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1 DIT, Kevin St. lectric Circuit Thery DT87/ Mesh Anlysis Summry In this sectin we use the frmt pprch fr slving circuit prblems using mesh nlysis. This technique genertes set f equtins, which cn be slved using determinnt lgebr nd the pplictin f Crmers rule. We will ssign the SAM DICTION fr ech current flw in ech brnch f prticulr circuit. We cn then write the lp equtin by inspectin. Let the sum f the impednces in the lp be psitive nd the mutul r trnsfer impednce, be negtive. The mutul r trnsfer impednce is tht impednce which is cmmn t tw lps. KVL sttes tht the sum f the e.f.m`s in clsed lp must be equl t the sum f the ptentil differences. If the e.m.f is in the sme directin s the current directin, it is given psitive e.m.f. If it is in the ppsite directin t the current directin, it is ssigned negtive sign. Cnsider the fllwing exmple f Tee netwrk fed frm tw DC surces in figure 9. Figure 9: Tee netwrk. The lp equtins re written pplying Kirchff s vltge lw And = I I I = I I I Nte the negtive sign in frnt f becuse the ssigned current is in the ppsite directin t the rise in ptentil. But by inspectin I = I = I I 4 Cpyright Pul Tbin 46
2 DIT, Kevin St. lectric Circuit Thery DT87/ Here the sum f the selfimpednces is nd the mutul impednce is. xpressed in mtrix frm: I = I 5 Use the fllwing rules t cn write the equtins by inspectin: Cnvert ll the current surces t vltge surces Nte the lctin f the mjr nde, Clculte the number f lps using the equtin N = b  n where N = number f independent lps, b = number f brnches, n = number f mjr ndes, Select ll lps nd drw line rund ech lp in clckwise directin yu culd g nticlckwise representing ech current. ch lp must cntin ne impednce r genertr which is nt in ny ther lp this ensures the equtins re independent, nd Apply KVL t the lp using the fllwing rule: A psitive sign is ssigned t ech e.m.f. if the rising vltge is in the sme directin s the ssigned current r negtive if in the ppsite directin. xercise Cnsider the circuit shwn in figure 0. Figure 0: Typicl circuit under investigtin. Lp 50 0 = i 4 j i j4 i 8 Lp = i j4 i 7 j i 0 j Lp 50 0 = i 8 i j i6 j Cpyright Pul Tbin 47
3 Determinnts DIT, Kevin St. lectric Circuit Thery DT87/ We will slve fr the individul currents t lter stge. We hve t cnsider determinnt lgebr in rder t slve fr the currents in the bve mtrix. A mtrix cnsists f elements rrnged in rws nd clumns. The first letter f the subscript f ech element tells yu the rw nd the secnd subscript tells yu the clumn in which the element is cntined. Fr exmple, cnsider by mtrix. A= 6 Determinnt f squre Mtrix The determinnt f squre mtrix A, is clculted s: A = 7 Cfctrs nd Minrs: If the determinnt f mtrix is bigger thn x then cfctrs must be used. The minr f n element A ij, f determinnt f rder n, is the determinnt f rder n btined by deleting the rw nd clumn tht cntins the given element. The minr f n element A i j is dented s: Fr the element the minr is Mij 8 A = M =  9 M = 0 And fr the element M The signed minr is = xmple Fr i =, nd j =, the cfctr is: j Mij = cfctr Mij = Mij Fr i =, nd j =, the cfctr is: Mij = Mij Cpyright Pul Tbin 48
4 DIT, Kevin St. lectric Circuit Thery DT87/ Cpyright Pul Tbin 49 Using the frmul we generte the fllwing mtrix with the sign f ech element given s: The vlue f determinnt: The vlue f determinnt A f rder n is the sum f the n prducts btined by multiplying ech element f chsen rw r clumn by its cfctr. A A A 4 vlute the fllwing mtrix rund the 5 0 rw. = [ ]  5[67]  0 = 7 Nte the brckets rund the negtive vlue. This is gd prctice. xmple Slve fr the currents in 5 = I = I xmple Clculte the vlue f current i in the circuit in figure using Thévenin's Therem Superpsitin Therem Mesh nlysis
5 DIT, Kevin St. lectric Circuit Thery DT87/ =0 Ω, =5 Ω, =5 Ω, X = j0 Ω, X =  j5 Ω, = nd =. 5 Figure : Slutin The Thévenin impednce Z TH = jx = j5. emember n idel vltge surce hs zer hm surce impednce nd my be replced by shrt circuit. The Thévenin vltge is: Figure : Applictin f Thévenin's therem. V = TH Cnvert ech vltge vectr in plr frm t rectngulr frm nd then dd ech cmpnent i.e. = = csθ jsinθ nd = 5 0 = csφ jsinφ = = 0cs60 jsin60 nd = 5 0 = 5cs0 jsin0 The Thévenin equivlent circuit, with the ld reinserted, is shwn in figure. Cpyright Pul Tbin 50
6 DIT, Kevin St. lectric Circuit Thery DT87/ Figure : Thévenin equivlent circuit with ld. Nte: In figure, mximum pwer is trnsferred t the ld when the ld is the cmplex cnjugte f the Thévenin impednce. Cpyright Pul Tbin 5
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