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7/25/2019 ECE225Midterm2 Sol
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Exercise 3-11 Determine the Th evenin-equivalent circuit at terminals (a , b ) in Fig. E3-11.
Solution:
(1) Open-circuit voltage
We apply node voltage method to determine open-circuit voltage:
V 1
2 4 +
V 1 V 23
= 0,
V 2 V 13
+ 3 +V 2
5 = 0.
Solution gives: V 2 = 3 . 5 V.Hence,
V Th = V oc = 3 . 5 V .
(2) Short-circuit current
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Because of the short circuit,V 2 = 0.
Hence at node V 1:
V 1
2 4 +
V 1
3 = 0
V 112
+ 13
= 4
V 1 = 24
5 V
I 1 = V 1
3 =
245 3
= 85
A ,
I sc = I 1 3 = 85
3 = 75
= 1 . 4 A
R Th = V Th I sc
= 3 . 5 1 . 4
= 2 . 5 .
Th evenin equivalent:
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
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Exercise 3-12 Find the Th evenin equivalent of the circuit to the left of terminals (a , b ) in Fig. E3-12, andthen determine the current I .
Figure E3-12
Solution: Since the circuit has no dependent sources, we will apply multiple steps of source transformation tosimplify the circuit.
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Across (a , b ),
V Th = V oc = 10 312 + 3
= 2 V
R Th = 3 12 + 0 . 6
= 3 123 + 12
+ 0 . 6 = 3
Hence,
I = 23 + 1 = 0 . 5 A .
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
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Exercise 3-13 Find the Norton equivalent at terminals (a , b ) of the circuit in Fig. E3-13.
Figure E3-13
Solution: Th evenin voltage
At node 1: I = 2 A .
Hence,V Th = V oc = 10 I 3 3 I = I = 2 V .
Next, we determine the short-circuit current:
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
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At node V 1:
2 3 I + V 1
10 +
V 1
3 = 0.
Also, I =
V 1
10 .
Hence,
2 3 I + I + 10
3 I = 0,
which gives
I = 1. 5 A , I 1 = 2 + 3 I I = 2 + 2 I = 5 A ,
I sc = 5 3 I = 5 4 . 5 = 0. 5 A .
R Th = V Th I sc
= 20 . 5
= 4 .
Norton circuit is:
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
7/25/2019 ECE225Midterm2 Sol
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Exercise 3-14 The bridge circuit of Fig. E3-14 is connected to a load R L between terminals (a , b ). Choose R L such that maximum power is delivered to R L . If R = 3 , how much power is delivered to R L?
Figure E3-14
Solution: We need to remove R L and then determine the Th evenin equivalent circuit at terminals (a , b ).Open-circuit voltage:
The two branches are balanced (contain same total resistance of 3 R ). Hence, identical currents will ow, namely
I 1 = I 2 = 243 R
= 8 R .
V oc = V a V b = 2 RI 1 RI 2 = RI 1 = R 8 R = 8 V .
To nd R Th , we replace the source with a short circuit:
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
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R 2 R = R 2 R R + 2 R
= 23
R
Hence, R Th =
4 R
3 ,
and the Th evenin circuit is
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
7/25/2019 ECE225Midterm2 Sol
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For maximum power transfer with R = 3 , R L should be
R L = 4 R
3 =
4 33
= 4 ,
and
P max = 2s
4 R L = 82
4 4 = 4 W .
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
7/25/2019 ECE225Midterm2 Sol
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Exercise 4-7 Express o in terms of 1 , 2 and 3 for the circuit in Fig. E4-7.
Figure E4-7
Solution: Starting from the output of the second stage and moving backwards towards the inputs,
o = 10 10 3
5 10 3
3 103
0 . 5 103 1 +
3 103
10 3 2 +
3 103
2 103 3
= 12 1 + 6 2 + 3 3 .
Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press
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Problem 4.10 In the circuit of Fig. P4.10, a bridge circuit is connected at the inputside of an inverting op-amp circuit.
(a) Obtain the Th evenin equivalent at terminals (a , b ) for the bridge circuit.(b) Use the result in (a) to obtain an expression for G = o/ s.(c) Evaluate G for R 1 = R 4 = 100 , R 2 = R 3 = 101 , and R f = 100 k .
o
s
R 2
b
a
R f
R 1
R 4 R 3
+
_
+ _
Figure P4.10: Circuit for Problem 4.10.
Solution: (a) The Th evenin equivalent circuit at (a , b ):
v s
R 2
b
a
R 1
R 4 R 3
+ _
i1
i2
+
_
v oc
s + i1( R 1 + R 2) = 0
ori1 =
s R 1 + R 2
.
Also, s + i2( R 3 + R 4) = 0
andi2 =
s R 3 + R 4
.
Th = oc = i 1 R 2 + i2 R 4
= s R 2 R 1 + R 2
+ s R 4
R 3 + R 4=
[ R 4( R 1 + R 2) R 2( R 3 + R 4)] s( R 1 + R 2)( R 3 + R 4)
. (1)
Suppressing s (by replacing it with a short circuit) leads to
R Th = ( R 1 R 2) + ( R 3 R 4)
= R 1 R 2 R 1 + R 2
+ R 3 R 4
R 3 + R 4=
R 1 R 2( R 3 + R 4) + R 3 R 4( R 1 + R 2)( R 1 + R 2)( R 3 + R 4)
.
(b) For the new circuit:
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v o
v Th
R f
R Th
+
_
+_
o = R f R Th
Th (inverting amplier) (3)
Inserting Eqs. (1) and (2) into (3) leads to
G = o
s = R f [ R 4( R 1 + R 2) R 2( R 3 + R 4)] R 1 R 2( R 3 + R 4) + R 3 R 4( R 1 + R 2)
(c) For R 1 = R 4 = 100 , R 2 = R 3 = 101 , and R f = 105 ,
G = 105[100 (100 + 101 ) 101 (100 + 101 )]
100 101 (100 + 101 ) + 100 101 (100 + 101 )
= 4.9505 5.
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Problem 4.11 Determine the output voltage for the circuit in Fig. P4.11 and specifythe linear range for s , given that V cc = 15 V and V 0 = 0.
Figure P4.11: Circuit for Problem 4.11.
v o
v sV 0
100 k
200 k 2 k
V cc = 15 V+
+ _
_
R 1 R 2
Inverting Amp
1
+ R 3
R 4
2
_
Solution: The given circuit is the same as the difference amplier circuit of Table4-3, with:
R 2 = 200 k , R 1 = 2 k , R 3 = 100 k , R 4 = , 1 = s , 2 = V 0 = 0.
Applying the difference amplier equation given by Eq. (4.41),
o = R 4
R 3 + R 4
R 1 + R 2 R 1
2 R 2 R 1
1
=
200 103
2 103
s = 100
s.
Since |( o)max | = 15 V, the linear range of s is
| s | 15100
= 150 mV ,
or 150 mV s 150 mV.
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Problem 4.23 For the circuit in Fig. P4.23, obtain an expression for voltage gainG = o/ s.
+_ 6 k
10 k
4 k
s
+_ 0
5 k
Figure P4.23: Circuit for Problem 4.23.
Solution: By voltage division,
p = s6
4 + 6 = 0.6 s .
n = p = 0.6 s. n s
10k +
n o5k
= 0.
Simplication leads to
o= 0.4
s.
Hence,G =
0
s= 0.4.
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Problem 4.24 Find the value of o in the circuit in Fig. P4.24.
+_ 0
2 A
5 V
2 k
6 k6 k
+_
4 k
Figure P4.24: Circuit for Problem 4.24.
Solution: Converting the input current source into a voltage source leads to
+_ o n 1
4 V
5 V
2 k6 k
6 k
+
_
4 k
+_
Fig. P4.24(a)
Apply nodal analysis:
p = n = 5 V.
@ n : n 1
6 k +
n o6 k
= 0,
@ 1 : 1 4
2 k +
1 o
4 k +
1 n
6 k = 0.
Simplify:
16k
( 1 + o) = 106 k
,
12k
+ 14k
+ 16k
1 14k
o = 176000
,
16k
16k
1112k
14k
1
o
=106k
176k
,
1
o =
327
387
.
o = 5. 429 V .
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Problem 4.36 Find the value of o in the circuit in Fig. P4.36.
4 6
2 n
+_ o
5 4
+_+_ 9 V3 V
p
Figure P4.36: Circuit for Problem 4.36.
Solution: By voltage division,
p = 9 46 + 4
= 3.6 V
n = p
n 35k
+ n 9
2k +
n o4k
= 0.
Solution gives o = 6.72 V .
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Problem 4.38 Determine o
and the power dissipated in RL
in the circuit of Fig. P4.38.
2
7
3 4 V
2 V
+_
4 2
5
+_
+ _
o
R L
3
R eq
a p
n
Figure P4.38: Circuit for Problem 4.38.
Solution: By voltage division,
p = 4 3k
2k + 3k =
125
= 2.4 V
n 27k
+ n a
5k = 0
n =
p =
2.4 V .
Solution gives a = 2.686 V .
The 2-k and 4-k output resistors are equivalent to a single resistor
R eq = 2 42 + 4
k = 86
k .
By voltage division,
o = a R eq
3k + R
eq
= 2.686 86 k
3k + 86 k
= 0.826 V .
P RL = 2o R L
= (0.826 )2
2k = 0.34 mW .
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Problem 4.52 Find the value of o in the circuit in Fig. P4.52.
+_
o+_
9 V 5 V 8 k3 k
4 k
3 k
6 k
+_
+_
8 k n2
a n1
p2
p1
i 1
i 2
i 3
Figure P4.52: Circuit for Problem 4.52.
Solution: For the rst stage:
p1 = 0 n1 = 0 , n1 9
8k +
n1 a6k
+ n1 o
3k = 0 ,
which simplies to8 o + 4 a + 27 = 0 . (1)
For the second stage:
p2 =5 83 + 8
=4011
V,
n2 = p2 =4011
V.
Since i n2 = 0, a = n2 = 4011 V.
Hence,
0 = 27 4 a
8
=18
27 4 4011
= 5 . 19 V .
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Exercise 5-9 Determine C eq and V eq(0) at terminals (a , b ) for the circuit in Fig. E5-9, given thatC 1 = 6 F, C 2 = 4 F and C 3 = 8 F, and the initial voltages on the three capacitors are 1(0) = 5 V and 2(0) = 3(0) = 10 V.
Figure E5-9
Solution:
C eq = C 1(C 2 C 3)C 1 + C 2 + C 3
= C 1(C 2 + C 3)C 1 + C 2 + C 3
= 6 10 6(4 10 6 + 8 10 6)
(6 + 4 + 8) 10 6 = 4 F ,
V eq(0) = 1(0) + 2(0) = 5 + 10 = 15 V .
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Problem 5.19 For the circuit in Fig. P5.19, nd C eq
at terminals (a , b ). Assume allinitial voltages to be zero.
Solution:
a
b
c
d
5 F 3 F 5 F
5 F3 F
6 F 6 F
a
b
5 F
6 F ( )13 + 13 16+ 65= F1
a
b
5 F
a
b
6 + = F65
365
C eq = 5 3655 + 365
= 180
61 = 2. 95 F
Figure P5.19
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Problem 5.20 Find C eq
at terminals (c , d ) in the circuit of Fig. P5.19.
Solution:
a
b
c
d
5 F 3 F 5 F
5 F3 F
6 F 6 F
c
d
5 F
5 F
6 F
c
d
5 F
5 F
c
d
( )13 + 13 16+ 65= F1
( )15 15 536+ + = 1.86 F1
6 + = F65
365
Figure P5.20
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Exercise 5-13 Determine L eq at terminals (a , b ) in the circuit of Fig. E5-13.
Figure E5-13
Solution:
Leq = 2 mH + ( 6 mH 12 mH )
= 2 + 6 12
6 + 12 mH
= 6 mH .
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Problem 5.30 All elements in Fig. P5.30 are 10-mH inductors. Determine Leq
.
Solution:
L
L
L L L
L
L
L
L
L 2 L2 L L eq
L eq
L
L
L L eq
L eq L eq = 2.5 L = 25 mH
( )1 L L212 L 12 L+ + =1
Figure P5.30
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