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Electrical circuit analysis 2012a mk

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N g u y ễn Côn g Phươn g gy g g Electric Circuit Theory Electric Circuit Theory Electrical Circuit Analysis Contents 1. Basic Elements Of Electrical Circuits 2. Basic Laws 3. Electrical Circuit Analysis 4. Circuit Theorems 5. Active Circuits 6. Capacitor And Inductor 7. First Order Circuits 8 SdOdCiit 8 . S econ d O r d er Ci rcu it s 9. Sinusoidal Steady State Analysis 10. AC Power Analysis 11 Three phase Circuits 11 . Three - phase Circuits 12. Magnetically Coupled Circuits 13. Frequency Response 14 The Laplace Transform Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 2 14 . The Laplace Transform 15. Two-port Networks Electrical Circuit Anal y sis 1. Branch current method 1. Branch current method 2. Node voltage method 3 Mesh current method 3 . Mesh current method Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 3 Branch current method (1) R 1 R 3 i 1 i 3 ab Ex. 1 R 2 R E J + i 2 i 4 + – v 1 + – v 3 v 2 – + v A B R 2 R 4 E 1 E 2 J + – + – + 2 v 4 – A B c Node a: i 1 + i 2 – i 3 = 0 i 1 + i 2 – i 3 = 0 i 1 1 2 3 Node b: i 3 – i 4 + J = 0 Loop A: – E 1 + R 1 i 1 – R 2 i 2 + E 2 = 0 i 3 – i 4 = –J R 1 i 1 – R 2 i 2 = E 1 – E 2 i 2 i 3 Loop B: – E 2 + R 2 i 2 + R 3 i 3 + R 4 i 4 = 0 R 2 i 2 + R 3 i 3 + R 4 i 4 = E 2 i 4 4 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (2) Ex. 1 R 1 R 3 i 1 i 3 ab A B R 2 R E J + i 2 i 4 + – v 1 + – v 3 v 2 – + v A B R 2 R 4 E 1 E 2 J + – + – + 2 v 4 – i 1 + i 2 – i 3 = 0 n = n 1 c i 3 – i 4 = –J R 1 i 1 – R 2 i 2 = E 1 – E 2 n KCL = n – 1 b +1 R 2 i 2 + R 3 i 3 + R 4 i 4 = E 2 n KVL = b – n + 1 5 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (3) Ex. 1 R 1 R 3 i 1 i 3 ab A B R 2 R E J + i 2 i 4 + – v 1 + – v 3 v 2 – + v A B R 2 R 4 E 1 E 2 J + – + – + 2 v 4 – i 1 + i 2 – i 3 = 0 1. Find: n KCL = n – 1 , a n d c i 3 – i 4 = –J R 1 i 1 – R 2 i 2 = E 1 – E 2 n KCL n ,a d n KVL = b – n + 1 2. Apply KCL at n KCL nodes 3 Apply KVL at n loops R 2 i 2 + R 3 i 3 + R 4 i 4 = E 2 3 . Apply KVL at n KVL loops 4. Solve simultaneous equations 6 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (4) n KCL = 4 – 1 = 3; n KVL = 6 – 4 + 1 = 3 a : i + i i 0 Ex. 2 J a : – i 1 + i 2 – i 6 = 0 b: i 1 – i 5 + i 3 + J = 0 c: – i 3 – i 4 + i 6 – J = R 1 R 3 E 1 i 1 i 3 + + – – b c + A: R 1 i 1 + R 5 i 5 + R 2 i 2 – e 1 = 0 B: R 3 i 3 + R 5 i 5 – R 4 i 4 = 0 C R i + R i + R i 0 0 A B + – R 2 R 4 R 5 1 i 2 i 4 i 5 a + – 1. Find: n KCL = n – 1 , a n d C : R 2 i 2 + R 6 i 6 + R 4 i 4 – e 6 = 0 C R E + + – – a d i n KCL n ,a d n KVL = b – n + 1 2. Apply KCL at n KCL nodes 3 Apply KVL at n nodes + – R 6 E 6 + – i 6 3 . Apply KVL at n KVL nodes 4. Solve simultaneous equations 7 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn Branch current method (5) i 1 i 2 R 1 R 2 E a Ex. 3 A B C 236 :0biii R 5 R E 6 E 1 J + – B 435 :0ciii i 5 i 6 i 3 i 4 14 :0 diiJ   R 3 R 4 R 6 E 3 d b c + – 3 4 14 :0 diiJ 11 55 4 4 1 : A Ri Ri Ri E 3 4 33 66 55 3 6 :BRiRiRiEE : CRiRiE   Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 8 66 22 6 : CRiRiE  Branch current method (6) + – R 1 E E i 1 i 3 Ex. 4 R 1 = 10Ω, R 2 = 20Ω, R 3 = 15Ω, E 1 = 30V, + – E 1 R 2 J R 3 E 3 i 2 E 3 = 45V, J = 2A. Find currents? 123 0iiiJ 123 2iii     R 2 J R 3 11 2 2 1 R iRi E 22 33 3 R iRiE 123 12 23 10 20 30 20 15 45 ii ii         123 12 3 ;;ii i      11 1 10 20 0 ;    1 21 1 30 20 0 ;     2 121 10 30 0 ;   3 11 2 10 20 30    Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 9 02015 45 20 15 04515 02045 Branch current method (7) + – R 1 E E i 1 i 3 R 1 = 10Ω, R 2 = 20Ω, R 3 = 15Ω, E 1 = 30V, Ex. 4 + – E 1 R 2 J R 3 E 3 i 2 E 3 = 45V, J = 2A. Find currents? 123 0iiiJ 123 2iii     R 2 J R 3 11 2 2 1 R iRi E 22 33 3 R iRiE 123 12 23 10 20 30 20 15 45 ii ii         11 1 10 20 0    20 0 1 1 1 1 1100 20 15 20 15 20 0    02015 20 15 20 15 20 0  1( 20 15 20 0) 10[1 15 20( 1)] 0[1 0 ( 20)( 1)]     650 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 10 1( 20 15 20 0) 10[1 15 20( 1)] 0[1 0 ( 20)( 1)]            650   [...]...  45  9 23 9.23  i3   3.62 A  15  Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 23 Electrical Circuit Analysis 1 Branch current method 2 Node voltage method 3 3 Mesh current method Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 24 Mesh current method (1) 3A Branch currents + + 2A – 5A 5A 3A Mesh currents Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn... 2350 R2 J R3 1350  i1   2.08 A  650  300   i2   0 46 A 0.46 650  2350   3.62 A i3   650  Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 11 Electrical Circuit Analysis 1 Branch current method 2 Node voltage method 3 3 Mesh current method Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 12 Node voltage method (1) i1 Ex 1 E2 c i1 –   E1  R1i1  (va ... 46 A 0.46 i  i  3.62 A 3 B i1  i A  2.08 A   i2  i A  iB  2  0 46 A 0.46 i  i  3.62 A 3 B Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 33 – Electrical Circuit Analysis 1 Branch current method 2 Node voltage method 3 3 Mesh current method Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 34 ... 1  E vd   3  R4  R3 1  E vd   J  1  R4  R1 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 21 Node voltage method (10) vc  0 Ex 4 –+ R2 a – + vb vd  J 0 R3 R4 vb  vd  E2  E1 E2 E1 b R1 vb d  vd  va  vd  E1   vb  E2  va  vd va  vb vb vd  i1  ; i2  ; i3  ; i4  R1 R2 R3 R4 R3 R4 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn c 22 Node voltage... i3 – i4 + J = 0 a i3 R3 R1 c Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 13 Node voltage method (2) i1 Ex 1 E2 c i2 –   E2  R2i2  (va  vc )  0 R4 J R2 a + v2 – vc  0 vac – E2 + E2  va  i2  R2 + v4 – +  E2  R2i2  vac  0 – i2 v1 R 2 + i4 – + + E1 b + v3 – + v1 – Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 a i3 R3 R1 c Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn... + R4)iB = E2 – R4J (R1+ R2)iA – R2iB = E1 – E2 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 27 Mesh current method (4) R1 Ex 1 R3 a v3 v1 J iB v4 R4 J + – E1 v2 R 2 E2 – + + iA b c R1iA + R2(iA – iB) = E1 – E2 R2(iB – iA) + R3iB + R4(iB + J) = E2 R1(iA ) + R2(iA – iB) = E1 – E2 R2(iB – iA)+ R3(iB ) + R4(iB + J) = E2 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 28 Mesh... E6  2 C i6 b R3 i A    iB i C i1  i A  J ; i2  iC ; i3  iB ; i4  i A ; i5  i A  iB ; i6  iC  iB Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 30 Mesh current method (7) R2 R1 Ex 4 J + iA iB – + – E1 E2 R1iA  R2iB  E1  E2 iA  iB   J Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 31 Mesh current method (8) Ex 5 R1 = 10Ω, R2 = 20Ω, R3 = 15Ω, E1 = 30V,... = 0 Node b: i3 – i4 + J = 0 a i3 R3 R1 R3 b + v3 – Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 15 Node voltage method (4) i1 Ex 1 – vc  0 + v4 – + E2 c b vb  i4  R4 + v4 vb  vc i4   R4 R4 – i2 v1 R 2 + i4 R4 J – + + E1 b + v3 – + v1 – Node a: i1 + i2 – i3 = 0 Node b: i3 – i4 + J = 0 a i3 R3 R1 v4 – i4 R4 c Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 16 Node...    vb  1  2  R1 R2 va  R1 R2 R3   R3     1   1 1  vb     va     vb  J   R3   R3 R4   E v E v v v v i1  1 a ; i2  2 a ; i3  a b ; i4  b R1 R2 R3 R4 Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn i1 i 2  i3 i4  18 Node voltage method (7) i1 Ex 1 E2 c i4 + v4 – R4 J – 5 6 6 7 – i2 v1 R 2 + + 4 E1 b + v3 – + v1 – + + 3 the reference node... Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 25 – Mesh current method (2) 6A ? –3A ? 6 – (–3) = 9A ? –3 – 8 = –11A ? 6A –3A 3A –6 – 2 = –8A ? 8A –8A ? ? ? 2A 5A ? –2A 2A ? 5A ? Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 26 Mesh current method (3) i1 Ex 1 a i3 R3 R1 + v3 – + v1 – – B: R2i2 + R3i3 + R4i4 = E2 i1 = iA i2 = iB – iA i3 = iB i4 = iB + J iB + E1 – i2 v1 R . g Electric Circuit Theory Electric Circuit Theory Electrical Circuit Analysis Contents 1. Basic Elements Of Electrical Circuits 2. Basic Laws 3. Electrical Circuit Analysis 4. Circuit Theorems 5 Active Circuits 6. Capacitor And Inductor 7. First Order Circuits 8 SdOdCiit 8 . S econ d O r d er Ci rcu it s 9. Sinusoidal Steady State Analysis 10. AC Power Analysis 11 Three phase Circuits 11 . Three - phase . Analysis 11 Three phase Circuits 11 . Three - phase Circuits 12. Magnetically Coupled Circuits 13. Frequency Response 14 The Laplace Transform Electrical Circuit Analysis - sites.google.com/site/ncpdhbkhn 2 14 . The

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