8/10/2019 EECE210 Q2 Sp14 Solution
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American
University of Beirut
Department
of Electrical
and Computer Engineering
EECE 2t0
-
Electric
Circuits
Qurz
2
Closed Book
-
No Programmable
Calculators -
90
minutes
TEST ID:2000
April8, 2014
Name:
l
8/10/2019 EECE210 Q2 Sp14 Solution
2/8
in
Fig.
1, find
power
aDsorr
--+
9+
I
)
l)
on)
,ll
Yq
)
3n
(,
)
t
?_
l.
absorbed by the2 A current source.
(b)
4w
(c)
12W
(d)
16w
(e)
None of the above
lvt
J
4
12.
;)
IAO
G
2h
---+
V,"r
=
(z
-?-L)
,s
=
-Vs
=-4V
?-
+ D
--VutZ
Lzh
vm = 8t,l
(abs
)
8+
4
+
?a
v*
Fig.
l:
Circuit
for Problem 1
2. If
Vs
=
36
V in
the circuit shown
in
Fig.
2, find the maximum
power
that can
be delivered to Rp.
(a)
9W
@r*
(c)
12W
(d)
0.7s
w
(e)
None
of
the above
12rz
BN
=
,{'r?
=7V
Fig. 2:
Circuit
for Problem
2
t
+sz
9o
*8)/f
/2
b2W
\,,
4v3
fi^
6(2
Kn
=(93
4O
=
352
652
If
Vs
=
8
V
in
the circuilshown
@srv
'E,yd
96v
t)
L
I
r.^*.;
8/10/2019 EECE210 Q2 Sp14 Solution
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3. In the
circuit
shown
in Fig.
3,
Rl
=
6 O,
R2
=
8
0,
and the
rnultiplying
constant
of
the
dependent
curent-controlled
voltage
source is
c
=
2 Y
lA.
Find the Th6venin
resistance
between terminals
a,b.
8a
G=-t,*ur*Zl,
a)
6O
@on
(c)
3O
(d)
so
ti,'"fQv'
(
2i,
I
it
=-V
+
=Y.Y*;e)
=
Y
r
Ei
.=
Eq
3+
6sz
(e)
None of the
above
Rrt
=
L/52
Fig. 3:
Circuit
of Problem 3
4. In the circuit
shown
in Fig.
4, the multiplying
constant of the dependent
voltage-conffolled cunent
source
is b
=
2
x
10-3
A/V.
Find
R1-
for maximum
power
transfer.
(a)
3.3
kf)
(b)
3.6
kCI
(c)
4.4
kf)
(tat)+.0
to
(e)
None
of
the above
1mA
Rru
Fig. 4: Circuit
of
Problem 4
. \\
=
VT
t*z
/
Vr
T
=
ZVt
=r3
(
I
Fn=
2ko
1ko
3ko
b.Vx
vr
=
tkQ
=)
ftr-r,
=
4Aa
T7
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5.
In the op-amp circuit
shown in Fig. 5, Vcc
=
12Y, Rl
=
20 kO,
R2
=
150 kQ, Ra
=
10 kO,
Rb
=
20
kO, Va
=
400 mY and
Vb
=
300 mV.
Find VR.
tt
3o
3,
eV
@)t,zu
(b)
2.60 v
(c)
5.13
V
(d)
1.s4
v
{n':Vp= 2o
xool
2o+lo
Vn=
2C
+ lo
xoc3
=
I
o12s
VR
:)
VR:
V
3o
e)
None
of
the above
2c
+
tSo
lSoh
nz
?oh Rl
loR
Ra
Va
o.AV
Vb
O,?Y
;
Fig. 5: Circuit
of
Problem
5
In
the
circuit shown
in
Fig.
6,
r,2
:
5
V
and
Vcc
=
16 V. Find the maximum
value
of
v1
that
can be
used
such
that the op-amp operates
in its
linear region.
5
_V,
3+
5
:)
lo-5\
-fo
=
o
=')
{"
=
lo-Sq
=
-lb
5-tr"
-n:) U-54r2-6=e
-(J'r:
z5
+
2ok
nu
(a)
4.8 V
(b)
s.6 v
(c)
6.0
V
@s.zv
(e)
None of the above
s
ko
5/gV
2s ko
q
=
lo
+
/e
5
Fig. 6: Circuit
of
Problem
6
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7.
Each
op-amp shown
in Fig.
kO.
Find
the
cunent
16.
(a)
54
mA
(b)
58
mA
$))38
mA
(d)
34
mA
(e)
None of the above
7
is operating
in the
linear region,
Vs
=
10
Y
Rr
=
20
kO,
and
Rr
=
4
V*
=
-4o
V5
=-LtoV
'
lo
Voz
=
-?o
Vot
=
$oV
To
=
*
?o
*
82-to
=
38
*ft
't
'fo
rg
'oV
l0
Ko
tloZ
:?
{30KA
I
l
Fig,
7:
Circuit
of Problem 7
8.
Find
the equivalent
capacitance
with respect to the
terminals a, b
in
the
network shown in
Fig.
8.
(a)
7
1tF
@t
uu
(c)
6
trrF
(d)
4
pF
(e)
None
of
the above
4fr
Fig.
8:
Circuit
of Problem 8
R.
-1
43
/xl7
=
8
Co-.
q
I
:)
cg
=
+
lt
5
q+8
=>
Cob=3+Z
=
S,,ttF
2,r
F
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9.
In the circuit
shown
in
Fig.
9,L1
=2H,Lz =
8
H, L3
=
equation around mesh I
(corresponding
to mesh-current I
(a)
6
-
24it
*
l4iz
-
4di1
ldt
-2di2/dt
*Tdlldt
=
0
(b)
6-
24\
*
l4h-
4dit
ldt
*
di2ldt
*5dlldr
=
0
@
-24it
*
t4b
-2dit
ldt
-2di2lclt
*4dhf
ctt
=
0
(d)
6
-
24i1*
t4l
-
8di1
I
dt
*
di2ldt
*Tdfuldt
=
0
(e)
None of the above
10
{}
5
H,
and
M
=
4 H.
Write the mesh-curent
-
6+zltit-
tq
tt
*
r#
(
i,
-L)
q,L
(ts-t,)
u
8* (tt-z-E)
Ji-
dt
q4-
(L:t-t'z)
-
o
da
r
+
-\.
)
-
6t
zvi,-tvQ
+
8f)
Z"lit
4T
I
alt3
=O
z't
oz
4t
4V
Fig.
9:
Circuit of Problem
9
10. The switch
in
the circuit shown in
Fig.
10 has been opened for
a
long time
and
is
closed at
/
:
0.
What
percentage
of the
initial
energy stored
in the inductor
has been dissipated
in
the
40
Q
resistor
when
,
:250
LLs.
(,i},rrr"
(b)
86.5Vo
(c)
39.3vo
(d)
95.0V0
(e)
None of the above
F
L
wtd
= t
LI;
_?E
+
Wr=*s*U-e_
L
'Z'40
t?
WB
=
t-e
Zor1il
vl(o)
)
)
-6
Sot
lo
dL
o,63L
63.1
al
/o
50
mA
40 O
,r(t)
/=0
f)roo
20
mH
i:
)
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11, InthecircuitshowninFig.
11,Cl
=41tF,C2=3[tF,C3=5trrF,andC4=5,uF,Thecapacitorsare
initially
charged
to
Va(0)
=
-5
Y
Vc(0)
=
-15
V
and
Vd(0)
=
54.4Y. The
switch
is
closed at t
=
0.
Find the total
energy delivered
to the
resistor R.
(a)
4710.4
ytl
@ruou.o
ut
(c)
2649.6
1.rI
(d)
5961.6
prJ
(e)
None
of the above
(a)
3.24
ttl
(b)
s.76
ttJ
(c)
7.29
1tI
@o'rt
(e)
None
of the
above
l'vwh
.Is
5
\f-
{L
-25oot
=
/o,5L
L
=
t
+
{+
cg4g
V,
=
V6+l/a
-Vg
=
v{B
=
u(o)
-ryhl
/
5
+
CI=
Hfo
-f
+g4'Ll
+
t5
--
6tl'VV
=
t
cuvo'=
3eo6'A7F
2'
C2
C4
tfi
Fig.
1l:
Circuit of Problem
1 1
12. In
the
circuit
shown
in
Fig.
12,
Is
=
1.4
mA.
The
switch
has been
in position
a
for
a
long time.
At
t
=
0,
the switch is moved
to
position
b. Find the total energy
delivered
to the 5 kQ resistor.
+
^r
vd
Vo
=
l,4t/Z
=
/6'8V
f
=
0'25Ylf
6^
l'6Y
lo3
=
4xlf
a
s
ue
=
16,8
e-.'oou
v
:)
W
te kQ,
*1
a
r"5
sk()
3ko
0.25 pt
?//8
Fig.
12:
Circuit of Problem
12
=
/,6fr52
V._
=J
o
Oo
_a
I o,5
-50oo
t
e
)+
lo.5'
.1
wt*
Sooo
$ooo
*
$DoD
=
A'uy(
8/10/2019 EECE210 Q2 Sp14 Solution
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13.
In
the
circuit
shown
in
Fig. 13, both switches
(1
and 2)
have
been closed
for
a long time and
are
opened at
t
=
0. Assume that Vs
=
300
V
Rs
=
10
kQ, Rl
=
25 k0, R2
=
25
kQ,
R3
=
50
k0,
and
C
=
40
1tF.
Find the value of the
voltage Y2 at t
=
1.2
s.
(a)
60.24Y
(b)
2s.1s v
/6)+s.rs v
Vru.,o
u
(e)
None
of the above
@t.r*o
(b)
0.s10
A
(c)
1.872
A
(d)
0.702
A
(e)
None of the above
Vs
3oov
T
IIL-
ZH
t
i(t)
=
?,ue
r
=?atl
Lo
,E+
=
?,qA
9u
{2
,fu)
Vo
=
3oo, Rul
;
lO+
Re-q
:)
Vs
=
tfloV
3
T=
4oxlo-6x/ftlo
=
lt
R*q
=
?s/6
/so
:
lokt)
-t,2
V2
=
t|Oe-t
V
=)
af
t=t'?s Vz=t5oo
=
49'$V
Fig.
13:
Circuit
of
Problem
13
14. In the circuit shown in
Fig.
14, the switch
has been
in position
a for a long time. At /
=
0, the
swjtch
moves instantaneously
to
position
b.
Assume that 11(0-)
:0,
Vs
=
300
Y
Rs
=
125
Q,
Rx
=
600 Q,
Ry
=
200 O, Ll
=
6
H,
and
L2=3
H.
Findthe
vaiue
of the curent
i2(t)
at
/=
4
ms.
9,o
t
Z
,fq
=
-
qtDt
=z.ye
-VNf
.
-/
1z0e'
v
[29{z
ns
-P
i
(|)
aF
Fig. 14: Circuit
of Problem 14
L1
6H
L.Gl
=
I
3
L2
nt
-qro
x
I
-t?zoe
4t
+
",
o
_
t,6
-,
=
1,60
+o,5
c
t=0
L2
ln
|ul,,rt
I
i,(0
I
Rx
6ue
(u*s)