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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
In Mesh 1: –4 + 400i + 300i – 300i – 1 = 0
1
1
2
In Mesh 2: 1 + 500i – 300i +2 – 2 = 0
2
Solving two equations
i1 = 5.923 mA and i2 = -2.846 mA
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
a. Define a clockwise mesh current i in the left-most mesh; a clockwise mesh current i
1
2
in the central mesh, and note that i can be used as a mesh current for the remaining
y
mesh.
Mesh 1: -10 + 7i1– 2i = 0
2
Mesh 2: -2i1 + 5i2 = 0
Mesh y: -2i2+ 9i = 0
y
Solving the three equations
i1 1.613A, iy=143.4mA
b. The power supplied by the 10 V source is (10)(i ) = 10(1.613) = 16.13 W
1
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
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SOLUTION:
By inspection, no current flows through the 2 Ω resistor, so i1= 0
Node A:
2=
Node B:
-2 =
VA
3
VB
6
+
+
VA −VB
VB
6
1
+
VB −VA
1
Solving, V = 0.8571 V and V = -0.8571 V
A
B
Thus v1 = 1.714V
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Two required nodal equations:
v
V1 −v2
2
v
3
v −v1
1
1
Node 1: 1= 1 +
Node 2: -3 = 2 + 2
Which modifies to
5v1-2v2 = 6
-v1+4v2=-9
Solving we find v1= 333.3mV
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us choose the bottom node as reference node, as then v will automatically become
x
a nodal voltage.
V
V −V
V −V
Node 1: 4 = 1 + 1 2 + 1 x
100
20
Vx −V1
Node 2: 10 -4-(-2) =
V
50
50
V −V
+
Vx −V2
40
V −V
Node 3: -2 = 2 + 2 x + 2 1
25
40
20
Simplifying
4 = 0.0800v – 0.0500v – 0.0200v
1
2
x
8 = -0.0200v – 0.02500v + 0.04500v
1
2
-2 = -0.0500v + 0.1150v – 0.02500v
1
2
Solving
v = 397.4 V
x
Chapter 3: Nodal and Loop Analysis Techniques
x
x
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The bottom node has the largest number of branch connections, so we choose that as
our reference node. This also makes v easier to find, as it will be a nodal voltage.
P
Working from left to right, we name our nodes 1, P, 2, and 3
Node1: 10 =
Node2: 0 =
V1
+
V1 −Vp
20
40
Vp −V1
Vp
40
Node 3: -2.5 + 2 =
+
100
V2 −Vp
50
V
Node 4: 5 – 2 = 3 +
200
Solving
60v - 20v = 8000 [1]
1
+
+
Vp −V2
50
V2 −V3
V3 −V2
10
10
P
-50v + 110 v - 40v = 0 [2]
1
P
2
- v + 6v - 5v = -25 [3]
P
2
3
-200v + 210v = 6000
2
3
Thus, v = 171.6 V
P
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
The supernode contains the 2-V source, nodes 1 and 2, and the 10-Ω resistor.Applying KCL to
the supernode as shown in the figure below gives,
2 = 𝑖1 + 𝑖2 + 7
Expressing 𝑖1 and𝑖2 in terms of the node voltages, we get,
2=
𝑣1 − 0
2
+
𝑣2 − 0
4
+7
=>
8 = 2𝑣1 + 𝑣2 + 28
Or,
𝑣2 = −20 − 2𝑣1
(i)
To get the relationship between 𝑣1 and 𝑣2 , we apply KVL to the circuit shown in figure below.
Going around the loop, we obtain,
−𝑣1 − 2 + 𝑣2 = 0
=>
𝑣2 = 𝑣1 + 2
(ii)
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Irwin, Engineering Circuit Analysis, 11e ISV
From (i) and (ii), we get,
𝑣1 = −7.333 𝑉
Chapter 3: Nodal and Loop Analysis Techniques
And
𝑣2 = −5.333 𝑉.
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We need to concern ourselves with the bottom part of this circuit only. Writing a single
nodal equation.
-4 + 2 = v/50
v = 100V
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
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Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCLto the two supernodes as
represented in figure below.
At supernode 1-2,
𝑖3 + 10 = 𝑖1 + 𝑖2
Expressing this in terms of the node voltages,
𝑣3 − 𝑣2
6
+ 10 =
𝑣1 − 𝑣4
5
+
𝑣1
2
Or,
5 𝑣1 + 𝑣2 − 𝑣3 − 2𝑣4 = 60
At supernode 3-4,
𝑖1 = 𝑖3 + 𝑖4 + 𝑖5
=>
Chapter 3: Nodal and Loop Analysis Techniques
𝑣1 − 𝑣4
3
𝑣 − 𝑣2
= 3
6
+
𝑣4
1
+
𝑣3
4
(i)
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Irwin, Engineering Circuit Analysis, 11e ISV
Or,
4𝑣1 + 2𝑣2 − 5𝑣3 − 16𝑣4 = 0
(ii)
We now apply KVL to the branches involving the voltage sources as shown in the figure below,
For loop 1,
−𝑣1 + 20 + 𝑣2 = 0
=>
𝑣1 − 𝑣2 = 20
(iii)
For loop 2,
−𝑣3 + 3 𝑣𝑥 + 𝑣4 = 0
But, 𝑣𝑥 = 𝑣1 − 𝑣4 , so that,
3𝑣1 − 𝑣3 − 2𝑣4 = 0
(iv)
For loop 3,
𝑣𝑥 − 3𝑣𝑥 + 6𝑖3 − 20 = 0
But, 6𝑖3 = 𝑣3 − 𝑣2 and𝑣𝑥 = 𝑣1 − 𝑣4 . Hence,
−2𝑣1 − 𝑣2 + 𝑣3 + 2𝑣4 = 20
(v)
We have four variables but 5 equations. Thus one extra equation can be used to check the results.
Solving equations from (i) to (v) using substitution and elimination method or using Cramer’s rule,
we get,𝑣1 = 26.67 𝑉
𝑣2 = 6.67 𝑉, 𝑣3 = 173.33 𝑉
𝑣4 = −46.67 𝑉
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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SOLUTION:
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SOLUTION:
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SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
The circuit in this example has three non-reference nodes. We assign voltages to the three
nodes as shown in the figure below and label the currents.
At node 1,
3 = i1 + ix
=>
𝑣 − 𝑣3
3= 1
4
𝑣1 − 𝑣2
+
2
Multiplying by 4 and rearranging terms, we get
3 v1 – 2 v2 + v3 = 12
(i)
At node 2,
ix= i2 + i3
=>
𝑣1 − 𝑣2
2
=
Multiplying by 8 and rearranging terms, we get,
Chapter 3: Nodal and Loop Analysis Techniques
𝑣2 − 𝑣3
8
+
𝑣2 − 0
4
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Irwin, Engineering Circuit Analysis, 11e ISV
-4v1 + 7 v2 – v3 = 0
(ii)
At node 3,
i1 + i2 = 2 ix
=>
𝑣1 − 𝑣3
4
𝑣 − 𝑣3
+ 2
8
𝑣 − 𝑣2
= 2( 1
2
)
Multiplying by 8, rearranging terms, and dividing by 3, we get,
2 v 1 – 3 v2 + v 3 = 0
(iii)
We have three simultaneous equations (i), (ii) and (iii) to solve to get the node voltages v1, v2
and v3. Thus, we find –
v1 = 4.8 V,
v2 = 2.4 V
Chapter 3: Nodal and Loop Analysis Techniques
and,
v3 = -2.4 V
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We begin by naming the top left node “1”, the top right node “2”, the bottom node of
the 6-V source “3” and the top node of the 2-Ω resistor “4.” The reference node has
already been selected, and designated using a ground symbol.
By inspection, v2 = 5V
Forming a supernode with nodes 1 and 3, we find
V
V −5
At the supernode:
-2 = 3 + 1
1
V4
10
V4 −5
At node 4:
2= +
2
4
Our supernode KVL equation, v1 − v3 = 6
Rearranging, simplifying and collecting terms,
v1 + 10v3 = −20 + 5 = −15
v1 − v3 = 6
Thus
v4 = 4.333V, v1 = 4.091V and v3 = −1.909V.
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
𝑉 −𝑉𝑐
𝐼1 = 𝑎
From the figure,
2
Assume that the currents are moving away from the node. Applying KCL at Node a,
𝑉𝑎
1
𝑉 −𝑉𝑐
+ 𝑎
2
𝑉 −2−𝑉𝑏
+ 𝑎
2
=4
(i)
Applying KCL at node b,
𝑉𝑏 +2−𝑉𝑎
2
𝑉 −𝑉𝑐
+ 𝑏
3
= 2𝐼1
(ii)
Applying KCL at node c,
𝑉𝑐 −𝑉𝑏
3
𝑉
+ 𝑐 = 𝐼1
=>
5
𝑉𝑐 −𝑉𝑏
3
𝑉
𝑉 −𝑉𝑐
5
2
+ 𝑐= 𝑎
Solving equations (i), (ii) and (iii), we get,
𝑉𝑎 = 4.303 𝑉
𝑉𝑏 = 3.87 𝑉
Chapter 3: Nodal and Loop Analysis Techniques
𝑉𝑐 = 3.33 𝑉
(iii)
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
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SOLUTION:
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SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us define the nodes as
Sol:
V −1.3
Node 1: -2 x 10-3 = 1 3
1.8 ×10
2,4 supernode
2.3 × 103 =
v4 − 1.3
v2 − v5
v4 − v5
v4
+
+
+
7.3 × 103 1 × 103 1.3 × 103 1.5 × 103
KVL equation: −v2 + v4 = 5.2
Node 5: 0 =
v5 −v2
1×103
+
v5 −v4
1.3×103
+
v5 −2.6
6.3×103
Simplifying and collecting terms,
14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275
Chapter 3: Nodal and Loop Analysis Techniques
[1]
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Irwin, Engineering Circuit Analysis, 11e ISV
-v + v4 = 5.2
[2]
-8.19 v – 6.3 v4 + 15.79 v5 = 3.38
[3]
2
2
Solving, we find the voltage at the central node is v = 3.460 V
4
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Define a voltage v at the top node of the current source I , and a clockwise mesh
x
2
current ib in the right-most mesh.
We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A.
Applying nodal analysis to the circuit,
vx − v1
I1 + I2 =
=1
6
so our requirement is I + I = 1. There is no constraint on the value of v1 other than
1
2
we are told to select a nonzero value
Thus we choose I1 = I2 = 500mA and v1 = 3.1415 V.
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
(*See Next Page)
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SOLUTION:
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Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
First, represent the resistor currents in terms of the node voltages as shown below,
Apply at KCL at node 1 to get,
𝑣1
50
𝑣 −𝑣2
+ 1
65
𝑣 −60
+ 1
80
=0
1
=>
(
=>
−
50
+
1
65
+
1
80
1
60
65
80
) 𝑣1 − ( ) 𝑣2 =
Apply KCL at node 2 to get
𝑣2 −𝑣1
65
𝑣 −60
+ 2
Solving, we get,
75
= 0.1
𝑣1 = 30.81𝑉
Chapter 3: Nodal and Loop Analysis Techniques
1
1
1
𝑣 + ( + ) 𝑣2 = 0.1
65 1
65
75
and
𝑣2 = 47.990𝑉
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
With the reference terminal already specified, we name the bottom terminal of the
3-mA source node “1,” the left terminal of the bottom 2.2-kΩ resistor node “2,” the
top terminal of the 3-mA source node “3,” the “+” reference terminal of the 9-V
source node “4,” and the “-” terminal of the 9-V source node “5.
Since we know that 1 mA flows through the top 2.2-kΩ resistor, v = -2.2 V.
5
Also, we see that v – v = 9, so that v = 9 – 2.2 = 6.8 V.
4
5
4
Proceeding with nodal analysis,
At node 1, −3 × 10−3 =
At node 2, 0 =
v2 −v1
2.2 ×103
3
+
V1
10×103
v2 −v3
+
4.7 ×103
3
At node 3, 1 × 10 + 3 × 10 =
V1 −V2
2.2×103
V3
3.3×103
+
Solving v1 = −8.614V, v2 = −3.909V, v3 = 6.143V
Chapter 3: Nodal and Loop Analysis Techniques
V3 −V2
4.7×103
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
If v = 0, the dependent source is a short circuit and we may redraw the circuit as
1
v
v − 96
40
20
At node 1: 1 + 1
v −V2
+ 1
10
Since v1 = 0
-2 = −
96
20
−
V2
10
So that V2 = -28V
Chapter 3: Nodal and Loop Analysis Techniques
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let𝑣𝑎 denote the node voltage at node a,𝑣𝑏 denotethe node voltage at node b, and 𝑣𝑐 denote
the nodevoltage at node c. Apply KCL at node a to obtain,
𝑣 −𝑣𝑏
−( 𝑎
𝑅1
𝑣 −𝑣𝑐
) + 𝑖1 − ( 𝑎
𝑅2
𝑣 −𝑣𝑏
) + 𝑖2 − ( 𝑎
𝑅5
)=0
Separate the terms of this equation that involve 𝑣𝑎 from the terms that involve 𝑣𝑏 and the terms
that
Involve𝑣𝑐 to obtain,
(
1
𝑅1
+
1
𝑅2
+
1
𝑅5
1
1
𝑅5
𝑅1
) 𝑣𝑎 − ( ) 𝑣𝑏 − (
+
1
𝑅2
) 𝑣𝑐 = 𝑖1 + 𝑖2
Apply KCL at node b to obtain,
𝑣 −𝑣𝑏
−𝑖1 + ( 𝑎
𝑅5
𝑣 −𝑣𝑐
)−( 𝑏
𝑅3
𝑣
) − ( 𝑏 ) + 𝑖3 = 0
𝑅4
Separate the terms of this equation that involve𝑣𝑎 from the terms that involve𝑣𝑏 and the terms
that involve 𝑣𝑐 to obtain,
1
1
𝑅5
𝑅3
− ( ) 𝑣𝑎 + (
+
1
𝑅4
+
1
𝑅5
1
) 𝑣𝑏 − ( ) 𝑣𝑐 = 𝑖3 − 𝑖2
𝑅3
Similarly, we can write the node equation at node c:
−(
1
𝑅1
+
1
𝑅2
1
1
𝑅3
𝑅1
) 𝑣𝑎 − ( ) 𝑣𝑏 + (
Chapter 3: Nodal and Loop Analysis Techniques
+
1
𝑅2
+
1
𝑅3
+
1
𝑅6
) 𝑣𝑐 = −𝑖1
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Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We apply KVL to the three meshes in turn. For mesh 1,
−24 + 10(𝑖1 − 𝑖2 ) + 12 (𝑖1 − 𝑖3 ) = 0
=>
11𝑖1 − 5𝑖2 − 6𝑖3 = 12
(i)
For mesh 2,
24𝑖2 + 4(𝑖2 − 𝑖3 ) + 10 (𝑖2 − 𝑖1 ) = 0
−5𝑖1 + 19𝑖2 − 2𝑖3 = 0
Or,
(ii)
For mesh 3,
4𝐼0 + 12(𝑖3 − 𝑖1 ) + 4(𝑖3 − 𝑖2 ) = 0
But at node A, 𝐼0 = 𝑖1 − 𝑖2 , so that
4(𝑖1 − 𝑖2 ) + 12(𝑖3 − 𝑖1 ) + 4(𝑖3 − 𝑖2 ) = 0
Or,
(iii)
−𝑖1 − 𝑖2 + 2𝑖3 = 0
Solving equations (i), (ii), and (iii) simultaneously, we get,
𝑖1 = 2.25 𝐴
𝑖2 = 0.75 𝐴
Chapter 3: Nodal and Loop Analysis Techniques
and,
𝑖3 = 1.5 𝐴
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Thus, 𝐼0 = 𝑖1 − 𝑖2 = 1.5 𝐴
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
The current through the 2 Ω resistor is i
1
Mesh 1:
5i – 3i = 0
Mesh 2:
–212 +8i –3i = 0
Mesh 3:
8i – 5i + 122 = 0
1
2
2
3
1
2
Solving, i = 20.52 A, i = 34.19 A and i = 6.121 A
1
2
3
The current through the 5 Ω resistor is i , or 6.121 A.
3
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SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Let us define four clockwise mesh currents. The top mesh current is labeled i . The
4
bottom left mesh current is labeled i , the bottom right mesh current is labeled i , and
1
3
the remaining mesh current is labeled i . Define a voltage “v ” across the 4-A current
2
4A
source with the “+” reference terminal on the left
Now i = 5 A and i = i4
3
a
MESH 1:
MESH 2:
-60 + 2i – 2i + 6i = 0
-6i + v + 4i – 4(5) = 0
MESH 4:
2i – 2i + 5i + 3i – 3(5) – v = 0
4
4
4A
2
1
4
4
4A
Also,
i2– i4 = 4
Solving these four equations
i = 16.83 A, i = 10.58 A, i = 6.583 A and v = 17.17 V.
1
2
4
4A
Thus, the power dissipated by the 2-Ω resistor is (i1-i4)2.2 = 210W
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
We begin our analysis by defining three clockwise mesh currents. We will call the top
mesh current i3, the bottom left mesh current i1, and the bottom right mesh current i2
Now
i =5A
[1]
1
i = -0.01 v
2
MESH 3:
[2]
1
50 i + 30 i – 30 i + 20 i – 20 i = 0 [3]
3
3
2
3
1
These three equations are insufficient, however, to solve for the unknowns. It would be
nice to be able to express the dependent source controlling variable v in terms of the
1
mesh currents. Returning to the diagram, it can be seen that KVL around mesh 1 will
yield
-v1 + 20i1 - 20i3 + 4v1=0
Or v = (20(5)/ 0.6 - 20 i3 / 0.6
[4]
1
Substituting Eq. [4] into Eq. [2] and then the modified Eq. [2] into Eq. [3], we find
20(5) – 30(-0.01)(20)(5)/0.6 + 30(-0.01)(20)i3/6 + 100i5 =0
Solving, we find that i3 = 555.6mA
Thus v = -1.481 A, i2 =1.481V and the power generated by the dependent voltage source
1
is
0.4 v (i – i ) = -383.9 W
1
2
1
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Note that meshes 1 and 2 form a supermesh since they have anindependent current source in
common. Also, meshes 2 and 3 formanother supermesh because they have a dependent current
source in common. The two supermeshes intersect and form a larger supermesh as
shown.Applying KVL to the larger supermesh,
2𝑖1 + 4𝑖3 + 8(𝑖3 − 𝑖4 ) + 6𝑖2 = 0
Or,
(i)
𝑖1 + 3𝑖2 + 6𝑖3 − 4𝑖4 = 0
For the independent current source, we apply KCL to node P
𝑖2 = 𝑖1 + 5
(ii)
For the dependent current source, we apply KCL to node Q
𝑖2 = 𝑖3 + 3𝐼0
But 𝐼0 = −𝑖4, hence,
𝑖2 = 𝑖3 − 3𝑖4
(iii)
Applying KVL in mesh 4,
2𝑖4 + 8(𝑖4 − 𝑖3 ) + 10 = 0
Or,
5𝑖4 − 4𝑖3 = −5
Chapter 3: Nodal and Loop Analysis Techniques
(iv)
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Solving equations (i) to (iv) simultaneously,
𝑖1 = −7.5 𝐴002,
𝑖2 = −2.5𝐴
Chapter 3: Nodal and Loop Analysis Techniques
𝑖3 = 3.93 𝐴,
𝑖4 = 2.1.43 𝐴
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SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
We define a clockwise mesh current i in the upper right mesh, a clockwise mesh
3
current i in the lower left mesh, and a clockwise mesh current i in the lower right mesh.
1
2
MESH 1:
MESH 2:
-6 + 6 i1 - 2 = 0
2 + 15 i – 12 i – 1.5 = 0
[1]
[2]
MESH 3:
i = 0.1 v
[3]
2
3
3
x
Eq. [1] may be solved directly to obtain i = 1.333 A.
1
It would help in the solution of Eqs. [2] and [3] if we could express the dependent source
controlling variable v in terms of mesh currents. Referring to the circuit diagram, we see
x
that v = (1)( i1) = i1, so Eq. [3] reduces to
i = 0.1 v = 0.1 i = 133.3 mA
3
x
As a result, Eq. [1] reduces to
i = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA
2
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Let us define currents i1, i2, i3, i4 in the left lower, left upper, right upper and right lower
loop respectively.
MESH 1:
–V + 9i – 2i – 7i = 0
z
1
2
4
MESH 2:
–2i + 7i – 5i = 0
MESH 3:
Vx - 5i2 + 8i – 3i = 0
MESH 4:
V – 7i – 3i + 10i = 0
1
2
3
3
y
1
4
3
4
Rearranging and setting i – i = 0, i – i = 0, i – i = 0 and i – i = 0,
1
2
2
3
1
4
4
3
9i - 2i -7i = V
1
2
4
z
-2i + 7i - 5i = 0
1
2
3
-5i2 + 8i3 – 3i = - V
4
x
-7i -3i + 10i = - V
1
3
4
y
Since i1 = i2 = i3 = i4, these equations produce
Vz=0
0=0
-Vx=0
-Vy=0
This is a unique solution. Therefore, the request that nonzero values be found cannot be
satisfied
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Starting with the left-most mesh and moving right, we define four clockwise mesh
currents i , i , i and i . We see that i = 2 mA.
1
2
3
4
MESH 2: -10 + 5000i + 4 + 1000i = 0
[1]
MESH 3: -1000i + 6 + 10,000 – 10,000i = 0 [2]
3
4
MESH 4: i = -0.5i
4
[3]
2
Reorganising, we find
5000 i + 1000 i = 6
[1]
9000 i – 10,000 i = -6
[2]
0.5 i + i = 0
[3]
2
3
3
2
4
4
On solving we find
i = 2mA, i = 1.5mA, i = -1.5mA and i = -0.75mA
1
The power generated by each source is:
P = 5000(i – i )(i ) = 5mW
2mA
1
2
1
P = 4 (-i ) = -6mW
4V
2
P = 6 (-i ) = 9mW
6V
3
P
= 1000 i (i – i ) = 4.5mW
P
= 10,000(i – i )(0.5 i) = -5.625mW
depV
depI
3
3
3
2
4
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
The circuit in Fig. 3.27 has four non-reference nodes, so we need fournode equations. This
implies that the size of the conductance matrixG, is 4 by 4. The diagonal terms of G, in Siemens,
are
𝐺11 = 0.2 + 0.1 = 0.3,
1
1
1
8
8
4
𝐺22 =
𝐺33 = + + = 0.5
1
5
1
1
8
1
+ +
1
1
1
8
2
1
= 1.325
𝐺44 = + + = 1.625
The off-diagonal terms are,
1
𝐺12 = − = −0.2,
𝐺13 = 𝐺14 = 0
5
1
1
𝐺21 = −0.2
𝐺23 = − = −0.125
𝐺24 = − = 1
𝐺31 = 0
𝐺32 = − = −0.125
𝐺34 = − = −0.125
𝐺41 = 0
𝐺42 = −1
𝐺43 = −0.125
8
1
1
1
8
8
The input current vector i has the following terms, in amperes
𝑖1 = 3
𝑖2 = −1 − 2 = −3
Thus the node-voltage equations are,
Chapter 3: Nodal and Loop Analysis Techniques
𝑖3 = 0
𝑖4 = 2 + 4 = 6
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0.3
−0.2
−0.2 1.325
[
0
−0.125
0
−1
0
−0.125
0.5
−0.125
Which can be solved to obtain 𝑣1 , 𝑣2 , 𝑣3 and 𝑣4
Chapter 3: Nodal and Loop Analysis Techniques
𝑣1
0
3
𝑣
−1
−3
2
][ ] = [ ]
−0.125 𝑣3
0
1.625 𝑣4
6
119
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
We have five meshes, so the resistance matrix is 5 by 5. The diagonalterms, in ohms, are:
𝑅11 = 5 + 2 + 2 = 9
𝑅22 = 2 + 4 + 1 + 1 + 2 = 10
𝑅33 = 2 + 3 + 4 = 9
𝑅44 = 1 + 3 + 4 = 8
𝑅55 = 1 + 3 = 4
The off-diagonal terms are:
𝑅12 = −2
𝑅13 = −2
𝑅14 = 0 = 𝑅15
𝑅21 = −2
𝑅23 = −4
𝑅24 = −1
𝑅31 = −2
𝑅32 = −4
𝑅34 = 0 = 𝑅35
𝑅41 = 0
𝑅42 = −1
𝑅43 = 0
𝑅45 = −3
𝑅51 = 0
𝑅52 = −1
𝑅53 = 0
𝑅54 = −3
The input voltage vector vhas the following terms in volts:
𝑣1 = 4
𝑣2 = 10 − 4 = 6
𝑣3 = −12 + 6 = −6
𝑣4 = 0
Thus, the mesh-current equations are:
Chapter 3: Nodal and Loop Analysis Techniques
𝑣5 = −6
𝑅25 = −1
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9
−2
−2
0
[0
−2
10
−4
−1
−1
−2
−4
9
0
0
4
0
0 𝑖1
6
−1 −1 𝑖2
0
0 𝑖3 = −6
0
8 −3 𝑖4
]
[
[
𝑖
]
−6]
−3 4
5
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
122
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Chapter 3: Nodal and Loop Analysis Techniques
123
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
124
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
125
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Chapter 3: Nodal and Loop Analysis Techniques
126
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
127
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Chapter 3: Nodal and Loop Analysis Techniques
128
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SOLUTION:
MESH 1: -7 + i1 – i2 = 0
MESH 2: i – i + 2i + 3i – 3i = 0
[2]
MESH 3: 3i – 3i + Xi3 +2i3 – 7 = 0
[3]
2
1
3
2
2
3
2
Grouping terms, we find that
i –i =7
[1]
1
2
-i1 + 6i2 -3i3 = 0
-3i + (5 + X)i = 7
2
3
[2]
[3]
Also i2 = 2.273A
Thus solving these equations
7−3i2 +5i3
x=
= 4.498Ω
i3
Chapter 3: Nodal and Loop Analysis Techniques
[1]
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SOLUTION:
We define a mesh current i in the left-hand mesh, a mesh current i in the top right
a
1
mesh, and a mesh current i in the bottom right mesh (all flowing clockwise).
The left-most mesh can be analysed separately to determine the controlling voltage va ,
as KCL assures us that no current flows through either the 1-Ω or 6-Ω resistor
Thus, -1.8 + 3ia – 1.5 + 2ia = 0, which may be solved to find ia = 0.66 A. Hence,
va = 3ia = 1.98 V
Forming one supermesh from the remaining two meshes, we may write:
-3 + 2.5 i + 3 i + 4 i = 0
1
2
2
And the supermesh KCL equation: i – i = 0.5 v = 0.5(1.98) = 0.99
2
1
a
Thus, we have two equations to solve:
2.5 i + 7 i = 3
1
2
-i1 + i2 = 0.99
Solving, we find that i = -413.7 mA and the voltage across the 2.5-Ω resistor
1
(Arbitrarily assuming the left terminal is the “+” reference) is 2.5 i1 = -1.034 V.
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
140
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
141
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SOLUTION:
Assigning clockwise currents in three meshes, we get,
From the figure,
𝐼𝑥 = 𝐼1
𝐼𝑦 = 𝐼2 − 𝐼3
But 𝐼3 = −1 𝐴, so
𝐼𝑦 = 𝐼2 + 1
Applying KVL to Mesh 1,
5 − 𝐼1 − 𝐼𝑦 − (𝐼1 − 𝐼2 ) = 0
=>
=>
𝐼1 = 2 𝐴
Applying KVL to Mesh 2,
−(𝐼2 − 𝐼1 ) + 𝐼𝑦 − 𝐼2 − 𝐼𝑥 − (𝐼2 − 𝐼3 ) = 0
Chapter 3: Nodal and Loop Analysis Techniques
−𝐼1 − 𝐼2 − 𝐼1 + 𝐼2 = −5 + 1
142
Irwin, Engineering Circuit Analysis, 11e ISV
=>
−𝐼2 + 𝐼1 + (𝐼2 + 1) − 𝐼2 − 𝐼1 − 𝐼2 + 𝐼3 = 0
=>
−2𝐼2 + 1 − 1 = 0
Chapter 3: Nodal and Loop Analysis Techniques
=>
𝐼2 = 0
143
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
144
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us define three clockwise mesh currents: i1 in the bottom left mesh, i2 in the top
mesh, and i3 in the bottom right mesh.
In mesh 1:
I1 = 5mA
Supermesh : i – i = 0.4 i
1
2
10
i – i = 0.4(i – i )
1
2
3
2
i – 0.6 i – 0.4 i = 0
1
Mesh 3:
2
3
-5000 i – 10000 i + 35000 i = 0
1
Simplify: 0.6 i + 0.4 i = 5×10-3
2
3
2
-10000i2 +35000i3 = 25
Solving, we find i = 6.6 mA and i = 2.6 mA. Since i = i – i , we find that
2
3
i = -4 mA
10
Chapter 3: Nodal and Loop Analysis Techniques
10
3
2
145
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us define four clockwise mesh currents, starting with i1 in the left-most mesh, then
i2, i3 and i4 moving towards the right.
Mesh 1:
-0.8ix + (2 + 5)i1 – 5i2 = 0
[1]
Mesh 2:
i =1A
[2]
2
Mesh 3:
(3 + 4) i – 3(1) – 4(i4) = 0
[3]
Mesh 4:
(4 + 3) i – 4 i – 5 = 0
[4]
3
4
3
Simplify and collect terms, noting that ix = i1 – i2 = i1 – 1
Thus, [3] and [4] become:
7i –4i =3
3
4
-4 i3 + 7i4 = 5
Solving, we find that i = 1.242 A and i = 1.424 A.
3
Chapter 3: Nodal and Loop Analysis Techniques
4
146
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Let us define four clockwise mesh currents i , i , i and i starting with the left-most
1
2
3
4
mesh and moving towards the right of the circuit.
At the 1,2 supermesh:
2000 i + 6000 i – 3 + 5000 i = 0
1
And
-3
i – i = 2×10
1
2
Also i4= -1mA
Solving, i1 = 1.923 mA and i2 = -76.92 μA
Thus, the voltage across the 2-mA source
v = -2000 i – 6000 (i – i ) = -15.85 V.
1
1
2
Chapter 3: Nodal and Loop Analysis Techniques
2
2
[1]
[2]
147
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
148
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
152
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Chapter 3: Nodal and Loop Analysis Techniques
153
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
159
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Chapter 3: Nodal and Loop Analysis Techniques
160
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
162
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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173
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174
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SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
175
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Chapter 3: Nodal and Loop Analysis Techniques
176
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SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
178
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
(See Next Page)
Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
181
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SOLUTION:
(See Next Page)
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Chapter 3: Nodal and Loop Analysis Techniques
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Chapter 3: Nodal and Loop Analysis Techniques
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SOLUTION:
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SOLUTION:
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Chapter 3: Nodal and Loop Analysis Techniques
189
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
190
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
191
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
192
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques
193
Irwin, Engineering Circuit Analysis, 11e ISV
SOLUTION:
Chapter 3: Nodal and Loop Analysis Techniques
194
Irwin, Engineering Circuit Analysis, 11e ISV
Chapter 3: Nodal and Loop Analysis Techniques