W
=
=
0
0
✓
3.4 A ) (A)
40에 1202
)
80KR1160 KR
¢
( b)
100r 11140R
150 KR
,
240이 113002
,
11160K에 180KM
8007.050111 700에 16002
d)
116KM
2 KR
b)
rm
133.33R
133.33R
紙科 1-5Tml
L.int 쎼一絅}
1-n
Late
에
(b)
-
"
하
i
墨沌毗
.
"
C)
34.291
TV
650r
4kt
hnr-towmrm-l.lt
t.IO
수
(d)
39.34.2
Lm
-
100KM
cnet.de/i=,iVj=핳쁧
3.13
m
=
V
Ios
Re
0.25A
V.
7.14
h)
=
肛仍 .
2.5V
→
M2111로
Rilargestresistance.no
=
Power
60 V
=
-
Re q
=
Ri
14V = 46V
笭一 ii. 2.03W
mm
A
3.20
→
Bq
n쵑늈
=
.
60 V
3.29 Re
)
=
q
301120t40 ⇒
14V
=
3.29
Req
6mA
.
=
3
-
E
=
쁪r
30000은
=
,
=
He
-
Mr
-
,
No
=
II.
二些
No
=
that is
Operation of thewltaged.irHer
Conrad to He 180 V source ?
종속전압원을 연결하면 왼쪽 전압 분배기에 의한 영향을 덜 받게 된다
i
120 V
20M res 伽孑 = 80V
effetdoesadd.mg/tedependent-wHage source
on
180
45mA
135V
Ref
=
40kt
=
3.29
3481-10쓰
=
b)
=
201E2
30kt Mk하
wltgeac.ms
Ri
3600 t R2
=
=
→
600t2
zvoltageac.ms
kt2%n
have
2.499 V
=
30r+쩻
Rim
=
180V
What
9.997R
=
3 R2
=
minimmpowerratlng-3_132-34_8.me
C)
=
nolond.ro/tage:l5VtiRz=l5V.ERtERz=60Vi.Ri=45V
MI
Target to Hugo Across
→
9.997S2
V0
R,
D)
is
.
440항 135V 二吐
.
v80
3.26
A)
H
,
450mA
18111 ( 2611-10R )
current
G3A-Ekg18R
→
=
12R
361m싸아
in The
j.int/I450rAl_l_.b)l5omA-36R=5.#C)roHageaCMSl
8
R
a.ws/or3.6V.
5.4 v
3.27
.
샹0.7표18예110마리에
¥
t 二止
A)
1
10V
d) 미쐥e
Req
vdkgeac.ms 47
i
1
(12st 18R ) 1110M
Root 7.5R
→
20211151117.5에 4St
-
( 16R1-4에 1130t → 12 R
b) current (41)
0.3쯔
=
C
)
4st 12 E2 H4 res or
Not
=
io
0.333A
n
=
=
E
10V
.
∴
V0
4V
=
Angela )
1.33V
=
幽升
20Ha씨에
=
d)
current ( 10에
미세 ( 10t)
=
∴ cumntao.rs )
3.33
wltage.ae)
e)
4V
-
-
3.2V대의
0.8V
=
0쁨 0.쁘 H 미쌰 (182)
=
J5A
=
0.8V
3.2V , current of
.
A 0.쁘
=
5A .int ia.ie 3mA
a)
in
Rain
in
""
150mV
=
5A
r
RA Il
b)
i. Req
C)
100mA
=
150m Vs
int 2G
ER
-
,
해
=
::::훃흆뿚
-
r
=
3mA
0.003A
→
Re
q
150mV
∴ Req
=
3쯔
RA
150mV
RA.com A -3nA -1.5¥
3.37
.su/ereading-5V50V-2omV=1mA.Rv
49햪
49퍞 고갱이다
4780.fr
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