Problem 4.1 Find I o in the circuit shown using linearity and the assumption that I o = 1mA. 4mA 4K Ω 4K Ω 4K Ω 2K Ω 4K Ω 12K Ω I 2 I 3 + + - V 1 1 o k I A = I 1 - Suggested Solution I T 4K Ω 4K Ω 4K Ω 2K Ω 4K Ω 12K Ω I 2 I 3 + + - V 1 1 o k I A = I 1 - V 2 4/K A 6 3 1 1 o 2 1 k k 4k 5 2 1 2 1 2 2 V2 7 3 2 3 4K + 12K 2 7/2mA 8 4mA 1mA x 7 I= (4 2 ) 6 , 4 16 1 o T o A V K K V I I I I mA V V KI V I mA I I I mA x I mA = + = = = = + = = + = = = ∴ = + = = ∴ = = If Then Then and Then So 2 mA
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Problem 4.1 Find Io in the circuit shown using linearity and the assumption that Io = 1mA.
4mA
4KΩ 4KΩ 4KΩ
2KΩ4KΩ12KΩ
I2 I3+ +
-
V1
1o kI A=I1-
Suggested Solution
IT
4KΩ 4KΩ 4KΩ
2KΩ4KΩ12KΩ
I2 I3+ +
-
V1
1o kI A=I1-
V2
4/K A
6 31 1o 2 1k k 4k52 1 2 1 22
V2 73 2 34K + 12K 2
7/2mA 84mA1mA x 7
I = (4 2 ) 6 ,4 16
1o
T
o
A V K K V II I I mA V V KI VI mA I I I mA
x I mA
= + = = =
= + = = + =
= = ∴ = + =
= ∴ = =
If Then
Then and
Then
So
2 mA
Problem 4.2 Find Io in the network shown using linearity and the assumption that Io=1mA.
6KΩ
2KΩ 3KΩ
2KΩ64V
6KΩIo
Suggested Solution
2
61 1 2 12
2 1 2 3 2 36
32 642 4 1
1 , 6 . 3 . 43 18 , 3 7
2 32 2
KV 4K
omA x
Io mA V V I mA I Io I mAV V KI V I mA I I I m
Vs V KI V x I mA
= = = = = + =
= + = = = = + =
= + = ∴ = ∴ = =
If
Then
A
4KΩ
2KΩ 3KΩ
2KΩ
I2
+
-
V264V
6KΩVs
I4 I3 +
-
V1
Io I1
Problem 4.3 Find Vo in the network shown using linearity and the assumption that Vo=1mV
I2
2KΩ 2KΩ
2KΩ2KΩ2KΩ6V
+
-
Vo
Suggested Solution
I2
R1 R2
R5V4R2V's
+
-
V'o=1mV
V1+ - V3 I3
I4 I5
V2
+
-
+
-
R4
' '1 15 4 3 45 2 223 3 3 2 3 2 2
1 2 3 1 1 1 1 2
6' 8
2 , 61
2 ' 3 1.52.5 5 ' 8
( )(1 ) 0.75 0.
V o V oR RT
VR
Vo VsVs V s m
5
R K Vs VI A I A I I IV I R mV V V V o mV I A
A
I I I A V I R mV V s V V mV
Vo m V Vo
µ µ µµ
µ
= Ω == = = = = + =
= = = + = = =
= + = = = = + =
= ⇒ = = =
All
75V
Problem 4.4 Find Vo in the circuit shown using linearity and the assumption that Vo=1V
12V
8K
4K6K
6K
2K
+
-
Vo
Figure P4.4
Suggested Solution
12V
8K
4K6K
6K
+
-
Vo
I1
Vx
2K
Vs
Is
I2
12 2 2
1 12
1 2
9 121
12(8 ) 4
0.3330.8333
(6 ) 9
1.33
VoKVxK
V VV Vo
Vo VI mA Vx I K VI mAIs I I mAVs Is K Vx V
Vo V
== = ⇒ = =
= =
= + == + =
= ⇒ =
Assume
Problem 4.5 In the network shown find Io using superposition
12V6KΩ
6KΩ6KΩ
6KΩ6mA
Io
Suggested Solution
6KΩ
6KΩ
6KΩ
6mAI'o
6KΩ
6mAI'o
9KΩ6KΩ
Zero the indep. voltage source
9 189 6 5' 0.006( )Ko K KI mA+= − = −
6KΩ
6KΩ
6KΩ
I''o
12V
6KΩ
Zero the indep. current souce
I
61256 6 (6 6 )
6 26 12 5
18 162( 5 5 5
'' ( )
' '' )
K K K K
Ko K K
o o o
I mA
I I mA
I I I mA mA
+ +
+
= + = −
= =
= =
+ = −
Problem 4.6 Find Io in the circuit shown using superposition
Io
6KΩ 2KΩ
2KΩ
2KΩ
12KΩ30V 30mA
Suggested Solution
I'o
6KΩ 2KΩ 2KΩ
12KΩ30V
I
Zero the indep. current source
30 6
6 12 ||6 183 , ' 1KK K K KI mA I o I mA+= = = =
I''o
6KΩ 2KΩ
2KΩ
2KΩ
12KΩ30mA
I1
Zero the indep. voltage source
41 4 2 6 ||12
618
0.03( ) 12
'' 0.012( ) 4
' '' 3
KK K K K
KK
I mA
I o mA
Io I o I o mA
+ += =
= − = −
= + =
Problem 4.7 In the network shown find Vo using superposition
1KΩ
1KΩ 1KΩ12V
1KΩ6mA Vo
+
-
Suggested Solution
1KΩ
1KΩ 1KΩ
1KΩ6mA V'o
+
-
Zero the indep. voltage source
2
2 2' 0.006( )(1 ) 3KK KV o k V+= =
1KΩ
1KΩ 1KΩ
1KΩV''o
+
-
Zero the indep. current source
12V
12
4'' (1 ) 3
' '' 0KV o K V
Vo V o V o V
−= = −
= + =
Problem 4.8 Find Vo in the network shown using superposition
3KΩ 3KΩ
Vo
+
-
3KΩ3KΩ9V
6KΩ
Suggested Solution
R1 R5
Vo
+
-
R4R29V
R3
12V
R3=6KΩ All other R=3KΩ
R5
Vo1
+
- R4
R2
R3R1 A
B
VAB
+
_9V
1
4 5
4 5 3
2 3 4 5
||01 ( || )
|| [ ( || )] 2.149( ) 3.75
( ) 0.75
ABAB
AB
RAB R RR R
AB R R R
R R R R R KV V
V V V+
+
= + == =
= =
Ω
R1
R5
Vo2
+
- R4
R2
R3
12V
C
D
5
4 3 1 2
02
01 02
|| [ ( || )] 2.1412( ) 5
4.25
CDCD
CD
RR R
R R R R R KV V
Vo V V V+
= + == − = −
= + = −
Ω
Problem 4.9 Find Io in the network shown using superposition.
6V3KΩ
2mA
6KΩ 3KΩ
2KΩ
Io
Suggested Solution
6V3KΩ2mA
3KΩ
2KΩ
Io2
Ic
C
D
6KΩ3KΩ
6KΩ
2KΩ
A
B
Io1
VAB
+
-
3KΩ
33 6
602 016 2
01 02
3 (6 || 2 ) 4.5 2 || (3 3 ) 1.52 ( ) 0.8 6( ) 1.2
( ) 0.6 / 2 0.6
1
ABCD AB
CD AB
K RABK R R KK ABK K
R K K K K R K K K KIc m mA V VI Ic mA I V K mA
Io I I
+ +
+
= + = Ω = + == = = =
= = = =
= + =
Ω
.2mA
Problem 4.10 Find Io in the network shown using superposition
1KΩ
3KΩ
2KΩ 2KΩ
6V
9mA
Io
Suggested Solution
1KΩ
3KΩ
2KΩ 2KΩ9mA
Io1
A
B
IA
101 1
3 (2 || 2 ) 49 ( ) 1.8
AB
AB
KA K R
R K K K KI I m m+
= + == = = A
Ω
1KΩ
3KΩ
2KΩ 2KΩ
6VI02
VCD
C
D
+
-
2
02 1 3
0 01 02
(1 3 ) || 2 1.336[ ] 2.4
0.6
2.4 , 0 2.4
CDCD
CD
CD
RCD R KVK K
R K K K KV VI mA
I I I mA I mA
+
+
= + = Ω= − = −
= − =
= + = =
Problem 4.11 Find Io in the network shown using superposition.
4mA
12V3KΩ
2KΩ
6KΩ
4KΩ
Io
Suggested Solution
302 3 Re 1
Re 1 6 (2 || 4 ) 7.334 [ ] 1.16K
K q
q K K KI m mA+
= + == =
K
4mA 3KΩ
6KΩ
4KΩ
I02
2KΩ
Req1
4mA12V
3KΩ
2KΩ
6KΩ
4KΩ
I01
+
-
Req2 Re 2
Re 2 2
01 9
0 01 02
Re 2 9 || 4 2.77 , 12[ ] 6.97
0.77
0.39
qq K
VxK
q K K k Vx
I mA
I I I mA
+= = Ω = =
= − = −
= + =
V
Problem 4.12 Find Io in the network shown using superposition
4mA
6V12KΩ
12KΩ
12KΩIo4mA
Suggested Solution
4mA
6V12KΩ
12KΩ
12KΩ
12KΩ
Io4mA
6V
12KΩ
12KΩ
12KΩ
12KΩ
I'o
6 312 12 ||(12 12 ) 10' K K K KI o m− −
+ += = A
4mA
6V12KΩ
12KΩ
12KΩ
12KΩ
4mA
12KΩ
12KΩ 12KΩ 12KΩ4mA
I1
I''o
121 12 12 ||12
1 12
0.003 0.008 110 10 2
4 ( ) 1.6
'' 0.8
' ''
KK K KI m mA
I o I mA
Io I o I o mA
+
−
= =
= =
= + = + =
Problem 4.13 Find Io in the circuit shown using superposition
6mA
12V
6KΩ
4KΩ
4KΩ 6KΩ
Io
Io
Suggested Solution
12V
6KΩ
4KΩ
4KΩ 6KΩ
Io
I'o
0.012
12' 1I o m= = A
6KΩ
4KΩ
4KΩ 6KΩ
I''o6mA 6KΩ
4KΩ4KΩ 6KΩ
I''o6mA
'' 3
' '' 2
I o m
Io I o I o mA
A= −
= + = −
Current splits equally:
Problem 4.14 Use superposition to find Io in the circuit shown
12V
6V
2KΩ
2KΩ
2KΩ
2KΩ
Io
2mA
Suggested Solution
12V
2KΩ
2KΩ
2KΩ
2KΩ
I'o
I'o = 12/4K = 3mA
2KΩ
2KΩ
2KΩ
2KΩ
I''o
2mA
Current Splits equallyI''o=1mA
6V
2KΩ
2KΩ
2KΩ
2KΩ
I'''o
I'''o = -6/4K = -3/2 mA
12V
Then Io=I'o + I''o + I'''o = 3mA + 1mA -3/2 mA = 5/2 mA
Problem 4.15 Find Io in the network shown using superposition
6V
6KΩ
6KΩ
6KΩ5mA
6KΩ
Io
Suggested Solution
6V
6KΩ
6KΩ
6KΩ
I'o
6 6
6 6 ||(6 6 ) 10' K K K K KI o A− −+ += =
6KΩ
6KΩ
6KΩ6KΩ
6KΩ
6KΩ
I''o
5mA
5mA6KΩ 6KΩ
I''o
6 16 6 6 ||6 2
3 25 5
'' 5 ( )( ) 1
' '' 1
KK K K KI o m mA
Io I o I o mA mA mA+ += =
= + = + =
Problem 4.16 Find Io in the network shown using superposition.
4mA
12V2KΩ
1KΩ1mA 2KΩ
Io
Suggested Solution
2KΩ
1KΩ 2KΩ
I03
12V
1KΩ 2KΩ
I01
2KΩ
1KΩ 2KΩ
I02
Io due to 12V source Io due to 2mA source Io due to 4mA source I01= -12/(1K+12K) = -4mA I02 = 2m[2K/(2K+3K)] = -1.33mA I03=0A Io = I01 + I02 + I03 = -5.33mA
Problem 4.17 Find Io in the network shown using superposition
Io
4KΩ
4KΩ
3KΩ
2KΩ
9V
6V
2mA
Suggested Solution
I01
4KΩ
4KΩ
3KΩ
2KΩ9V
Io due to 9V source
6V 4KΩ
3KΩ
2KΩI02
Io due to 6V source (redrawn)
9 601 024 42.25 1.5K KI mA I mA−= = = = −
I03
4KΩ
4KΩ
3KΩ
Io due to 2mA source
2KΩ
2mA
03 03
01 02 03
(4 ) 0 0K I IIo I I I
0.75Io mA
= → == + +
=
Problem 4.18 Find Vo in the network shown using superposition
Problem 4.68 Given the linear circuit shown, it is known that when a 2-KΩ load is connected to the terminals A-B, the load current is 10mA. If a 10-KΩ load is connected to the terminals the load current is 6mA. Find the current in a 20KΩ load.
A
B
Rth
Voc
Suggested Solution
Voc
Rth
I
RL
A
B
VAB
+
-
(Rth + RL) I = Vocif RL = 2KΩ , I = 10mA => Voc = 20 + 0.01Rth
if RL = 10KΩ , I = 6mA => Voc = 60 + 0.006Rth
yield Voc = 120V and Rth = 10KΩ
If RL = 20KΩ , I = Voc / (RL + Rth)
I = 4 mA
Problem 4.69 If an 8-KΩ load is connected to the terminals of the network shown, VAB = 16V. If a 2-KΩ load is connected to the terminals VAB = 8V. Find VAB if a 20KΩ load is connected to the terminals.
Problem 4FE-2 Find the value of the load RL in the network shown that will achieve maximum power transfer, and determine the value of the maximum power.
1KΩ1KΩ
RL
+ -Vx
12V 2Vx
Suggested Solution
1KΩ1KΩ
RL = 2/7 K
+ -V'x
12V 2Vx
1KΩ1KΩ
Isc
+ -V''x
12V 2V''x
+
-Voc
I
- 12 + 2K I + 1K I + 2V'x = 0V'x = 2K I
Voc = 12 - 2K I = 66/7 V
Vx = 12Isc = 12 / 2K + 24 / 1K = 30mA
Rth = Voc / Isc = 2/7 K
66/7 V
2/7 KP = [(60/7)/(4/7 K)]2 (2/7 K)
P = 64.3 mW
Problem 4FE-3 Find the value of RL in the network shown for maximum power transfer to this load.