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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
If 𝐼0 = 1𝐴 then 𝑉1 = 3 + 5 𝐼0 = 8 𝑉 and 𝐼1 =
𝑉1
4
= 2 𝐴.ApplyingKCL at node 1 gives
𝐼2 = 𝐼1 + 𝐼0 = 3 𝐴
𝑉2 = 𝑉1 + 2𝐼2 = 8 + 6 = 14 𝑉,
𝐼2 =
𝑉2
7
=2𝐴
Applying KCL at node 2 gives
𝐼4 = 𝐼3 + 𝐼2 = 5 𝐴
Therefore,𝐼𝑠 = 5 𝐴.This shows that assuming𝐼0 = 1 𝐴 gives𝐼𝑠 = 5 𝐴.The actual source current of 15 A
will give 𝐼0 = 3 𝐴 as the actual value.
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Open circuit the 4 A source. Then,
(7 + 2) || (5 + 5) = 4.737 Ω, we can calculate v1’ =1.4.737 = 4.737V
To find the total current flowing through the 7 Ω resistor, we first determine the total
voltage v1 by continuing our superposition procedure. The contribution to v1 from the 4 A
source is found by first open-circuiting the 1 A source, then noting that current division
yields:
5
20
4
=
= 1.053A
5 + (5 + 7 + 2) 19
V1’’ = (1.053)(9) = 9.477V
Hence v1 = 14.21V
We may now find the total current flowing downward through the 7 Ω resistor as
14.21/7 = 2.03 A.
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Short-circuit the 10 V source
Note that 6 || 4 = 2.4 Ω. By voltage division, the voltage across the 6 Ω resistor is then
2.4
4×
= 1.778V
3 + 2.4
So that i1’ = 0.2693A
Short-circuit the 4 V source
Note that 3 || 6 = 2 Ω. By voltage division, the voltage across the 6 Ω resistor is then
2
−10 × = −3.33V
6
i1’’= -5.556mA
a. I = i1’ + i1’’
= -259.3mA
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Taking one source at a time:
The contribution from the 24-V source may be found by shorting the 45-V source and opencircuiting the 2-A source. Applying voltage division
vx‘ = 24
20
10+20+45||30
= 24
20
10+20+18
= 10V
We find the contribution of the 2-A source by shorting both voltage sources and
applying current division:
20
vx’’ = 24 10+20+18 = 8.333V
Finally, the contribution from the 45-V source is found by open-circuiting the 2-A
source and shorting the 24-V source.
Defining v across the 30-Ω resistor with the “+” reference on top:
30
0=
v30
v30
v30 − 45
+
+
20
10 + 20
45
solving, v = 11.25 V and hence v ”’ = -11.25(20)/(10 + 20) = -7.5 V
30
x
Adding the individual contributions, we find that v = v ’ + v ” + v ”’ = 10.83 V
x
Chapter 5: Additional Analysis Techniques
x
x
x
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
We find the contribution of the 4-A source by shorting out the 100V source and
analysing the resulting circuit:
4=
V1′
+
20
V1′
V ′1 −V′
10
V ′ −V ′1
4i1’ = 30 + 10
Where i1 ' = V '/ 20
1
Simplifying & collecting terms, we obtain
30 V ' – 20 V' = 800
1
-7.2 V ' + 8 V' = 0
1
Solving, we find that V' = 60 V. Proceeding to the contribution of the 60-V source, we
analyse the following circuit after defining a clockwise mesh current i flowing in the left
a
mesh and a clockwise mesh current i flowing in the right mesh.
b
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
30 ia – 60 + 30 ia – 30 ib = 0
ib = -0.4i1" = +0.4ia
Solving, we find that ia = 1.25 A and so V" = 30(ia – ib) = 22.5 V
Thus, V = V' + V" = 82.5 V.
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Since there are two sources, let
𝑣 = 𝑣1 + 𝑣2
Where 𝑣1 and𝑣2 are the contributions due to the 6-V voltage sourceand the 3-A current source,
respectively. To obtain 𝑣1 we set the currentsource to zero, as shown in the figure below.
Applying KVL to the loop, we get,
12𝑖1 − 6 = 0
Thus,
=>
𝑖1 = 0.5 𝐴
𝑣1 = 4𝑖1 = 2 𝑉
To get 𝑣2 we set the voltage source to zero, as shown.
Using current division,
8
𝑖3 = 4+8 3 = 2 𝐴
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Hence,
𝑣2 = 4𝑖3 = 8 𝑉
And we find,
𝑣 = 𝑣1 + 𝑣2 = 10 𝑉
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
The circuit involves a dependent source, which must be leftintact. We let,
𝑖0 = 𝑖0𝑎 + 𝑖0𝑏
Where𝑖0𝑎 and 𝑖0𝑏 are due to the 4-A current source and 20-V voltagesource respectively. To obtain𝑖0𝑎
we turn off the 20-V source so that we have the circuit shown below.
We apply mesh analysis in order toobtain 𝑖0𝑎 . For loop 1,
𝑖1 = 4 𝐴
(i)
−3𝑖1 + 6𝑖2 − 𝑖3 − 5𝑖0𝑎 = 0
(ii)
For loop 2,
For loop 3,
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
−5𝑖1 − 𝑖2 + 10𝑖3 + 5𝑖𝑜𝑎 = 0
(iii)
𝑖3 = 𝑖1 − 𝑖0𝑎 = 4 − 𝑖0𝑎
(iv)
But at node 0,
Solving equations (i)-(iv), we get,
52
𝑖0𝑎 = 17 𝐴
To obtain 𝑖𝑜𝑏 we turn off the 4-A current source so that the circuitbecomes as that shown below.
For loop 4, KVL gives,
6𝑖4 − 𝑖5 − 5𝑖0𝑏 = 0
(v)
−𝑖4 + 10𝑖5 − 20 + 5𝑖0𝑏 = 0
(vi)
And for loop 5,
But, 𝑖5 = −𝑖0𝑏 . Thus, using this and equations (v) and (vi) we get,
Hence,
𝑖0 = 𝑖𝑜𝑎 + 𝑖0𝑏 = −0.4706 𝐴
Chapter 5: Additional Analysis Techniques
60
𝑖0𝑏 = − 17 𝐴
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
In this case, we have three sources. Let
𝑖 = 𝑖1 + 𝑖2 + 𝑖3
Where, 𝑖1 , 𝑖2 and 𝑖3 are due to the 12-V, 24-V, and 3-A sources respectively. To get 𝑖1 , consider the circuit
shown below.
Combining the 4 Ω (on the right-hand side) in series with 8 Ω gives 12 Ω. The 12 Ω inparallel with 3 Ω
gives4Ω as shown in the figure above. Thus,
To get 𝑖2 , consider the circuit shown below,
Chapter 5: Additional Analysis Techniques
𝑖1 =
12
6
=2𝐴
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Irwin, Engineering Circuit Analysis, 11e, ISV
Applying mesh analysisgives,
16𝑖𝑎 − 4𝑖𝑏 + 24 = 0
=>
4𝑖𝑎 − 𝑖𝑏 = −6
7𝑖𝑏 − 4𝑖𝑎 = 0
=>
𝑖𝑎 = 4 𝑖𝑏
(ii)
7
(i)
𝑖2 = 𝑖𝑏 = −1 𝐴
From (i) and (ii), we get,
To get𝑖3 , consider the circuit shown below.
Using nodal analysis gives,
𝑣2
8
+
𝑣2 −𝑣1
4
𝑣2 −𝑣1
4
=
𝑣1
4
=3
=>
3𝑣2 − 2𝑣1 = 24
(iii)
𝑣1
3
=>
𝑣2 =
10
𝑣
3 1
(iv)
+
Substituting (iv) into (iii) leads to 𝑣1 = 3 and
𝑖3 =
Thus,
𝑣1
3
=1𝐴
𝑖 = 𝑖1 + 𝑖2 + 𝑖3 = 2 𝐴
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
One approach to this problem is to write a set of mesh equations, leaving the voltage source
and current source as variables which can be set to zero.
We first rename the voltage source as V . We next define three clockwise mesh currents in
x
the bottom three meshes: i , i and i . Finally, we define a clockwise mesh current i in the
1
y
4
top mesh, noting that it is equal to –4 A
Our general mesh equations are then:
-V + 18i – 10i = 0
x
1
y
–10i + 15i – 3i = 0
1
y
4
–3i + 16i – 5i = 0
y
4
3
Set V = 10 V, i = 0. Our mesh equations then become
x
3
18i – 10iy’ = 10
1
–10i + 15iy’ – 3i = 0
1
4
– 3iy + 16i = 0
4
Solving, iy’ = 0.6255 A.
Set V = 0 V, i = – 4 A. Our mesh equations then become
x
3
18i – 10iy’’ = 0
1
–10i + 15iy’’ – 3i = 0
1
4
-3iy’’ + 16i4 = -20
Solving,iy’’ = –0.4222 A
Thus, iy = iy’ + iy’’ = 203.3mA
Chapter 5: Additional Analysis Techniques
3
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
.
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Shorting the 14 V source, we find that RTH = 10 || 20 + 10 = 16.67 Ω.
Next, we find V by determining V (recognising that the right-most 10 Ω resistor carries no
TH
OC
current, hence we have a simple voltage divider):
10 + 10
VTH = Voc = 14
= 9.33V
10 + 10 + 10
Thus, our Thevenin equivalent is a 9.333 V source in series with a 16.67 Ω resistor, which is
in series with the 5 Ω resistor of interest.
9.333
Now, I5Ω = 5+16.67 = 0.4307A
Thus, P5Ω = (0.4307)2.5 = 927.5mW
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
First
R = 10 mV/ 400 μA = 25 Ω
TH
Then
R = 110 V/ 363.6×10-3 A = 302.5 Ω
TH
Increased current leads to increased filament temperature, which results in a higher
resistance (as measured). This means the Thévenin equivalent must apply to the specific
current of a particular circuit – one model is not suitable for all operating conditions (the
light bulb is nonlinear).
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
a. Shorting out the 88-V source and open-circuiting the 1-A source, we see looking
into the terminals x and x' a 50-Ω resistor in parallel with 10 Ω in parallel with (20 Ω + 40
Ω), so
R = 50 || 10 || (20 + 40) = 7.317 Ω
TH
Using superposition to determine the voltage V across the 50-Ω resistor, we find
xx'
Vxx ′ = VTH
50||(20 + 40)
40
= 88
+ 1 (50||10)[
]
10 + [50|| 20 + 40 ]
40 + 20 + (50| 10
= 69.27V
b. Shorting out the 88-V source and open-circuiting the 1-A source, we see looking
into the terminals y and y' a 40-Ω resistor in parallel with [20 Ω + (10 Ω || 50 Ω)]:
R = 40 || [20 + (10 || 50)] = 16.59 Ω
TH
Using superposition to determine the voltage V across the 1-A source, we find
yy'
V𝑦𝑦 ′ = VTH = 1. R TH + 88
27.27
40
[
]
10 + 27.27 20 + 40
= 59.52V
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
a. Removing terminal c, we need write only one nodal equation
𝑣𝑦 𝑣𝑦 − 𝑣𝑥
3= +
2
3
which may be solved to yield V = 4 V. Therefore, Vab = VTH = 2 – 4 = 2V
b
R
TH
= 12 || 15 = 6.667 Ω. We may then calculate I as I = V / R
N
= -300 mA (arrow pointing upwards)
b. Removing terminal a, we again find R
TH
N
TH
TH
= 6.667 Ω, and only need write a single
nodal equation; in fact, it is identical to that written for the circuit above, and we once again
find that V = 4 V. In this case, V = V = 4 – 5 = -1 V, so I = -1/ 6.667
b
TH
= –150 mA (arrow pointing upwards)
Chapter 5: Additional Analysis Techniques
bc
N
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
We inject a current of 1 A into the port (arrow pointing up), select the bottom terminal
as our reference terminal, and define the nodal voltage V across the 200-Ω resistor.
x
Then, 1 = V / 100 + (V – V )/ 50 [1]
1
1
x
-0.1 V = V / 200 + (V – V )/ 50 [2]
1
x
x
1
which may be simplified to
3 V – 2 V = 100 [1]
1
x
16 V + 5 V = 0 [2]
1
x
Solving, we find that V = 10.64 V, so R
1
TH
= V / (1 A) = 10.64 Ω.
1
Since there are no independent sources present in the original network, I = 0.
N
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
85
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e, ISV
SOLUTION:
Set the independent sources equal to zero. This leads to thecircuit shown below, from which we find
𝑅𝑁 .Thus,𝑅𝑁 = 4 Ω
To find we short-circuit terminals 𝑎 and 𝑏, as shown below.
We ignore the 5Ω resistor because it has been short-circuited.Applying mesh analysis, we obtain,
𝑖1 = 2 𝐴,
20𝑖2 − 4𝑖1 − 12 = 0
From these equations,, we get,
𝑖2 = 1𝐴 = 𝑖𝑠𝑐 = 𝐼𝑁
Chapter 5: Additional Analysis Techniques
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Thus, the Norton equivalent circuit is as,
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Since V = VTh= Vab = Vx, we apply KCL at the node a and obtain,
30−𝑉𝑇ℎ
12
=
𝑉 𝑇ℎ
60
+ 2𝑉𝑇ℎ
=>
𝑉𝑇ℎ = 1.19 𝑉
=>
𝑉𝑋 = 0.4762 𝑉
=>
𝐼𝑁 =
To find RTh , consider the circuit below.
At node a, KCL gives
2𝑉𝑋 +
𝑉𝑋
60
+
𝑉𝑋
12
𝑅𝑇ℎ =
𝑉𝑋
1
= 0.4762 Ω
=1
Thus, 𝑉𝑇ℎ = 1.19 𝑉, 𝑅𝑇ℎ = 𝑅𝑁 = 0.4762 Ω, 𝐼𝑁 = 2.5 𝐴
Chapter 5: Additional Analysis Techniques
𝑉 𝑇ℎ
𝑅𝑇ℎ
= 2.5 𝐴
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
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SOLUTION:
To find we set the independent voltage source equal to zero andconnect a voltage source𝑣0 of 1V to the
terminals. We obtain the circuit shown below.
We ignore the 5-Ω resistor because it is short-circuited. Also due to the short circuit,the 4-Ω resistor, the
voltage source, and the dependent current sourceare all in parallel. Hence,𝑖𝑥 = 0. At node a,
𝑖0 =
𝑅𝑁 =
𝑣0
𝑖0
1𝑉
= 0.2 𝐴
5Ω
= 5Ω
To find 𝐼𝑁 we short-circuit terminals aand band find the current𝑖𝑠𝑐 as indicated in figure below.
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The 4-Ωresistor, the 10-V voltage source, the 5-Ω resistor, and the dependentcurrent source are all in
parallel. Hence,
𝑖𝑥 =
10
4
= 2.5𝐴
At node a, KCL gives,
𝑖𝑠𝑐 =
10
5
+ 2𝑖𝑥 = 7 𝐴 = 𝐼𝑁
Chapter 5: Additional Analysis Techniques
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SOLUTION:
Chapter 5: Additional Analysis Techniques
106
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Chapter 5: Additional Analysis Techniques
107
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SOLUTION:
Chapter 5: Additional Analysis Techniques
108
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Chapter 5: Additional Analysis Techniques
109
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Chapter 5: Additional Analysis Techniques
110
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SOLUTION:
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SOLUTION:
For RTh, consider the circuit shown below,
𝑅𝑇ℎ = 1 + 4
3 + 2 + 5 = 3.857 Ω
For VTh, consider the circuit shown below,
Applying KVL gives,
=>
10 − 24 + 𝑖 3 + 4 + 5 + 2 = 0, 𝑜𝑟 𝑖 = 1𝐴
𝑉𝑇ℎ = 4𝑖 = 4 𝑉
(b) For RTh, consider the circuit shown below,
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𝑅𝑇ℎ = 5
2 + 3 + 4 = 3.214 Ω
To get VTh, consider the circuit shown below,
At the node, KCL gives,
24−𝑣0
9
+2 =
𝑣0
,
5
=>
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𝑣0 = 15 𝑉
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SOLUTION:
First we note the three current sources are in parallel, and may be replaced by a single current
source having value 5 – 1 + 3 = 7 A, arrow pointing upwards. This source is in parallel with
the 10Ω resistor and the 6Ω resistor. Performing a source transformation on the current
source and 6 Ω resistor, we obtain a voltage source (7)(6) = 42 V in series with a 6 Ω resistor
and in series with the 10 Ω resistor
By voltage division, v = 42(10)/16 = 26.25 V.
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SOLUTION:
We can ignore the 1-kΩ resistor, at least when performing a source transformation on
this circuit, as the 1-mA source will pump 1 mA through whatever value resistor we place
there. So, we need only combine the 1 and 2 mA sources (which are in parallel once we
replace the 1-kΩ resistor with a 0-Ω resistor). The current through the 5.8-kΩ resistor is then
simply given by voltage division:
4.7
i = 3 × 10−3
= 1.343mA
4.7 + 5.8
2.
3
The power dissipated by the 5.8-kΩ resistor is then i 5.8×10 = 10.46 mW.
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SOLUTION:
(1)(47) = 47 V. (20)(10) = 200 V. Each voltage source “+” corresponds to its
corresponding current source’s arrow head
Using KVL on the simplified circuit above,
3
3
47 + 47×10 I – 4 I + 13.3×10 I + 200 = 0
1
1
1
3
Solving, we find that I = -247/ (60.3×10 – 4) = -4.096 mA.
1
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SOLUTION:
(2 V1)(17) = 34V1
Analysing the simplified circuit above,
34 V – 0.6 + 7 I + 2 I + 17 I = 0 [1]
1
V1 = 2 I [2]
Substituting, we find that I = 0.6/ (68 + 7 + 2 + 17) = 6.383 mA. Thus
V = 2 I = 12.77 mV
1
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SOLUTION:
Doing source transformation
12
= 1.333 mA
9000
9k || 7k = 3.938 kΩ
→ (1.333 mA)(3.938 kΩ) = 5.249V
3
5.249/ 473.938×10 = 11.08 μA
473.9 k || 10 k = 9.793 kΩ. (11.08 mA)(9.793 kΩ) = 0.1085 V
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3
Ix = = 0.1085/ 28.793×10 = 3.768 μA
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SOLUTION:
We obtain a 5v /4 A current source in parallel with 4 Ω, and a 3 A current source in parallel
3
with 2 Ω. We now have two dependent current sources in parallel, which may be combined
to yield a single –0.75v current source (arrow pointing upwards) in parallel with 4 Ω.
3
Selecting the bottom node as a reference terminal, and naming the top left node V and the
x
top right node V , we write the following equations:
y
𝑣𝑥 𝑣𝑥 − 𝑣𝑦
−0.75𝑣3 = +
4
3
𝑣𝑦 𝑣𝑦 − 𝑣𝑥
3= +
2
3
𝑣3 = 𝑣𝑦 − 𝑣𝑥
Solving we find that v3 = −2V
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SOLUTION:
We first transform the current and voltage sources to obtain the circuit,
Combining the 2Ω and 4Ω resistors in series andtransforming the 12-V voltage source gives us,
We nowcombine the 6Ω and 8Ω resistors in parallel to get 2Ω. We alsocombine the 2-A and 4-A current
sources to get a 2-A source. Thus,by repeatedly applying source transformations, we obtain the circuit,
We use current division to get, 𝑖 = 0.4 𝐴 and 𝑣0 = 8 𝑖 = 3.2 𝑉
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SOLUTION:
The circuit involves a voltage-controlled dependent currentsource. We transform this dependent
current source as well as the 6-Vindependent voltage source as shown,
The 18-V voltagesource is not transformed because it is not connected in series with anyresistor. The
two 2Ω resistors in parallel combine to give a 1Ω resistor, which is in parallel with the 3-A current
source. The currentsource is transformed to a voltage source as shown in the figure below. Notice that
the terminals for 𝑣𝑥 are intact.
Applying KVL around theloop gives,
−3 + 5𝑖 + 𝑣𝑥 + 18 = 0
Applying KVL to the loop containing only the 3-V voltage source, the 1-Ω resistor, and𝑣𝑥 yields
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(i)
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−3 + 𝑖 + 𝑣𝑥 = 0
Using (i) and (ii), we get,
=>
𝑣𝑥 = 7.5 𝑉
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𝑣𝑥 = 3 − 𝑖
(ii)
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SOLUTION:
We need to find the Thévenin equivalent resistance of the circuit connected to R , so
L
we short the 20-V source and open-circuit the 2-A source; by inspection, then
R = 12 || 8 + 5 + 6 = 15.8 Ω
TH
Analyzing the original circuit to obtain V and V with R removed
1
V = 20 8/ 20 = 8 V; V = -2 (6) = -12 V
1
We define V
2
TH
= V – V = 8 + 12 = 20 V. Then
1
2
PRL max = (Vth )2/4RL = 400/4.15.8 =6.329W
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L
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SOLUTION:
We need to find 𝑅𝑇ℎ and the 𝑉𝑇ℎ across the terminals 𝑎 − 𝑏. To get 𝑅𝑇ℎ , we use the circuit
shown below, and obtain 𝑅𝑇ℎ = 9 Ω
To get𝑉𝑇ℎ we consider the circuit,
Applying meshanalysis gives,
−12 + 18𝑖1 − 12𝑖2 = 0,
=>
𝑖2 = −2 𝐴
2
Solving for𝑖1 we get 𝑖1 = − 3 𝐴. Applying KVL around the outer loopto get𝑉𝑇ℎ across terminals ab, we obtain,
−12 + 6𝑖1 + 3𝑖2 + 𝑉𝑇ℎ = 0
For maximum power transfer
𝑅𝐿 = 𝑅𝑇ℎ = 9 Ω
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=>
𝑉𝑇ℎ = 22 𝑉
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And the maximum power is
𝑉2
𝑝𝑚𝑎𝑥 = 4𝑅𝑇ℎ = 13.44 𝑉
𝐿
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SOLUTION:
We need RThand V at terminals a and b. To find RTh and VTh, we insert a 1-mA source at terminals
a and b as shown below,
Assume that all resistances are in k ohms, all currents are inmA, and all voltages are in volts. At
node a,
1=
𝑣𝑎
40
+
𝑣𝑎 +120𝑣0
,
10
Or
40 = 5𝑣𝑎 + 480𝑣0
(i)
The loop on the left side has no voltage source. Hence,𝑣0 = 0. From (i), 𝑣𝑎 = 8 𝑉
1
To get VTh, consider the original circuit. For the left loop,𝑣0 = 4 × 8 = 2 𝑉
For the right loop,𝑣𝑅 = 𝑉𝑇ℎ = −192 𝑉
The resistance at the required resistor is𝑅 = 𝑅𝑇ℎ = 8 𝑘Ω
𝑝=
2
𝑉𝑇ℎ
4𝑅𝑇ℎ
=
−192 2
4×8×10 3
= 1.152 𝑊
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SOLUTION:
We need the Thevenin equivalent across the resistor R.To find RTh, consider the circuit below,
=>
10 kΩ
40 kΩ = 8 kΩ
𝑣
𝑣
𝑣
1 + 3𝑣0 = 301 + 301 = 151
8
But, 𝑣0 = 30 𝑣1 , hence, 15 +
𝑅𝑇ℎ =
𝑣1
1
And
8 kΩ + 22 kΩ = 30 kΩ
=>
15 + 45𝑣0 = 𝑣1
45×8
𝑣1 , which
30
leads to 𝑣1 = 1.3636 𝑉
= −1.3636 𝑘Ω
RThbeing negative indicates an active circuit and if you now make R equal to 1.3636 kΩ, then the
active circuit will actually try to supply infinitepower to the resistor. Thecorrect answer is
therefore:
𝑝𝑅 =
2
𝑉 𝑇ℎ
1363.6
−1363 .6+1363.6
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=
𝑉 𝑇ℎ 2
1363.6
0
=∞
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SOLUTION:
IN =2.5A
20i2 = 80
I = 2A
By current division
RN
2 = 2.5
R N + 20
Solving RN = RTH = 80Ω
Thus VTH = VOC = 2.5 X 80 = 200V
Pmax = Vth2/4Rth = 125W
RL=Rth=80Ω
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