loop analysis the second systematic technique to determine
TRANSCRIPT
The second systematic technique to determine all currents and voltages in a circuit
IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUITAND THEN IT USES OHM’S LAW TO COMPUTE NECESSARY VOLTAGES
THERE ARE SITUATION WHERE NODE ANALYSIS IS NOT AN EFFICIENT TECHNIQUEAND WHERE THE NUMBER OF EQUATIONS REQUIRED BY THIS NEW METHOD ISSIGNIFICANTLY SMALLER
LOOP ANALYSIS
+
-
+
-
1R 2R
3RV18
V12
2RV 1RV
3RV
Apply node analysis to this circuit
There are 4 non reference nodes
There is one super node
There is one node connected to thereference through a voltage source
We need three equations to compute all node voltages
…BUT THERE IS ONLY ONE CURRENT FLOWING THROUGH ALL COMPONENTS AND IF THAT CURRENT IS DETERMINED ALL VOLTAGES CAN BE COMPUTED WITH OHM’S LAW
I
STRATEGY:1. Apply KVL(sum of voltage drops =0)
0][18][12321
RRRVVVVV
2. Use Ohm’s Law to expressvoltages in terms of the “loop current.”
0][18][12 321 IRVIRIRV
RESULT IS ONE EQUATION IN THE LOOP CURRENT!!!
SHORTCUT
Skip this equation
Write this one
directly
3V2V1V
4V
LOOPS, MESHES AND LOOP CURRENTS
EACH COMPONENTIS CHARACTERIZEDBY ITS VOLTAGEACROSS AND ITSCURRENT THROUGH
A LOOP IS A CLOSED PATH THAT DOES NOTGO TWICE OVER ANY NODE.THIS CIRCUIT HAS THREE LOOPS
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
fabcdef
A MESH IS A LOOP THAT DOES NOT ENCLOSEANY OTHER LOOP.fabef, ebcde ARE MESHES
A LOOP CURRENT IS A (FICTICIOUS) CURRENTTHAT IS ASSUMED TO FLOW AROUND A LOOP
fabef ebcde
1I
2I
3I
CURRENTS LOOP ARE321
,, III
A MESH CURRENT IS A LOOP CURRENT ASSOCIATED TO A MESH. I1, I2 ARE MESHCURRENTS
CLAIM: IN A CIRCUIT, THE CURRENT THROUGHANY COMPONENT CAN BE EXPRESSED IN TERMSOF THE LOOP CURRENTS
1
2 3
4
56
7
A BASIC CIRCUIT
ab c
def
FACT: NOT EVERY LOOP CURRENT IS REQUIREDTO COMPUTE ALL THE CURRENTS THROUGHCOMPONENTS
1I
3I3
1
31
II
II
III
cb
eb
fa
CURRENTS LOOP
TWO USING
32
21
31
III
III
III
cb
eb
fa
EXAMPLES THE DIRECTION OF THE LOOPCURRENTS IS SIGNIFICANT
FOR EVERY CIRCUIT THERE IS A MINIMUMNUMBER OF LOOP CURRENTS THAT ARENECESSARY TO COMPUTE EVERY CURRENTIN THE CIRCUIT.SUCH A COLLECTION IS CALLED A MINIMALSET (OF LOOP CURRENTS).
FOR A GIVEN CIRCUIT LETB NUMBER OF BRANCHESN NUMBER OF NODES
THE MINIMUM REQUIRED NUMBER OF LOOP CURRENTS IS
)1( NBL
MESH CURRENTS ARE ALWAYS INDEPENDENT
AN EXAMPLE
2)16(7
6
7
L
N
BTWO LOOP CURRENTS AREREQUIRED.THE CURRENTS SHOWN AREMESH CURRENTS. HENCE THEY ARE INDEPENDENT ANDFORM A MINIMAL SET
KVL ON LEFT MESH
REPLACING AND REARRANGING
THESE ARE LOOP EQUATIONS FOR THECIRCUIT
DETERMINATION OF LOOP CURRENTS
KVL ON RIGHT MESH
2 4 5 30
Sv v v v
USING OHM’S LAW
1 1 1 2 1 2 3 1 2 3
4 2 4 5 2 5
, , ( )
,
v i R v i R v i i R
v i R v i R
IN MATRIX FORM
1 2 3 3 11
3 3 4 5 22
S
S
R R R R vi
R R R R vi
WRITE KVL ON EACH MESH
WRITE THE MESH EQUATIONS
IDENTIFY ALL VOLTAGE DROPS 4Rv
1Rv
2Rv
5Rv
3Rv
BOOKKEEPINGBRANCHES = 8NODES = 7
LOOP CURRENTS NEEDED = 2
02211 RSRS vvvv :MESH TOP
033452 RSRRR vvvvv :BOTTOM
USE OHM’S LAW
11Ri
52Ri
42Ri
221 )( Rii
32Ri
AND WE ARE TOLD TOUSE MESH CURRENTS!THIS DEFINES THE LOOPCURRENTS TO BE USED
+
-
+ -
V1
V2R1
R2 R3
R4R5
WRITE THE MESH EQUATIONS
1I
2I
DRAW THE MESH CURRENTS. ORIENTATIONCAN BE ARBITRARY. BUT BY CONVENTIONTHEY ARE DEFINED CLOCKWISE
NOW WRITE KVL FOR EACH MESH AND APPLYOHM’S LAW TO EVERY RESISTOR.
0)( 51221111 RIRIIRIV
AT EACH LOOP FOLLOW THE PASSIVE SIGN CONVENTION USING LOOP CURRENT REFERENCE DIRECTION
0)( 21242322 RIIRIRIV
DEVELOPING A SHORTCUT
WHENEVER AN ELEMENTHAS MORE THAN ONELOOP CURRENT FLOWINGTHROUGH IT WE COMPUTENET CURRENT IN THE DIRECTION OF TRAVEL
SHORTCUT: POLARITIES ARE NOT NEEDED.APPLY OHM’S LAW TO EACH ELEMENT AS KVLIS BEING WRITTEN
1I @KVL
2I @KVL
396
12612
21
21
kIkI
kIkIREARRANGE
add and 2/*
mAIkI 5.0612 22
mAIkIkI4
561212 121
EXPRESS VARIABLE OF INTEREST AS FUNCTIONOF LOOP CURRENTS
21 IIIO
1I @KVL
2I @KVL
1IIO NOW
THIS SELECTION IS MORE EFFICIENT
996
12612
21
21
kIkI
kIkIREARRANGE
substract and 2/*
3/*
mAIkI4
31824 11
LEARNING EXAMPLE: FIND Io USING LOOP ANALYSII
AN ALTERNATIVE SELECTION OF LOOP CURRENTS
A PRACTICE EXAMPLEIF THE CIRCUIT CONTAINS ONLY INDEPENDENTSOURCE THE MESH EQUATIONS CAN BE WRITTEN“BY INSPECTION”
THE RIGHT HAND SIDE IS THE ALGEBRAIC SUMOF VOLTAGE SOURCES AROUND THE LOOP(VOLTAGE RISES - VOLTAGE DROPS)
MUST HAVE ALL MESH CURRENTS WITH THESAME ORIENTATION
IN LOOP K
THE COEFFICENT OF Ik IS THE SUM OF RESISTANCES AROUND THE LOOP.
THE COEFFICIENT OF Ij IS THE SUM OFRESISTANCES COMMON TO BOTH k AND j ANDWITH A NEGATIVE SIGN.
LOOP 1 12612 21 kIkI
LOOP 2 396 21 kIkI
LOOP 1 kkI 641 oft coefficien
02 I oft coefficien
kI 63 oft coefficien
LOOP 2 01 I oft coefficien
kkI 392 oft coefficien
kI 33 oft coefficien
][6 V RHS
][6 VRHS
Loop 31 2 3
(6 ) (3 ) (3 6 12 ) 0k I k I k k k I
1. DRAW THE MESH CURRENTS
1I 2I
2. WRITE MESH EQUATIONS
MESH 1 ][32)242( 21 VkIIkkk
MESH 2 )36()62(2 21 VVIkkkI DIVIDE BY 1k. GET NUMBERS FOR COEFFICIENTSON THE LEFT AND mA ON THE RHS
][982
][328
21
21
mAII
mAII
add and 4/*
][3330 2 mAI ][
5
336 2 VkIVO
LEARNINGEXTENSION
3. SOLVE EQUATIONS
12k
6k
4k4k
2k
12V
9V
WRITE THE MESH EQUATIONS
BOOKKEEPING: B = 7, N = 4
1I
2I
3I
4I 1. DRAW MESH CURRENTS
2. WRITE MESH EQUATIONS. USE KVL
0)(61212 311 IIkVkI :1 MESH
0)(4)(412 3242 IIkIIkV :2 MESH
0)(4)(69 2313 IIkIIkV :3 MESH
02)(49 424 kIIIkV :4 MESH
EQUATIONS BY INSPECTION
VkIkI 12618 31
VkIkIkI 12448 432
VkIkIkI 91046 321
VkIkI 964 42
CHOOSE YOUR FAVORITE TECHNIQUETO SOLVE THE SYSTEM OF EQUATIONS
CIRCUITS WITH INDEPENDENT CURRENT SOURCES
THERE IS NO RELATIONSHIP BETWEEN V1 ANDTHE SOURCE CURRENT! HOWEVER ...
MESH 1 CURRENT IS CONSTRAINED
MESH 1 EQUATION mAI 21
MESH 2
VkIkI 282 21 “BY INSPECTION”
][2
96
4
3
8
2)2(222 VkIVmA
k
VmAkI O
CURRENT SOURCES THAT ARE NOT SHAREDBY OTHER MESHES (OR LOOPS) SERVE TO DEFINE A MESH (LOOP) CURRENT AND REDUCE THE NUMBER OF REQUIRED EQUATIONS
TO OBTAIN V1 APPLY KVL TO ANY CLOSEDPATH THAT INCLUDES V1
KVL
ANALYSISMESH USING COMPUTE OV
TWO MESH CURRENTS ARE DEFINED BY CURRENTSOURCES
mAImAI 24 21 MESH 3
“BY INSPECTION” VkIkIkI 31242 321
mAk
mAkmAkVI
4
1
12
)2(4)4(233
KVL FOR Vo
USE KVL TOCOMPUTE Vo
LEARNING EXAMPLE
WE ACTUALLY NEED THE CURRENT ON THE RIGHT MESH. HENCE, USE MESH ANALYSIS
1I2I
MESH 1: mAI 41
MESH 2: 06)(4][5 212 kIIIkV
mAmAmAI 1144510 2
][5
336 2 VkIVO
1I
2I
MESH 1: mAI 41
MESH 2: 0124 21 kIkI
mAI3
4
12
162
][86 2 VkIVO
LEARNING EXTENSIONS
+
-
+
VO
_
6k
4k
2k
2k
IS
VS
IS = 2mA, VS = 6V
Determine VO
Problem 3.46 (6th Ed)
1. Select loop currents.
In this case we use meshes. We note that the current source could define one mesh.
1I
2I
3I
2. Write loop equations.
Loop 1SII 1
Loop 2 0)(2)(4 1232 IIkIIkVS
Loop 3 0)(26)(4 13323 IIkkIIIk
Since we need to compute Vo it is efficient to solve for I3 only.
HINT: Divide the loop equations by 1k. Coefficientsbecome numbers and voltage source becomes mA.
mAII28
323421028 33
VkIVO7
486 3
3 non-reference nodes. 3 meshes
One current source, one super node
SELECTING THE SOLUTION METHOD
BOTH APPROACHES SEEM COMPARABLE. CHOOSELOOP ANALYSIS
We use the fact that I1 = Is
])[46(21
46 132 mAIk
VII S
Loop 2
mAIII S 42124 32 Loop 3
2/*
eqs add and 3/*
CURRENT SOURCES SHARED BY LOOPS - THE SUPERMESH APPROACH
1. SELECT MESH CURRENTS
2. WRITE CONSTRAINT EQUATION DUE TOMESH CURRENTS SHARING CURRENT SOURCES
mAII 432
3. WRITE EQUATIONS FOR THE OTHER MESHES
mAI 21
4. DEFINE A SUPERMESH BY (MENTALLY)REMOVING THE SHARED CURRENT SOURCE
SUPERMESH
5. WRITE KVL FOR THE SUPERMESH
0)(1)(2216 131223 IIkIIkkIkI
NOW WE HAVE THREE EQUATIONS IN THREE
UNKNOWNS. THE MODEL IS COMPLETE
CURRENT SOURCES SHARED BY MESHES - THE GENERAL LOOP APPROACH
THE STRATEGY IS TO DEFINE LOOP CURRENTS THAT DO NOT SHARE CURRENT SOURCES -
EVEN IF IT MEANS ABANDONING MESHES
FOR CONVENIENCE START USING MESH CURRENTSUNTIL REACHING A SHARED SOURCE. AT THATPOINT DEFINE A NEW LOOP.
IN ORDER TO GUARANTEE THAT IF GIVES ANINDEPENDENT EQUATION ONE MUST MAKE SURETHAT THE LOOP INCLUDES COMPONENTS THATARE NOT PART OF PREVIOUSLY DEFINED LOOPS
A POSSIBLE STRATEGY IS TO CREATE A LOOPBY OPENING THE CURRENT SOURCE
mAI
mAI
4
2
2
1
THE LOOP EQUATIONS FOR THE LOOPS WITHCURRENT SOURCES ARE
THE LOOP EQUATION FOR THE THIRD LOOP IS
0)(1)(2)(21][6 13123233 IIkIIIkIIkkIV
THE MESH CURRENTS OBTAINED WITH THISMETHOD ARE DIFFERENT FROM THE ONESOBTAINED WITH A SUPERMESH. EVEN FORTHOSE DEFINED USING MESHES.
-
+
1R
2R
3R
4R
1SI 2SI
3SISV
RESISTORS ACROSS VOLTAGESFIND
Three independent current sources.Four meshes.One current source shared by twomeshes.
For loop analysis we notice...
Careful choice of loop currents should make only one loop equationnecessary. Three loop currents canbe chosen using meshes and notsharing any source.
1I 2I
3I
)( 43111 IIIRV
)( 1222 IIRV
SOLVE FOR THE CURRENT I4.USE OHM’S LAW TO C0MPUTE REQUIREDVOLTAGES
Now we need a loop current that doesnot go over any current source and passes through all unused components.
HINT: IF ALL CURRENT SOURCES ARE REMOVEDTHERE IS ONLY ONE LOOP LEFT
4I
11 sII
22 SII
33 SII
MESH EQUATIONS FOR LOOPS WITHCURRENT SOURCES
0)()()( 3441341243 IIRIIIRIIRVSKVL OF REMAINING LOOP
)( 4233 IIRV
)( 4344 IIRV
4V
1V
2V
3V
-
+
1R
2R
3R
4R
1SI 2SI
3SISV
A COMMENT ON METHOD SELECTIONThe same problem can be solved by node analysis but it requires 3 equations
1V
2V3V
4V
0
0
0
1
14
1
41
1
43
2
13
3
233
21
2
31
2
R
VV
R
VI
R
VV
R
VV
R
VVI
IIR
VV
VV
S
S
SS
S
CIRCUITS WITH DEPENDENT SOURCES Treat the dependent source as thoughit were independent.Add one equation for the controllingvariable
k
VI
mAI
X
2
4
2
1
SOURCESBY DETERMINED
CURRENTS MESH
0)(1)(21 4313 IIkIIkkIx :3 MESH
012)(1)(1 2434 VIIkIIk :4 MESH
)(2 1324 IIkVIII xx
VARIABLESGCONTROLLIN
COMBINE EQUATIONS. DIVIDE BY 1k
122
823
0
4
432
432
321
1
III
III
III
I
122
823
0
4
432
432
321
1
III
III
III
I
SOLVE USING MATLAB
PUT IN MATRIX FORM
12
8
0
4
2110
2310
0111
0001
4
3
2
1
I
I
I
I
» R=[1,0,0,0; %FIRST ROW
1,1, -1, 0; %SECOND ROW
0,1,3,-2; %THIRD ROW
0,-1,-1,2] %FOURTH ROW
R =
1 0 0 0
1 1 -1 0
0 1 3 -2
0 -1 -1 2
DEFINE THE MATRIX
DEFINE THE RIGHT HAND SIDE VECTOR
» V=[4;0;8;12]
V =
4
0
8
-12
Since we divided by1k the RHS is mA andall the coefficientsare numbers
SOLVE AND GET THE ANSWER » I=R\V
I =
4
-6
-2
-10
The answers are in mA
>> is the MATLAB prompt. Whatfollows is the command entered
LEARNING EXTENSION: Dependent Sources
We treat the dependent source as one more voltage source
and solve...
The selection of loop currentssimplifies expression for Vx and computation of Vo.
USING LOOP CURRENTSUSING MESH CURRENTS
MESH 1 0)(422 211 IIkkIVx LOOP 1 04)(22 121 kIIIkVx
MESH 2 0)(463 122 IIkkI LOOP 2 063)(22 221 kIIIkVx
NOW WE EXPRESS THE CONTROLLING VARIABLE IN TERMS OF THE LOOP CURRENTS
14kIVx )(4 21 IIkVx
3104
042
21
21
kIkI
kIkI REPLACE AND REARRANGE
386
066
21
21
kIkI
kIkI
mAImAI 5.1,3 21 mAImAI 5.1,5.1 21 SOLUTIONS
][96 2 VkIVO
NOTICE THE DIFFERENCE BETWEEN MESH CURRENT I1 AND LOOP CURRENT I1 EVENTHOUGH THEY ARE ASSOCIATED TO THESAME PATH
Find Vo
DEPENDENT CURRENT SOURCE. CURRENT SOURCES NOT SHARED BY MESHES
We treat the dependent source as a conventional source
Then KVL on the remaining loop(s)
And express the controlling variable,Vx, in terms of loop currents
Equations for meshes with current sources
WE ARE ASKED FOR Vo. WE ONLY NEEDTO SOLVE FOR I3
REPLACE AND REARRANGE
mAIIIIkV
kIV
x
x42
)(4
221
21
1
mAIkIkI8
11238 323
][4
336 3 VkIVO
1I
DRAW MESH CURRENTS
2I
WRITE MESH EQUATIONS.
0)(4212 122 IIkkI :2 MESH
CONTROLLING VARIABLE IN TERMS OF LOOP CURRENTS
2IIx
0)(422 211 IIkkIkIx :1 MESH
REPLACE AND REARRANGE
1264
066
21
21
kIkI
kIkI
SOLVE FOR I2
mAIkI 6122 22
][122 2 VkIVO
In the following we shall solve using loopanalysis two circuits that had previously beensolved using node analysis
This is one circuit.we recap first the node analysisapproach and then we solve usingloop analysis
LEARNING EXAMPLE FIND THE VOLTAGE Vo
AT SUPER NODE
1 22
XV V V
2 3 1 32 14
2 01 1 1 1
V V V VV V VmA
k k k k
3 2 3 1
3@ : 2 0
1 1
V V V VV mA
k k
4 4@ : 4V V V
CONTROLLING VARIABLE2X
V V
1k
1k
SOLVE EQUATIONS NOW
1
1 3
1 3
3
2 2 6
2 2
X
X
X
V V
V V V V
V V V V
VARIABLE OF INTEREST1 3O
V V V
RECAP OF NODE ANALYSIS
USING LOOP ANALYSISDETERMINE Vo
1I
2I
3I
START SELECTION USING MESHES
4I
SELECT A GENERAL LOOP TO AVOIDSHARING A CURRENT SOURCE
1
3
1: 2
3 : 2
Loop I mA
Loop I mA
Write loop equations
2 2 32: 2 1 1 ( ) 0
XLoop V kI k I I
4 3 1 44 : 4 1 ( ) 2 1 0
XLoop V k I I I V kI
Controlling variable:1 3 4
1 ( )X
V k I I I
2 4
2 4
4
2 2 61 , 2
4 8
kI kII mA I mA
kI
21
OV kIVariable of Interest
LEARNING EXAMPLE Find the current Io
2 2@ : 12V V V
3 3@ : 2
XV V V
4 1
1 3 4 3 4 51 2 4
@super node:
6 (constraint eq.)
2 01 1 1 1 1
X
V V V
V V V V V VV V VI
k k k k k
5 4 5
5@ : 2 0
1 1X
V V VV I
k k
1 2
4
CONTROLLING VARIABLES
1
X
X
V V V
VI
k
7 eqs in 7 variables
VARIABLE OF INTEREST 5
1O
VI
k
RECAP OF NODE ANALYSIS
Find the current Io using mesh analysis
Write loop/mesh equations
1 1 2 1 4Loop1: 1 1 ( ) 1 ( ) 0kI k I I k I I
2 1 2 5Loop 2: 1 ( ) 6 1 ( ) 0k I I V k I I
3Loop 3: 2
XI I
4 1Loop 4: 12 1 ( ) 2 0
XV k I I V
5 2 5 6Loop 5: 1 ( ) 1 ( ) 2 0
Xk I I k I I V
6 3 6 6 5Loop 6: 1 ( ) 1 1 ( ) 0k I I kI k I I
1
5 6
Controlling variables
1X
X
V kI
I I I
8 eqs in 8 unknowns
Variable of interest:6O
I I
Select mesh currents