Chapter 2 Resistive Circuits - Seoul National Universityengineering.snu.ac.kr/lecture_pdf/EE/2005 -...
Transcript of Chapter 2 Resistive Circuits - Seoul National Universityengineering.snu.ac.kr/lecture_pdf/EE/2005 -...
Chapter 2Resistive Circuits
1. Solve circuits by combining resistances in Series and Parallel.
2. Apply the Voltage-Division and Current-Division Principles.
3. Solve circuits by the Node-Voltage Technique.
4. Solve circuits by the Mesh-Current Technique.
5. Find Thévenin and Norton Equivalents.
6. Apply the Superposition Principle.
7. Understand the Wheatstone Bridge.
Goal
Resistance in Series
v2=R2iv1=R1i v3=R3iOhm’s Law
Using KVL, v = v1 + v2 + v3
v = (R1 + R2 + R3)i
Req = R1 + R2 + R3
Resistance in Parallel
v=R2iv=R1i v=R3iOhm’s Law
Using KCL, i = i1 + i2 + i3
i = (1/R1 + 1/R2 + 1/R3)v
Req = 1/R1 + 1/R2 + 1/R3
Example
Series Parallel
Series
Network Analysis using Series/Parallel Equivalents
1. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.
2. Redraw the circuit with the equivalent resistance for the combination found in step 1.
3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often end with a single source and a single resistance.
4. Solve for the currents and voltages in the final equivalent circuit. Transfer Results to Backward until Original Circuit
Network Analysis : The process of determining the Current, Voltage & Power for each element given in Circuit & Element Values
Step 1,2
Step 3
Step 1,2 & 3
Step 4
Voltage Division
total321
111 v
RRRRiRv
++== total
321
222 v
RRRRiRv
++==
Of the total voltage, the fraction that appears across a given resistance in a series circuit is the ratio of the given resistance to total series resistance Voltage Division Principle
Application of the Voltage-Division Principle
V5.1
156000200010001000
1000
total4321
11
=
×+++
=
+++= v
RRRRRv
Generalized Voltage Divider
total
in
ZVI =
21 ZZZ +=total
21
22 ZZ
ZVZIV+
== inout
Current Division
total21
2
11 i
RRR
Rvi
+== total
21
1
22 i
RRR
Rvi
+==
For two resistance in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to sum of the two resistance (Two Resistors Case) Current Division Principle
Application of the Current-Division Principle
Ω=+×
=+
= 2060306030
32
32eq RR
RRR
A10152010
20
eq1
eq1 =
+=
+= siRR
Ri
Position Transducer Based on Voltage-Division Principle
θKRR
Rvv so =+
=21
2
Typical Knob in Electrical System based on Analog Signalsuch as Amplifier, Adjustable Light, Volume
Series Sources
• Ideal independent voltage sources in series add algebraically
I+ - + - + - + -
I V1 V2 V3Vn
- VR +
R
Note : Parallel Voltage Sources are not Resolvable. WHY?
Parallel Sources
• Ideal independent current sources in parallel add algebraically
In
I3I2I1
R
+
V
_
IT
•Note : Series current sources are not resolvable. WHY?
Example
• R2 and R3 are effectively open circuited and therefore can be omitted• R7 and R8 are short circuited, and can be omitted• I1 and I2 are Algebraically added because they are in Series.• V1 and V2 are Algebraically added because they are in Parallel• R1 and R4 are in Series & R5 & R6 are in Parallel
Although they are veryimportant concepts,
series/parallel equivalents andthe current/voltage division
principles are not sufficient tosolve all circuits.
Wye - Delta Transformations
• Use Y to ∆ transformation
- Need to find equivalent resistance to determine current. - HOW?- (They are not in series, not in parallel)
Equating Resistance's• Resistance between X - Y• In ∆ ⇒ Ra // (Rb + Rc)
• In Y ⇒ R1 + R3
21)()( RR
RRRRRRR
cab
cabXZ +=
+++
=
RbRa
Rc
X
Y Z
R1 R2
R3
X
Y
Z
31)()( RR
RRRRRRR
cba
cbaXY +=
+++
=
32)()( RR
RRRRRRR
bac
bacYZ +=
+++
=
Similarly
Above 3 Equations are Linearly Solvable between R1 , R2 , R3 & Ra , Rb , Rc
X
Y Z
Y
X Z
∆ Y
With Voltage Source With Dependent Source
∆ Y Transformation : Not Enough for All Circuits
Node-Voltage Analysis
1. Select Reference Node2. Assign Node Voltages
If the reference node is chosen at one end of an independent voltage source,one node voltage is known at the start, and fewer need to be computed.
3. Write Network Equations. - First, use KCL to write current equations for nodes and supernodes.
(Write as many current equations as you can without using all of the nodes.) - If not have enough equations because of voltage sources between nodes ,
(use KVL to write additional equations)- If the circuit contains dependent sources,
Find expressions for the controlling variables in terms of the node voltages.Substitute into the network equationsObtain equations having only the node voltages as unknowns.
4. Solve for the Node Voltages. (Equation in standard form) 5. Calculate currents or voltages of interest.
with the node voltages from Step 4
Select Reference Node & Assign Node Voltage
Node : Two More Circuit Elements are Joined TogetherReference Node : Node with Known & Referencing Voltage Ground Symbol : Zero Voltage
- Typically Reference Node is set to be Ground Symbol- Negative Polarity to Reference Node
Assign Element Unknown Voltages
12 vvvy −= 32 vvvx −=
- Assign Node Voltage across R1, R2 & R3- Use KVL In R2 In R3
13 vvvz −=In R1
Writing KCL Equations with Node Voltages
03
32
4
2
2
12 =−
++−
Rvv
Rv
Rvv
03
23
5
3
1
13 =−
++−
Rvv
Rv
Rvvsvv =1
3 Unknown Node Voltages 3 Equations
At Node 1 At Node 2
At Node 3
KCL
1133
22
Solution of Equations
Solving KCL equation as Standard form# of Unknown Node Voltage = # of Equation
2222121
1212111
ivgvgivgvg
=+=+
3333232131
2323222121
1313212111
ivgvgvgivgvgvgivgvgvg
=++=++=++
With 2 Unknown Node Voltages With 3 Unknown Node Voltage
Generally
ijij ivg =
Solution of Linear Simultaneous Equation gives Node Voltage
Calculation of Remaining Node Voltages & Currents
Tips on KCL Equation Writing
03
32
4
2
2
12 =−
++−
Rvv
Rv
Rvv
At Node 2
- Writing KCL Node not Connected to Voltage Source (note : Supernode) - Direction of Current from Node : Positive
(Node Voltage of Interest – Node Voltage Surrounding ) / Resistance = Current Output from Node of Interest to Node Surrounding- Sum of All Currents from Surrounding Nodes
Node 2 Node 1 Node 2 Ground Node 2 Node 3
02
21
1
1 =+−
+ siRvv
Rv 0
4
32
3
2
2
12 =−
++−
Rvv
Rv
Rvv
siRvv
Rv
=−
+4
23
5
3
KCL Equations with a Current Source
v2-v1 v2-v3
Node 1 Node 3Node 2
KCL Equations with 2 Current Sources
Node 1 02
21
1
31 =−−
+−
aiRvv
Rvv
Node 2 04
32
3
2
2
12 =−
++−
Rvv
Rv
Rvv
Node 3 01
13
4
23
5
3 =+−
+−
+ biRvv
Rvv
Rv
Different choice of Reference Node
Node 1 010520
21131 =−
++− vvvvv
Node 2 05
1010
3212 =−
++− vvvv
Node 3 051020
23313 =−
++− vvvvv
VvVvVv
45.4573.7227.27
3
2
1
−=−=−=
Aix 909.0=
Same Logic regardless of Reference Node
KCL Equations with Voltage & Current Sources
Node 1 01210
5121 =−−
+− vvv
Node 2 0510
105
1222 =−
+−
+vvvv
VvVv 129.632.10 21 ==
Same Logic for Current & Voltage Sources Containing Circuit
Node 3 : Not Writing KCL on , because of Voltage Source
Circuits with Voltage Sources
( ) ( ) 01515
3
2
4
2
1
1
2
1 =−−
++−−
+R
vRv
Rv
Rv
“The Net Current flowing through any closed surface must equal to Zero”Supernode : Combination of Node including Voltage Sources
If Write KCL equations with All Node & Supernodes
When a Voltage Source is connected between Nodes, KCL can not be written at Nodesbecause both Current & Resistance are not defined in Voltage Source Above Circuit can not Write KCL, because All node are connected to Voltage Sources
( ) ( ) 01515
3
2
4
2
1
1
2
1 =−−
−−−−
−−R
vRv
Rv
Rv
Supernode 1 Supernode 2
Supernode 1
Supernode 2Dependant EqnsUse KVL at Supernode
( ) ( ) 01515
3
2
4
2
1
1
2
1 =−−
++−−
+R
vRv
Rv
Rv
010 21 =+−− vv
Solution by Supernode
KCL
KVL
Supernode
Independent Equation!! Solvable
at Supernode
Write KVL at Supernode Write KCL for Super node
010 21 =++− vv
13
32
2
31
1
1 =−
+−
+R
vvR
vvRv
04
3
3
23
2
13 =+−
+−
Rv
Rvv
Rvv
14
3
1
1 =+Rv
Rv
Supernode Example
KVL at Supernode
KCL around Supernode
KCL at Node 3
KCL at Ground
Same Equation!
Node-Voltage Analysis with a Dependent Source
First, write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source.
xs iiR
vv 21
21 +=−
03
32
2
2
1
12 =−
++−
Rvv
Rv
Rvv
024
3
3
23 =++−
xiRv
Rvv
1
3
2
Substitution yields
3
23
1
21 2R
vviR
vvs
−+=
−
03
32
2
2
1
12 =−
++−
Rvv
Rv
Rvv
023
23
4
3
3
23 =−
++−
Rvv
Rv
Rvv
Next, find an expression for the controlling variable ix in terms of the node voltages.
3
23
Rvvix
−=
Solvable!
Example with Dependant Voltage Source
050 21 =++− vv.v x 13 vvvx −=Supernode (Node 1 & 2) KVL
KCL
01
13
3
23
4
3 =−
+−
+R
vvR
vvRv
siRvv
Rvv
Rv
=−
+−
+3
32
1
31
2
1
Node 3
Ground siRv
Rv
=+4
3
2
1311 ,, vvvforSolution
Mesh-Current Analysis
1. Define the Mesh Currents (usually a clockwise direction)If necessary, redraw the network without crossing conductors or elements.
2. Write Network equations ( # of equations = # of mesh currents).
- Use KVL for meshes with no current sources. - If current sources, write expressions for their currents in terms of the mesh currents. - If common current source to two meshes, Use KVL for Supermesh.- If dependent sources,
Find expressions for the controlling variables Substitute into the network equations,Obtain equations having only the mesh currents as unknowns.
3. Solve for the mesh currents (Standard form equation)
4. Calculate any other currents or voltages of interest.
Basic Rule for Mesh Current
KVL for vA, R1 & R3 AviRiR =+ 3311
KVL for vB, R2 & R3 BviRiR −=+− 2233
KCL for Node 1321 iii +=
Node-Voltage Analysis Rearrange
AviiRiR =−+ )( 21311
BviRiiR =+−− 22213 )(
When Several Mesh Currents flow through one Elements,the Current in that Element to be Algebraic Sum of the Mesh Current
Choosing the Mesh Currents
Only for Planar Network : Network without Crossing Element
Mesh : Loop of Elements Mesh Current : Current flowing through MeshDefine Current Direction Clockwise for ConsistencyExample, Current flow in R2 to Left = i3 - i1
Current flow in R3 to Up = i2 - i1
Writing Equations to Solve for Mesh Currents
If Network with only resistances and independent voltage sources,write required equations by following each current around its mesh &applying KVL.
( ) 024123 =++− BviRiiR
( ) 031132 =−+− BviRiiR
( ) ( ) 021312 =−−+− As viiRiiRMesh 1
Mesh 2
Mesh 3
( ) ( ) 021441211 =−−+−+ AviiRiiRiR
( ) ( ) 032612425 =−+−+ iiRiiRiR
( ) ( ) 043823637 =−+−+ iiRiiRiR
( ) ( ) 034814243 =−+−+ iiRiiRiR
Example of Mesh Current with Voltage Source
Solution of Network Equation & Find Voltage at Node
Example of Mesh-Current Analysis
Mesh 1 20i1 + 10(i1 – i2) – 150 = 0
Mesh 2 10(i2 – i1)+ 15 i2 + 100 = 0
30i1 - 10(i1 – i2) = 150 -10 i1+ 25 i2 = -100
In standard form i1 = 4.231Ai2 = -2.308A
Current through R2 downward = i1 – i2 = 6.539A
Mesh Currents in Circuits with Current Sources
A common mistake made by beginning students isto assume that the voltages across current sources are zero.
A21 =i
0105)(10 212 =++− iiiKVL to Mesh 2
Combine meshes 1 and 2 into a Supermesh. Write KVL equation around the periphery of meshes 1 & 2
( ) ( ) 01042 32311 =+−+−+ iiiii( ) ( ) 0243 13233 =−+−+ iiiii
512 =− ii
Mesh Currents with Voltage & Current Sources
KVL for Mesh 1 & 2 is impossible,because Voltage across 5A is unknown
Mesh 3
Current Source 5A
( ) 0100510 212 =−+− iii
Ai 51 −=
01020105 21 =−++ ii
Aii 112 =−
Supermesh
Example of Mesh-Current Analysis
026420 221 =+++− iii
124iivx −=
22ivx =
Mesh-Current with Dependant Source
Supemesh
Dependant Source
- Write Equations exactly the same as Independent Source- Express Dependant Variable with Mesh Current- Substitute into Network Equation
Thévenin Equivalent Circuits
=
Real Complex CircuitEasy UnderstandableEquivalent Circuit
a
a
b
a
b
Thévenin Equivalent Circuits
ocvVt =
sc
oc
ivRt =
Example of Thévenin Equivalent Circuits
ARR
vi s 1.021
1 =+
=
ViRvoc 51.05012 =×==
ARvi s
sc 15.010015
1
===
Ω=== 3.3315.05
sc
oct i
vR
To find Open Circuit Voltage To find Short Circuit Current
Ω=+
=+
= 3.3350100
5000
501
1001
1eqR Rt equals Req with Zero Voltage!
Example of Thévenin Equivalent Circuits
Aioc 5=
ViRvoc 5051011 =×==
ARR
RAisc 1521
1 =+
=
Ω=== 50150
sc
oct i
vR
To find Open Circuit Voltage To find Short Circuit Current
ioc
voc i1
isc
Req = 10 + 40 = 50Ω Rt equals Req with Zero Current!
Finding the Thévenin Resistance Directly
If No Dependant Source,When zeroing a Voltage source, it becomes a Short circuitsWhen Zeroing a Current Source, it becomes a Open Circuit
In Zeroing Independent Source, Replace a Voltage Source with Short Circuit Replace a Current Source with Open Circuit.
When the Source is Zeroed, the resistance seen from the Circuit terminals is equal to the Thévenin Resistance
Short Circuit
Direct Finding the Thévenin Resistance
b) Zeroing SourceΩ=
+=
+= 4
201
51
111
1
21 RR
Req
c) Short Circuit
02 =isciii +=+ 21 2 A
Rvi s 4
520
11 === AiSC 6=
ViRV sctt 2464 =×==
Exercise 2.24
Find Thévenin Resistance
Zero
Rt=30Ω
Zero
Ω=+
+= 14
201
51
110eqR
Ω=+
+= 10
201
51
161eqR Ω=102
eqR Ω=+
= 5
101
101
1eqR
Zero Zero
Thévenin Resistance with Dependant Source
102 oc
xxvii =+
Can not use Zeroing Source Technique due to dependant source
Open Circuit : Node-Voltage Tech
KCL at Node 1KVL Loop ocx vi −=105
Vvoc 57.8=
Short Circuit
AAix 25
10== Aii xsc 63 ==
Ω=== 43.1657.8AV
ivR
sc
oct
Norton-Equivalent Circuit Analysis
Norton Equivalent Circuit : Independent Current Source +Thévenin Resistance
In is isc in Thévenin Equivalent Analysis
Step-by-step Thévenin/Norton-Equivalent-Circuit Analysis
1. Perform two of these:a. Determine the open-circuit voltage Vt = voc.b. Determine the short-circuit current In = isc.c. Zero the sources and find the Thévenin resistance Rt
looking back into the terminals.2. Use the equation Vt = Rt In to compute the remaining value.3. Thévenin equivalent : a voltage source Vt in series with Rt .4. Norton equivalent : a current source In in parallel with Rt .
0154 321
=+
+−
+RR
vR
vv ococxKCL at Node
KVL at Loopococx vv
RRRv 25.0
32
3 =+
=
Vvoc 62.4=
Current along R2 & R3 is Zero : vx = 0
ocs
sc vRvi 75.0
2015
1
===
Ω=== 15.675.062.4
sc
oct i
vR
Example of Norton-Equivalent-Circuit Analysis
Maximum Power TransferThe load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.
Lt
tL RR
Vi+
= LL RipL
2=( )2
2
Lt
LtL RR
RVp+
=
( ) ( )( )
024
222
=+
+−+=
Lt
LtLtLtt
L
L
RRRRRVRRV
dRdp
At Power Maximum
Lt RR =t
tmaxL R
Vp4
2
=
SUPERPOSITION PRINCIPLE
The total response is the sum of the responses to each of the Independent Sources acting individually with Other
Independent Source Zeroed. When Zeroed, Current Source become Open and Voltage Source become short.
nT rrrr +++= .......21
In equation form, this is
In Circuit with Resistance, Linear Dependant Source, Independent Source
Superposition Principle
Response : Voltage or Current
221
1sx
TsT iKiRv
Rvv
=++−
2Rvi T
x =
2121
211
121
2ssT i
KRRRRRv
KRRRRv
+++
++=
1121
21 sv
KRRRRv++
=
2121
212 siKRRR
RRv++
=
Example of Superposition PrinciplevT is Interesting
KCL at Top
If is2 is set to Zero, Response due to vs1 :
If vs1 is set to Zero, Response to is2 :
21 vvvT +=In General, dependant sources do not contribute a separate term to the total
response , and we must zero dependant source in applying superposition
WHEATSTONE BRIDGE
The Wheatstone bridge is used by mechanical and civil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings.
31
2 RRRRx =
Wheatstone Bridge
42
4
31
3
RRVRV
RRVRV ba +
=+
=
( ) ( )4231
4123
42
4
31
3
RRRRRRRRV
RRVR
RRVRV
+⋅+−
=
+−
+=∆
4123@ RRRRVoltageZero =
ba VVV −=∆ Va = Voltage @ aVb = Voltage @ b
DC Wheatstone bridgeV : Bridge Supply VoltageD : Voltage Detector
GTh
ThG RR
VI+
=
42
42
31
31
RRRR
RRRRRTh +
++
=
( )( )4231
4123
RRRRRRRRVVTH ++
−=
Bridge Circuits(Galvanometer)
Galvanometer BridgeV : Bridge Supply VoltageG : Voltage Detector
G
Lead Compensation for Remote Sensor Contact Resistance of each Resistor
Lead Compensation
( ) ( )L
L
RRRRRRRRRRVV2
)2(
4231
4123
++⋅++−
=∆
LRRR 244 +→Lead Problem
- RL makes ∆V change- Environmental Effect
Temperature, Stress,Vapor- Transient effect
( ) ( )4231
4123
RRRRRRRRVV+⋅+
−=∆
RL
RL
( ) ( )4231
4123
RRRRRRRRVV+⋅+
−=∆
g
Lead Compensated Circuit
Lead Compensation by Power Lead ExtensionConnection (3) : g→c
No Change on ∆VConnection (1) & (2) : R3 → R3 + RL , R4 → R4 + RL
Same Condition : R3 & R4 are identically changedFinally No Change in Bridge
RL
RL
RL
( ) 54254 & RRRRR >>+>>
( )5
542
54 IRRRR
RRVVb +++
+=
( )5
542
54
31
3 IRRRR
RRVRR
VRV −++
+−
+=∆
Current Balance Bridge
1. Splitting R4→ R4 +R52. Current on R5 <<R4 to make ∆V null
by adjusting magnitude and polarity of current I manually or electronically
To make null bridge circuit
axC VVV +=
baxbC VVVVVV −+=−=∆
042
4
31
3 =+
−+
+RR
VRRR
VRVx
( ) 05542
54
31
3 =−++
+−
++ IR
RRRRRV
RRVRVx
Potential Measurement
Make ∆V zero (null condition)
Or current balance combination
05 =− IRVx
Bridge off-null voltage ∆V(a) nonlinear for large-scale changes in resistance(b) nearly linear at small ranges of resistance change
Linearity of Bridge Circuit
( ) ( )4231
4123
RRRRRRRRVV+⋅+
−=∆