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1 MeshCurrent Method The meshcurrent s analog of the nodeoltage method. We sole for a new set of arables, mesh currents, that automatcally satsfy KCLs. As such, meshcurrent method reduces crcut soluton to wrtng a bunch of KVLs. Note: Meshcurrent method only works for planar crcuts: crcuts that can be drawn on a plane (lke on a paper) wthout any elements or connectng wres crossng each other as shown below. Note that n some cases a crcut that looks nonplanar can be made nto a planar crcut by mong some of the connectng wres (see fgure) Wres crossng Ths crcut than can be reduced to a planar crcut A Nonplanar Crcut Meshcurrent method s best explaned n the context of example crcut below. A mesh s defned as a closed path (a loop) that contans no closed path wthn t. Mesh current s the current that crculates n the mesh.e., a) f an element s located on a sngle mesh (such as R 1, R 2, s1,and s2 ) t carres the same current as the mesh current, b) If an element s located on the boundary of two meshes (such as R 3 ), t wll carry a current that s the algebrac sum of the the two mesh currents: 1 1 s1 R 1 R R s2 3 = = 2 1 In ths way, KCLs are automatcally satsfed. In addton, as we can wrte current n each element n terms of mesh currents, we can use characterstcs of element to wrte the MAE140 Notes, Wnter

2 oltage across each element n terms on mesh currents. Therefore, we need only to wrte KVLs n terms of mesh currents. In the crcut aboe, KVLs ge: Mesh 1: R 1 1 R 3 ( 1 2 ) s1 =0 (R 1 R 3 ) 1 R 3 2 = s1 Mesh 2: R 3 ( 2 1 )R 2 2 s1 =0 R 3 1 (R 2 R 3 ) 2 = s2 or n matrx form, [ R1 R 3 R 3 R 3 R 2 R 3 ] [ 1 2 ] [ s1 = s2 ] R = s Whch s smlar n form to matrx equaton found for nodeoltage method. s the array of mesh currents (unknowns), s s the array of ndependent oltage sources, and R s the resstance matrx and s symmetrc. The dagonal element, R jj, s the sum of resstance around mesh j and the offdagonal elements, R jk, are the sum of resstance shared by meshes j and k. Example Fnd and. Usng meshcurrent method: 1 2Ω 6 Ω 2 Mesh 1: ( 1 2 ) 16 = 0 Mesh 2: ( 2 1 ) 9=0 { = =3 16 V 9 V 1 2 3Ω 6 V { 1 =2A 2 =1A The problem unknowns, and can now be found from the mesh currents: = 1 2 =1A =2 1 =4V Mesh currents method for crcuts wth current sources Because of characterstcs of a current source does not specfy ts oltage, we hae to modfy meshcurrent method. Ths s best seen n the example below: MAE140 Notes, Wnter

3 From the crcut, we note: If a current source s located on only one mesh (1A ICS n the crcut), the mesh current can be drectly found from the current source and we do not need to wrte any KVL: 1 = 1 A 1 A 1 2Ω 2 10 V 2A 2Ω 3 6Ω If a current source s located on the boundary between two meshes (2A ICS n the crcut), KVL on these meshes (mesh 2 or 3 n the aboe crcut) contan the oltage across the 2A ICS whch s unknown. We need two equatons to substtue for the two KVLs on meshes 2 and 3 that are not useful now. The frst one s found from the characterstcs of the current source (ts current should be 2 A): 3 2 =2A The second equaton can be found by notng that KVL can be wrtten oer any closed loop. We defne a supermesh as the combnaton of two meshes whch hae a current source on ther boundary as shown n the fgure. Whle KVL on mesh 2 or on mesh 3 both nclude the oltage across the 2A current source that s unknown, KVL on the supermesh does not nclude that: Supermesh 2&3: 2( 2 1 )2( 3 1 ) = =10 1 =1 2 3 = = =6 1 =1 2 = 3 2 2( 3 2) 8 3 = 10 whch results n 1 =1A, 2 = 1 A,and 3 =1A, 1 A Note that we could hae used 1 =1and 2 = 3 2 equatons drectly on the meshes as shown n the fgure and wrote only the KVL on the supermesh: 2Ω 1 = 1 2Ω 2 = V 2A 3 6Ω supermesh 2&3: 2[( 3 2) (1)) 2[ 3 (1)] = 0 3 =1A 2 =1 2=1 A MAE140 Notes, Wnter

4 Recpe for MeshCurrent Method 1. Checkf crcut s planar. 2. Identfy meshes, mesh currents, and supermeshes. a) Rearrange the crcut f possble to poston current source on a sngle mesh. b) Use characterstc equatons of ICS to fnd mesh currents and reduce the number of unknowns. 3. Wrte KVL at each mesh and supermesh. 4. Sole for mesh currents. 5. Calculate problem unknowns from mesh currents. If you need to calculate the oltage across a current source you may hae to wrte KVL around a mesh contanng the current source. 6. For consstency and elmnaton of errors, always markall mesh currents n clockwse drecton and wrte down KVLs n the same drecton. Comparson of Nodeoltage and Meshcurrent methods Nodeoltage and meshcurrent are powerful methods that smplfy crcut analyss substantally. They are methods of choce n almost all cases (except for ery smple crcuts or specal crcuts). Examnaton of the crcut can also tell us whch of the two methods are best suted for the crcut at hand. We always want to reduce the crcut equatons nto the smallest number of equatons n smallest number of unknowns. The number of equatons from nodeoltage method, N NV and mesh current method, N MC are gen by: N NV = N node 1 N VS N MC = N mesh N CS where N VS and N CS are numbers of oltage and current sources, respectely. Thus, always nspect the crcut, fnd N VS and N CS, and proceed wth the method that results n the smallest number of equatons to sole. Note: You need to checkto ensure that the crcut s a planar crcut. If t s not one cannot use meshcurrent method and should use nodeoltage method. MAE140 Notes, Wnter

5 Addte propertes and Superposton In solng any lnear crcut, we always end up wth a set of smultaneous lnear equatons of the form A x = s A : x : s : matrx of resstances or conductances array of crcut arables, and/or, (unknowns) array of ndependent sources Lnear algebra tells that f we know the soluton to A x = s 1 to be x 1 (.e., A x 1 = s 1 ) and f we know the soluton to A x = s 2 to be x 2 (.e., A x 2 = s 2 ), then the soluton to A x = s 1 s 2 s x = x 1 x 2 because: A x 1 A x 2 = s 1 s 2 A (x 1 x 2 )=s 1 s 2 In a lnear crcut ths property means that: Addte property of lnear crcut or Prncple of Superposton: If a lnear crcut s dren by more than one ndependent source, the response of the crcut can be wrtten as the sum of the responses of the crcut to nddual sources wth all other sources klled (.e., ther strength set to zero.) Note that kllng a source does not mean remoeng t. s s = s for any klled = 0 for any = s klled = 0 for any for any MAE140 Notes, Wnter

6 Example: Fnd by superposton. Because we hae two ndependent sources, we frst kll the current source to arre at crcut a and then we kll the oltage source to arre at crcut b. By superposton, = a b Crcut a s a oltage dder crcut and a can be wrtten down drectly as 10 Ω 5Ω 3 A 15 V 10 Ω a = = 5 V Crcut b s a current dder crcut and current can be wrtten down drectly as a 5Ω 10 Ω 15 V = 1/51/10 3=2A 1/5 b = 5 = 10 V b 5Ω 3 A Thus, = a b =5 10 = 5 V. Note: Usng superposton results n slghtly smpler crcuts (one element s replaced wth ether a short or open crcut) but more crcuts. In general superposton requres more work than nodeoltage or meshcurrent methods. Superposton s used: a) If sources are fundamentally dfferent (e.g., dc and ac sources as we see later). In ths case superposton may be the only choce, b) If crcut s repette (see example 312 text book) such that crcuts resultng from applyng superposton lookdentcal and, thus, we need only to sole one crcut. MAE140 Notes, Wnter

7 Reducton of twotermnal subcrcuts to Theenn form Recall Theenn and Norton forms and the R T fact that they are equalent. The conenton s to wrte the characterstcs of N Theenn/Norton forms wth acte sgn conenton: T RN = T R T = N R Theenn Form Norton Form N Equalent f R =R T N and = R T N T We used the equalency of Norton and Theenn forms n crcut reducton. Recall our dscusson of equalent elements and subcrcuts. We can replace any twotermnal subcrcut wth another one as long as they hae the same characterstcs. We wll show below that the characterstcs of any twotermnal element contanng lnear elements s n Theenn form. Frst let s examne how to fnd characterstcs of a twotermnal element. In order to fnd the characterstcs of a resstor n the lab, we connect a oltage sources (wth adjustable strength) to ts termnal, change the strength of the oltage of the source and Subcrcut measure the current flowng through the resstor. After a suffcent number of pars of and are measured, we can plot the result and deduce = R. We can perform smlar, but mathematcal, experment to fnd the characterstcs of a twotermnal element. Attach a oltage source across ts termnal wth a strength. Sole the crcut and calculate current Subcrcut whch wll be n terms of ( characterstcs!). Alternately, we can attach a current source wth strength to the subcrcut and sole for n terms as s shown. Ths s the general method to calculate the characterstcs of a twotermnal element. Suppose we used nodeoltage method and assgn the ground as shown. After wrtng all of nodeoltage equatons, we wll get: G = s G : : s : Conductance matrx Array of node oltages Array of ndependent current sources If we choose as node no. 1 to be the poste termnal of the subcrcut, the nodeoltage array wll be a column ector, =[, 2, 3,..] wth oltage as ts frst element. The MAE140 Notes, Wnter

8 current source array s also wll hae as ts frst element. Now, f we sole the aboe matrx equaton and denotng the nerse matrx of G as G 1,weget: 2... n = G 1 11 G G1n 1 G 1 21 G G2n G 1 n1 G 1 n2... G 1 nn s2... sn The frst row of ths matrx equaton reduces to = C 1 C 2 where C 1 and C 2 are two constants (Snce all G and s are constants). C 1 should be a resstance (call t R T )and C 2 should be a oltage (call t V T ). Thus: the characterstcs of any twotermnal element contanng lnear elements s n the Theenn form of = T R T. Next, consder the twotermnal subcrcut = 0 R T = 0 and ts Theenn equalent (they hae exactly same characterstcs). Let the cur T Subcrcut = rent = 0 and calculate,.e., calculate the oc = oc oltage across the termnals of the subcrcut whle the termnal are open crcut. Ths oltage s called the open crcut oltage, oc.examnaton of the Theenn form shows that f =0, T = oc. Next, consder the twotermnal subcrcut and ts Norton equalent (they hae exactly same characterstcs). Let the oltage = 0 and calculate,.e., calculate the current whle the subcrcut termnals are shorted. Ths current s called the short crcut current, sc. Examnaton of the Norton form shows that f =0, N = sc. Subcrcut = sc = 0 N R N = sc Lastly, examnaton of the matrx equaton aboe shows that the Theenn resstance depends of conductance matrx only. Thus, f one klls all of the sources n the subcrcut, the remanng crcut should be equalent to the Theenn s resstance. The aboe bold/underlned statements consttute the Theenn s Theorem Fndng Equalent Theenn/Norton Forms: Three methods are aalable: Method 1: Use source transformaton and crcut reducton to reduce the crcut to a Theenn/Norton form. Ths s a cumbersome method, does not always work, and should be used only on smple crcuts. MAE140 Notes, Wnter = 0

9 Method 2: Drectly fnd characterstcs of the subcrcut by attachng a current source or a oltage source to the crcut as dscussed aboe. Ths method always work. The drawback s that the crcut has to be soled analytcally. Method 3: Compute two of the followng quanttes by solng the approprate crcuts: T = oc, N = sc, and R T by kllng the sources. The thrd parameter s found from T = R T N. Ths s the best method and wth a few exceptons, always work. Example: Fnd Theenn equalent of ths subcrcut: 5 Ω 20Ω Method 1: Source transformaton and crcut reducton 5 Ω 5A 5 Ω 20Ω 20Ω 4 Ω 8 Ω 8A 32V 32V Method 2: Drectly fnd characterstcs. We attach a oltage source wth strength to the output termnals as shown. Assume s known and sole for. Crcut has 4 nodes and two oltage sources, so the number of equaton for nodeoltage method, N NV =4 1 2=1. Crcut has three meshes and 1 current source, so the number of equatons for meshcurrent method s N MC =3 1 = 2. So, better to do nodeoltage method. 5 Ω 1 20Ω MAE140 Notes, Wnter

10 =160.5 = = 4 4 T =32V R T =8Ω = = 0 =32 8 ( characterstcs!) Method 3: Theenn s Theorem: fnd two of the followng three: R T, oc,and sc. a) Fnd R T by kllng the sources 5 Ω 5 Ω 20Ω 20Ω 5 20 = R = 44 = 8Ω T b) Fnd T = oc (set =0) Usng nodeoltage method (note that the oltage drop across the 4 Ω resstor s zero. oc 25 oc 0 3= oc 100 oc 60 = 0 5 Ω 20 Ω oc = 0 oc T = oc =32V c) Fnd N = sc (set =0) Usng nodeoltage method: =0 =0 5 Ω 20Ω 2 = 0 sc 2 =16V N = sc = 2 4 =4A MAE140 Notes, Wnter

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