RESISTIVE CIRCUITS OHM’S LAW - DEFINES THE SIMPLEST PASSIVE ELEMENT: THE RESISTOR.
Chapter 4 Methods of Analysis of Resistive Circuits
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Transcript of Chapter 4 Methods of Analysis of Resistive Circuits
Chapter 4
Methods of Analysis of Resistive Circuits
• Node voltage analysis - with independent source : current source / voltage source - with dependent source
• Mesh current analysis - with independent source : voltage source / current source
- with dependent source
Resistive Circuit Analysis
Figure 4.2-1(a) A circuit with three nodes.(b) The circuit after the nodes have been labeled and a reference node has been selected and marked.(c) Using voltmeters to measure the node voltages.
Node Voltage Analysis – with current sources
To write a set of node equations, we do two things:
1. Express element currents as functions of the node voltages.
2. Apply KCL at each of the nodes of the circuits, except for the
reference node.
Figure 4.2-2Node voltages, v1 and v2 , and element voltage, va, of a circuit element.
Figure 4.2-3 Node voltages, v1 and v2, and element voltage, v1 - v2, of a (a) generic circuit element, (b) voltage source, and (c) resistor.
21 vvva
Node Voltage Analysis – with current sources
Figure 4.2-4(a) A circuit with three resistors.(b) The resistor voltages expressed as functions of the node voltages.(c) The resistor currents expressed as functions of the node voltages.
Node Voltage Analysis – with current sources
12 R
vv
R
vi baaS
KCL
31 R
v
R
vv bba
If
15.04 baa vvv
5.01bba vvv
AiRRR S 4,5.0,1 321 node (a)
node (b)
Example 4.2-1
Node Voltage Analysis – with current sources
node (a)
node (b)
0510
4
baa vvv
045
R
vvv bba
Example 4.2-2
Example 4.2-3
Node Voltage Analysis – with current sources
05
22
11
R
vvi
R
vvi
R
vv bacacanode (a)
node (b)
node (c)
03435
2
i
R
v
R
vv
R
vvi bcbba
21215521
111111iiv
RRv
Rv
RRR cba
2335435
11111iiv
Rv
RRRv
R cba
16321321
1111111iv
RRRRv
Rv
RR cba
Exercise 4.2-1
Exercise 4.2-2
Node Voltage Analysis – with current sources
32
1
2
1
4
1
ba vv
43
1
2
1
2
1
ba vv
node (a)
node (b)
Figure 4.3-1Circuit with an independent voltage sourceand an independent current sources.
Figure 4.3-2Circuit with a supernode that incorporate va and vb.
Node Voltage Analysis – with current and voltage sources
A supernode consists of two nodes connected by an
independent
or a dependent voltage source
Sa vv 23 R
vv
R
vi abbS
23 R
vv
R
vi SbbS
32
332
RR
vRiRRv SSb
Sba vvv bSa vvv
SbbS iR
v
R
vv
21
21
221
RR
vRiRRv SSb
Example 4.3-1
Example 4.3-2
Node Voltage Analysis – with current and voltage sources
Method #1
Apply KCL to the Supernode
Method #2
Vva 4
Vvv cb 8
212126
cbab vvvv
8 cb vv
Method #1
Method #2 : apply KCL to the Supernode
Vva 1212 ab vv
65.1 avi
Vvb 0
03
5.3 bvi
635.35.1 ab vv
360.2 ba vv
35.3
65.1 ba vv
36
0.2 ba vv
Example 4.3-3
Exercise 4.3-1 Exercise 4.3-2
Node Voltage Analysis – with current and voltage sources
Let Supernode be the 10V source
Apply KCL to the Supernode
524010
bcba vvvv
Vvb 12
10 ca vv
530
220
ba vv10 ba vv
10
12
403
ab vv8 ba vv
Example 4.4-1
Example 4.4-2
Example 4.4-3
Node Voltage Analysis – with dependent sources
6ba
x
vvi
24
6
833 bb
xc
vviv
KCL at node (b)
32
6
8 cbb vvv
Vvb 7 Vvc 5.0
aaxba vvvvv 444
KCL at supernode
aaaba vvvvv
4
3
10
5
41043
Vva 4 Vvv ab 205
ab vv 5
2010
6 baa
a vvi
v
KCL at node (b)
ab iv
520
0
20
12 ba
vi
KCL at node
(a)
Vvb 10
Exercise 4.4-1
Exercise 4.4-2
Node Voltage Analysis – with dependent sources
12
4
128
6 aaa
baa
a ivi
vvi
v
Vva 0
8
9ai
KCL at node (a)
Viv ab 5.44
015
4
20
6
aaa vvv
Figure 4.5-1Nonplanar circuit with a crossover.
Figure 4.5-2Circuit with four meshes.Each mesh is identified by dashed lines.
- Loop
- Mesh : 다른 loop 를 포함하고 있지 않은 loop (cross over 가 없는 circuit 에만 적용
가능 )
Mesh Current Analysis – with independent voltage source
Figure 4.5-3(a) A circuit with two meshes(b) Inserting ammeters to measure the mesh currents.
Figure 4.5-4 Mesh currents, i1 and i2,and element current, i1 – i2, of a(a) generic circuit element,(b) current source, and (c) resistor.
321 ii
Mesh Current Analysis – with independent voltage source
To write a set of mesh equations, we do two things:
1. Express element voltages as functions of the mesh
currents.
2. Apply KVL to each of the meshes of the circuits.
Figure 4.5-5
Mesh Current Analysis – with independent voltage source
21311 iiRiRvs
22213 iRiiR
Figure 4.5-6
sviRiRR 24141 )(
0)( 35252414 iRiRRRiR
gviRRiR 35325 )(
Mesh #1
Mesh #2
Mesh #1
Mesh #2
Mesh #3
21411 iiRiRvs
012432522 iiRiiRiR
0)( 23533 gviiRiR
)8(123)36( 21 ii
8)63(3 21 ii
Exercise 4.5-1
Mesh Current Analysis – with two independent voltage sources
812336 211 iii
06338 212 iii
Figure 4.6-2Circuit with an independent current sourcecommon to both meshes.
Figure 4.6-1Circuit with an independent voltage source and an independent current source.
sii 2
sab vviR 11
abviRR 232 )(
12 iiis
Mesh Current Analysis – with current and voltage source
Mesh #1
Mesh #2
Mesh #1 sviRiRR 22121 )(
21
21 RR
iRvi ss
sviRRiR 23211
ss viiRRiR 13211
Example 4.6-1
Figure 4.6-4 Circuit with a supermesh that incorporates mesh 1 and mesh 2, indicated by the dashed line.
Mesh Current Analysis – with current and voltage source
10121 abviiR
0)( 33132 abviRiiR
41 i
Mesh #2
Mesh #3
532 ii
A Supermesh is one larger mesh created from two meshes that have an independent or dependent current source in common.
SupermeshMesh #3
102)(3)(1 23231 iiiii
Current source
10451 321 iii
0631 321 iii
521 ii
0)(32)(1 23313 iiiii
SupermeshMesh #3
Example 4.6-2
Mesh Current Analysis – with current and voltage source
0129 1 vi
5.121 ii
Method #1
Method #2
063 22 vii
1299 21 ii 012639 221 iii
1299 21 ii012639 221 iii
Exercise 4.6-1
Exercise 4.6-2
Mesh Current Analysis – with current and voltage source
Ai4
31
04)23(9 122 iii
Ai3
4
9
122
Vv 433
4
0)3(6315 ii
Example 4.7-1
Mesh Current Analysis – with dependent source
21 iiia
KVL to mesh #1
02432 1 i
12 4
5ii
032 2 mviKVL to mesh #2
25 iia
Ai4
31 Ai
16
152
Vivm 3016
153232 2
Example 4.7-2
Mesh Current Analysis – with dependent source
VAia 2.72.7
21 iiia
KVL to mesh #1
KVL to mesh #2
03610 1 i Ai 6.31
Ai 8.12 02.74 2 i
AVii
Ai
i
AiA a
a
a /48.16.3
2.7
21
Example 4.8-1
Figure 4.8-1
Node Voltage & Mesh Current Analysis – comparison
Figure 4.8-2
0122 23313 iiRiii
Ai 11
2R
Ai 32
Ai 5.03
KVL to mesh #3
Example 4.8-2
Figure 4.8-4
Node Voltage & Mesh Current Analysis – comparison
Figure 4.8-3
R
vvv 331 22
Vv 161
8R
Vvv 1821
KCL at node 3
Vv 22
Vv 163
Mesh Current Analysis – MATLAB
0)( 12131411 viiRiRiR 0)( 21322225 iiRviRiR
1231341 viRiRRR 2232513 viRRRiR
123131 viRiRaRR p
223213 1 viRRRaiR p
Node Voltage and Mesh Current Analysis – PSpice
Example 4.10-1
Example 4.11-1
Example 4.11-2
Node Voltage and Mesh Current Analysis – MATLAB
Figure 4.12-1Proposed circuit for measuring and displaying the angular position of the potentiometer shaft.
Potentiometer Angle Display
Figure 4.12-2Circuit diagram containing models of the power supplies, voltmeter, and potentiometer.
Figure 4.12-3 The redrawn circuit showing the mode v1.
kvo /1.0 Vk ov101.0ovwhere
Angle varies from to
180 180Potentiometer varies from to
0 1
Potentiometer Angle Display
Figure 4.12-3 The redrawn circuit showing the mode v1.
0)1(
)15(15
2 21
p
i
p
ii
RaR
v
aRR
v
M
v
RRR 21
ppp
pi RRRMRaRaRR
RRaRMv
2121
21
2)1(
15)12(2
Assumption MRp 2 KR 20
Then, ppp RRMRaRaRR 22)1(
pp
i RR
aRv
2
15)12(
180
15
2
V
RR
Rv
p
pi
180
5.7
180
15
2
1 VVviLet kRkR p 10,5
4.2)1.0(5.7
180b
180
5.7 Vbbvv io
ree
voltVb
deg1.0
180
5.7
Figure 4.12-4 The final designed circuit.
Potentiometer Angle Display
ppp
pi RRRMRaRaRR
RRaRMv
2121
21
2)1(
15)12(2
150 9167.02
1
360
150
aSuppose
24.6
1052210)9167.01(5109167.05
15)19167.02(102
kkMkkkk
kMvi
98.14)24.6()4.2( ov8.14910 ov
Problems P4.2-4 / P4.2-7 / P4.3-5 / P4.3-11 /
P4.4-7 / P4.4-10 / P4.4-17 / P4.5-6
/
P4.6-10 / P4.7-10 / P4.8-2
Homework #4