Post on 21-Dec-2015
description
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
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
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
-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
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
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
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
. 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
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
, 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
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
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
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
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
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