NODAL AND LOOP ANALYSIS TECHNIQUES. Develop systematic techniques to determine all the voltages and currents in a circuit NODE ANALYSIS

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1 NODAL AND LOOP ANALYSS TECHNQUES LEANNG GOALS NODAL ANALYSS LOOP ANALYSS Develop systematic techniques to determine all the voltages and currents in a circuit NODE ANALYSS One of the systematic ways to determine every voltage and current in a circuit The variables used to describe the circuit will be Node oltages -- The voltages of each node with respect to a pre-selected reference node

2 T S NSTUCTE TO STAT THE PESENTATON WTH A ECAP OF A POBLEM SOLED BEFOE USNG SEES/ PAALLEL ESSTO COMBNATONS COMPUTE ALL THE OLTAGES AND CUENTS N THS CCUT k k k SECOND: BACKTACK USNG KL, KCL OHM S OHM' S : a KCL : OHM' S : b k * OTHE OPTONS... FST EDUCE TO A SNGLE LOOP CCUT k * KCL : 5 OHM'S : C k * 5 b k a () 9

3 THE NODE ANALYSS PESPECTE S S S a KL a a a KL a b b b 5 KL c c EFEENCE 5 b WHAT S THE PATTEN??? THEE AE FE NODES. F ONE NODE S SELECTED AS EFEENCE, THEN THEE AE FOU OLTAGES WTH ESPECT TO THE EFEENCE NODE. 5 b c ONCE THE OLTAGES AE KNOWN, THE CUENTS CAN BE COMPUTED USNG OHM S LAW. THEOEM: F ALL NODE OLTAGES WTH ESPECT TO A COMMON EFEENCE NODE AE KNOWN, THEN ONE CAN DETEMNE ANY OTHE ELECTCAL AABLE FO THE CCUT. v v m v N v DLL QUESTON ca A GENEAL EW THE EFEENCE DECTON FO CUENTS S ELEANT v ' v i' USNG THE LEFT-GHT EFEENCE DECTON THE OLTAGE DOP ACOSS THE ESSTO MUST HAE THE POLATY SHOWN ' ii OHM'S LAW v i v m PASSE SGN CONENTON ULES! N F THE CUENT EFEENCE DECTON S EESED... THE PASSE SGN CONENTON WLL ASSGN THE EESE EFEENCE POLATY TO THE OLTAGE ACOSS THE ESSTO ' v OHM'S LAW i v N m

4 DEFNNG THE EFEENCE NODE S TAL THE STATEMENT S MEANNGLES S UNTL THE EFEENCE PONT S DEFNED. BY CONENTON THE GOUND SYMBOL SPECFES THE EFEENCE PONT. ALL NODE OLTAGES AE MEASUED WTH ESPECT TO THAT EFEENCE PONT? THE STATEGY FO NODE ANALYSS S a b c. DENTFY ALL NODES AND SELECT A EFEENCE NODE. DENTFY KNOWN NODE OLTAGES. AT EACH NODE WTH UNKNOWN OLTAGE, WTE A KCL EQUATON (e.g.,sum OF CUENT LEANG a : a s a a b 9k b : 5 b a b b c k k c : 6 c b c 9k k 5 EFEENCE. EPLACE CUENTS N TEMS OF NODE OLTAGES AND GET ALGEBAC EQUATONS N THE NODE OLTAGES... SHOTCUT: SKP WTNG THESE EQUATONS... AND PACTCE WTNG THESE DECTLY

5 WHEN WTNG A NODE EQUATON... AT EACH NODE, ONE CAN CHOSE ABTAY DECTONS FO THE CUENTS a b a b c d d c AND SELECT ANY FOM OF KCL. WHEN THE CUENTS AE EPLACED N TEMS OF THE NODE OLTAGES, THE NODE EQUATONS THAT ESULT AE THE SAME O EQUALENT. CUENTS LEANG a b b d b c CUENTS NTO NODE a b b d b c a b a b c ' d d ' ' CUENTS LEANG c ' ' ' b a b d c b CUENTS NTO NODE ' ' ' b a b d c b WHEN WTNG THE NODE EQUATONS, WTE THE EQUATON DECTLY N TEMS OF THE NODE OLTAGES. BY DEFAULT, USE KCL N THE FOM SUM-OF-CUENTS-LEANG =. THE EFEENCE DECTON FO THE CUENTS DOES NOT AFFECT THE NODE EQUATON. CCUTS WTH ONLY NDEPENDENT SOUCES HNT: THE FOMAL MANPULATON OF EQUATONS MAY BE SMPLE F ONE USES CONDUCTANCES NSTEAD OF NODE USNG NODE i A v v v WTH CONDUCTANCES EODENG TEMS EODENG TEMS i A Gv G( v v) THE MODEL FO THE CCUT S A SYSTEM OF ALGEBAC EQUATONS THE MANPULATON OF SYSTEMS OF ALGEBAC EQUATONS CAN BE EFFCENTLY DONE USNG MAT ANALYSS.

6 EAMPLE WTE THE KCL NODE WE SUALZE THE CUENTS LEANG AND WTE THE KCL EQUATON EPEAT THE POCESS AT NODE i v v v v O SUALZE CUENTS GONG NTO NODE ANOTHE EAMPLE OF WTNG NODE EQUATONS B B 5mA A MAK THE NODES (TO NSUE THAT NONE AE MSSNG) A 8k k 8k k C SELECT AS EFEENCE k 8k 8k B 5mA k WTE KCL AT EACH NODE N TEMS OF A A NODE A B B

7 A MODEL S SOLED BY MANPULATON OF EQUATONS AND USNG MAT ANALYSS NUMECAL MODEL LEANNG EAMPLE USE GAUSSAN ELMNATON THE NODE EQUATONS i A ma, i B k, ma THE MODEL EPLACE ALUES AND SWTCH NOTATON TO UPPE CASE ALTENATE MANPULATON ADD EQS [ ] * /k * / GHT HAND SDE S OLTS. COEFFS AE NUMBES * / (and add equations) 6[ ] SOLUTON USNG MAT ALGEBA PLACE N MAT FOM AND DO THE MAT ALGEBA... USE MAT ANALYSS TO SHOW SOLUTON PEFOM THE MAT MANPULATONS A Adj( A) A FO THE ADJONT EPLACE EACH ELEMENT BY TS COFACTO SAMPLE 8k k

8 AN EAMPLE OF NODE ANALYSS earranging v COULD WTE EQUATONS BY v CONDUCTANCES CONNECTED TO NODE CONDUCTANCES BETWEEN & CONDUCTANCES BETWEEN & CONDUCTANCES BETWEEN & WTNG EQUATONS BY NSPECTON FO CCUTS WTH ONLY NDEPENDENT SOUCES THE MAT S ALWAYS SYMMETC THE DAGONAL ELEMENTS AE POSTE THE OFF-DAGONAL ELEMENTS AE NEGATE Conductances connected to node Conductances between and Conductances between and Conductances between and ALD ONLY FO CCUTS WTHOUT DEPENDENT SOUCES

9 LEANNG : ma : ma k BY NSPECTON k k k k ma USNG KCL ma LEANNG ETENSON 6mA Node : ma 6mA 6 : 6mA k N MOST CASES THEE AE SEEAL DFFEENT WAYS OF SOLNG A POBLEM 8mA k (6 ma) ma k (6 ma) ma k NODE EQS. BY NSPECTON 6mA k ma k 6 Once node voltages are known k k CUENTS COULD BE COMPUTED DECTLY USNG KCL AND CUENT DDE!!

10 CCUTS WTH DEPENDENT SOUCES CCUTS WTH DEPENDENT SOUCES CANNOT BE MODELED BY NSPECTON. THE SYMMETY S LOST. A POCEDUE FO MODELNG WTE THE NODE EQUATONS USNG DEPENDENT SOUCES AS EGULA SOUCES. FO EACH DEPENDENT SOUCE WE ADD ONE EQUATON EPESSNG THE CONTOLLNG AABLE N TEMS OF THE NODE OLTAGES. NUMECAL EAMPLE LEANNG EAMPLE v v k k v v ma k k * / k * / v i o v i A v v v v MODEL FO CONTOLLNG AABLE v i o EPLACE AND EAANGE v v v v i A [ ] ADDNG THE EQUATONS 5 [ ] 5 LEANNG EAMPLE: CCUT WTH OLTAGE-CONTOLLED CUENT EPLACE AND EAANGE WTE NODE EQUATONS. TEAT DEPENDENT SOUCE AS EGULA SOUCE CONTNUE WTH GAUSSAN ELMNATON... O USE MAT ALGEBA EPESS CONTOLLNG AABLE N TEMS OF NODE OLTAGES FOU EQUATONS N OU UNKNOWNS. SOLE USNG FAOTE TECHNQUE.

11 USNG MATLAB TO SOLE THE NODE EQUATONS k, k, i [ A/ ] A ma, i k, B ma, DEFNE THE COMPONENTS OF THE CCUT DEFNE THE MAT G Entries in a row are separated by commas (or plain spaces). ows are separated by semi colon» =;=;=; =; %resistances in Ohm» ia=.;ib=.; %sources in Amps» alpha=; %gain of dependent source» G=[(/+/), -/, ; %first row of the matrix -/, (/+alpha+/), -(alpha+/); %second row, -/, (/+/)], %third row. End in comma to have the echo G = DEFNE GHT HAND SDE ECTO» =[ia;-ia;ib]; %end in ";" to skip echo SOLE LNEA EQUATON» =G\ % end with carriage return and get the echo = LEANNG ETENSON: FND NODE OLTAGES EAANGE AND MULTPLY BY k [ ] * / and add eqs. NODE EQUATONS : ma k k : O CONTOLLNG AABLE (N TEMS ON NODE OLTAGES) O k EPLACE ma k k k k k

12 LEANNG ETENSON FND THE OLTAGE O NODE EQUATONS x x ma k * / x O O * /k k k x O [ ] x x NOTCE EPLACEMENT OF DEPENDENT SOUCE N TEMS OF NODE OLTAGE [ ] O [ ] CCUTS WTH NDEPENDENT OLTAGE SOUCES nodes plus the reference. n principle one needs equations... but two nodes are connected to the reference through voltage sources. Hence, those node voltages are known!!! Hint: Each voltage source connected to the reference node saves one node equation One more example. Only one KCL is necessary k [ ] k SOLNG THE EQUATONS 6[ ] ( 6[ ] ) ( THESE AE THE EMANNG TWO NODE EQUATONS ).5[ ]

13 THE SUPENODE TECHNQUE We will use this example to introduce the concept of a SUPENODE SUPENODE Conventional node analysis requires all currents at a 6 ma S S k eqs, unknowns...panic!! The current through the source is not related to the voltage of the source. S Math solution: add one equation 6[ ] Efficient solution: enclose the source, and all elements in parallel, inside a surface. Apply KCL to the surface!!! ma ma k 6 The source current is interior to the surface and is not required. We STLL need one more equation. 6[ Only eqs in two unknowns!!! ] ALGEBAC DETALS The Equations () () 6mA ma k 6[ ] /k Solution. Eliminate denominators in Eq().Multiply by... [ ] 6[ ]. Add equations to eliminate [ ] [ ]. Use Eq() to compute 6[ ] [ ]

14 s S FND THE NODE OLTAGES AND THE POWE SUPPLED BY THE OLTAGE SOUCE s S k, [ ], s k [ ma], s 6[ ma] [ ] ma k k * /k [ ] adding: 6[ ] [ ] TO COMPUTE THE POWE SUPPLED BY OLTAGE SOUCE WE MUST KNOW THE CUENT THOUGH T. 6mA 8mA k k P [ ] 8[ ma] 6mW BASED ON PASSE SGN CONENTON, THE POWE S ECEED BY THE SOUCE!! LEANNG EAMPLE FND O SUPENODE, 6 SUPENODE KNOWN NODE OLTAGES CONSTANT

15 LEANNG ETENSON SUPENODE 6 SOUCES CONNECTED TO THE EFEENCE CONSTANT EQUATON S NOT NEEDED FO O SUPENODE k k k ( ) k 6 * / 5 8 OHM'S LAW and add * / k O. 8mA k WTE THE NODE EQUATONS supernode dentify all nodes, select a reference and label nodes. Nodes connected to reference through a voltage source oltage sources in between nodes and possible supernodes EQUATON BOOKKEEPNG: KCL@ _, KCL@ supernode, constraints equations and one known node Supernodes can be more complex KCL@_ 5 (Careful not to omit any current) CONSTANTS DUE TO OLTAGE SOUCES S 5 S 5 S 5 EQUATONS N FE UNKNOWNS.

16 CCUTS WTH DEPENDENT SOUCES PESENT NO SGNFCANT ADDTONAL COMPLETY. THE DEPENDENT SOUCES AE TEATED AS EGULA SOUCES. WE MUST ADD ONE EQUATON FO EACH CONTOLLNG AABLE LEANNG EAMPLE FND O OLTAGE SOUCE CONNECTED TO EFEENCE : KCL@ 6 x k k CONTOLLNG AABLE N TEMS OF NODE OLTAGES k * / 6 O ma k x 6 k EPLACE

17 SUPE NODE WTH DEPENDENT SOUCE OLTAGE SOUCE CONNECTED TO EFEENCE SUPENODE CONSTANT KCL AT SUPENODE x 6 CONTOLLNG AABLE N TEMS OF NODES x ( 6) * /k CUENT CONTOLLED OLTAGE SOUCE CONSTANT DUE TO SOUCE k CONTOLLNG AABLE N TEMS OF NODES x k k KCL AT SUPENODE ma ma [ ] * / and add 8[ ] O ma k x k x k

18 An example with dependent sources k k S k x a x nodes are connected to the reference through voltage sources DENTFY AND LABEL NODES EPESS CONTOLLNG AABLE N TEMS OF NODE OLTAGES S a x k S k k v k ( a has units of [olt/amp] EPLACE x N (8 a) S k * a k EPLACE N KCL ) What happens when a=8? S ( a a / ) LEANNG EAMPLE FND THE OLTAGE : k AT SUPE NODE ma k k k k k k : ma CONTOLLNG AABLE SOLE EQUATONS NOW AABLE OF NTEEST O 6

19 LEANNG EAMPLE Find the current o FND NODES AND SUPE node: 6 (constraint eq.) 5 k k k k k 5 : 5 k k CONTOLLNG AABLES k 7 eqs in 7 variables AABLE OF NTEEST O 5 k LOOP ANALYSS The second systematic technique to determine all currents and voltages in a circuit T S DUAL TO NODE ANALYSS - T FST DETEMNES ALL CUENTS N A CCUT AND THEN T USES OHM S LAW TO COMPUTE NECESSAY OLTAGES THEE AE STUATON WHEE NODE ANALYSS S NOT AN EFFCENT TECHNQUE AND WHEE THE NUMBE OF EQUATONS EQUED BY THS NEW METHOD S SGNFCANTLY SMALLE

20 Apply node analysis to this circuit 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 THEE S ONLY ONE CUENT FLOWNG THOUGH ALL COMPONENTS AND F THAT CUENT S DETEMNED ALL OLTAGES CAN BE COMPUTED WTH OHM S LAW STATEGY:. Apply KL (sum of voltage drops =) [ ] 8[ ]. Use Ohm s Law to express voltages in terms of the loop current. [ ] 8[ ] ESULT S ONE EQUATON N THE LOOP CUENT!!! SHOTCUT Skip this equation Write this one directly LOOPS, MESHES AND LOOP CUENTS a b 7 c f 6 e 5 d A BASC CCUT EACH COMPONENT S CHAACTEZED BY TS OLTAGE ACOSS AND TS CUENT THOUGH A LOOP S A CLOSED PATH THAT DOES NOT GO TWCE OE ANY NODE. THS CCUT HAS THEE LOOPS. fabef ebcde abcdefa A MESH S A LOOP THAT DOES NOT ENCLOSE ANY OTHE LOOP. fabef, ebcde AE MESHES. A LOOP CUENT S A (FCTTOUS) CUENT THAT S ASSUMED TO FLOW AOUND A LOOP.,, AE LOOP CUENTS A MESH CUENT S A LOOP CUENT ASSOCATED TO A MESH., AE MESH CUENTS. CLAM: N A CCUT, THE CUENT THOUGH ANY COMPONENT CAN BE EPESSED N TEMS OF THE LOOP CUENTS. EAMPLES a f FACT: NOT EEY LOOP CUENT S EQUED TO COMPUTE ALL THE CUENTS THOUGH COMPONENTS. a b e b c b 7 c f 6 e 5 d A BASC CCUT THE DECTON OF THE LOOP CUENTS S SGNFCANT. USNG TWO LOOP CUENTS a f be bc FO EEY CCUT THEE S A MNMUM NUMBE OF LOOP CUENTS THAT AE NECESSAY TO COMPUTE EEY CUENT N THE CCUT. SUCH A COLLECTON S CALLED A MNMAL SET (OF LOOP CUENTS).

21 FO A GEN CCUT LET: B NUMBE OF BANCHES N NUMBE OF NODES THE MNMUM EQUED NUMBE OF LOOP CUENTS S L B ( N ) DETEMNATON OF LOOP CUENTS KL ON LEFT MESH KL ON GHT MESH MESH CUENTS AE ALWAYS NDEPENDENT AN EAMPLE USNG OHM S LAW v i, v i, v ( i i ) v i, v i 5 5 EPLACNG AND EAANGNG B 7 N 6 L 7 (6 ) TWO LOOP CUENTS AE EQUED. THE CUENTS SHOWN AE MESH CUENTS. HENCE THEY AE NDEPENDENT AND FOM A MNMAL SET. N MAT FOM i v S i v 5 S THESE AE LOOP EQUATONS FO THE CCUT WTE THE MESH EQUATONS v i BOOKKEEPNG BANCHES = 8 NODES = 7 LOOP CUENTS NEEDED = AND WE AE TOLD TO USE MESH CUENTS! THS DEFNES THE LOOP CUENTS TO BE USED DENTFY ALL OLTAGE DOPS WTE KL ON EACH MESH v i TOP MESH: v BOTTOM: v USE OHM S LAW v ( i i v ) S v vs v v 5 v5 v vs v i i 5

22 DEELOPNG A SHOTCUT WTE THE MESH EQUATONS WHENEE AN ELEMENT HAS MOE THAN ONE LOOP CUENT FLOWNG THOUGH T, WE COMPUTE NET CUENT N THE DECTON OF TAEL 5 DAW THE MESH CUENTS. OENTATON CAN BE ABTAY, BUT BY CONENTON THEY AE DEFNED CLOCKWSE. NOW WTE KL FO EACH MESH AND APPLY OHM S LAW TO EEY ESSTO. AT EACH LOOP, FOLLOW THE PASSE SGN CONENTON USNG LOOP CUENT EFEENCE DECTON. ( ) 5 ( ) LEANNG EAMPLE: FND o USNG LOOP ANALYS AN ALTENATE SELECTON OF LOOP CUENTS SHOTCUT: POLATES AE NOT NEEDED. APPLY OHM S LAW TO EACH ELEMENT AS KL S BENG WTTEN EAANGE k 9k NOW O * / and add k 6. 5mA 5 k ma EPESS AABLE OF NTEEST AS FUNCTON OF LOOP CUENTS O EAANGE THS SELECTON S MOE EFFCENT k * / 9k 9 * / and substract k 8 ma

23 F THE CCUT CONTANS ONLY NDEPENDENT SOUCES, THE MESH EQUATONS CAN BE WTTEN BY NSPECTON. A PACTCE EAMPLE MUST HAE ALL MESH CUENTS WTH THE SAME OENTATON N LOOP K THE COEFFCENT OF k S THE SUM OF ESSTANCES AOUND THE LOOP. LOOP coefficient of coefficient of coefficient of k HS 6[ ] THE GHT HAND SDE S THE ALGEBAC SUM OF OLTAGE SOUCES AOUND THE LOOP (OLTAGE SES - OLTAGE DOPS) THE COEFFCENT OF j S THE SUM OF ESSTANCES COMMON TO BOTH k AND j AND WTH A NEGATE SGN. LOOP coefficient of coefficient of 9k k coefficient of k HS 6[ ] LOOP k LOOP 9k Loop (6 k ) ( k ) (k k ) LEANNG ETENSON. DAW THE MESH CUENTS. WTE MESH EQUATONS MESH ( k k k) k [ ] MESH k (k ) (6 ). SOLE EQUATONS 8 [ ma] 8 9[ ma] * / [ ma] O and add DDE BY k. GET NUMBES FO COEFFCENTS ON THE LHS AND ma ON THE HS. [ ] 5

24 WTE THE MESH EQUATONS k k k k. DAW MESH CUENTS 9 BOOKKEEPNG: B = 7, N =. WTE MESH EQUATONS. USE KL MESH : k ( ) MESH : k( ) k( ) MESH : 9 ( ) k( ) MESH : k( ) k 9 CHOOSE YOU FAOTE TECHNQUE TO SOLE THE SYSTEM OF EQUATONS EQUATONS BY NSPECTON 8k 8k k k k k 9 k 9 CCUTS WTH NDEPENDENT CUENT SOUCES KL THEE S NO ELATONSHP BETWEEN AND THE SOUCE CUENT! HOWEE... MESH CUENT S CONSTANED MESH EQUATON MESH BY NSPECTON ma k 8k k (ma) 9 ma O [ 8k CUENT SOUCES THAT AE NOT SHAED BY OTHE MESHES (O LOOPS) SEE TO DEFNE A MESH (LOOP) CUENT AND EDUCE THE NUMBE OF EQUED EQUATONS ] TO OBTAN, APPLY KL TO ANY CLOSED PATH THAT NCLUDES

25 LEANNG EAMPLE COMPUTE O USNG MESH ANALYSS KL FO o TWO MESH CUENTS AE DEFNED BY CUENT SOUCES ma ma MESH USE KL TO COMPUTE o BY NSPECTON k(ma) k( ma) k k k k ma LEANNG ETENSONS WE ACTUALLY NEED THE CUENT ON THE GHT MESH. HENCE, USE MESH ANALYSS. MESH : ma MESH : 5[ ] k( ) 5mA ma ma O [ ] 5 MESH : ma MESH : k k O 6 ma 8[ ]

26 CUENT SOUCES SHAED BY LOOPS - THE SUPEMESH APPOACH. WTE CONSTANT EQUATON DUE TO MESH CUENTS SHANG CUENT SOUCES ma. WTE EQUATONS FO THE OTHE MESHES ma. DEFNE A SUPEMESH BY (MENTALLY) EMONG THE SHAED CUENT SOUCE. SELECT MESH CUENTS 5. WTE KL FO THE SUPEMESH 6 k k k( ) k ( ) SUPEMESH NOW WE HAE THEE EQUATONS N THEE UNKNOWNS. THE MODEL S COMPLETE. CUENT SOUCES SHAED BY MESHES - THE GENEAL LOOP APPOACH THE STATEGY S TO DEFNE LOOP CUENTS THAT DO NOT SHAE CUENT SOUCES. EEN F T MEANS ABANDONNG MESHES FO CONENENCE, STAT USNG MESH CUENTS UNTL EACHNG A SHAED SOUCE. AT THAT PONT, DEFNE A NEW LOOP. N ODE TO GUAANTEE THAT T GES AN NDEPENDENT EQUATON, ONE MUST MAKE SUE THAT THE LOOP NCLUDES COMPONENTS THAT AE NOT PAT OF PEOUSLY DEFNED LOOPS. A POSSBLE STATEGY S TO CEATE A LOOP BY OPENNG THE CUENT SOUCE. THE LOOP EQUATONS FO THE LOOPS WTH CUENT SOUCES AE ma ma THE LOOP EQUATON FO THE THD LOOP S 6[ ] k k( ) k( ) k ( ) THE MESH CUENTS OBTANED WTH THS METHOD AE DFFEENT FOM THE ONES OBTANED WTH A SUPEMESH. EEN FO THOSE DEFNED USNG MESHES.

27 FND OLTAGES ACOSS ESSTOS S S S For loop analysis we notice S 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. Now we need a loop current that does not go over any current source and passes through all unused components. HNT: F ALL CUENT SOUCES AE EMOED THEE S ONLY ONE LOOP LEFT SOLE FO THE CUENT. USE OHM S LAW TO COMPUTE EQUED OLTAGES ( ) MESH EQUATONS FO LOOPS WTH CUENT SOUCES s S S KL OF EMANNG LOOP S ( ) ( ) ( ) ( ) ( ) ( ) A COMMENT ON METHOD SELECTON The same problem can be solved by node analysis but it requires equations S S - + S S S S S S S

28 CCUTS WTH DEPENDENT SOUCES Treat the dependent source as though it were independent. Add one equation for the controlling variable COMBNE EQUATONS. DDE BY k. MESH CUENTS DETEMNED BY SOUCES ma k MESH : MESH : x k x k( ) k ( ) k( ) k ( ) CONTOLLNG AABLES x k( ) 8 SOLE USNG MATLAB 8 PUT N MAT FOM Since we divided by k, the HS is ma and all the coefficients are numbers. 8 >> is the MATLAB prompt. What follows is the command entered. DEFNE THE MAT» =[,,,; %FST OW,, -, ; %SECOND OW,,,-; %THD OW,-,-,] %FOUTH OW = DEFNE THE GHT HAND SDE ECTO» =[;;8;] = 8 - SOLE AND GET THE ANSWE» =\ The answers are in ma =

29 LEANNG ETENSON: Dependent Sources Find o USNG MESH CUENTS USNG LOOP CUENTS We treat the dependent source as one more voltage source x LOOP k( ) k MESH k k( ) MESH 6 k( ) and solve... x k LOOP k( ) x NOW WE EPESS THE CONTOLLNG AABLE N TEMS OF THE LOOP CUENTS x k( ) k k k k ma,. 5mA NOTCE THE DFFEENCE BETWEEN MESH CUENT AND LOOP CUENT EEN THOUGH THEY AE ASSOCATED WTH THE SAME PATH EPLACE AND EAANGE O SOLUTONS 9[ ] x k 8k.5mA,. 5mA The selection of loop currents simplifies expression for x and computation of o. DEPENDENT CUENT SOUCE. CUENT SOUCES NOT SHAED BY MESHES WE AE ASKED FO o. WE ONLY NEED TO SOLE FO EPLACE AND EAANGE 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 [ ] Then KL on the remaining loop(s) And express the controlling variable, x, in terms of loop currents

30 DAW MESH CUENTS WTE MESH EQUATONS. MESH : k x k k( ) MESH : k k( ) CONTOLLNG AABLE N TEMS OF LOOP CUENTS x EPLACE AND EAANGE k SOLE FO k 6mA O k [ ] 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.

31 LEANNG EAMPLE FND THE OLTAGE o ECAP OF NODE : k AT SUPE NODE ma k k k k k ma k : CONTOLLNG AABLE SOLE EQUATONS NOW 6 AABLE OF NTEEST O DETEMNE o USNG LOOP ANALYSS Write loop equations Loop : ma Loop : ma Loop : k k( ) Loop : k( ) k Controlling variable: k( ) k k 6 ma, ma k 8 ariable of nterest O k STAT SELECTON USNG MESHES SELECT A GENEAL LOOP TO AOD SHANG A CUENT SOUCE

32 LEANNG EAMPLE Find the current o ECAP OF NODE node: 6 (constraint eq.) 5 k k k k k 5 : 5 k k CONTOLLNG AABLES k AABLE OF NTEEST 7 eqs in 7 variables O 5 k Find the current o using mesh analysis Write loop/mesh equations Select mesh currents Loop: k k( ) k( ) Loop : k ( ) 6 k ( ) 5 Loop : Loop : k( ) Loop 5: k ( ) k ( ) Loop 6: k ( ) k k ( ) Controlling variables k 8 eqs in 8 unknowns 5 6 ariable of interest: O 6

33 APPLCATON MANUAL SPEED CONTOL FO DC MOTO Battery i Speed is proportional to applied voltage, M Potentiometer Arm is movable and allows to change and keeping their sum constant: pot Assumption speed i (no load) battery Potentiometer is not affected by connection to amplifier Circuit diagram [rpm] K K M M M speed [rpm] K M battery Design Example A k A Analysis of solution O Model using node analysis S An 8-volt source is to be used in conjunction with two standard resistors to design a voltage divider that will output 5 when connected to a μa load. While keeping the consumed power as low as possible, we wish to minimize the error between the actual output and the required 5 volts. P Standard= Off-the-shelf available only in certain values O s O A, 5, 8 O O S O O S O Find, eg..: 5 O 5 or : O Design equations O O.98 esistor values should be high Why not use very large values? Use trial and error. Pick Highest possible, determine and analyze resulting circuit. 5 7 k 5 / k O Closest k Are there other factors that we should consider; e.g., is o really constant? Tolerances?

34