Engineering Circuit Analysis Solutions 7ed Hayt_[Upload by R1LhER
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Engineering Circuit Analysis, 7 th Edition Chapter Two Solutions 10 March 2006 1. (a) 12 μs (d) 3.5 Gbits (g) 39 pA (b) 750 mJ (e) 6.5 nm (h) 49 kΩ (c) 1.13 kΩ (f) 13.56 MHz (i) 11.73 pA PROPRIETARY MATERIAL . © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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Transcript of Engineering Circuit Analysis Solutions 7ed Hayt_[Upload by R1LhER
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Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
1. (
a)
12 s
(d
) 3.
5 G
bits
(g)
39 p
A
(b
) 75
0 m
J (e
) 6.
5 nm
(h)
49 k
(c)
1.13
k
(f)
13.5
6 M
Hz
(i
) 11
.73
pA
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
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nies
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imite
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str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
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duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
2.
(a)
1 M
W
(e)
33 J
(i)
32 m
m
(b
) 12
.35
mm
(f)
5.3
3 nW
(c)
47. k
W
(g)
1 ns
(d)
5.46
mA
(h
) 5.
555
MW
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
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nies
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imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
3. (
a)
()
74
5.7
W40
0 H
p =
1
hp 2
98.3
kW
(b
) 12
ft =
12
in2.
54 c
m1
m(1
2 ft
)3.
658
m1
ft1
in10
0 cm
=
(c
) 2.
54 c
m =
25.4
mm
(d
) (
)105
5 J
67 B
tu =
1
Btu
70.
69
kJ
(e)
285.
410
-15
s =
285.
4 fs
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
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raw
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nies
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imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
4.
(15
V)(
0.1
A)
= 1
.5 W
= 1
.5 J
/s.
3
hrs
runn
ing
at th
is p
ower
leve
l equ
ates
to a
tran
sfer
of
ener
gy e
qual
to
(1.5
J/s
)(3
hr)(
60 m
in/ h
r)(6
0 s/
min
) =
16.
2 kJ
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
5.
Mot
or p
ower
= 1
75 H
p (a
) W
ith
100%
eff
icie
nt m
echa
nica
l to
elec
tric
al p
ower
con
vers
ion,
(1
75 H
p)[1
W/ (
1/74
5.7
Hp)
] =
130
.5 k
W
(b) R
unni
ng f
or 3
hou
rs,
Ene
rgy
= (
130.
510
3 W
)(3
hr)(
60 m
in/h
r)(6
0 s/
min
) =
1.4
09 G
J (c
) A
sin
gle
batt
ery
has
430
kW-h
r ca
paci
ty. W
e re
quir
e (1
30.5
kW
)(3
hr)
= 3
91.5
kW
-hr
ther
efor
e on
e ba
tter
y is
suf
fici
ent.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
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imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
6.
The
400
-mJ
puls
e la
sts
20 n
s.
(a)
To
com
pute
the
peak
pow
er, w
e as
sum
e th
e pu
lse
shap
e is
squ
are:
400 Ene
rgy
(mJ)
t (ns
) 20
The
n P
= 4
001
0-3 /
201
0-9
= 2
0 M
W.
(b
) At 2
0 pu
lses
per
sec
ond,
the
aver
age
pow
er is
P a
vg =
(20
pul
ses)
(400
mJ/
puls
e)/(
1 s)
= 8
W.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
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imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
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rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
7.
T
he 1
-mJ
puls
e la
sts
75 f
s.
(a)
To
com
pute
the
peak
pow
er, w
e as
sum
e th
e pu
lse
shap
e is
squ
are:
1
Ene
rgy
(mJ)
t (fs
) 75
The
n P
= 1
10-
3 /75
10-
15 =
13.
33 G
W.
(b
) A
t 100
pul
ses
per
seco
nd, t
he a
vera
ge p
ower
is
P avg
= (
100
puls
es)(
1 m
J/pu
lse)
/(1
s) =
100
mW
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
8.
T
he p
ower
dra
wn
from
the
batt
ery
is (
not q
uite
dra
wn
to s
cale
):
5 7
1724
P (W
)
10
6
t (m
in)
(a
) T
otal
ene
rgy
(in
J) e
xpen
ded
is
[6(5
) +
0(2
) +
0.5
(10)
(10)
+ 0
.5(1
0)(7
)]60
= 6
.9 k
J.
(b
) The
ave
rage
pow
er in
Btu
/hr
is
(690
0 J/
24 m
in)(
60 m
in/1
hr)
(1 B
tu/1
055
J) =
16.
35 B
tu/h
r.
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
9.
The
tota
l ene
rgy
tran
sfer
red
duri
ng th
e fi
rst 8
hr
is g
iven
by
(1
0 W
)(8
hr)(
60 m
in/ h
r)(6
0 s/
min
) =
288
kJ
The
tota
l ene
rgy
tran
sfer
red
duri
ng th
e la
st f
ive
mi n
utes
is g
iven
by
30
0 s
0
10 +
10
30
0t
dt
=
300
2
0
10+
10
=
60
0t
t
1.5
kJ
(a)
The
tota
l ene
rgy
tran
sfer
red
is 2
88 +
1.5
=
28
9.5
kJ
(b)
The
ene
rgy
tran
sfer
red
in th
e la
st f
ive
min
utes
is 1
.5 k
J PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
10. T
otal
ch
arge
q =
18t
2
2t4
C.
(a
) q(
2 s)
= 4
0 C
. (b
) To
find
the
max
imum
cha
rge
with
in 0
t
3
s, w
e ne
ed to
take
the
fir
st a
nd
seco
nd d
eriv
itiv
es:
dq
/dt =
36t
8
t3 =
0, l
eadi
ng to
roo
ts a
t 0,
2.1
21 s
d2
q/dt
2 =
36
24
t2
subs
titu
ting
t =
2.1
21 s
into
the
expr
essi
on f
or d
2 q/d
t2 , w
e ob
tain
a v
alue
of
1
4.9,
so
that
this
roo
t rep
rese
nts
a m
axim
u m.
T
hus,
we
find
a m
axim
um c
harg
e q
= 4
0.5
C a
t t =
2.1
21 s
.
(c)
The
rat
e of
cha
rge
accu
mul
atio
n at
t =
8 s
is
dq/d
t| t =
0.8 =
36(
0.8)
8
(0.8
)3 =
24.
7 C
/s.
(d
) S
ee F
ig. (
a) a
nd (
b).
00.
51
1.5
22.
53
-20
-10010203040506070
time
(s)
q (C)
00.
51
1.5
22.
53
-150
-100-5
0050
i (A
)
time (t)
(a)
(b)
PR
OP
RIE
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RY
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ion
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rse
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arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Tw
o So
luti
ons
10
Mar
ch 2
006
11.
Ref
erri
ng to
Fig
. 2.6
c,
0
A
,
3 2-
0
A,
3
2-
)(
35
1
>
+ e
1 =
'4 =
v1/
100
+ (
v1 -
v2)
/20
+ (
v1 -
vx)
/50'
; >
> e
2 =
'10
- 4
- (-
2) =
(vx
- v
1)/5
0 +
(vx
- v
2)/4
0';
>>
e3
= '-
2 =
v2/
25 +
(v2
- v
x)/4
0 +
(v2
- v
1)/2
0';
>>
a =
sol
ve(e
1,e2
,e3,
'v1'
,'v2'
,'vx'
);
>>
a.v
1 an
s =
82
200/
311
>>
a.v
2 an
s =
57
200/
311
>>
a.v
x an
s =
12
3600
/311
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
4. W
e se
lect
the
botto
m n
ode
as o
ur r
efer
ence
term
inal
and
def
ine
two
noda
l vol
tage
s:
R
ef.
Nex
t, w
e w
rite
the
two
requ
ired
nod
al e
quat
ions
: N
ode
1:
11
21
23
vv
v =
+
Nod
e 2:
2
23
13
vv
v
=+
1
Whi
ch m
ay b
e si
mpl
ifie
d to
: 5v
1
2v2
= 6
an
d
-v1
+ 4
v 2 =
-9
Solv
ing,
we
find
that
v1
= 3
33.3
mV
. PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
5.
We
begi
n by
sel
ectin
g th
e bo
ttom
nod
e as
the
refe
renc
e te
rmin
al, a
nd d
efin
ing
tw
o no
dal v
olta
ges
VA a
nd V
B, a
s sh
own.
(N
ote
if w
e c
hoos
e th
e up
per
rig
ht
nod
e,
v 1 b
ecom
es a
nod
al v
olta
ge a
nd f
alls
dir
ectly
out
of
the
solu
tion.
)
VA
VB
Ref
.
We
note
that
aft
er c
ompl
etin
g no
dal a
naly
sis,
we
can
find
v1
as v
1 =
VA
VB.
A
t nod
e A
: A
AV
VV
B
105
4
=+
[1
]
A
t nod
e B
: B
BV
VV
(6)
85
A
=
+ [
2]
Si
mpl
ifyi
ng,
3V
A
2V
B =
40
[
1]
8
VA +
13V
B =
240
[2
] S
olvi
ng,
VA =
43.
48 V
and
VB =
45.
22 V
, so
v 1 =
1.
740
V.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
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tors
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rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
6.
By
insp
ectio
n, n
o cu
rren
t flo
ws
thro
ugh
the
2
resi
stor
, so
i 1 =
0.
W
e ne
xt d
esig
nate
the
botto
m n
ode
as t
he r
efer
ence
term
inal
, and
def
ine
VA a
nd
V
B a
s sh
own:
VA
V
B
Ref
.
At n
ode
A:
AA
BV
VV
2
[1]
31
=+
A
t nod
e B
: B
BB
AV
VV
V2
[6
61
2]
=
++
N
ote
this
yie
lds
VA a
nd V
B, n
ot v
1, d
ue to
our
cho
ice
of r
efer
ence
nod
e. S
o, w
e o
btai
n v 1
by
KV
L: v
1 =
VA
VB.
Si
mpl
ifyi
ng E
qs. [
1] a
nd [
2],
4
VA
3V
B =
6
[1]
3V
A +
4V
B =
6
[2]
Solv
ing,
VA =
0.8
571
V a
nd V
B =
-0.
8571
V, s
o v 1
= 1
.714
V.
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
7.
The
bot
tom
nod
e ha
s th
e la
rges
t num
ber
of b
ranc
h co
nnec
tions
, so
we
choo
se th
at a
s
our
refe
renc
e no
de.
Thi
s al
so m
akes
vP e
asie
r to
fin
d, a
s it
will
be
a no
dal v
olta
ge.
W
orki
ng f
rom
left
to r
ight
, we
nam
e ou
r no
des
1, P
, 2, a
nd 3
. N
OD
E 1
: 10
=
v1/
20
+ (
v 1
vP)/
40
[1]
N
OD
E P
: 0
= (
v P
v1)
/ 40
+ v
P/ 1
00 +
(v P
v
2)/ 5
0 [2
] N
OD
E 2
: -2
.5 +
2 =
(v 2
v
P)/
50
+ (
v 2
v3)
/ 10
[3]
NO
DE
3:
5
2 =
v3/
200
+ (
v 3
v2)
/ 10
[4
] Si
mpl
ifyi
ng,
60
v 1 -
20v
P
=
800
0 [1
] -
50v 1
+ 1
10 v
P -
40v
2
= 0
[2
]
- v P
+
6v 2
- 5
v 3
=
-25
[3
]
-20
0v2
+ 2
10v 3
= 6
000
[4]
Sol
ving
,
v P =
171
.6 V
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
8.
The
logi
cal c
hoic
e fo
r a
refe
renc
e no
de is
the
botto
m n
ode,
as
then
vx
wil
l au
tom
atic
ally
bec
ome
a no
dal v
olta
ge.
NO
DE
1:
4 =
v1/
100
+ (
v 1
v2)
/ 20
+ (
v 1
vx)
/ 50
[1
]
NO
DE
x:
10
4
(-2
) =
(v x
v
1)/ 5
0 +
(v x
v
2)/ 4
0
[2]
NO
DE
2:
-2 =
v2
/ 25
+ (
v 2
vx)
/ 40
+ (
v 2
v1)
/ 20
[3
]
Sim
plif
ying
,
4
= 0
.080
0v1
0
.050
0v2
0.0
200v
x [1
]
8
= -
0.02
00v 1
0
.025
00v 2
+ 0
.045
00v x
[2]
-2
= -
0.05
00v 1
+ 0
.115
0v2
0.02
500v
x [3
] S
olvi
ng,
v x =
397
.4 V
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
9.
Des
igna
te th
e no
de b
etw
een
the
3-
and
6-
resi
stor
s as
nod
e X
, and
the
righ
t-ha
nd
node
of
the
6-
resi
stor
as
node
Y. T
he b
otto
m n
ode
is c
hose
n as
the
refe
renc
e no
de.
(a
) W
r itin
g th
e tw
o no
dal e
quat
ions
, the
n N
OD
E X
: 1
0 =
(v X
2
40)/
3 +
(v X
v
Y)/
6
[1]
N
OD
E Y
:
0 =
(v Y
v
X)/
6 +
vY/ 3
0 +
(v Y
6
0)/ 1
2 [2
]
Sim
plif
ying
, -1
80 +
144
0 =
9 v
X
3 v
Y [
1]
10
800
=
- 3
60 v
X +
612
vY
[2]
Sol
ving
,
v X =
181
.5 V
an
d
v Y =
124
.4 V
T
hus,
v 1
= 2
40
vX =
5
8.50
V
and
v 2
= v
Y
60
=
64.
40 V
(b)
The
pow
er a
bsor
bed
by th
e 6-
re
sist
or is
(vX
vY)2
/ 6
= 5
43.4
W
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
10.
Onl
y on
e no
dal e
quat
ion
is r
equi
red:
A
t the
nod
e w
here
thre
e re
sist
ors
join
,
0.02
v 1
=
(vx
5
i 2) /
45 +
(v x
1
00)
/ 30
+ (
v x
0.2
v3)
/ 50
[1
] T
his,
how
ever
, is
one
equa
tion
in f
our
unkn
owns
, the
oth
er th
ree
resu
lting
fro
m th
e pr
esen
ce o
f th
e de
pend
ent s
ourc
es.
Thu
s, w
e re
quir
e th
ree
addi
tiona
l equ
atio
ns:
i 2
= (
0.2
v 3 -
vx)
/ 50
[2]
v 1 =
0.2
v3
- 1
00
[3]
v 3 =
50i
2
[4]
Sim
plif
ying
,
v 1
0.2
v 3
= -
100
[3]
v3
+
50
i 2
= 0
[4
]
vx
+
0.2
v 3
5
0 i 2
=
0
[2]
0.07
556v
x
0.02
v 1
0.0
04v 3
0
.111
i 2 =
33.
33
[1]
Solv
ing,
we
find
that
v1
= -
103.
.8 V
a
nd
i 2 =
-37
7.4
mA
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
11. I
f v 1
= 0
, the
dep
ende
nt s
ourc
e is
a s
hort
cir
cuit
and
we
may
red
raw
the
circ
uit a
s:
A
t NO
DE
1:
4
- 6
= v
1/ 4
0 +
(v 1
9
6)/ 2
0 +
(v 1
V
2)/ 1
0 S
ince
v 1
= 0
, th
is s
impl
ifie
s to
-2
=
-9
6 / 2
0 -
V2/
10
so
that
V
2 =
-2
8 V
.
20
10
40
+
v 1
= 0
-
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
12.
We
choo
se th
e bo
ttom
nod
e as
gro
und
to m
ake
calc
ulat
ion
of i 5
eas
ier.
The
left
-mos
t no
de is
nam
ed
1, t
he to
p no
de is
nam
ed
2, t
he c
entr
al n
ode
is n
amed
3
and
the
no
de b
etw
een
the
4-
and
6-
resi
stor
s is
nam
ed
4.
N
OD
E 1
: -
3 =
v1/
2 +
(v 1
v
2)/ 1
[
1]
N
OD
E 2
: 2
= (
v 2
v1)
/ 1 +
(v 2
v
3)/ 3
+ (
v 2
v4)
/ 4
[2]
N
OD
E 3
: 3
= v
3/ 5
+ (
v 3
v4)
/ 7 +
(v 3
v
2)/ 3
[3
]
NO
DE
4:
0 =
v4/
6 +
(v 4
v
3)/ 7
+ (
v 4
v2)
/ 4
[4]
R
earr
angi
ng a
nd g
roup
ing
term
s,
3v1
2
v 2
=
-6
[1]
-12
v 1 +
19v
2
4v 3
3v4
=
24
[2]
35
v 2 +
71v
3
15v 4
=
315
[3]
-42
v 2
24v
3 +
94v
4 =
0
[4]
So
lvin
g, w
e fi
nd th
at v
3 =
6.7
60 V
and
so
i 5
= v
3/ 5
= 1
.352
A.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
13.
We
can
redr
aw th
is c
ircu
it an
d el
imin
ate
the
2.2-
k r
esis
tor
as n
o cu
rren
t flo
ws
th
roug
h it:
At N
OD
E 2
: 7
10-3
5
10
-3
=
(v2
+ 9
)/ 4
70 +
(v 2
v
x)/ 1
010
-3
[1]
A
t NO
DE
x: 5
10
-3
0.2
v 1
=
(vx
v 2
)/ 1
010
3
[2]
T
he a
dditi
onal
equ
atio
n re
quir
ed b
y th
e pr
esen
ce o
f th
e de
pen d
ent s
ourc
e an
d th
e fa
ct
that
its
cont
rolli
ng v
aria
ble
is n
ot o
ne o
f th
e no
dal v
olta
ges:
v 1 =
v2
v
x
[3]
Elim
inat
ing
the
vari
able
v1
and
grou
ping
term
s, w
e ob
tain
:
10,4
70 v
2
470
vx
=
89,
518
and
1999
v2
1
999
v x
=
50
So
lvin
g, w
e fi
nd
v
x =
8.
086
V.
9
V
7 m
A
5 m
A
0.2
v 1
10 k
47
0
+ v
1 -
v x
v 2
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
14.
We
need
con
cern
our
selv
es w
ith th
e bo
ttom
par
t of
this
cir
cuit
only
. W
ritin
g a
sing
le
nod
al
equa
tion,
-4
+ 2
=
v/
50
W
e fi
nd th
at
v
= -
100
V.
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
15.
We
choo
se th
e bo
ttom
nod
e as
the
refe
renc
e te
rmin
al. T
hen:
N
ode
1:
11
22
1
vv
v2
+
=
[1]
Nod
e 2:
2
32
12
44
12
4
vv
vv
vv
=
++
[2]
Nod
e 3:
3
23
34
2 [
3]
25
10
vv
vv
v
=
++
Nod
e 4:
4
34
42
610
4
vv
vv
v
0
=+
+ [
4]
Nod
e 5:
5
57
12
1
vv
v
=
[5]
+ N
ode
6:
66
76
8
52
10
vv
vv
v
=
++
1 [
6]
Nod
e 7:
7
57
67
82
[7]
1
24
vv
vv
vv
=
++
Nod
e 8:
8
86
87
610
4
vv
vv
v
=
++
0 [
8]
N
ote
that
Eqs
. [1-
4] m
ay b
e so
lved
inde
pend
ently
of
Eqs
. [5-
8].
Sim
plif
ying
,
to
yiel
d
12
12
34
23
4
23
4
32
4
[1]
47
216
[
2]
58
20
[3]
156
310
[4
]
vv
vv
vv
vv
vv
vv
=
+
=
+
=
+
=
1 2 3 4
3.37
0 V
7.05
5 V
7.51
8 V
4.86
9 V
v v v v
= = = = a
nd
57
67
8
56
78
67
8
32
2
[5]
85
10
[6]
4
27
8
[7]
615
310
[
8]
vv
vv
vv
vv
vv
vv
=
=
+
=
+=
to
yiel
d
5 6 7 8
1.68
5 V
3.75
9 V
3.52
7 V
2.43
4 V
v v v v
= = = =
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
16.
We
choo
se th
e ce
nter
nod
e fo
r ou
r co
mm
on te
rmin
al, s
ince
it c
onne
cts
to th
e la
rges
t nu
mbe
r of
bra
nche
s. W
e na
me
the
left
no
de
A,
the
top
node
B
, th
e ri
ght n
ode
C
, a
nd th
e bo
ttom
nod
e D
. W
e ne
xt f
orm
a s
uper
node
bet
wee
n no
des
A a
nd B
.
At t
he s
uper
node
: 5
= (
VA
VD)/
10
+ V
A/ 2
0 +
(V
B
VC)/
12.
5 [1
] A
t nod
e C
:
VC =
150
[2]
At n
ode
D:
-1
0 =
VD/ 2
5 +
(V
D
VA)/
10
[3]
O
ur s
uper
node
-rel
ated
equ
atio
n is
VB
VA =
100
[4
] Si
mpl
ifiy
ing
and
grou
ping
term
s,
0.15
VA +
0.0
8 V
B -
0.0
8 V
C
0.1
VD
= 5
[1]
V
C
=
150
[2]
-25
VA
+ 3
5 V
D =
-2
500
[3]
- V
A
+
VB
=
10
0 [
4]
Solv
ing,
we
find
that
VD =
-63
.06
V.
Sinc
e v 4
= -
VD,
v 4
= 6
3.06
V.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
17.
Cho
osin
g th
e bo
ttom
nod
e as
the
refe
renc
e te
rmin
al a
nd n
amin
g th
e le
ft n
ode
1,
the
ce
nter
nod
e 2
an
d th
e ri
ght n
ode
3,
we
next
for
m a
sup
erno
de a
bout
nod
es 1
and
2,
enc
ompa
ssin
g th
e de
pend
ent v
olta
ge s
ourc
e.
A
t the
sup
erno
de,
5
8 =
(v 1
v
2)/ 2
+ v
3/ 2
.5
[1]
A
t nod
e 2,
8 =
v2
/ 5 +
(v 2
v
1)/ 2
[2]
O
ur s
uper
node
equ
atio
n is
v1
- v
3 =
0.8
vA
[3]
Sin
ce
v A =
v2,
we
can
rew
rite
[3]
as
v 1
v
3 =
0.8
v 2
Sim
plif
ying
and
col
lect
ing
term
s,
0.5
v1
- 0
.5 v
2 +
0.4
v3
=
-3
[1]
-0.5
v1
+ 0
.7 v
2
=
8
[2
]
v
1 -
0.8
v2
- v
3 =
0
[3
]
(a)
Solv
ing
for
v 2 =
vA, w
e fi
nd th
at
v A
= 2
5.91
V
(b)
The
pow
er a
bsor
bed
by th
e 2.
5-
resi
stor
is
(v3)
2 / 2
.5 =
(-0
.454
6)2 /
2.5
= 8
2.66
mW
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
18.
Sele
ctin
g th
e bo
ttom
nod
e as
the
refe
renc
e te
rmin
al, w
e na
me
the
left
nod
e 1
, th
e m
iddl
e no
de
2 a
nd th
e ri
ght n
ode
3.
NO
DE
1:
5 =
(v 1
v
2)/ 2
0 +
(v 1
v
3)/ 5
0
[1]
NO
DE
2:
v 2
= 0
.4 v
1
[2
] N
OD
E
3:
0.01
v1
= (
v 3
v2)
/ 30
+ (
v 3
v1)
/ 50
[3]
Si
mpl
ifyi
ng a
nd c
olle
ctin
g te
rms,
we
obta
in
0.0
7 v 1
0
.05
v 2
0.02
v3
=
5
[1]
0.
4 v 1
v 2
=
0 [2
] -0
.03
v 1
0.0
3333
v2
+ 0
.053
33 v
3 =
0
[3]
Si
nce
our
cho
ice
of
ref
eren
ce te
rm
inal
mak
es th
e c
ontr
ollin
g va
ria
ble
of b
oth
de
pend
ent s
ourc
es a
nod
al v
olta
ge, w
e ha
ve
no n
eed
for
an a
dditi
onal
equ
atio
n as
we
mig
ht h
ave
expe
cted
.
Solv
ing,
we
find
that
v
1 =
148
.2 V
, v2
= 5
9.26
V, a
nd v
3 =
120
.4 V
.
The
pow
er s
uppl
ied
by th
e de
pend
ent c
urre
nt s
ourc
e is
ther
efor
e
(0.0
1 v 1
)
v 3
=
177.
4 W
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
19.
At n
ode
x:
v x/ 4
+ (
v x
vy)
/ 2 +
(v x
6
)/ 1
= 0
[1
]
At n
ode
y:
(vy
kv
x)/ 3
+ (
v y
vx)
/ 2
=
2
[2]
O
ur a
dditi
onal
con
stra
int i
s th
at v
y =
0, s
o w
e m
ay s
impl
ify
Eqs
. [1]
and
[2]
: 1
4 v x
=
48
[1]
-2k
v x
- 3
vx
=
12
[2]
Si
nce
Eq.
[1]
yie
lds
vx
=
48/1
4 =
3.4
29 V
, we
find
that
k =
(1
2 +
3 v
x)/ (
-2 v
x)
=
-3.2
50
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
20.
Cho
osin
g th
e bo
ttom
nod
e jo
inin
g th
e 4-
re
sist
or, t
he 2
-A c
urre
nt s
ourc
ee a
nd th
e
4-V
vol
tage
sou
rce
as o
ur r
efer
ence
nod
e,
we
next
nam
e th
e ot
her
node
of
the
4-
resi
stor
nod
e 1
, a
nd
the
node
join
ing
the
2-
r esi
stor
and
the
2-A
cur
rent
sou
rce
node
2.
F
inal
ly, w
e cr
eate
a s
uper
node
with
nod
es
1 a
nd
2.
A
t the
sup
erno
de:
2
= v
1/ 4
+ (
v 2
4)/
2
[1]
Our
rem
aini
ng e
quat
ions
: v 1
v
2 =
3
0.
5i1
[2
] an
d
i 1 =
(v 2
4
)/ 2
[3]
E
quat
ion
[1]
sim
plif
ies
to
v 1
+ 2
v2
= 0
[1
] C
ombi
ning
Eqs
. [2]
and
[3,
4
v 1
3 v
2 =
8
[4
]
Solv
ing
thes
e la
st tw
o eq
uatio
ns, w
e fi
nd th
at v
2 =
727
.3 m
V.
Mak
ing
use
of E
q. [
3],
we
ther
efor
e fi
nd th
at
i 1 =
1.63
6 A
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
21.
We
firs
t num
ber
the
node
s as
1, 2
, 3, 4
, and
5 m
ovin
g le
ft to
rig
ht. W
e ne
xt s
elec
t no
de 5
as
the
refe
renc
e te
rmin
al.
To
sim
plif
y th
e an
alys
is, w
e fo
rm a
sup
erno
de f
rom
no
des
1, 2
, and
3.
At th
e su
pern
ode,
-4
8 +
6 =
v1/
40
+ (
v 1
v3)
/ 10
+ (
v 3
v1)
/ 10
+ v
2/ 5
0 +
(v 3
v
4)/ 2
0 [1
]
Not
e th
at s
ince
bot
h en
ds o
f th
e 10
-
resi
stor
are
con
nect
ed to
the
supe
rnod
e, th
e re
late
d te
rms
canc
el e
ach
othe
r ou
t, an
d so
cou
ld h
ave
been
igno
red.
At n
ode
4:
v 4
= 2
00
[2]
Supe
rnod
e K
VL e
quat
ion:
v 1
v
3 =
400
+ 4
v 20
[3
] W
here
the
cont
rolli
ng v
olta
ge
v
20 =
v3
v 4
= v
3
200
[4]
Thu
s, E
q. [
1] b
ecom
es -
6 =
v1/
40
+ v
2/ 5
0 +
(v 3
2
00)/
20
or,
mor
e si
mpl
y,
4
= v
1/ 4
0 +
v2/
50
+ v
3/ 2
0 [1
]
and
Eq.
[3]
bec
omes
v 1
5
v3
= -
400
[3
]
Eqs
. [1
], [
3],
and
[5]
are
not
suf
fici
ent,
how
ever
, as
we
hav
e fo
ur u
nkno
wns
. At t
his
poin
t we
need
to s
eek
an a
dditi
onal
equ
atio
n, p
ossi
bly
in te
rms
of v
2. R
efer
ring
to th
e ci
rcui
t, v 1
- v
2 =
40
0
[5]
Rew
ritin
g as
a m
atri
x eq
uatio
n,
=
400
400
-
4
0
1-
1
5-
0
120
1
501
401
321 vvv
Solv
ing,
we
find
that
v 1
= 1
45.5
V, v
2 =
-25
4.5
V, a
nd v
3 =
109
.1 V
. Sin
ce v
20 =
v3
20
0, w
e fi
nd th
at
v 20
= -
90.9
V.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
22.
We
begi
n by
nam
ing
the
top
left
nod
e 1
, th
e to
p ri
ght n
ode
2,
the
botto
m n
ode
of
the
6-V
sou
rce
3
and
the
top
node
of
the
2-
resi
stor
4.
T
he
refe
renc
e no
de h
as
alre
ady
been
sel
ecte
d, a
nd d
esig
nate
d us
ing
a gr
ound
sym
bol.
By
insp
ectio
n,
v 2 =
5 V
.
Form
ing
a su
pern
ode
with
nod
es 1
& 3
, we
find
At t
he s
uper
node
:
-2 =
v3/
1 +
(v 1
5
)/ 1
0
[1]
At n
ode
4:
2 =
v4/
2 +
(v 4
5
)/ 4
[2]
Our
sup
erno
de K
VL e
quat
ion:
v 1
v
3 =
6
[3]
Rea
rran
ging
, sim
plif
ying
and
col
lect
ing
term
s,
v 1 +
10
v 3 =
-20
+ 5
= -
15
[1
] an
d v 1
- v
3 =
6
[2]
Eq.
[3]
may
be
dire
ctly
sol
ved
to o
btai
n
v4
= 4
.333
V.
Solv
ing
Eqs
. [1]
and
[2]
, we
find
that
v 1 =
4.0
91 V
and
v
3 =
-1.
909
V.
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
23.
We
begi
n by
sel
ectin
g th
e bo
ttom
nod
e as
the
refe
renc
e, n
amin
g th
e no
des
as s
how
n be
low
, and
for
min
g a
supe
rnod
e w
ith n
odes
5 &
6.
B
y in
spec
tion,
v4
= 4
V.
By
KV
L,
v3
v 4
= 1
so
v 3
= -
1 +
v4
= -
1 +
4
or
v
3 =
3 V
.
At t
he s
uper
node
,
2 =
v6/
1 +
(v 5
4
)/ 2
[1
]
At n
ode
1,
4
= v
1/ 3
ther
efor
e,
v
1 =
12
V.
At n
ode
2,
-4
2
= (
v 2
3)/
4
Solv
ing,
we
find
that
v 2 =
-21
V
Our
sup
erno
de K
VL
equ
atio
n is
v5
- v
6 =
3
[2]
Solv
ing
Eqs
. [1]
and
[2]
, we
find
that
v 5 =
4.6
67 V
a
nd
v 6 =
1.6
67 V
. T
he p
ower
sup
plie
d by
the
2-A
sou
rce
ther
efor
e is
(v 6
v
2)(2
) =
45
.33
W.
4 A
2 A
1 V
4 V
3 V
4
3
2
1
v 2 v 1
v 3v 4
v 5v 6
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
24.
We
begi
n by
sel
ectin
g th
e bo
ttom
nod
e as
the
refe
renc
e, n
amin
g ea
ch n
ode
as s
how
n
belo
w, a
nd f
orm
ing
two
diff
eren
t sup
erno
des
as in
dica
ted.
By
insp
ectio
n,
v7
= 4
V
an
d v
= (
3)(4
) =
12
V.
1 A
t nod
e 2:
-4
2 =
(v 2
v
3)/ 4
or
v 2
-v 3
= -
24
[1
]
At t
he 3
-4 s
uper
node
:
0 =
(v
v
32)
/ 4 +
(v
v
)/ 6
or
-6
v +
6v
45
23
+ 4
v 4
4v
= 0
[2
] 5
At n
ode
5:
0
= (
v
v5
4)/ 6
+
(v
4
)/ 7
+ (
v
v5
56)
/ 2
or
-14v
+ 6
8v4
5
42v
= 4
8 [3
] 6
At t
he 6
-8 s
uper
node
: 2
= (
v
v6
5)/ 2
+ v
8/ 1
or
-
v +
v5
6 +
2v 8
= 4
[4
] 3-
4 su
pern
ode
KV
L e
quat
ion:
v 3
- v
4 =
-1
[5
] 6-
8 su
pern
ode
KV
L e
quat
ion:
v 6
v
= 3
[6]
8 R
ewri
ting
Eqs
. [1]
to [
6] in
mat
rix
form
,
=
3
1-
4
48
0
24-
1-
1
0
0
0
0
0
0
0
1-
1
0
2
1
1-
0
0
0
0
42-
68
14-
0
0
0
0
4-
4
6
6-
0
0
0
0
1-
1
865432 vvvvvv
Solv
ing,
we
find
that
v 2
= -
68.9
V, v
3 =
-44
.9 V
, v =
-43
.9 V
, v =
-7.
9 V
, v =
700
mV
, v =
-2.
3 V
. 4
56
8 T
he p
ower
gen
erat
ed b
y th
e 2-
A s
ourc
e is
ther
efor
e (
v 8
v)(
2) =
13
3.2
W.
6
v 1
v 2
v 3
v 4
v 5
v 6
v 7
v 8
Volta
ges i
n vo
lts.
Cur
rent
s in
ampe
res.
Resi
stan
ces
in o
hms.
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
25.
Wit
h th
e re
fere
nce
term
inal
alr
eady
spe
cifi
ed, w
e na
me
the
bott
om te
rmin
al o
f th
e
3-m
A s
ourc
e no
de
1,
the
left
term
inal
of
the
bott
om 2
.2-k
re
sist
or n
ode
2,
the
to
p te
rmin
al o
f th
e 3-
mA
sou
rce
node
3,
th
e +
re
fere
nce
term
inal
of
the
9-V
so
urce
nod
e 4
, a
nd th
e -
te
rmin
al o
f th
e 9-
V s
ourc
e no
de
5.
Sinc
e w
e kn
ow th
at 1
mA
flo
ws
thro
ugh
the
top
2.2-
k r
esis
tor,
v
= -
2.2
V.
5
Als
o, w
e se
e th
at v
4
v =
9, s
o th
at v
54
= 9
2
.2 =
6.8
V.
Pro
ceed
ing
with
nod
al a
naly
sis,
A
t nod
e 1:
-3
10
-3 =
v1/
10
103
+
(v1
v 2
)/ 2
.2
103
[1
] A
t nod
e 2:
0
= (
v 2
v1)
/ 2.2
10
3 +
(v 2
v
3)/ 4
.7
103
[2
] A
t nod
e 3:
1
103
+ 3
10
3 =
(v 3
v
2)/ 4
.7
103
+ v
3/3.
310
3 [3
] So
lvin
g,
v 1 =
-8.
614
V, v
= -
3.90
9 V
and
v
23
= 6
.143
V.
N
ote
that
we
coul
d al
so h
ave
mad
e us
e of
the
supe
rnod
e ap
proa
ch h
ere.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
26.
Mes
h 1:
4
+ 4
00i
+ 3
00i
11
30
0i2
1
= 0
or
70
0i1
30
0i =
5
2
M
esh
2: 1
+ 5
00i 2
30
0i +
2
2 =
0
or
300i
+ 5
00i
11
2 =
3.
2 S
olvi
ng,
i =
5.9
23 m
A
and
i
= -
2.84
6 m
A.
12
PRO
PR
IET
AR
Y M
AT
ER
IAL
. 2
007
Th
e M
cGra
w-H
ill
Com
pani
es, I
nc.
Lim
ited
dist
rib
utio
n pe
rmitt
ed o
nly
to
teac
hers
and
edu
cato
rs f
or c
ours
e pr
epar
atio
n. I
f yo
u ar
e a
stud
ent u
sing
this
Man
ual,
you
are
usin
g it
with
out p
erm
issi
on.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
27.
(a)
Def
ine
a cl
ockw
ise
mes
h cu
rren
t i 1
in th
e le
ft-
mos
t mes
h; a
clo
ckw
ise
mes
h c
urre
nt
i in
the
cent
ral m
esh,
and
not
e th
at
i2
y ca
n be
use
d as
a m
esh
curr
ent f
or th
e r
emai
ning
mes
h.
M
esh
1: -
10 +
7i
2i
12
= 0
Mes
h 2:
-2i
+ 5
i =
0
12
Mes
h y:
-
2i +
9i
2y
= 0
Sol
ve th
e re
sult
ing
mat
rix
equa
tion
:
t
o fi
nd th
at i
1 2
72
010
25
00
02
90
yi i i
=
1 =
1.6
13 A
, and
i y =
143
.4 m
A.
(b
) T
he p
ower
sup
plie
d by
the
10 V
sou
rce
is (
10)(
i 1) =
10(
1.61
3) =
16.
13 W
. PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
28.
Def
ine
thre
e m
esh
curr
ents
as
show
n:
(a)
The
cur
rent
thro
ugh
the
2
resi
stor
is i 1
.
Mes
h 1:
5i 1
3i
2 =
0
or
5i
3i
=
0
1
2
M
esh
2:
212
+8i
3i
21
= 0
or -
3i1
+8i
2
=
212
Mes
h 3:
8i 3
5i
2 +
122
= 0
or
5i
+ 8
i =
12
2 2
3
Sol
ving
, i
= 2
0.52
A, i
= 3
4.19
A a
nd i
12
3 =
6.1
21 A
.
(b)
The
cur
rent
thro
ugh
the
5
res i
stor
is i 3
, or
6.1
21 A
.
***
Not
e: s
ince
the
prob
lem
sta
tem
ent d
id
not s
peci
fy a
dir
ecti
on, o
nly
the
curr
ent
m
agni
tude
is r
elev
ant,
and
its s
ign
is a
rbitr
ary.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
29.
We
begi
n by
def
inin
g th
ree
cloc
kwis
e m
esh
curr
ents
i,
i1
2 an
d i 3
in th
e le
ft-
mos
t,
cent
ral,
and
righ
t-m
ost m
eshe
s, r
espe
ctiv
ely.
The
n,
(a)
N
ote
that
i x =
i 2
i 3.
Mes
h 1:
i 1 =
5 A
(by
insp
ectio
n)
Mes
h 3:
i 3 =
2
A (
by in
spec
tion)
M
esh
2:
25i 1
+ 7
5i2
20
i 3 =
0, o
r, m
akin
g us
e of
the
abov
e,
1
25
+ 7
5i +
40=
0
so th
at i
= 1
.133
A.
22
Thu
s,
i x =
i 2
i 3 =
1.1
33
(2
) =
3.1
33 A
.
(b)
The
pow
er a
bsor
bed
by th
e 25
r
esis
tor
is
P 25
= 2
5 (i 1
i 2
)22
= 2
5 (5
1
.133
) =
373
.8 W
. PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
30.
Def
ine
thre
e m
esh
curr
ents
as
show
n. T
hen,
4
0i
Mes
h 1:
2
+ 8
0i1
2
30i 3
=
0
M
esh
2:
40i
1 +
70i
2
=
0
M
esh
3:
30i
1
+70
i =
0
3
Sol
ving
,
1 2 3
8040
302
4070
00
300
700
i i i
=
w
e fi
nd th
at i 2
= 2
5.81
mA
and
i 3 =
19.
35 m
A.
Thu
s, i
= i 3
i 2
=
6.46
mA
. PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
31.
Mov
ing
from
left
to r
ight
, we
nam
e th
e bo
ttom
thre
e m
eshe
s, m
esh
1,
mes
h 2
,
and
mes
h 3
. I
n ea
ch o
f th
ese
thre
e m
eshe
s w
e de
fine
a
cloc
kwis
e cu
rren
t. T
he
rem
aini
ng m
esh
curr
ent i
s cl
earl
y 8
A.
We
may
then
wri
te:
ME
SH 1
: 12
i 1 -
4 i
=
100
2
ME
SH 2
: -4
i +
9 i
-
3 i
= 0
1
23
ME
SH 3
:
-3
i +
18
i =
-80
2
3
Sol
ving
this
sys
tem
of
thre
e (i
ndep
ende
nt)
equa
tions
in th
ree
unkn
owns
, we
find
that
i 2 =
i x =
2.7
91 A
.
PR
OP
RIE
TA
RY
MA
TE
RIA
L.
200
7 T
he
McG
raw
-Hil
l C
ompa
nies
, Inc
. L
imite
d di
str
ibut
ion
perm
itted
onl
y to
te
ache
rs a
nd e
duca
tors
for
cou
rse
prep
arat
ion.
If
you
are
a st
uden
t usi
ng th
is M
anua
l, yo
u ar
e us
ing
it w
ithou
t per
mis
sion
.
-
Eng
inee
ring
Cir
cuit
Ana
lysi
s, 7
th E
diti
on
Cha
pter
Fou
r So
luti
ons
10
Mar
ch 2
006
32.
We
defi
ne f
our
cloc
kwis
e m
esh
curr
ents
. The
top
mes
h cu
rren
t is
labe
led
i 4. T
he
bott
om le
ft m
esh
curr
ent i
s la
bele
d i,
the
botto
m r
ight
mes
h cu
rren
t is
labe
led
i1
3, a
nd
the
rem
aini
ng m
esh
curr
ent i
s la
bele
d i.
Def
ine
a vo
ltage
v
24A
ac
ross
the
4-A
cur
rent
so
urce
with
the
+
refe
renc
e te
rmin
al o
n th
e le
ft.
B
y in
spec
tion,
i 3
= 5
A
and
ia
= i
. 4
ME
SH 1
: -6
0 +
2i
2i
+ 6
i =
0
or
2i1
44
1
+
4i 4
= 6
0
[1]
M
ESH
2:
-6i 4
+ v
4A +
4i 2
4(
5) =
0
o
r
4i2
- 6i
+ v
44A
= 2
0 [2
]
ME
SH 4
: 2i
2
i +
5i
+ 3
i
3(5)
v
41
44
4A =
0
or
-2i
+
10i
- v
