RESISTIVE CIRCUITS MULTI NODE/LOOP CIRCUIT ANALYSIS

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Oliver Anderson
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1 RESSTE CRCUTS MULT NODE/LOOP CRCUT ANALYSS

2 DEFNNG THE REFERENCE NODE S TAL 4 THESTATEMENT 4 S MEANNGLES UNTL THE REFERENCE PONT S DEFNED BY CONENTON THE GROUND SYMBOL SPECFES THE REFERENCE PONT. ALL NODE OLTAGES ARE MEASURED WTH RESPECT TO THAT REFERENCE PONT?

3 THE STRATEGY FOR NODE ANALYSS S a b c. DENTFY ALL NODES AND SELECT A REFERENCE NODE. DENTFY KNOWN NODE a : a s a a b 9k 6k b : 4 5 b a b b c k 4k c : 6 c b c 9k k 5 REFERENCE. AT EACH NODE WTH UNKNOWN OLTAGE WRTE A KCL EQUATON (e.g.,sum OF CURRENT LEANG ) 4. REPLACE CURRENTS N TERMS OF NODE OLTAGES AND GET ALGEBRAC EQUATONS N THE NODE OLTAGES... SHORTCUT: SKP WRTNG THESE EQUATONS... AND PRACTCE WRTNG THESE DRECTLY

4 EXAMPLE WRTE THE KCL NODE WE SUALZE THE CURRENTS LEANG AND WRTE THE KCL EQUATON REPEAT THE PROCESS AT NODE i v v v v R4 R OR SUALZE CURRENTS GONG NTO NODE

5 6mA Node : ma 6mA 6 : 6mA 6k k N MOST CASES THERE ARE SEERAL DFFERENT WAYS OF SOLNG A PROBLEM 8mA k (6 ma) ma k 6k 6k (6 ma) 4mA k 6k NODE EQS. BY NSPECTON k ( ) ( 6)mA 6k k ( ) 6 ma Once node voltages are known k 6k k CURRENTS COULD BE COMPUTED DRECTLY USNG KCL AND CURRENT DDER!!

6 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 Only one KCL is necessary 6k k [ ] k 4 THESE ARE THE REMANNG TWO NODE EQUATONS 6[ ] SOLNG THE EQUATONS ( ) 6[ ] (.5[ ] )

7 THE SUPERNODE TECHNQUE SUPERNODE S Conventional analysis requires all currents at a node 4 ma S S k eqs, unknowns...panic!! The current through the source is not related to the voltage of the source Math solution: add one equation 6[ ] Efficient solution: enclose the source, and all elements in parallel, inside a surface. Apply KCL to the surface!!! 4mA 6k k 6 ma The source current is interior to the surface and is not required We STLL need one more equation 6[ ] Only eqs in two unknowns!!!

8 ALGEBRAC DETALS The Equations () () Solution 6mA 6k k 6[ ] 4mA. Eliminate denominators in Eq(). Multiply by... 4[ ] 6[ ]. Add equations to eliminate [ ] [ ]. Use Eq() to compute 6[ ] 4[ ]

9 Find alue of O SUPERNODE 4 6 SOURCES CONNECTED TO THE REFERENCE 4 CONSTRANT EQUATON SUPERNODE 6 k S NOT NEEDED FOR O ( 4) k k k * / k * / and add 5 8 OHM'S LAW O. 8mA k

10 Apply node analysis to this circuit - R R R R R - 8 There are 4 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 R 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 4 Skip this equation Write this one directly

11 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 4 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 b e b c b 7 c 4 f 6 e 5 d A BASC CRCUT THE DRECTON OF THE LOOP CURRENTS S SGNFCANT USNG TWO LOOP CURRENTS a f be bc 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).

12 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 4 5 USNG OHM S LAW v i R, v i R, v ( i i ) R v i R, v i R 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

13 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 4 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 R4 ( ) R

14 EXAMPLE: FND o 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 * / and add k 6. 5mA 5 k 6k ma 4 EXPRESS ARABLE OF NTEREST AS FUNCTON OF LOOP CURRENTS O NOW O REARRANGE THS SELECTON S MORE EFFCENT k 6k * / 6k 9k 9 * / and substract 4k 8 ma 4

15 . DRAW THE MESH CURRENTS. WRTE MESH EQUATONS MESH k 4k 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] * / 4 and add [ ma] O 6k [ ] 5

16 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) ma O 6k 8k 4 9 [ ] 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

WRTE EQUATONS FOR THE OTHER MESHES ma 4.

17 CURRENT SOURCES SHARED BY LOOPS - THE SUPERMESH APPROACH. WRTE CONSTRANT EQUATON DUE TO MESH CURRENTS SHARNG CURRENT SOURCES 4mA. WRTE EQUATONS FOR THE OTHER MESHES ma 4. 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

18 FND OLTAGES R 4 R S S 4 R ACROSS S R RESSTORS 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. 4 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 S SOLE FOR THE CURRENT 4. USE OHM S LAW TO CMPUTE REQURED OLTAGES MESH EQUATONS FOR LOOPS WTH CURRENT SOURCES s S S KL OF REMANNG LOOP R ( 4 ) R ( 4 ) R4 ( 4 ) R ( 4 ) R ( ) R ( 4 )

LOOP ANALYSIS. determine all currents and Voltages in IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUIT

LOOP ANALYSIS. determine all currents and Voltages in IT IS DUAL TO NODE ANALYSIS - IT FIRST DETERMINES ALL CURRENTS IN A CIRCUIT LOOP ANALYSS The second systematic technique to determine all currents and oltages 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