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RESISTIVE CIRCUITS
•MULTI NODE/LOOP CIRCUIT ANALYSIS
DEFINING THE REFERENCE NODE IS VITAL
SMEANINGLES IS4V VSTATEMENT THE 1 =UNTIL THE REFERENCE POINT IS DEFINED
BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.
−
+V4
ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT
+
−V2
_____?12 =V
−+ 12V
THE STRATEGY FOR NODE ANALYSIS1. IDENTIFY ALL NODES AND SELECT
A REFERENCE NODE
2. IDENTIFY KNOWN NODE VOLTAGES
3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION
(e.g.,SUM OF CURRENT LEAVING =0)
0:@ 321 =++− IIIVa
4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES
0369
=−
++−
kVV
kV
kVV baasa AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
REFERENCE
SV aV bV cV
0:@ 543 =++− IIIVb
0:@ 65 =+− IIVc
0943
=−
++−
kVV
kV
kVV cbbab
039
=+−
kV
kVV cbc
SHORTCUT: SKIP WRITING THESE EQUATIONS...
AND PRACTICE WRITING
THESE DIRECTLY
EXAMPLE
@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
03
12
4
122 =
−+
−+−
Rvv
Rvvi
OR VISUALIZE CURRENTS GOING INTO NODE
WRITE THE KCL EQUATIONS
mA6
1I2I
3I
⇒=++− 036
6:@ 222 k
Vk
VmAV 2 12V V=
CURRENTS COULD BE COMPUTED DIRECTLYUSING KCL AND CURRENT DIVIDER!!
1
2
3
83 (6 ) 2
3 66 (6 ) 4
3 6
I mAkI mA mA
k kkI mA mA
k k
= −
= =+
= =+
IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM
NODE EQS. BY INSPECTION
( ) ( )mAVVk
62021
21 +−=+
( ) mAVkk
V 631
610 21 =⎟
⎠⎞
⎜⎝⎛ ++
1 16V V⇒ = −
kVI
kVI
kVI
3622
32
21
1 ===
Once node voltages are known
11@ : 2 6 0
2VV mA mAk+ + =
Node analysis
3 nodes plus the reference. In principle one needs 3 equations...
…but two nodes are connected tothe reference through voltage sources. Hence those node voltages are known!!!
…Only one KCL is necessary
012126
12322 =−
+−
+kVV
kVV
kV
][5.1][640)()(2
22
12322
VVVVVVVVV
=⇒==−+−+
EQUATIONSTHESOLVING
Hint: Each voltage sourceconnected to the referencenode saves one node equation ][6
][12
3
1
VVVV
−==
THESE ARE THE REMAININGTWO NODE EQUATIONS
Conventional analysis requires all currents at a node
@V_1 06
6 1 =++− SIk
VmA
@V_2 012
4 2 =++−k
VmAIS
2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the source
Math solution: add one equation
][621 VVV =−
Efficient solution: enclose the source, and all elements in parallel, inside a surface.
Apply KCL to the surface!!!
04126
6 21 =+++− mAk
Vk
VmA
The source current is interiorto the surface and is not required
We STILL need one more equation
][621 VVV =−Only 2 eqs in two unknowns!!!
SUPERNODE
THE SUPERNODE TECHNIQUE
SI
][6
04612621
21
VVV
mAmAk
Vk
V
=−
=+−+
(2)
(1)
Equations The
ALGEBRAIC DETAILS
...by Multiply Eq(1). in rsdenominato Eliminate 1.Solution
][6][242
21
21
VVVVVV
=−=+
][10][303 11 VVVV =⇒=2Veliminateto equations Add2.
][4][612 VVVV =−=2Vcompute to Eq(2) Use 3.
VVVV4
6
4
1
−== SOURCES CONNECTED TO THE
REFERENCE
SUPERNODE
CONSTRAINT EQUATION VVV 1223 =−
KCL @ SUPERNODE
02
)4(212
6 3322 =−−
+++−
kV
kV
kV
kV
k2/*
VVV 223 32 =+VVV 1232 =+− addand 3/*VV 385 3 =
mAk
VIO 8.32
3 == LAW SOHM'
OIV FORNEEDED NOT IS 2
OI of Value Find
+-
+-
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][12 321 =−+++− RRR VVVVV
2. Use Ohm’s Law to expressvoltages in terms of the “loop current.”
0][18][12 321 =++++− IRVIRIRVRESULT IS ONE EQUATION IN THE LOOP CURRENT!!!
SHORTCUT
Skip this equation
Write this onedirectly
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
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 LOOPTWOUSING
32
21
31
III
III
III
cb
eb
fa
+=
−=
−−=EXAMPLES THE DIRECTION OF THE LOOP
CURRENTS 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(767
=−−===
LNB
TWO LOOP CURRENTS AREREQUIRED.THE CURRENTS SHOWN AREMESH CURRENTS. HENCE THEY ARE INDEPENDENT ANDFORM A MINIMAL SET
KVL ON LEFT MESH
REPLACING AND REARRANGING
DETERMINATION OF LOOP CURRENTS
KVL ON RIGHT MESH
2 4 5 3 0Sv 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 Rv i R v i R
= = = −
= =
+-
+ -
V1
V2R1
R2 R3
R4R5
WRITE THE MESH EQUATIONS
1I2I
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
39612612
21
21
−=+−=−
kIkIkIkIREARRANGE
addand 2/*mAIkI 5.0612 22 =⇒=
mAIkIkI4561212 121 =⇒+=
EXPRESS VARIABLE OF INTEREST AS FUNCTIONOF LOOP CURRENTS
21 IIIO −=
1I@KVL
1IIO = NOWTHIS SELECTION IS MORE EFFICIENT
99612612
21
21
=+=+
kIkIkIkIREARRANGE
substractand 2/*3/*
mAIkI431824 11 =⇒=
EXAMPLE: FIND Io
AN ALTERNATIVE SELECTION OF LOOP CURRENTS
2I@KVL
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
mAIImAII
=+−−=−
addand 4/*][3330 2 mAI = ][
5336 2 VkIVO ==
3. SOLVE EQUATIONS
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”
][296
43
82)2(2
22 VkIVmAk
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
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
-+
1R2R
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
KVL OF REMAINING LOOP
)( 4233 IIRV −=
−
+
4V
−+ 1V
−
+
2V
−+ 3V
0)()()( 3441341243 =++−++−+ IIRIIIRIIRVS