LOOP ANALYSIS. The second systematic technique to determine all currents and voltages in a circuit


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1 LOOP ANALYSS The second systematic technique to determine all currents and voltages in a circuit T S DUAL TO NODE ANALYSS  T FRST DETERMNES ALL CURRENTS N A CRCUT AND THEN T USES OHM S LAW TO COMPUTE NECESSARY OLTAGES THERE ARE STUATON WHERE NODE ANALYSS S NOT AN EFFCENT TECHNQUE AND WHERE THE NUMBER OF EQUATONS REQURED BY THS NEW METHOD S SGNFCANTLY SMALLER
2 Apply node analysis to this circuit +  R R R R R R There are non reference nodes There is one super node There is one node connected to the reference through a voltage source We need three equations to compute all node voltages BUT THERE S ONLY ONE CURRENT FLOWNG THROUGH ALL COMPONENTS AND F THAT CURRENT S DETERMNED ALL OLTAGES CAN BE COMPUTED WTH OHM S LAW STRATEGY:. Apply KL (sum of voltage drops =). Use Ohm s Law to express voltages in terms of the loop current. [ ] 8[ ] R R R [ ] R R 8[ ] R RESULT S ONE EQUATON N THE LOOP CURRENT!!! SHORTCUT Skip this equation Write this one directly
3 LOOPS, MESHES AND LOOP CURRENTS a b 7 c f 6 e 5 d A BASC CRCUT EACH COMPONENT S CHARACTERZED BY TS OLTAGE ACROSS AND TS CURRENT THROUGH A LOOP S A CLOSED PATH THAT DOES NOT GO TWCE OER ANY NODE. THS CRCUT HAS THREE LOOPS fabef ebcde fabcdef A MESH S A LOOP THAT DOES NOT ENCLOSE ANY OTHER LOOP. fabef, ebcde ARE MESHES A LOOP CURRENT S A (FCTCOUS) CURRENT THAT S ASSUMED TO FLOW AROUND A LOOP,, ARE LOOP CURRENTS A MESH CURRENT S A LOOP CURRENT ASSOCATED TO A MESH., ARE MESH CURRENTS CLAM: N A CRCUT, THE CURRENT THROUGH ANY COMPONENT CAN BE EXPRESSED N TERMS OF THE LOOP CURRENTS EXAMPLES FACT: NOT EERY LOOP CURRENT S REQURED TO COMPUTE ALL THE CURRENTS THROUGH COMPONENTS a a f be bc b 7 c f 6 e 5 d A BASC CRCUT THE DRECTON OF THE LOOP CURRENTS S SGNFCANT USNG LOOP a f b e b c TWO CURRENTS FOR EERY CRCUT THERE S A MNMUM NUMBER OF LOOP CURRENTS THAT ARE NECESSARY TO COMPUTE EERY CURRENT N THE CRCUT. SUCH A COLLECTON S CALLED A MNMAL SET (OF LOOP CURRENTS).
4 FOR A GEN CRCUT LET B NUMBER OF BRANCHES N NUMBER OF NODES THE MNMUM REQURED NUMBER OF LOOP CURRENTS S L B ( N ) MESH CURRENTS ARE ALWAYS NDEPENDENT AN EXAMPLE DETERMNATON OF LOOP CURRENTS KL ON LEFT MESH KL ON RGHT MESH v v v v S 5 USNG OHM S LAW v i R, v i R, v ( i i ) R v i R, v i R 5 5 REPLACNG AND REARRANGNG B 7 N 6 L 7 (6 ) TWO LOOP CURRENTS ARE REQURED. THE CURRENTS SHOWN ARE MESH CURRENTS. HENCE THEY ARE NDEPENDENT AND FORM A MNMAL SET N MATRX FORM R R R R i v S R R R R i v 5 S THESE ARE LOOP EQUATONS FOR THE CRCUT
5 WRTE THE MESH EQUATONS v R BOOKKEEPNG BRANCHES = 8 NODES = 7 LOOP CURRENTS NEEDED = i R AND WE ARE TOLD TO USE MESH CURRENTS! THS DEFNES THE LOOP CURRENTS TO BE USED DENTFY ALL OLTAGE DROPS v R i R v R ( i i R v R ) v R5 i R i R 5 WRTE KL ON EACH MESH TOP MESH: v BOTTOM: v USE OHM S LAW S vr vs vr R vr5 vr vs vr
6 DEELOPNG A SHORTCUT WRTE THE MESH EQUATONS +  R +  R R WHENEER AN ELEMENT HAS MORE THAN ONE LOOP CURRENT FLOWNG THROUGH T WE COMPUTE NET CURRENT N THE DRECTON OF TRAEL R 5 R DRAW THE MESH CURRENTS. ORENTATON CAN BE ARBTRARY. BUT BY CONENTON THEY ARE DEFNED CLOCKWSE NOW WRTE KL FOR EACH MESH AND APPLY OHM S LAW TO EERY RESSTOR. AT EACH LOOP FOLLOW THE PASSE SGN CONENTON USNG LOOP CURRENT REFERENCE DRECTON R ( ) R R5 R R ( ) R
7 LEARNNG EXAMPLE: FND o USNG LOOP ANALYS AN ALTERNATE SELECTON OF LOOP CURRENTS SHORTCUT: POLARTES ARE NOT NEEDED. APPLY OHM S LAW TO EACH ELEMENT AS KL S BENG WRTTEN REARRANGE k 6k 6k 9k */ k 6. 5mA k 6k and add 5 ma EXPRESS ARABLE OF NTEREST AS FUNCTON OF LOOP CURRENTS O NOW REARRANGE O THS SELECTON S MORE EFFCENT k 6k */ 6k 9k 9 */ and substract k 8 ma
8 F THE CRCUT CONTANS ONLY NDEPENDENT SOURCE THE MESH EQUATONS CAN BE WRTTEN BY NSPECTON A PRACTCE EXAMPLE MUST HAE ALL MESH CURRENTS WTH THE SAME ORENTATON N LOOP K THE COEFFCENT OF k S THE SUM OF RESSTANCES AROUND THE LOOP. LOOP coefficient of coefficien t of coefficient of k 6k 6k RHS 6[ ] THE RGHT HAND SDE S THE ALGEBRAC SUM OF OLTAGE SOURCES AROUND THE LOOP (OLTAGE RSES  OLTAGE DROPS) THE COEFFCENT OF j S THE SUM OF RESSTANCES COMMON TO BOTH k AND j AND WTH A NEGATE SGN. LOOP k 6k LOOP k 9k 6 Loop LOOP coefficient of coefficient of 9k k coefficient of k RHS 6[ ] (6 k) ( k) (k 6k k)
9 LEARNNG EXTENSON. DRAW THE MESH CURRENTS. WRTE MESH EQUATONS MESH k k k) k [ ] ( k (k 6k) (6 MESH ) DDE BY k. GET NUMBERS FOR COEFFCENTS ON THE LEFT AND ma ON THE RHS. SOLE EQUATONS 8 [ ma] 8 9[ ma] * / [ ma] O and add 6k [ ] 5
10 WRTE THE MESH EQUATONS k k k k. DRAW MESH CURRENTS 6k 9 BOOKKEEPNG: B = 7, N =. WRTE MESH EQUATONS. USE KL MESH : k 6k( ) MESH : k( ) k( ) MESH : 9 6k( ) k( ) MESH : k( ) k 9 CHOOSE YOUR FAORTE TECHNQUE TO SOLE THE SYSTEM OF EQUATONS EQUATONS BY NSPECTON 8k 6k 8k k k 6k k k 9 k 6k 9
11 CRCUTS WTH NDEPENDENT CURRENT SOURCES KL THERE S NO RELATONSHP BETWEEN AND THE SOURCE CURRENT! HOWEER... MESH CURRENT S CONSTRANED MESH EQUATON MESH ma BY NSPECTON k 8k k (ma) 9 ma O 6k [ 8k ] CURRENT SOURCES THAT ARE NOT SHARED BY OTHER MESHES (OR LOOPS) SERE TO DEFNE A MESH (LOOP) CURRENT AND REDUCE THE NUMBER OF REQURED EQUATONS TO OBTAN APPLY KL TO ANY CLOSED PATH THAT NCLUDES
12 LEARNNG EXAMPLE COMPUTE O USNG MESH ANALYSS KL FOR o TWO MESH CURRENTS ARE DEFNED BY CURRENT SOURCES ma ma MESH USE KL TO COMPUTE o BY NSPECTON k(ma) k( ma) k k k k ma
13 LEARNNG EXTENSONS WE ACTUALLY NEED THE CURRENT ON THE RGHT MESH. HENCE, USE MESH ANALYSS MESH : ma MESH : [ ] k( ) 6k 5 5mA ma ma O 6k [ ] 5 MESH : ma MESH : k k O 6 ma 6k 8[ ]
14 Problem.6 (6th Ed). Write loop equations. S S +  Determine O k k k S = ma, S = 6. Select loop currents. 6k SELECTNG THE SOLUTON METHOD + O nonreference nodes. meshes One current source, one super node BOTH APPROACHES SEEM COMPARABLE. CHOOSE LOOP ANALYSS n this case we use meshes. We note that the current source could define one mesh. _ Loop S Loop S k( ) k( ) Loop k( ) 6k k( ) Since we need to compute o it is efficient to solve for only. HNT: Divide the loop equations by k. Coefficients become numbers and voltage source becomes ma. Loop Loop We use the fact that = s S (6 )[ ma] * / k */ and add eqs 6 S ma O 6k 7 ma
15 CURRENT SOURCES SHARED BY LOOPS  THE SUPERMESH APPROACH. WRTE CONSTRANT EQUATON DUE TO MESH CURRENTS SHARNG CURRENT SOURCES ma. WRTE EQUATONS FOR THE OTHER MESHES ma. DEFNE A SUPERMESH BY (MENTALLY) REMONG THE SHARED CURRENT SOURCE. SELECT MESH CURRENTS 5. WRTE KL FOR THE SUPERMESH 6 k k k( ) k ( ) SUPERMESH NOW WE HAE THREE EQUATONS N THREE UNKNOWNS. THE MODEL S COMPLETE
16 CURRENT SOURCES SHARED BY MESHES  THE GENERAL LOOP APPROACH THE STRATEGY S TO DEFNE LOOP CURRENTS THAT DO NOT SHARE CURRENT SOURCES  EEN F T MEANS ABANDONNG MESHES FOR CONENENCE START USNG MESH CURRENTS UNTL REACHNG A SHARED SOURCE. AT THAT PONT DEFNE A NEW LOOP. N ORDER TO GUARANTEE THAT F GES AN NDEPENDENT EQUATON ONE MUST MAKE SURE THAT THE LOOP NCLUDES COMPONENTS THAT ARE NOT PART OF PREOUSLY DEFNED LOOPS A POSSBLE STRATEGY S TO CREATE A LOOP BY OPENNG THE CURRENT SOURCE THE LOOP EQUATONS FOR THE LOOPS WTH CURRENT SOURCES ARE ma ma THE LOOP EQUATON FOR THE THRD LOOP S 6 [ ] k k( ) k( ) k ( ) THE MESH CURRENTS OBTANED WTH THS METHOD ARE DFFERENT FROM THE ONES OBTANED WTH A SUPERMESH. EEN FOR THOSE DEFNED USNG MESHES.
17 S ( FND OLTAGES R R S S R ACROSS S R RESSTORS  + S Now we need a loop current that does not go over any current source and passes through all unused components. HNT: F ALL CURRENT SOURCES ARE REMOED THERE S ONLY ONE LOOP LEFT MESH EQUATONS FOR LOOPS WTH CURRENT SOURCES s S S KL OF REMANNG LOOP For loop analysis we notice... Three independent current sources. Four meshes. One current source shared by two meshes. Careful choice of loop currents should make only one loop equation necessary. Three loop currents can be chosen using meshes and not sharing any source. SOLE FOR THE CURRENT. USE OHM S LAW TO CMPUTE REQURED OLTAGES R R ( ) R ( ) R ( ) ( ) ) ) R ( R ) ( R
18  + R R R R S S S S A COMMENT ON METHOD SELECTON The same problem can be solved by node analysis but it requires equations R R R R R R S S S S S
19 CRCUTS WTH DEPENDENT SOURCES Treat the dependent source as though it were independent. Add one equation for the controlling variable k ma X BY SOURCES DETERMNED CURRENTS MESH ) ( ) ( k k k x MESH : ) ( ) ( k k MESH : ) ( k x x CONTROLLN G ARABLES COMBNE EQUATONS. DDE BY k 8
20 SOLE USNG MATLAB 8 PUT N MATRX FORM Since we divided by k the RHS is ma and all the coefficients are numbers 8 >> is the MATLAB prompt. What follows is the command entered DEFNE THE MATRX» R=[,,,; %FRST ROW,, , ; %SECOND ROW,,,; %THRD ROW,,,] %FOURTH ROW R = DEFNE THE RGHT HAND SDE ECTOR» =[;;8;] = 8  SOLE AND GET THE ANSWER» =R\ The answers are in ma =
21 LEARNNG EXTENSON: Dependent Sources Find o USNG MESH CURRENTS USNG LOOP CURRENTS We treat the dependent source as one more voltage source x LOOP k( ) k MESH k k( ) 6 k( ) MESH x k LOOP k( ) 6k x NOW WE EXPRESS THE CONTROLLNG ARABLE N TERMS OF THE LOOP CURRENTS x k k k( ) and solve... x k k k ma,. 5mA REPLACE AND REARRANGE 6k 6k O SOLUTONS 6k 9[ ] 6k 8k.5mA,. 5mA NOTCE THE DFFERENCE BETWEEN MESH CURRENT AND LOOP CURRENT EEN THOUGH THEY ARE ASSOCATED TO THE SAME PATH The selection of loop currents simplifies expression for x and computation of o.
22 DEPENDENT CURRENT SOURCE. CURRENT SOURCES NOT SHARED BY MESHES WE ARE ASKED FOR o. WE ONLY NEED TO SOLE FOR REPLACE AND REARRANGE x x k k( ) 8k k ma ma 8 We treat the dependent source as a conventional source Equations for meshes with current sources O 6k [ ] Then KL on the remaining loop(s) And express the controlling variable, x, in terms of loop currents
23 DRAW MESH CURRENTS WRTE MESH EQUATONS. MESH MESH : k k( ) CONTROLLNG ARABLE N TERMS OF LOOP CURRENTS x : k x k k( ) REPLACE AND REARRANGE 6k 6k k 6k SOLE FOR k 6mA O k [ ]
24 n the following we shall solve using loop analysis two circuits that had previously been solved using node analysis This is one circuit. we recap first the node analysis approach and then we solve using loop analysis
25 LEARNNG EXAMPLE FND THE OLTAGE o RECAP OF NODE : k AT SUPER NODE X ma k k k k k k : ma CONTROLLNG ARABLE X SOLE EQUATONS NOW X 6 X X ARABLE OF NTEREST O
26 DETERMNE o USNG LOOP ANALYSS Write loop equations Loop : ma Loop : ma Loop : k k( ) X Loop : k( ) k X Controlling variable: k( ) X k k 6 ma, ma k 8 ariable of nterest O k START SELECTON USNG MESHES SELECT A GENERAL LOOP TO AOD SHARNG A CURRENT SOURCE
27 LEARNNG EXAMPLE Find the current o RECAP OF NODE : node: 6 (constraint eq.) 5 X k k k k k 5 : 5 X k k CONTROLLNG ARABLES X X k 7 eqs in 7 variables ARABLE OF NTEREST O 5 k
28 Find the current o using mesh analysis Write loop/mesh equations Select mesh currents Loop: k k( ) k( ) Loop : k( ) 6 k( ) 5 Loop : X Loop : k( ) X Loop 5: k( ) k( ) X Loop 6: k( ) k k( ) Controlling variables X X k 8 eqs in 8 unknowns 5 6 ariable of interest: O 6
+ + +   This circuit than can be reduced to a planar circuit
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