circuit analysis method. topic node-voltage method mesh-current method source of embodiment...
TRANSCRIPT
CIRCUIT ANALYSIS CIRCUIT ANALYSIS METHODMETHOD
CIRCUIT ANALYSIS CIRCUIT ANALYSIS METHODMETHOD
TOPICTOPIC•Node-Voltage Method•Mesh-current Method•Source of embodiment
principle•Thevenin’s Circuit•Norton’s Circuit•Maximum Power Transfer•Superposition Principle
INTRODUCTION TO NODE-VOLTAGE
METHOD
•Base on Kirchhoff’s Current Law
•Important step: select one node as a reference.
Example:Node-Voltage method
•Previous circuit, set node 3 as a reference. By using Kirchhoff’s Current law at node 1,
251
100 2111 VVVV
Node-voltage equation at node2
2102
0 212
VVV
•Solve previous equation
VV
VV
91.1011
120
09.911
100
2
1
NODE-VOLTAN METHOD THAT CONTAIN
DEPENDENT SOURCE
•If the circuit contains dependent source, the node-voltage equation imposed by the presence of the dependent source.
Find power that absorb by 5Ω Resistor using node-voltagemethod.
•Circuit have 3 node. •Need 2 node-voltage
equations.•Summing the currents away
from node 1 generates the equation,
05202
20 2111
VVVV
•Summing the currents away from node 2 yields
02
8
1052212
iVVVV
•These two node-voltage equations contain three unknowns, namely, V1, V2 and iø . To eliminate iø, we must express this controlling current in terms of the node-voltage,
521 VV
i
•Substituting this relationship into the node 2 equation simplifies the two node-voltage equations
06.1
102.075.0
21
21
VV
VV
•Solving for V1 and V2 gives,
V1 =16V V2 = 10V
•Then,
•Power absorb by 5Ω resistor
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 are three essential nodes in this circuit, which means that two simultaneous equations are needed.
•There is only one unknown node-voltage V2, but V1 =100V.
•Solution of this circuit thus involves only a single node-voltage equation at node 2.
055010
212 VVV
Have V1 =100V, and solvedV2 =125V.
SUPERNODESUPERNODE
•A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.
Example:supernode
•Select node:
•Node voltage equation at node 2 and node 3
0505
212
iVVV
04100
3 iV
•Add previous equation
04100505
3212 VVVV
•Previous equation could get when use supernode concept at node 2 and node 3.
Supernode
•From 5Ω resistor
04100505
3212 VVVV
•Pink equation was equal to green equation.
•Using supernode at node 2 and 3 make it simple to analyse the circuit.
•Have V1 =50V and V3 can be describe with V2,
iVV 1023
5
502 V
i
•Replace V1 =50, V3 and iø, pink equation become
1410500
10
100
1
5
1
50
12
V
VV
V
60
15)25.0(
2
2
•Insert V2
Ai 25
5060
VV 8020603
INTRODUCTION TO INTRODUCTION TO MESH-CURRENT MESH-CURRENT
METHODMETHOD
•One mesh mean a loop that no others loop inside.
•This mesh-current method used Kirchhoff’s voltage law to find current each mesh.
Example:Mesh-current
•From Kirchhoff’s law
(1)
(2)
321 iii
33222
33111
RiRiV
RiRiV
•Use i3 from equation (1) and insert to equation (2)
323111 )( RiRRiV )( 322312 RRiRiV
• Mesh-current circuit with mesh current ia and ib.
•Using KVL at those two mesh
311 RiiRiV baa
232 RiRiiV bab
•After ia and ib known, then we can calculate voltage at power at each resistor.
MESH-CURRENT METHOD THAT HAVE DEPENDENT SOURCE
•When circuit have dependent source, mesh-current equation will have constant value related to dependent source.
Example:mesh method with dependent source
• Find power that obserb by 4Ω resistor using mesh-current method.
•From Kirchhoff’s voltage law
iiiii
iiiii
iiii
154200
4150
20550
2313
32212
3121
•Have
•Insert equation iø to related equation,
31 iii
321
321
321
9450
41050
2052550
iii
iii
iii
•By using Cramer law, i2 and i3 can be calculated as below,
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 that absorb by 4Ω resistor
W
Rip
16
4)2628( 2
2
SPECIAL CASE (SUPERMESH)
•When a branch of current source can be remove and use supermesh concept (current source assume as open circuit)
• Assume that current source as open circuit
•Supermesh equation
06450
23100
ac
bcba
ii
iiii
cba iii 65950
•Mesh-current equation for mesh 2
cb
bab
ii
iii
2
1030
•Known
ic –ia= 5A By using Cramer law at those three equation, value for those three mesh current could be calculated.
SOURCE SOURCE TRANSFORMATIONSTRANSFORMATIONS
•Source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.
Source Transformation
Example:Source transformation
•When resistor R=0, terminal a-b become close circuit. Beginning, close circuit current should be same. Therefore,
s
ss R
VI
•Close circuit current for second circuit was Is. Therefore,
ps R
VsI
•When resistor R = ∞, these circuit become open circuit.
•From first circuit, we have Vab =Vs . Therefore, it was voltage for open circuit.
psab RIV
•Vab for those two circuit should be same. Therefore, Vs = Is Rp .
•Replace Is
ps
ps
ss
RR
RR
VV
•Summarize for Source transformation
Tetapkan
s
ss R
VI
sp RR
MethodAfterBefore
Tetapkan
pss RIV
ps RR
Before After Method
THEVENIN EQUIVALENT CIRCUIT
•Introduced in 1883 by M. Leon Thevenin (1857-1926), a French telegraph engineer.
• Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh where VTh
is the open-circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent
sources are turned off.
•This theorem usually used to replace large sequence part (complex) with one simple equivalent circuit. This simple circuit makes voltage, current and circuit power could be calculated easily.
Thevenin equivalent circuit
•Thevenin voltage, VTh = open circuit voltage for origin circuit.
•When load decrease until zero, circuit become close circuit and current become:
sc
ThTh
Th
Thsc i
VR
R
Vi
Example
•Step 1: node-voltage equation for open circuit:
ThVVV
VV
32
03205
25
1
11
• Step 2: replace close circuit at a-b terminal
•Node voltage equation for close circuit:
VV
VVV
16
04
3205
25
2
222
84
32
sc
ThTh i
VR
Aisc 44
16
Close circuit current:
Thevenin resistance
Thevenin equivalent circuit
Norton equivalent circuit
•In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed Norton’s theorem.
•This equivalent circuit have one independent source that parallel
with one resistor.
•Norton equivalent circuit could have from Thevenin equivalent circuit by source transformation.
ExampleStep 1: Source transformation
Step 2: Combine source and parallel resistors
Step 3: Source transformation, Series resistors combined, producing the Thevenin equivalent circuit
Step 4: Source transformation and Producing the Norton equivalent circuit
Norton equivalent circuit
TOPICTOPIC• Node-Voltage method• Mesh-current method• Source transformation
principle• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer
principle• Superposition principle
MAXIMUM POWER TRANSFER
•Power system designed to provide power to load at high-efficiency and decrease power loss when delivered to load. Therefore, we need to decrease source resistance and delivering resistance.
•Definition for Maximum power transfer tell that power that transfer from one source was represent by Thevenin equivalent circuit become max when load resistor RL and Thevenin resistor RTh was
ExampleExample
•Power absorb by resistor RL
LLTh
Th
L
RRR
V
Rip2
2
•Differentiate p with RL
4
22 2
LTh
LThLLThTh
L RR
RRRRRV
dR
dp
•Differential was zero and p become maximum
•Solve
)(22LThLLTh RRRRR
LTh RR
•Therefore, for maximum power transfer, RL must equal with RTH .
•Maximum power transfer equation:
L
Th
L
LTh
R
V
R
RVp
42
2
2
2
SUPERPOSITION 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 current through) that element due to each independent source acting alone.
Step to Apply Step to Apply Superposition PrincipleSuperposition Principle
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the independent sources.
1. Independent voltage source become close circuit with zero volt.
2. Independent current source become open circuit.
3. If dependent source exist, it should be active while superposition process.
REMEMBER!
ExampleExample
•Step 1: turn off current source
•Use voltage divider law to calculate V0 :
Vk
kV 54
1020
•Step 2: turn off voltage source
•Use current divider law to calculate V0 ,
VkmV
mAmk
ki
2)2)(1(
1)2(4
2
0
0
Total V0 :
V0 =2+5=7V.
Question 1
Answer•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
•Supermesh:
•Mesh 3:
0)(510 321 III
125510
012555
23
233
II
III
•Current source
•known
021 2VII
)(5 320 IIV
•Replace V0
01011
)(10
321
3221
III
IIII
•By using Cramer law
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
•Open circuit voltage:
•Node-voltage equation for Voc
VV
V
VV
VV
oc
oc
ococ
ococ
10
202
0424
0222
24
• Thevenin resistance:
5422THR
• Get:
VV 88.6)10(16
110
Question 4
• Close circuit current
AI sc 64
123
• Norton resistance
RTH = 4Ω
• Norton equivalent circuit:
VV 18)3(612460
Question 5
•Turn off current source
VV 412
2241
0
•Turn off voltage source
VV 4212
46110
•Total V0
VVV 0110
10
Question 6
Node-voltage equation:
020
80
10
5
2003 000
ViVV
20
800 V
iKnown:
•Get:
V0 =50V
•Finally we get:
V0 =50V