Post on 30-Jun-2020
2012/9/17
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Basic Laws
•Ohm’s Law (resistors)•Nodes, Branches, and Loops•Kirchhoff’s Laws•Series Resistors and Voltage Division•Parallel Resistors and Current Division•Wye-Delta Transformations•Applications
Ohm’s Law•Resistance R is represented by
•Ohm’s Law:
AR
Rv+
_
i
1 = 1 V/A
Cross-sectionarea A
Meterialresistivity
ohm
Riv
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Resistors
0Riv == R = 0v = 0
+
_
i
R = v
+
_
i = 0
0Rv
limiR
==∞→
Variable resistor Potentiometer (pot)
Open circuitShort circuit
Nonlinear Resistors
i
v
Slope = R
v
i
Slope = R(i) or R(v)
Linear resistor Nonlinear resistor
•Examples: lightbulb, diodes•All practical resistors may exhibit certain
nonlinear behavior.
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Conductance and PowerDissipation
•Conductance G is represented by
vi
RG
11 S = 1 = 1 A/V
siemens mho
Gi
Gvivp
Rv
Riivp
vGi
22
22
===
===
=
positive R : power absorption (+)
negative R: power generation (-)
Nodes, Branches, & Loops•Branch: a single element (R,
C, L, v, i)
•Node: a point of connectionbetween branches (a, b, c)
•Loop: a closed path in acircuit (abca, bcb, etc)–An independent loop contains
at least one branch which isnot included in other indep.loops.
–Independent loops result inindependent sets of equations.
+_
a
c
b
+_
c
ba
redrawn
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ContinuedElements in parallelElements in series
•Elements in series–(10V, 5)
• Elements in parallel–(2, 3, 2A)
•Neither–((5/10V), (2/3/2A))
10V
5
2 3 2A+_
Kirchhoff’s Laws•Introduced in 1847 by German physicist G. R.
Kirchhoff (1824-1887).
•Based on conversation of charge and energy.
•Two laws are included,Kirchhoff’s current law (KCL) andKirchhoff’s votage law (KVL).
•Combined withOhm’s law, we have apowerful set of tools for analyzing resistivecircuits.
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KCL
i1i2
in
0211
nn
N
niiii
•Assumptions–The law of conservation of charge–The algebraic sum of charges within a system
cannot change.
•Statement–The algebraic sum of currents entering a node
(or a closed boundary) is zero.
Proof of KCL
proved)(KCLanyfor0)()(
anyfor0)(it.onstoredbetoallowednotisCharge
object.physicalanotisnodeA
)()(
)()(1
ttidt
tdqttq
dttitq
titi
TT
T
TT
n
N
nT
i1i2
in
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Example 1
i1
i3i2
i4
i5
leaving,entering,
52431
54321 0)-()-(
TT
T
ii
iiiii
iiiiii
Example 2
321
312
IIII
IIII
T
T
I1 I2 I3
ITIT
321 IIIIS
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Case with A Closed Boundary
cancelled.arecurrentsbranchInternal
0
0
0
1111
baab
nbn
mam
jbj
iai
iiii
ii
ii
a
Treat the surfaceas a big node
leavingentering ii
b
ia1
ib1
KVL
01
m
M
mv
•Assumption–The principle of conservation of energy
•Statement–The algebraic sum of all voltage drops (or rises)
around a closed path (or loop) is zero.
v1+ _ v2+ _ vm+ _
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Proof of KVL
proved)(KVLanyfor0)(0)(
anyfor0)()()()()(
anyforconstant)(givesenergyofonconservatitheofprincipleThe
)()()(
)()(
1
1
ttvti
ttitvtitvdt
tdwttw
dttitvtw
tvtv
T
Tm
M
m
T
T
TT
m
M
mT
v1+ _ v2+ _ vM+ _
i
Example 1
41532
54321 0
vvvvv
vvvvvvRT
v4v1
v5
+_ +_
+_
v2+ _ v3+ _
Sum of voltage drops = Sum of voltage rises
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Example 2
321
321 0
VVVV
VVVV
ab
ab
V3
V2
V1
Vab
+_
+_
+_
+
_
a
b
Vab+_
+
_
a
b321 VVVVS
Example 3Q: Find v1 and v2.Sol:
V12,V8A4205
03220(2),Eq.into(1)Eq.ngSubstituti
(2)020givesKVLApplying
(1)3,2,lawsOhm'From
21
21
21
vvii
ii
vv
iviv
v1+ _
v2
+
_
20V
2
3+_ i
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Example 4Q: Find currents and voltages.Sol:
(3)8330
03830
0301,looptoKVLAppying
(2)0givesKCL,nodeAt
(1)6,3,8,lawsOhm'By
21
21
21
321
332211
ii
ii
vv
iiia
iviviv
V6V,6V,24
A1A,3A2gives(2)Eq.(5),Eq.&(3)Eq.By
(5)236(1),Eq.By
(4)02,looptoKVLAppying
321
312
2323
2332
vvv
iii
iiii
vvvv
v1+ _
30V
8
3+_
i1
6+
_v3
i3
i2
Loop 1 Loop 2
a
+
_v2
b
Example 5
Q: Find vo.Sol:
V5,A1053035
(2),Eq.into(1)Eq.ngSubstituti
(2)0235givesKVLApplying
(1)5,10,lawsOhm'From
o
oxx
ox
viii
vvv
iviv
i
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Series Resistors
(5)
,Let
(4)or
(3)(2),Eq.&(1)Eq.By
(2)0KVL,Applying
(1),,lawsOhm'By
21eq
eq
21
2121
21
2211
RRR
iRvRR
vi
RRivvv
vvv
iRviRv
v1+ _
v
R1
+_
i
v2+ _
R2a
b
v +_
i
v+ _
Reqa
b
Voltage Division
vRR
vRR
RiRv
vRR
vRR
RiRv
eq
2
21
222
eq
1
21
111
v1+ _
v
R1
+_
i
v2+ _
R2a
b
v +_
i
v+ _
Reqa
b
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Continued
vRR
vRRR
Rv
GGGG
RRRR
eq
n
N21
nn
N21eq
N21eq
1111
v +_
i
v+ _
Reqa
b
v1+ _
v
R1
+_
i
v2+ _
R2a
b
vN+ _
RN
Parallel Resistors
(5)or
(4)111
(3)11
(2),Eq.&(1)Eq.By(2)
,nodeatKCLApplying
(1),or
,lawsOhm'By
21
21
21eq
eq2121
21
22
11
2211
RRRR
R
RRR
Rv
RRv
Rv
Rv
i
iiia
Rv
iRv
i
RiRiv
eq
i a
b
R1+_ R2v
i1 i2
i a
b
Req or Geq+_v v
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Current Division
iGG
GRRiR
Rv
i
iGG
GRRiR
Rv
i
RRRiR
iRv
21
2
21
1
22
21
1
21
2
11
21
21eq
i a
b
R1+_ R2v
i1 i2
i a
b
Req or Geq+_v v
Continued
iGG
iGGG
Gi
GGGG
RRRR
eq
n
N21
nn
N21eq
N21eq
1111
i a
b
Req or Geq+_v v
i a
b
R1+_ R2v
i1 i2
RN
iN
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iGG
iGGG
Gi
GGGG
RRRR
eq
n
N21
nn
N21eq
N21eq
1111
vRR
vRRR
Rv
GGGG
RRRR
eq
n
N21
nn
N21eq
N21eq
1111
Brief Summary
v1+ _v
R1
+_
i
v2+ _
R2a
b
vN+ _
RN i a
b
R1+_ R2vi1 i2
RN
iN
Example
Req
6 35
8
2
4 1
Req
26
8
2
4
Req 2.48
4
Req 14.4
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How to solve the bridge network?
•Resistors are neitherin series nor inparallel.•Can be simplified by
using 3-terminalequivalent networks.
R1
R2 R3R4
R5R6
Wye (Y)-Delta () Transformations
R3
R1 R21
2
3
4
R3
R1 R2
3
4
1
2
Rb
Rc1
2
3
4
RaRb
Rc1
2
3
4
Ra
Y T
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to Y Conversion
cba
bac
cab
RRRRRRR
RRRRRRR
RRRRRRR
//)()Y(
//)()Y(
//)()Y(same.thebemustbehaviorterminal-Two
323434
211313
311212
cba
ba
cba
ac
cba
cb
RRRRR
R
RRRRR
R
RRRRR
R
3
2
1
R3
R1 R2
3
4
1
2Y
Rb
Rc1
2
3
4
Ra
Y-Transformations
3
133221
2
133221
1
133221
RRRRRRR
R
RRRRRRR
R
RRRRRRR
R
c
b
a
cba
ba
cba
ac
cba
cb
RRRRR
R
RRRRR
R
RRRRR
R
3
2
1
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Example
Rab
12.5
15
510
30
20
a
b
Rab
12.5
15
17.570 30
a
b
35
Rab7.292
10.521
a
b
Rab9.632
a
b
Applications: Lighting Systems
0N21 ... Vvvv NV
vvv 0N21 ...
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Applications: DC Meters
Parameters:IFS: full-scale currentRm: meter resistance
Continued
RR V R
RR A
R
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Continued
elementmeter
elementmeter
VVII
elementmeter
elementmeter
VVII
Voltmeters
mnFSFS RRIV range-Single
mFSFS
mFSFS
mFSFS
RRIV
RRIV
RRIV
33
22
11
range-Multiple
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Ammeters
nmnFSM RRRII :range-Single
333
222
111
:range-Multiple
RRRII
RRRII
RRRII
mFSM
mFSM
mFSM