Chapter 3
METHODS OF ANALYSIS
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3.1 Introduction
Having understood the fundamental laws of circuit theory (Ohm’s law and
Kirchhoff’s laws), we are now prepared to apply these laws to develop two
powerful techniques for circuit analysis: nodal analysis, which is based on
a systematic application of Kirchhoff’s current law (KCL), and mesh
analysis, which is based on a systematic application of Kirchhoff’s voltage
law (KVL). The two techniques are so important that this chapter should
be regarded as the most important in the book. Students are therefore
encouraged to pay careful attention. With the two techniques to be
developed in this chapter, we can analyze almost any circuit by obtaining a
set of simultaneous equations that are then solved to obtain the required
values of current or voltage.
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3.2 Nodal Analysis
• Nodal analysis provides a general procedure for analyzing circuits using
node voltages as the circuit variables. Choosing node voltages instead of
element voltages as circuit variables is convenient and reduces the
number of equations one must solve simultaneously
• Nodal analysis is also known as the node-voltage method.
• In nodal analysis, we are interested in finding the node voltages.
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3.2 Nodal Analysis
𝐽3ሶ
• Consider the circuit in Figure. 3.1:
At node 1, applying KCL (K1) gives:
Y3
−𝐼1ሶ + 𝐼3ሶ − 𝐼4ሶ + 𝐽1ሶ − 𝐽3ሶ = 0
⟺ −𝐼1ሶ + 𝐼3ሶ − 𝐼4ሶ = −𝐽1ሶ + 𝐽3ሶ (1)
At node 2, applying KCL (K1) gives:
−𝐼2ሶ − 𝐼3ሶ + 𝐼4ሶ + 𝐽2ሶ + 𝐽3ሶ = 0
⟺ −𝐼2ሶ − 𝐼3ሶ + 𝐼4ሶ = −𝐽2ሶ − 𝐽3ሶ (2)
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Y4
1
𝐼1ሶ
𝐽1ሶ
𝐼3ሶ
2
𝐼2ሶ
𝐼4ሶ
Y1
Y2
3
Figure. 3.1
𝐽2ሶ
3.2 Nodal Analysis
Since 𝜑ሶ 1 , 𝜑ሶ 2 , 𝜑ሶ 3 are potential at node1, 2 and 3
The voltage across each the admittance are:
𝜑ሶ 1 −𝜑ሶ 3
ሶ
➢ 𝑌1 : 𝜑ሶ 1 − 𝜑ሶ 3 → 𝐼1 =
= (𝜑ሶ 1 −𝜑ሶ 3 ). 𝑌1
𝑍1
Note that: Current flows from a
higher potential to a lower
potential in a Impedance (resistor)
𝜑ሶ ℎ𝑖𝑔ℎ𝑒𝑟 − 𝜑ሶ 𝑙𝑜𝑤𝑒𝑟
𝐼ሶ =
𝑍
𝐽3ሶ
𝜑ሶ 2 −𝜑ሶ 3
ሶ
➢𝑌2 : 𝜑ሶ 2 − 𝜑ሶ 3 → 𝐼2 =
= (𝜑ሶ 2 −𝜑ሶ 3 ). 𝑌2
𝑍2
Y3
𝜑ሶ 2 −𝜑ሶ 1
ሶ
➢𝑌3 : 𝜑ሶ 2 − 𝜑ሶ 1 → 𝐼3 =
= (𝜑ሶ 2 −𝜑ሶ 1 ). 𝑌3
𝑍3
➢𝑌4 : 𝜑ሶ 1 − 𝜑ሶ 2 → 𝐼4ሶ =
𝜑ሶ 1 −𝜑ሶ 2
𝑍4
Y4
1
= (𝜑ሶ 1 −𝜑ሶ 2 ). 𝑌4
𝐼1ሶ
𝐽1ሶ
2
𝐼2ሶ
𝐼4ሶ
Y1
Y2
3
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𝐼3ሶ
𝐽2ሶ
3.2 Nodal Analysis
Node 3 is chosen as the reference node 3 and have zero potential (𝜑ሶ 3 = 0)
Eq(1): −𝐼1ሶ + 𝐼3ሶ − 𝐼4ሶ = −𝐽1ሶ + 𝐽3ሶ
⇔ −𝜑ሶ 1 . 𝑌1 + (𝜑ሶ 2 −𝜑ሶ 1 ). 𝑌3 − (𝜑ሶ 1 −𝜑ሶ 2 ). 𝑌4 = −𝐽1ሶ + 𝐽3ሶ
⇔ 𝜑ሶ 1 . 𝑌1 − (𝜑ሶ 2 −𝜑ሶ 1 ). 𝑌3 + (𝜑ሶ 1 −𝜑ሶ 2 ). 𝑌4 = +𝐽1ሶ − 𝐽3ሶ
⇔ (𝑌1 +𝑌3 + 𝑌4 ). 𝜑ሶ 1 − 𝑌3 + 𝑌4 . 𝜑ሶ 2 = 𝐽1ሶ − 𝐽3ሶ (3)
Eq(2): −𝐼2ሶ − 𝐼3ሶ + 𝐼4ሶ = −𝐽2ሶ − 𝐽3ሶ
⇔ (𝑌2 +𝑌3 + 𝑌4 ). 𝜑ሶ 2 − 𝑌3 + 𝑌4 . 𝜑ሶ 1 = 𝐽2ሶ + 𝐽3ሶ (4)
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Y3
𝐽1ሶ
Y4
1
𝐼1ሶ
⇔ −𝜑ሶ 2 . 𝑌2 − (𝜑ሶ 2 −𝜑ሶ 1 ). 𝑌3 + (𝜑ሶ 1 −𝜑ሶ 2 ). 𝑌4 = −𝐽2ሶ − 𝐽3ሶ
⇔ 𝜑ሶ 2 . 𝑌2 + (𝜑ሶ 2 −𝜑ሶ 1 ). 𝑌3 − (𝜑ሶ 1 −𝜑ሶ 2 ). 𝑌4 = +𝐽2ሶ + 𝐽3ሶ
𝐽3ሶ
𝐼3ሶ
2
𝐼2ሶ
𝐼4ሶ
Y1
Y2
𝐽2ሶ
3
We solve Eqs. (3) and (4)
to obtain the node
voltages 𝜑ሶ 1 and 𝜑ሶ 1
3.2 Nodal Analysis
Procedure of Nodal Analysis:
❑ Step 1: The first step in nodal analysis is selecting a node as the reference. The
reference node is commonly called the ground since it is assumed to have zero
potential.
❑Step 2: All the node voltages with respect to the ground from all the principal
nodes should be labelled except the reference node
❑Step 3: The nodal equations at all the principal nodes except the reference node
should have a nodal equation. The nodal equation is obtained from Kirchhoff’s
current law and then from Ohm’s law.
❑ Step 4: Solve the resulting simultaneous equations to obtain the unknown node
voltages.
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3.2 Nodal Analysis
At node i, applying KCL (K1) gives:
𝑌𝑖𝑖 × 𝜑ሶ 𝑖 − σ𝑛𝑗=1 𝑌𝑖𝑗 × 𝜑ሶ 𝑗 = σ𝑛𝑖=1 ±𝐽𝑖ሶ (i ≠ 𝑗)
✓ 𝑌𝑖𝑖 : Sum of the admittances of the impedance connected to the node i
✓ 𝜑ሶ 𝑖 : The voltage at the node i
✓ 𝑌𝑖𝑗 : is the total admittance of the impedance joining node i to node j.
✓ 𝜑ሶ 𝑗 : the voltage at the node j
✓σ𝑛𝑖=1 ±𝐽𝑖ሶ : Algebraic sum current source entering or leaving the node i
✓ +𝐽𝑖ሶ : if current source entering the node i
✓ −𝐽𝑖ሶ : if current source leaving the node i
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3.2 Nodal Analysis
Example 3.1: For the circuit shown in Fig. 3.2
a. Find the node voltages
b. Calculate the power supplied by each source.
c.
Check the power balance in the circuit
Ans:
𝜑1 = 4.8𝑉
𝜑2 = 2.4𝑉
𝜑3 = −2.4𝑉
Figure. 3.2
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3.2 Nodal Analysis
Example 3.2: By nodal analysis,find io in the
circuit in Fig.3.3
Fig. 3.3
Practice problem 3.2: By nodal analysis,
find 𝑉1ሶ , 𝑉ሶ2 in the circuit in Fig.3.4
Fig. 3.4
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3.2 Nodal Analysis
❑ We now consider how voltage sources affect nodal analysis.Consider
the following three possibilities:
➢ Case 2: If a voltage source is connected between the reference node
and a nonreference node, we simply set the voltage at the nonreference
node equal to the voltage of the voltage source
➢ Case 2: If the voltage source (dependent or independent) is connected
➢between two nonreference nodes, the two nonreference nodes form a
generalized node or supernode; we apply both KCL (K1) and KVL (K2)
to determine the node voltages
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3.2 Nodal Analysis
❑ We now consider how voltage sources affect nodal analysis.Consider the
following three possibilities:
➢ Case 1: if the voltage source in series with a impedance (or resistor) with an
equivalent network which has a current source in parallel with a impedance
(or resistor)
➢Case 2: If a voltage source is connected between the reference node and a
nonreference node, we simply set the voltage at the nonreference node equal
to the voltage of the voltage source
➢ Case 3: If the voltage source (dependent or independent) is connected
between two nonreference nodes, the two nonreference nodes form a
generalized node or supernode; we apply both KCL (K1) and KVL (K2) to
determine the node voltages
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3.2 Nodal Analysis
➢Case 3: We use the circuit in Fig. 3.5 (a) for illustration
✓Step 1: We write voltages equation between two node 2 and node 3 in Fig.3.5 (a)
𝑣2 − 𝑣3 = 5
Fig. 3.5(a)
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3.2 Nodal Analysis
➢Case 3: We use the circuit in Fig. 3.5 (a) for illustration
✓Step 2: Any element is connected between node-2 and node-3 can be
removed and circuit is redrawn and applying KCL as Fig. 3.5 (b)
Fig. 3.5(a)
Fig. 3.5(b)
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3.2 Nodal Analysis
➢Case 3: We use the circuit in Fig. 3.5(a) for illustration
✓Step 3: We now rewrite node voltages equation for node 2 and node 3 in Fig.
3.7 (b) as
At node 1:
1
𝑉2
2
1
6
At node 3: 𝑉3
+
1
8
1
𝑉1 ( )
2
−
+
1
4
− 𝑉1 ( ) = 0 (2)
= 0 (1)
1
4
Adding Eqs. (1) to (2) gives
1
𝑉2
2
+
1
8
+
1
𝑉3
6
+
1
4
−
1
𝑉1
2
+
1
4
=0
Fig. 3.5(b)
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3.2 Nodal Analysis
Example 3.3: For the circuit shown in Fig.3.6. Find 𝐼1ሶ , 𝐼2ሶ using nodal
analysis.
𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟
𝐼1ሶ = 0.97∠78.110 𝐴;
5Ω 𝜑ሶ 1 2Ω 𝜑ሶ 2
𝐼2ሶ = 0.74∠69.20 𝐴;
𝐼1ሶ
j12Ω
𝐼2ሶ
4Ω
0

0
10 V
𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑠𝑜𝑢𝑟𝑐𝑒
2- 30 A
0
9Ω
-j4Ω
𝜑ሶ 3 = 0
Fig. 3.6
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3.2 Nodal Analysis
Example 3.4: Use nodal analysis to determine voltage v1, v2, and v3 in the
circuit in Fig.3.7
Fig. 3.7
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3.2 Nodal Analysis
Example 3.5: Use nodal analysis to determine voltage v1, v2, v3 and v4 in the
circuit in Fig.3.8
Fig. 3.8
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3.2 Nodal Analysis
Example 3.6: For the circuit shown in Fig.3.9, find
a. 𝑉1ሶ , 𝑉ሶ2 , and 𝑉1ሶ using nodal analysis
b. Power of each element
c. Check the power balance in the circuit
Practice problem 3.6: For the circuit shown in Fig.3.10, find
Fig. 3.9
a. 𝑉1ሶ , 𝑉ሶ2 , using nodal analysis
b. Power of each element
c. Check the power balance in the circuit
Fig 3.10
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3.2 Nodal Analysis
Example 3.7: For the circuit shown in Fig. 3.11. Find voltage each node using
nodal analysis
Ans: 𝜑1 = 2𝑉, 𝜑2 = −2𝑉, 𝜑3 = 5𝑉, 𝜑4 = 2𝑉
1Ω
𝜑1 = 2𝑉
𝜑2
2Ω
+
U1
2V
-
I1
1Ω 𝜑3
I3
9A
𝐼
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𝜑4
(𝜑4 −𝜑3 )
2U1
𝜑5 = 0
3I2
-+
I2
2Ω
Fig. 3.11
3.2 Nodal Analysis
Practice problem 3.7: For the circuit in Fig.3.12. Fin𝑑 𝑉1ሶ , 𝑉2 using nodal
analysis thế nút.
𝑉1ሶ = 18,97∠18.430 𝑉;
𝑉ሶ2 = 13,91∠198,30 𝑉;
Fig. 3.12
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3.3 Mesh Analysis
Mesh analysis provides another general procedure for analyzing circuits,
using mesh currents as the circuit variables. Using mesh currents instead of
element currents as circuit variables is convenient and reduces the number of
equations that must be solved simultaneously. Recall that a loop is a closed
path with no node passed more than once. A mesh is a loop that does not
contain any other loop within it
Nodal analysis applies KCL(K1) to find unknown voltages in a given
circuit, while mesh analysis applies KVL(K2) to find unknown currents.
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3.3 Mesh Analysis
❑ Step to Determine Mesh Current
- In Mesh analysis, Identify all meshes of the circuit
- Assign mesh current 𝐼1ሶ , 𝐼2ሶ , … 𝐼𝑛ሶ to the n meshes
- The direction of the mesh current is arbitrary (clockwise or
counterclockwise)
- The current passing through an element is the algebraic sum of all mesh
currents passing through that element
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3.3 Mesh Analysis
Z1
Consider the circuit in Fig.3.13
- There are 2 meshes , assign mesh
ሶ , 𝐼𝑚𝑙2
ሶ
current 𝐼𝑚𝑙1
to the 2 meshes
Z2
𝐼1ሶ
𝐸ሶ1
ሶ
- The branch current 1: 𝐼1ሶ = 𝐼𝑚𝑙1
𝐼2ሶ
ሶ
𝐼𝑚𝑙1
Z3
𝐼3ሶ
ሶ
- The branch current 2: 𝐼2ሶ = −𝐼𝑚𝑙2
ሶ
ሶ
- The branch current 3: 𝐼3ሶ = 𝐼𝑚𝑙1
− 𝐼𝑚𝑙2
Chương 3 - Nguyễn Ngọc Thiêm
Fig. 3.13
ሶ
𝐼𝑚𝑙2
𝐸ሶ 2
3.3 Mesh Analysis
Z1
- Applying KVL(K2) to loop 1:
𝐼1ሶ 𝑍1 + 𝐼3ሶ 𝑍3 = 𝐸ሶ1 (1)
Z2
𝐼1ሶ
𝐸ሶ1
- Applying KVL(K2) to loop 2
𝐼2ሶ
ሶ
𝐼𝑚𝑙1
Z3
ሶ
𝐼𝑚𝑙2
𝐼3ሶ
−𝐼2ሶ 𝑍2 − 𝐼3ሶ 𝑍3 = −𝐸ሶ 2 (2)
Fig. 3.13
ሶ , 𝐼𝑚𝑙2
ሶ ,
Replace each branch current 𝐼1ሶ , 𝐼2ሶ , 𝐼3ሶ equal each mesh curent 𝐼𝑚𝑙1
respectively, we obtain
ሶ 𝑍1 + (𝐼𝑚𝑙1
ሶ −𝐼𝑚𝑙2
ሶ )𝑍3 = 𝐸ሶ1
1 ⟺ 𝐼𝑚𝑙1
ሶ 𝑍2 − (𝐼𝑚𝑙1
ሶ −𝐼𝑚𝑙2
ሶ )𝑍3 = −𝐸ሶ 2
2 ⟺ 𝐼𝑚𝑙2
Chương 3 - Nguyễn Ngọc Thiêm
𝐸ሶ 2
3.3 Mesh Analysis
Z1
ሶ 𝑍1 + (𝐼𝑚𝑙1
ሶ −𝐼𝑚𝑙2
ሶ )𝑍3 = 𝐸ሶ1
1 ⟺ 𝐼𝑚𝑙1
ሶ (𝑍1 +𝑍3 ) − 𝐼𝑚𝑙2
ሶ 𝑍3 = 𝐸ሶ1 (3)
⟺ 𝐼𝑚𝑙1
Z2
𝐼1ሶ
𝐸ሶ1
ሶ 𝑍2 − (𝐼𝑚𝑙1
ሶ −𝐼𝑚𝑙2
ሶ )𝑍3 = −𝐸ሶ 2
2 ⟺ 𝐼𝑚𝑙2
ሶ (𝑍2 +𝑍3 ) − 𝐼𝑚𝑙1
ሶ 𝑍3 = −𝐸ሶ 2 (4)
⟺ 𝐼𝑚𝑙2
ሶ
ሶ
Solving Eqs (3) and (4), we get 𝐼𝑚𝑙1
and 𝐼𝑚𝑙2
Chương 3 - Nguyễn Ngọc Thiêm
𝐼2ሶ
ሶ
𝐼𝑚𝑙1
Z3
𝐼3ሶ
Fig. 3.13
ሶ
𝐼𝑚𝑙2
𝐸ሶ 2
3.3 Mesh Analysis
❑ Step to Determine Mesh Current
- Determine the number of mesh currents required in a circuit (if a circuit has
n nodes, d branches, and l meshes: l = n-d+1)
- The direction of the mesh current
- Assign mesh current 𝐼1ሶ , 𝐼2ሶ , … 𝐼𝑛ሶ to the n meshes
- Apply KVL to each of the n meshes
ሶ 𝑍𝑖𝑖 + σ ±𝐼𝑚𝑙𝑗
ሶ 𝑍𝑖𝑗 = σ ±𝐸ሶ 𝑖 (i ≠j)
- Applying KVL to mesh i, we obtain: 𝐼𝑚𝑙𝑖
Chương 3 - Nguyễn Ngọc Thiêm
3.3 Mesh Analysis
ሶ : Mesh current in the i mesh
✓ 𝐼𝑚𝑙𝑖
✓ 𝑍𝑖𝑖 : Sum of the impedances in the i mesh
ሶ 𝑍𝑖𝑖 + σ ±𝐼𝑚𝑙𝑗
ሶ 𝑍𝑖𝑗 = σ ±𝐸ሶ 𝑖 (i ≠j)
𝐼𝑚𝑙𝑖
✓𝑍𝑖𝑗 : Sum of the impedances of the common impedance (or resistors) in the i and j meshes
ሶ : Mesh current in the j mesh
✓ 𝐼𝑚𝑙𝑗
✓σ ±𝐸ሶ 𝑖 : Algebraic sum of the voltage sources across the i mesh ( positive signs (+) are
assigned to those voltage source having a polarity such that the mesh current passes from the
negative to the positive terminal. A negative sign (−) is assigned to those voltage source
having a polarity such that the mesh current passes from the positive to the negative
terminal.)
ሶ 𝑍𝑖𝑗 : if direction of the mesh current 𝐼𝑚𝑙𝑗
ሶ
ሶ
✓+𝐼𝑚𝑙𝑗
in the same direction mesh current 𝐼𝑚𝑙𝑖
ሶ 𝑍𝑖𝑗 : if direction of the mesh current 𝐼𝑚𝑙𝑗
ሶ
ሶ
✓− 𝐼𝑚𝑙𝑗
in the opposite direction mesh current 𝐼𝑚𝑙𝑖
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3.3 Mesh Analysis
Example 3.8: Applying mesh analysis to find 𝐼𝐴 , 𝐼𝐵 , 𝐼𝐶 and the power each
source in the circuit of Fig.3.14
36V
4Ω
7
25
𝐼𝐴 = 2𝐴, 𝐼𝐵 = 𝐴, 𝐼𝐶 =
𝐴
6
6
𝑃12𝑉 =24W, 𝑃36𝑉 = 150W
𝐼3
𝐼2
𝐼𝐶
2Ω
5Ω
𝐼4
4Ω
𝐼1
12V
𝐼𝐴
10Ω
𝐼5
Fig. 3.14
Chương 3 - Nguyễn Ngọc Thiêm
𝐼𝐵
𝐼6
20Ω
3.3 Mesh Analysis
+ 𝑈ሶ 𝑥 −.
Example 3.9: Find 𝐼 ሶ in the
circuit of Fig.3.15 Ans: 𝐼 ሶ =
0.5Ω
𝐼1ሶ
𝐼ሶ
4∠00 (𝑚𝑉)
20∠590 𝑚𝐴
-+
.
𝐼1ሶ
(0.5-j0.8)Ω
6𝑈ሶ 𝑥
𝐼2ሶ
(2-j)Ω
Fig 3.15
Practice problem 3.9: Solve for io in
Fig 3.16
Fig.3.16 using mesh analysis
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3.3 Mesh Analysis
❑ Mesh Analysis with Current Sources
Applying mesh analysis to circuits containing current sources (dependent or
independent) may appear complicated. But it is actually much easier than what we
encountered in the previous section, because the presence of the current sources
reduces the number of equations. Consider the following three possible cases.
➢Case 1: if the voltage source in parallel with a impedance (or resistor) with an
equivalent network which has a voltage source in series with a impedance (or
resistor)
➢Case 2: When a current source exists only in one mesh, then 𝐼𝐾ሶ = ±𝐽 ሶ
- where: 𝐼𝐾ሶ is a mesh current in the K mesh, 𝐽 ሶ is current source, +𝐽:ሶ if if direction of
the mesh current𝐼𝐾ሶ in the same direction current source 𝐽,ሶ −𝐽:ሶ if if direction of the
mesh current 𝐼𝐾ሶ in the opposite direction current source 𝐽 ሶ
Chương 3 - Nguyễn Ngọc Thiêm
3.3 Mesh Analysis
❑ Mesh Analysis with Current Sources
➢Case 3: When a current source exists between two meshes: Consider the circuit in
Fig. 3.17(a), for example. We create a supermesh by excluding the current source
and any elements connected in series with it, as shown in Fig. 3.17(b). Thus
supermesh
Z1
𝐸ሶ1
Z2
𝐽1ሶ
ሶ
𝐼𝑚𝑙𝑖
Z4
ሶ
𝐼𝑚𝑙𝑘
ሶ
𝐼𝑚𝑙𝑗
Z3
Z1
𝐸ሶ 2
𝐸ሶ1
Z2
ሶ
𝐼𝑚𝑙𝑗
ሶ
𝐼𝑚𝑙𝑖
Z4
Z3
Z5
ሶ
𝐼𝑚𝑙𝑘
Fig. 3.17(a)
Chương 3 - Nguyễn Ngọc Thiêm
Z5
Fig. 3.17(b)
𝐸ሶ 2
3.3 Mesh Analysis
❑ Mesh Analysis with Current Sources
Case 3: A supermesh results when two meshes have a (dependent or independent)
current source in common.
applying KVL to the supermesh in Fig. 3.23(b) gives
ሶ 𝑍1 + 𝐼𝑚𝑙𝑗
ሶ 𝑍2 + 𝑍2 𝐼𝑚𝑙𝑗
ሶ − 𝐼𝑚𝑙𝑘
ሶ
ሶ − 𝐼𝑚𝑙𝑘
ሶ
𝐼𝑚𝑙𝑖
+ 𝑍4 𝐼𝑚𝑙𝑖
= 𝐸ሶ1 − 𝐸ሶ 2
Z1
𝐸ሶ1
Z2
ሶ
𝐼𝑚𝑙𝑗
ሶ
𝐼𝑚𝑙𝑖
Z4
Fig. 3.17(b)
Chương 3 - Nguyễn Ngọc Thiêm
supermesh
ሶ
𝐼𝑚𝑙𝑘
Z3
Z5
𝐸ሶ 2
3.3 Mesh Analysis
❑ Mesh Analysis with Current Sources
➢Case 3: One additional constraint equations are necessary. These can be
determined by the requirement that the algebraic sum of the mesh currents passing
through a current source must equal the current provided by the source. Thus, we
obtain:
Z1
ሶ − 𝐼𝑚𝑙𝑗
ሶ
𝐼𝑚𝑙𝑖
= 𝐽1ሶ
𝐸ሶ1
Z2
𝐽1ሶ
ሶ
𝐼𝑚𝑙𝑖
Z4
ሶ
𝐼𝑚𝑙𝑘
ሶ
𝐼𝑚𝑙𝑗
𝐸ሶ 2
Z3
Z5
Fig. 3.17(a)
Chương 3 - Nguyễn Ngọc Thiêm
3.3 Mesh Analysis
Example 3.10: For the circuit shown in Fig.3.18. Find 𝑢0 (𝑡) using mesh
analysis, given that 𝑡 𝑗 𝑡 = 10 cos 2𝑡 + 300 (𝐴)
Ans: 𝑈0 𝑡 = 6.4676 cos 2𝑡 + 440 (𝑉)
1
2F
1Ω
1Ω
2
+
Current source
j(t)
1Ω
2H
resistor
parallel
Fig. 3.18
Chương 3 - Nguyễn Ngọc Thiêm
U0(t)
2Ω
-
i0(t)
3.3 Mesh Analysis
Practice problem 3.10: Find current 𝐼0ሶ in the circuit in Fig.3.19 using mesh
analysis
current source exists only in one mesh
Ans: 𝐼0ሶ = 1,194∠65,450 𝐴
Fig. 3.19
Chương 3 - Nguyễn Ngọc Thiêm
3.3 Mesh Analysis
Example 3.11: Find I1, I2, I3 in the circuit in Fig.3.20 using analysis
𝐼1 = 3𝐴, 𝐼2 = 8𝐴, 𝐼3 = 6𝐴
Current source exists only in one mesh
Ta có: 𝐼𝑐 = 2
2A
−𝐼𝑎 + 𝐼𝑏 = 5
current source exists
between two meshes
I1
4Ω
38V
𝐼𝑐
I2
1Ω
𝐼𝑏
𝐼𝑎
5A
Fig. 3.20
Chương 3 - Nguyễn Ngọc Thiêm
I3
3Ω
3.3 Mesh Analysis
Exmple 3.12: Solve for V0 in the circuit in Fig.3.21 using mesh analysis.
Ans:𝑉ሶ0 = 9,756∠222,320 𝑉
Current source exists
between two meshes
Current source exists
only in one mesh
Fig. 3.21
Chương 3 - Nguyễn Ngọc Thiêm
3.3 Mesh Analysis
Practice problem 3.12: Solve for 𝐼 0ሶ in the circuit in Fig. 3.22 using mesh
analysis.
Ans: 𝐼0ሶ = 1,465∠38,480 𝐴
Fig. 3.22
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
i1
The transformer is an electrical device designed
on the basis of the concept of magnetic
coupling. It uses magnetically coupled coils to
transfer energy from one circuit to another.
+
u1
-
i2
+
u2
-
Transformers are key circuit elements.
They are used in electronic circuits such as radio and television receivers for
such purposes as impedance matching, isolating one part of a circuit from
another, and again for stepping up ordown ac voltages and currents.
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
When two inductors (or coils) are in a close
i1
proximity to each other, the magnetic flux caused +
by current in one coil links with the other coil,
there by inducing voltage in the latter. This
u1
-
i2
𝚿𝟏𝟐
𝚿𝟐𝟐
+
u2
𝚿𝟏𝟏
𝚿𝟐𝟏
-
phenomenon is known as mutual inductance
Consider two coils with self –inductances 𝐿1 and 𝐿1 that are in close
proximity with each other. Coil 1 has 𝑁1 turns, while coil 2 has 𝑁2 turns.
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
The magnetic flux Ψ1 emanating from coil 1 has two components: Ψ1 = Ψ11 + Ψ12
✓ Ψ11 = 𝐿1 𝑖1 links only coil 1 is caused by the current 𝑖1 flowing in coil 1
✓Ψ12 = ±𝑀12 𝑖2 links coil 2 is caused by the current 𝑖1 flowing in coil 1
➢ Hence: Ψ1 = Ψ11 + Ψ12 = 𝐿1 𝑖1 ± 𝑀𝑖2
The magnetic flux emanating from coil 2 has two components: Ψ2 = Ψ22 + Ψ21
✓ Ψ22 = 𝐿2 𝑖21 : links only coil 2 is caused by the current 𝑖2 flowing in coil 2
✓ Ψ21 =±𝑀21 𝑖1 : links coil 1 is caused by the current 𝑖2 flowing in coil 2
➢Hence: Ψ2 = Ψ22 + Ψ21 = 𝐿2 𝑖2 ± 𝑀𝑖1
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
✓𝑀12 = 𝑀21 = 𝑀 as the mutual inductance between the two coils, measured in
henrys (H)
✓𝑘 =
𝑀
𝐿1 .𝐿2
is the couple coefficient, 0 < 𝑘 < 1
The voltage in coil 1: 𝑢1 =
𝑑Ψ1
𝑑𝑡
The voltage in coil 2: 𝑢2 =
𝑑Ψ2
𝑑𝑡
=
𝑑Ψ11
𝑑𝑡
𝑑Ψ12
+
𝑑𝑡
=
𝑑Ψ22
𝑑𝑡
𝑑Ψ21
𝑑𝑡
+
=
𝑑𝑖1
±𝐿1
𝑑𝑡
±
𝑑𝑖2
𝑀
𝑑𝑡
(3.4a)
=
𝑑𝑖2
±𝐿2
𝑑𝑡
𝑑𝑖1
±𝑀
𝑑𝑡
(3.4b)
✓ where :
-
𝑑𝑖1
𝐿1
𝑑𝑡
-
𝑑𝑖2
𝐿2
𝑑𝑡
is the self-voltage induced in coil 1,
𝑑𝑖2
𝑀
𝑑𝑡
is the voltage induced in coil 1
is the self-voltage induced in coil 2,
𝑑𝑖1
𝑀
𝑑𝑡
is the voltage induced in coil 2
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
❑ The self-induced
𝑑𝑖1
±𝐿1
𝑑𝑡
and
𝑑𝑖2
±𝐿2
𝑑𝑡
whose polarity is determined by the
reference direction of the current and the reference polarity of the voltage
❑ The polarity of mutual voltage
𝑑𝑖2
±𝑀
𝑑𝑡
and
𝑑𝑖1
±𝑀
𝑑𝑡
is not easy to determine
➢ We apply the dot convention in circuit analysis. By this convention, a dot is
placed in the circuit at one end of each of the two magnetically coupled coils
to indicate the direction of the magnetic flux if current enters that dotted
terminal of the coil.
➢The dots are used along with the dot convention to determine the polarity of
the mutual voltage
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
➢A dot illustrated in Fig.3.23
The dot convention is stated as follows:
- If a current enters the dotted terminal of one
coil, the reference polarity of the mutual voltage
in the second coil is positive at the dotted
terminal of the second coil.
Fig.3.23
- If a current leaves the dotted terminal of one coil, the reference polarity of
the mutual voltage in the second coil is negative at the dotted terminal of the
second coil
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
Application of the dot convention is illustrated in the four pairs of mutually
coupled coils in Fig. 3.24
i2
i1
M
+
u1
L1
-
L2
u2 u1
Hình 1
M
+ +
+
L1
L2
-
i2
i1
i2
i1
u2
-
M
+
u1
L1
-
Fig. 3.24
Chương 3 - Nguyễn Ngọc Thiêm
u2
Hình 3
Hình 2
+
L2
i2
i1
M
+
u1
L1
-
+
L2
u2
Hình 4
3.4 Mutual Inductane
We can write Eqs (3.a) and (3.4b) in the frequency domain ( phasor domain)
𝑈ሶ 1 = ±𝑗𝑋𝐿1 . 𝐼1ሶ ± 𝑗𝑋𝑀 . 𝐼2ሶ
𝑈ሶ 2 = ±𝑗𝑋𝐿2 . 𝐼2ሶ ± 𝑗𝑋𝑀 . 𝐼1ሶ
where: 𝑋𝐿1 = 𝜔𝐿1 Ω , 𝑋𝐿2 = 𝜔𝐿2 Ω , 𝑋𝑀 = 𝜔𝑀 Ω
𝐼1ሶ
+
u1
-
⟺
M
i1
i2
L1
L2
+
𝑈ሶ 1
jXL1
-
Fig 3.25
Chương 3 - Nguyễn Ngọc Thiêm
𝑈ሶ 2
jXM
+
u2
𝐼2ሶ
+
jXL2
-
3.4 Mutual Inductane
Example 3.13: Determine 𝑈ሶ 1 , 𝑈ሶ 2 , 𝑈ሶ 3 in the circuit in Fig.3.26
+
𝐼1ሶ
𝑈ሶ 1
-
𝑗𝜔𝐿1
𝑗𝜔𝑀31
𝑗𝜔𝑀12
𝐼2ሶ
𝐼3ሶ
𝑗𝜔𝐿2
𝑗𝜔𝑀23
𝑈ሶ 2
Fig. 3.26
Chương 3 - Nguyễn Ngọc Thiêm
𝑗𝜔𝐿3
𝑈ሶ 3
3.4 Mutual Inductane
Example 3.14: determine mesh equations in the circuit of Fig.3.27
𝐼2ሶ
𝑈ሶ 1
𝑈ሶ 2
+
−
Fig. 3.27
Chương 3 - Nguyễn Ngọc Thiêm
𝐼1ሶ
−
+
3.4 Mutual Inductane
Example 3.15: Solve for 𝑉ሶ0 in Fig.3.28
𝑈ሶ 1
𝑈ሶ 2 +
+ −
−
𝑈ሶ 1
Practice problem 3.15:
Calculate the phasor current
𝐼1ሶ and 𝐼2ሶ in the circuit of
Fig.3.29
Fig.3.28
Ans: 𝑉ሶ0 = −𝑗0,6 𝑉
Fig3.29
Ans: 𝐼1ሶ = 13,01∠ −49,390 𝐴
𝐼2ሶ = 12,91∠14,040 𝐴
Chương 3 - Nguyễn Ngọc Thiêm
3.4 Mutual Inductane
Example 3.16: Determine the
phasor current 𝐼1ሶ and 𝐼2ሶ in the
circuit of Fig. 3.30
Practice problem 3.16:
Determine the phasor current 𝐼1ሶ
and 𝐼2ሶ in the circuit of Fig. 3.31
Fig. 3.31
Fig. 3.30
Ans: 𝐼1ሶ = 20,3∠3,5 A,
𝐼2ሶ = 8,693∠190 A,
0
Ans 3.19: 𝐼1ሶ = 2,15∠86,560 A,
𝐼2ሶ = 3,23∠86,560 A,
Chương 3 - Nguyễn Ngọc Thiêm
3.5 Maximum Power Transfer
In many practical situations, a circuit is designed to provide power to a load. While for
electric utilities, minimizing power losses in the process of transmission and
distribution is critical for efficiency and economic reasons, there are other applications
in areas such as communications where it is desirable to maximize the power
delivered to a load. We now address the problem of delivering the maximum power to
a load when given a system with known internal losses. It should be noted that this
will result in significant internal losses greater than or equal to the power delivered to
the load.
Chương 3 - Nguyễn Ngọc Thiêm
3.5 Maximum Power Transfer
Consider the circuit in Fig.3.32. We assume that the voltage source 𝐸ሶ = 𝐸𝑚 ∠𝜑𝑒 and
the impedance source 𝑍𝑠 = 𝑅𝑠 + 𝑗𝑋𝑠 to provide power to a load with the load
impedance 𝑍𝐿 = 𝑅𝐿 + 𝑗𝑋𝐿 . We now address the problem of delivering the maximum
power to a load when given a system with known internal losses
RS
The current through the load is:
𝐸ሶ
𝐸ሶ
𝐼ሶ =
=
𝑍𝑠 + 𝑍𝐿 𝑅𝑠 + 𝑗𝑋𝑠 + 𝑅𝐿 + 𝑗𝑋𝐿
The average power delivered to the load is:
1
2
𝑅
𝐸
1
𝐿 𝑚
2
2
𝑃 = 𝑅𝐿 𝐼𝑚 =
2
(𝑅𝑠 + 𝑅𝐿 )2 +(𝑋𝑠 + 𝑋𝐿 )2
Chương 3 - Nguyễn Ngọc Thiêm
jXs
.
I
jXL
.
E
RL
Fig 3.32
3.5 Maximum Power Transfer
• Our objective is to adjust the load parameters 𝑍𝐿 = 𝑅𝐿 + 𝑗𝑋𝐿 so that P is
maximum. To do this we set
• Setting
𝜕𝑃
𝜕𝑋𝐿
And setting
𝑅𝐿 =
𝜕𝑃
𝜕𝑋𝐿
and
𝜕𝑃
𝜕𝑋𝐿
RS
= 0.
jXs
.
I
jXL
= 0 gives : 𝑋𝐿 = −𝑋𝑆 (3.5a)
𝜕𝑃
𝜕𝑋𝐿
.
E
RL
= 0 results in
Fig.3.32
𝑅𝑆2 + (𝑋𝐿 + 𝑋𝑆 )2 (3.5b)
Combining Eqs(3.5a) and (3.5b) leads to the conclusion that for
maximum average power tranfor, 𝑍𝐿 must be selected so that
𝑋𝐿 = −𝑋𝑆 and 𝑅𝐿 = 𝑅𝑆 or 𝑍𝐿 = 𝑅𝐿 + 𝑗𝑋𝐿 = 𝑅𝑆 − 𝑗𝑋𝐿 = 𝑍𝑆∗
Chương 3 - Nguyễn Ngọc Thiêm
3.5 Maximum Power Transfer
For maximum average power transfer, the load impedance ZL must be equal to the
complex conjugate of the source impedance ZS.
The maximum average power as:
𝑃𝑚𝑎𝑥
2
𝐸𝑚
=
8𝑅𝑠
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
3.6.1 Thevenin’s Theorem
It often occurs in practice that a particular element in a circuit is variable
(usually called the load) while other elements are fixed. As a typical example,
a household outlet terminal may be connected to different appliances
constituting a variable load. Each time the variable element is changed, the
entire circuit has to be analyzed all over again. To avoid this problem,
Thevenin’s theorem provides a technique by which the fixed part of the
circuit is replaced by an equivalent circuit
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
3.6.1 Thevenin’s Theorem
According to Thevenin’s theorem, the linear circuit in Fig. 3.33(a) can be replaced by
that in Fig. 3.33(b). (The load may be asingle impedance or resistor or another
circuit.) The circuit to the left of the terminals a-b in Fig. 3.33(b) is known as the
Thevenin equivalent circuit
𝐼ሶ
𝐼ሶ
𝑈ሶ
⇔
𝑈ሶ 𝑇ℎ
𝑈ሶ
Fig. 3.33
Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source 𝑈ሶ 𝑇ℎ in series with a impedance ZTh,
where 𝑈ሶ 𝑇ℎ is the open-circuit voltage at the terminals and ZTh is the input or equivalent
impedance at the terminals when the independent sources are turned off.
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
3.6.2 Norton’s Theorem
In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an
American engineer at Bell Telephone Laboratories, proposed a similar theorem
Norton’s theorem states that a linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a current source 𝐼𝑁ሶ in parallel with a impedance
ZN, where 𝐼𝑁ሶ is the short-circuit current through the terminals and ZN is the
input or equivalent resistance at the terminals when the independent sources are
turned off
𝐼ሶ
𝑈ሶ
⇔
ሶ
𝐼𝑁ሶ = 𝐼𝑠𝑐
Fig. 3.34
Chương 3 - Nguyễn Ngọc Thiêm
𝑈ሶ ሶ
𝑈
3.6 Thevenis’s Theorem and Norton’s Theorem
3.6.2 Norton’s Theorem
Observe the close relationship between Norton’s and Thevenin’s
𝐼𝑁ሶ =
𝑈ሶ 𝑇ℎ
𝑍𝑇ℎ
(3.6)
This is essentially source transformation. For this reason, source transformation
is often called Thevenin-Norton transformation. The Thevenin and Norton
equivalent circuits are related by a source transformation.
Since 𝑈ሶ 𝑇ℎ , 𝐼𝑁ሶ and 𝑍𝑇ℎ are related according to Eq.(3.6), to determine the
Thevenin or Norton equivalent circuit requires that we find:
• The open-circuit voltage 𝑈ሶ 𝑜𝑐 = 𝑈ሶ ℎ𝑚 = 𝑈ሶ 𝑇ℎ across terminals a and b
ሶ = 𝐼𝑁ሶ = 𝐼𝑛𝑔𝑎𝑛
ሶ
• The short-circuit current 𝐼𝑠𝑐
𝑚𝑎𝑐ℎ at terminals a and b
• The equivalent at terminals a and b
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
❑ To apply this idea in finding the Thevenin impedance (𝑍𝑇ℎ ) or resistance
(𝑅𝑇ℎ ), we need to consider two cases
• Case 1: If the network has no dependent source, we turn off all independent
sources. 𝑍𝑇ℎ or 𝑅𝑇ℎ is the input impedance or resistance of the network
looking between terminals a and b, as shown in Fig
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
❑ To apply this idea in finding the Thevenin impedance (𝑍𝑇ℎ ) or resistance (𝑅𝑇ℎ ), we
need to consider two cases
• Case 2: If the network has dependent sources, we turn off all independent sources.
As with superposition, dependent sources are not to be turned off because they are
controlled by circuit variables. Find the open-circuit voltage 𝑈ሶ 𝑜𝑐 = 𝑈ሶ ℎ𝑚 = 𝑈ሶ 𝑇ℎ
ሶ = 𝐼𝑁ሶ = 𝐼𝑛𝑔𝑎𝑛
ሶ
across terminals a and b and find the short- circuit current 𝐼𝑠𝑐
𝑚𝑎𝑐ℎ at
terminals a and b. Then 𝑍𝑇ℎ =
𝑈ሶ 𝑜𝑐
ሶ
𝐼𝑠𝑐
=
𝑈ሶ 𝑇ℎ
ሶ
𝐼𝑁
𝑈ሶ 𝑜𝑐 = 𝑈ሶ ℎ𝑚 = 𝑈ሶ 𝑇ℎ
Chương 3 - Nguyễn Ngọc Thiêm
ሶ = 𝐼𝑁ሶ
𝐼𝑠𝑐
3.6 Thevenis’s Theorem and Norton’s Theorem
Example 3.17: Find the Thevenin equivalent circuit of the circuit shown in
Fig.3.35, to the left of the terminals a-b. Then find the current through RL = 6 Ω,
and RL = 36Ω (Ans: RTh = 4 Ω, UTh = 30 V, 3A, 0,75 A)
Fig. 3.35
Fig. 3.36
Practice problem 3.17: Using Thevenin’s theorem, find the equivalent
circuit to the left of the terminals in the circuit in Fig. 3.36. Then find I
(Ans: RTh = 4 Ω, UTh = 6 V, 1,5A)
Chương 3 - Nguyễn Ngọc Thiêm
3.6 Thevenis’s Theorem and Norton’s Theorem
Example 3.18: Find the Thevenin equivalent of the circuit in Fig. 3.37
(Ans: RTh = 6 Ω, UTh = 20 V)
Fig. 3.37
Fig. 3.38
Practice problem 3.18: Find the Thevenin equivalent circuit of the circuit
in Fig. 3.38 to the left of the terminals a and b
(Ans: RTh = 0,44 Ω, UTh = 5,33 V)
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3.6 Thevenis’s Theorem and Norton’s Theorem
Example 3.19: Obtain the Thevenin equivalent at terminals a-b of the circuit in
Fig.3.39
Fig. 3.39
Fig. 3.40
Practice problem 3.19: Obtain the Thevenin equivalent at terminals a-b
of the circuit in Fig.3.40
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3.6 Thevenis’s Theorem and Norton’s Theorem
Exmaple 3.20: Find the Thevenin equivalent of the circuit in Fig.3.41 as seen
from terminals a-b.
Fig.3.41
Fig.3.42
Practice problem 3.20: Determine the Thevenin equivalent of the circuit in
Fig. 3.42 as seen from the terminals a-b.
Ans: 𝑍𝑇𝐻 = 12,166∠136,30 Ω, 𝑉𝑇𝐻 = 7,35∠72.90 𝑉,
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3.6 Thevenis’s Theorem and Norton’s Theorem
Example 3.21: Cho In the circuit of Fig.3.43
a. Find the Thevenin equivalent and Norton of the circuit to the left of the
terminals a and b
b. Find the load impedance 𝑍𝐿 that bsorbs the maximum average power
Fig. 3.43
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3.6 Thevenis’s Theorem and Norton’s Theorem
Practice problem 3.21: in the circuit in Fig .3.44
a. Find the Thevenin equivalent and Norton of the circuit to the left of the
terminals a and b
b. Find the load impedance 𝑍𝑇 that bsorbs the maximum average power
5Ω
.
j10Ω
I
a
-j4Ω
20∠900 V (hd)
ZT
3Ω
Fig. 3.44
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b
3.7 Superpositon Principle
The superposition principle states that the voltage across (or current through)
an element in a linear circuit is the algebraic sum of the voltages across (or
currents through) that element due to each independent source acting alone
❑ Step to Apply Superposition Principle:
Step 1. Turn off all independent sources except one source. Find the output
(voltage or current) due to that active source using nodal or mesh analysis.
Step 2. Repeat step 1 for each of the other independent sources.
Step 3. Find the total contribution by adding algebraically all the contributions
due to the independent sources.
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3.7 Superpositon Principle
Example 3.22: Find v0(t) in the circuit in Fig.3.45 using the superposition
theorem
Fig.3.45
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3.7 Superpositon Principle
Practice problem 3.22: For the circuit shown in Fig.3.46. Find v0 using
superposition
Fig.3.46
𝑣0 𝑡 = 4,631 sin 5𝑡 − 81,120 + 1,051cos(10𝑡 − 86,240 )
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