CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad.
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Transcript of CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad.
CIRCUIT ANALYSIS METHODS
Chapter 3
Mdm shahadah ahmad
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
INTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHODINTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHOD
• Use KCL.
• Important step: select one of the node as reference node
• Then define the node voltage in the circuit diagram.
Node-voltage example
• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.
• The node-voltage equation for node 1 is,
• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.
• The node-voltage equation for node 1 is,
251
100 2111 VVVV
• Node-voltage equation of node 2,
2102
0 212
VVV
• Solving for V1 and V2 yeilds
VV
VV
91.1011
120
09.911
100
2
1
THE NODE-VOLTAGE METHOD AND DEPENDENT SOURCES
• If the circuit contains dependent sources, the node-voltage equations must be supplemented with the constraint equation imposed by the presence of the dependent sources.
example…
Use the node-voltage method to find the power dissipated in the 5Ω resistor.
• The circuit has 3 node. • Thus there must be 2 node-voltage
equation.• Summing the currents away from node 1
generates the equation,
05202
20 2111
VVVV
• Summing the current away from node 2 yields,
02
8
1052212
iVVVV
• As written, these two equations contain three unknowns namely V1, V2 and iØ.
• To eliminate iØ, express the current in terms of node-voltage,
521 VV
i
• Substituting this relationship into the node 2 equation,
06.1
102.075.0
21
21
VV
VV
• Solving for V1 and V2 gives,
VV 161 VV 102
• Then,
Ai 2.15
1016
W
Rip
2.7
544.12
SPECIAL CASE
• When a voltage source is the only element between two essential nodes, the node-voltage method is simplified.
Example…
• There is three essential nodes, so two simultaneous equation are needed.
• Only one unknown node voltage, V2 where as V1=100V.
• Therefore, only a single node-voltage equation is needed which is at node 2.
055010
212 VVV
Using V1 =100V, thus V2=125V.
SUPERNODESUPERNODE
• When a voltage source is between two essential nodes, those nodes can be combine to form a supernode (voltage sourse is assume as open circuit).
Supernode example…
• Nodes chosen,
• Node-voltage equation for node 2 and 3,
0505
212
iVVV
04100
3 iV
• Summing both equation,
04100505
3212 VVVV
Above equation can be generates directly using supernode approach
Supernod
• Starting with resistor 5Ω branch and moving counterclockwise around the supernode,
04100505
3212 VVVV
• Using V1 =50V and V3 as a function of V2,
iVV 1023
5
502 V
i
• Substituded into the node-voltage equation,
1410500
10
100
1
5
1
50
12
V
VV
V
60
15)25.0(
2
2
• Using V2 value, gives
Ai 25
5060
VV 8020603
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
INTRODUCTION OF MESH-INTRODUCTION OF MESH-CURRENT METHODCURRENT METHOD
• A mesh is a loop with no loop inside it.
• A mesh current is the current that exist only in the perimeter of a mesh.
• Mesh-current method use KVL to generates equation for each mesh.
Mesh-current example…
• Mesh-current circuit with mesh current ia and ib.
• Use KVL on both mesh,
311 RiiRiV baa
232 RiRiiV bab
• Solving for ia and ib, and you can compute any voltages or powers of interest.
THE MESH-CURRENT METHOD AND DEPENDENT SOURCES
• If the circuit contains dependent sources, the mesh-current equations must be supplemented by the appropriate constraint equations.
Example…
• Use the mesh-current method to determine the power dissipated in the 4Ω resistor.
• Using KVL,
iiiii
iiiii
iiii
154200
4150
20550
2313
32212
3121
• But
• Substituting into the mesh-current equation,
31 iii
321
321
321
9450
41050
2052550
iii
iii
iii
• Using Cramer rule, the values of i2 and i3 can be determine,
945
4105
20525
905
405
205025
2i
45
5254
95
202510
94
2055
95
4550
2i
Ai
i
26125
32505001250625
3250
)125(4)125(10)125(5
)65(50
2
2
A
i
28125
3500125
45
10550
125
045
0105
50525
3
• Power dissipated by 4Ω resistor is
W
Rip
16
4)2628( 2
2
SPECIAL CASE (SUPERMESH)
• When a branch includes a current source, the mesh-current method can be simplified.
• To create a supermesh, remove the current source from the circuit by simply avoiding the branch when writing the mesh-current equations.
• Supermesh equation,
06450
23100
ac
bcba
ii
iiii
cba iii 65950
• Mesh 2 equation,
cb
bab
ii
iii
2
1030
• From the circuit,
ic –ia= 5A• Using Cramer rule, the three
mesh current can be obtain.
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
SOURCE SOURCE TRANSFORMATION TRANSFORMATION
• Source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa.
Sorce transformation
Example…
• Source transformation procedure
s
ss R
VI
sp RR
From To methodUse,
pss RIV
ps RR
From To method
Use,
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
THEVENIN EQUIVALENT THEVENIN EQUIVALENT CIRCUITCIRCUIT
• Thevenin equivalent circuit consist of an independent voltage source, VTh in series with a resistor RTh.
Thevenin equivalent circuit
ThV
ThRa
b
• Thevenin voltage, VTh = open circuit voltage in the original circuit.
• Thevenin resistance, RTh is the ratio of open-circuit voltage to the short-circuit current.
sc
ThTh
Th
Thsc i
VR
R
Vi
Example…
V25
5a
b
20
4
A3
1V
abV
• Step 1: node-voltage equation for open-circuit:
ThVVV
VV
32
03205
25
1
11
• Step 2: short-circuit condition at terminal a-b
V25
5 a
b
20
4
A3
2V scI
• Node-voltage equation for short-circuit:
VV
VVV
16
04
3205
25
2
222
84
32
sc
ThTh I
VR
AI sc 44
16
Short-circuit current:
Thevenin resistance:
Thevenin equivalent circuit
V32
8a
b
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
NORTON EQUIVALENT NORTON EQUIVALENT CIRCUITCIRCUIT
• A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance.
• Can be derive from a Thevenin equivalent circuit simply by making a source transformation.
• Norton current, IN = the short-circuit current at the terminal of interest.
• Norton resistance, RN = Thevenin resistance, RTh
Example
V25
5a
b
20
4
A3
Step 1: Source transformation
A5 5
a
b
20
4
A3
Step 2: Parallel sources and parallel resistors combined
a
b
4
A8 4
Step 3: Source transformation, series resistors combined, producing the Thevenin equivalent circuit
V32
8a
b
THEVENINEQUIVALENT
CIRCUIT
Step 4: Source transformation, producing the Norton equivalent circuit
a
b
A4 8
NORTONEQUIVALENT
CIRCUIT
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
MAXIMUM POWER MAXIMUM POWER TRANSFERTRANSFER
• Two basic types of system:– Emphasizes the efficiency of the power
transfer– Emphasizes the amount of power
transferred.
• Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL.
• Maximum power transfer occurs when RL=RTh.
Example…Example…
ThV
ThR a
b
LRi
• Power dissipated by resistor RL
LLTh
Th
L
RRR
V
Rip2
2
• Derivative of p with repect to RL
4
22 2
LTh
LThLLThTh
L RR
RRRRRV
dR
dp
• Derivative is zero and p is maximum when
)(22LThLLTh RRRRR
LTh RR
• The maximum power transfer occurs when the load resistance, RL = RTh
• Maximum pwer transfer delivered to RL:
L
Th
L
LTh
R
V
R
RVp
42
2
2
2
CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
PRINSIP PRINSIP SUPERPOSISISUPERPOSISIPRINSIP PRINSIP SUPERPOSISISUPERPOSISI
• In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are activate.
Step of Superposition principle
1. Deactivated all the sources and only remain one source at one time. Do circuit analysis to find voltages or currents.
2. Repeat step 1 for each independent sources.
3. Sum the resulting voltages or currents.
1. Independent voltage source will become short-circuit with 0Ω resistance.
2. Independent current source will become open-circuit.
3. Dependent sources are never deactivated when applying superposition.
REMEMBER!!!
Example…Example…
• Step 1: deactivated all sources except
voltage source
• V0 is calculated using voltage divider:
Vk
kV 54
1020
• Step 2: Deactivated all sources except
current source
• V0 is calculated by using current divider:
VkmV
mAmk
ki
2)2)(1(
1)2(4
2
0
0
V0 =2+5=7V.
• Step 3: Sum all the resulting voltages:
Question 1 (node-voltage)
• Calculate the value of Io
Solution
• Node 1:
622
3
4221
21
211
VV
VVV
• Node 2:
44
5
2
44
1
2
1
2
1
2
4422
21
21
2212
VV
VV
VVVV
846.1625.1
36
4
6
45
21
21
23
21
23
2
V
AI 923.02
846.10
Question 2 (mesh-current)
• Determine the value of currents, I1, I2 and I3.
• Supermesh:
• Mesh 3:
0)(510 321 III
125510
012555
23
233
II
III
• Dependent current source
• Vo
021 2VII
)(5 320 IIV
• Substitute V0
01011
)(10
321
3221
III
IIII
• Use Cramer rule
10111
1050
5510
10110
105125
550
1I
A1625
625
111
505
101
1005
1011
10510
1011
55125
• Current I2:
A
I
21625
13125625
101
510125
625
1001
101250
5010
2
• Current I3:
A
I
23625
14375625
101
510125
625
0111
12550
0510
3
Question 3 (thevenin)
• Open-circuit voltage, Voc:
• Node-voltage equation for Voc
VV
V
VV
VV
oc
oc
ococ
ococ
10
202
0424
0222
24
• Thevenin resistance, RTh:
5422THR
• Thevenin equivalent circuit:
VV 88.6)10(16
110
Question 4 (norton)
• Open-circuit current, Isc:
AI sc 64
123
• Norton resistance, RN:
RN = 4Ω
• Norton equivalent circuit:
VV 18)3(612460
Question 5 (superposition)
• Use superposition principle to determine the voltage Vo.
• Deactivated current source
V412
224V0
• Deactivated voltage source
V4212
46)2(iV oo
• Summing the voltage V0
V8VV 00
Question 6 (node-voltage)
• Determine the value of Vo.
node-voltage equation:
020
80
10
5
2003 000
ViVV
20
800 V
iCurrent iΔ:
• Thus:
V0 =50V