ee602 circuit analysis
DESCRIPTION
Ee602 Network AnalysisTRANSCRIPT
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UNIT 1
NETWORK ANALYSIS
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COURSE LEARNING OUTCOME :
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Apply basic circuit theorems in solving alternating current (AC) in electrical circuit problems. (C3)
Use appropriate table and formula to determine the equation and series in order electrical circuit problems. (C3)
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NETWORK ANALYSIS 1.1 Understand the application of network analysis
theorems to AC circuits:
1.1.1 Apply mesh analysis to AC networks. 1.1.2 Apply nodal analysis to AC networks. 1.1.3 Apply the superposition theorem to AC networks. 1.1.4 Apply Thevenin's and Norton's theorems to AC
networks. 1.1.5 Determine Delta-Wye and Wye-Delta conversions
in AC networks.
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A general expression for impedance , Z
Z = R ± jX Ω R = resistance , X = reactance
Element Impedance Resistance
Reactance
Resistor ZR = R R 0
Inductor ZL = jL 0 L
Capacitor ZC = 1 / jC
= -j/C0 -1 / C
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Steps to Analyze AC Circuits
1. Transform the circuit to the phasor domain or frequency domain
2. Solve the problem using circuit techniques (Nodal, Mesh,…etc)
3. Transform the resulting phasor to the time domain
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MESH ANALYSIS7
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an equation in terms of the loop currents.
4. Solve the resulting system of linear equations.
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Example 28 Determine I0 for the circuit below using mesh analysis
Ans: 6.12 144.78 A
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Answer :9
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Example 410
Find V0 for the circuit below.
Ans: 9.754 -137.69 V
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Exercise 211
Find I0 for the circuit below using mesh analysis.
Ans: 1.194 65.45 A
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NODAL ANALYSIS12
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.
4. Solve the resulting linear equations.
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NODAL ANALYSIS13
40
80j 15j 25Ao05 A03 o
• Based on KCLAssigns unknown voltages to all its essential node
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Step 1: Choose a Reference Node
j80 Ω5 A
40 Ω
-j15 Ω 25 Ω3 A
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Select a reference node. Mark the reference node with the earth sign ┴ or downward arrow ↓.A reference node is the node from where all the other node voltages join.
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Step 2: Assign Unknown Node Voltages
5 A 3 A
40
80j 15j 25
15
Remember : between 2 different nodes, there should be at least 1 element
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Step 2: Assign Unknown Node Voltages16
j80 Ω5 A
40 Ω
-j15 Ω 25 Ω3 A
V1 V2
Assign node voltages at the marked essential nodes.
V3
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Step 3: Perform KCL at the Selected Nodes
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• KCL is performed with the current going out of the node as positive (i.e. currents going out are added, going in are subtracted)
• Unknown currents assume to leave node
(1)
(2)
… and so on until all the simultaneous equations are performed for all unknowns
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Step 4: Solve the equations18
(1)
(2)
V1 = ? V3 = ?V2 = ?
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Example 119
Find ix in the circuit below using nodal analysis.
Ans: 7.59 108.4 A
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Example 221
Compute V1 and V2 in the circuit below using nodal analysis
Ans: 25.78 -70.48 V, 31.41 -87.18 V
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Exercise 123
Using nodal analysis, find v1 and v2 in the circuit below
Ans: 11.34 60.04 V, 33.07 57.15 V
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Circuit with Dependent Sources24
If a dependent source is present in the circuit, we need to come up with a constraint equation imposed by the presence of the dependent source.
The constraint equation is an equation describing the dependent term (of the dependent source) in terms of node voltages or values.
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Circuit with Dependent Sources25
In this case, the dependent source is the 8io voltage source. The dependent term is io. From Ohm's Law, we obtain the constraint equation:
521 vv
io
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Nodal or Mesh Analysis??26
Go for the analysis that will result in lesser number of simultaneous equations.
Compare the number of node-voltage equations to the number of mesh-current equations required.
The one that is less represents the analysis that would be the better choice.
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Example 827
Obtain current I0 in figure below using Mesh analysis :
Ans: 1.465 38.48 A
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Source Transformation &
Superposition Theorem (Independent &
dependant source)
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Source transformation : replacing a
voltage source vs in series with a
impedance Z by a current source is in
parallel with a impedance Z, or vice versa.
Source Transformation
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Source Transformation
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Example 1
Calculate Vx in the circuit of figure below using the concept of source transformation.
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Superposition Theorem
Superposition : the voltage across (or current through) an element in a linear circuits is the
algebraic sum of the voltage across (or current through) that element due to each independent
source acting alone.Current Source open circuit(0 A)
Voltage Source short circuit (0 V)
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Superposition Theorem
Step to apply:
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source.
2. Repeat step 1 for each other independent sources.
3. Find the total contribution by adding algebraically all the contribution due to the independent source.
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EXAMPLE 1
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Step 1: Eliminate one of the source :
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Step 2 : Calculate the current/voltage needed
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Example 2 :38
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Example 3
Calculate Io in the circuit of figure shown below
using the superposition theorem.
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SOLUTIONS :43
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TRY THIS!44
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Example Superposition to dependent source :
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SOLUTION
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SOLUTION - CONT
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Any two-terminal network containing voltage of current sources can be replaced by a equivalent circuit consisting of a voltage to the open circuit voltage of the original circuit in series with the measured back into the original circuit.
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Thevenin’s Theorem
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Thevenin’s Theorem
Step :1. The portion of the original network
considered as the load is removed or imagined to be removed.
2. The open circuit voltage is calculated3. The thevenin resistance is calculated
looking back into the network. 4. The equivalent circuit is drawn, the load
reconnected, and the load current determined.
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Thevenin Equivalent
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Example Thevenin 1:51
Find the Thevenin equivalent at terminals a–b of the circuit below.
Zth =12.4 – j3.2 Ω VTH = 18.97<-51.57º V
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Norton Theorem
Norton Theorem may be stated as the linear network containing resistance and energy sources, can be replaced by an equivalent circuit consisting of a current source ( IN ). In parellel with resistance ( RN )
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Norton Theorem
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NN
Th
sc
ocTh
scN
ocTh
RI
V
i
vR
iI
vV
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Q & A
OUTCOMES ?
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