1. 1 ECE 221 Electric Circuit Analysis I Chapter 10 Circuit Analysis 3 Ways Herbert G. Mayer, PSU Status 10/21/2015

2. 2 Syllabus Goal Sample Problem 1 Solve by Substitution KCL Using Cramer’s Rule Solve by Node Voltage Method Solve by Mesh Current Method Conclusion Problem 1 Same for Problem 2

3. 3 Goal We’ll analyze simple circuits, named Sample Problem 1 and Sample Problem 2 With various constant voltage sources and resistors Goal is to compute branch currents i 1, i 2, and i 3 First by using conventional algebraic substitution, applying Kirchhoff’s Laws; we’ll need 3 equations Secondly, we use the Node Voltage Method Thirdly we compute fictitious currents i a and i b, using the Mesh Current Method Any method may apply Cramer’s Rule to conduct the arithmetic computations, once the equations exist

4. 4 Problem 1

5. 5 Circuit for Sample Problem 1

6. 6 Solve Problem 1 Via KCL, KVL Using Arithmetic Substitution

7. 7 Sample Problem 1: 3 Equations KCL at node n1: (1)i1=i2+ i3 KVL in the left mesh, labeled ia: (2)R1*i1 + R3*i3 - v1=0 KVL in the right mesh, labeled ib: (3)R2*i2 + v2 - R3*i3=0 (3)’i3 = (R2*i2)/R3 + v2/v3

8. 8 Solve Problem 1 Arithmetic Substitution   in (2) R1*(i2+i3) + R3*i3= v1 R1*i2 + R1*i3 + R3*i3= v1 R1*i2 + i3*(R1+R3)= v1 R1*i2 + (R2*i2 + v2)*(R1+R3)/R3= v1... i2*(R1+R2*(R1+R3)/R3) = v1-v2*(R1+R3)/R3... i2*(100+2*400/3) = 10 - 20*(400/300) i2 = -45.45 mA

9. 9 Solve Problem 1 Arithmetic Substitution i3=i2 * R2/R3 + v2/R3 = -0.0303+0.066667 i3= 0.03636 A i3 = 36.36 mA i1=i2 + i3 i1 = -9.09 mA

10. 10 Solve Problem 1 Via KCL, KVL Using Cramer’s Rule

11. 11 Solve Problem 1 Using Cramer’s Rule i1=i2+i3 R1*i1 + R3*i3 - v1=0 R2*i2 + v2 - R3*i3=0 Normalized: i1 – i2 - i3= 0 R1*i1 + 0 + R3*i3= v1 0 + R2*i2- R3*i3= -v2

12. 12 Cramer’s Characteristic Determinant Normalize i1, i2, i3 positions in matrix |1-1 -1||0| Δ =|R1 0 R3|,R =|v1| |0R2-R3||-v2| |1 -1 -1| Δ=|100 0 300| |0 200-300| | 1 -1 1| S=|-1 1-1| | 1 -1 1|

13. 13 Cramer’s Characteristic Determinant Δ= 1| 0300|-100|-1 -1| + 0 | 200 -300|| 200 -300| Δ= 1*( 0 – 60,000 ) - 100*( 300 + 200 ) Δ= -60k - 50k Δ= -110,000

14. 14 Numerator Determinant N1, and i1 | 0 -1-1| N(i1) = N 1 =| 10 0 300| |-20 200 -300| N 1 = -10| -1-1| -20|-1 -1| | 200 -300|| 0300| N 1 =-10 * (300+200) -20 * (-300 ) N 1 =-10*500 + 6,000 N 1 =1,000 i1=1,000 / -110,000 i1 = -0.00909 A =-9.09 mA

15. 15 Numerator Determinant N2, and i2 |1 0-1| N(i2) = N 2 =| 100 10300| | 0-20 -300| N 2 = 1| 10 300| -100|0 -1| |-20-300|| -20 -300| N 2 = -3,000 + 6,000 -100 * ( 0 - 20 ) N 2 = 3,000 + 2,000 = 5,000 i2 = 5,000 / -110,000 i2 = -0.04545 A =-45.45 mA

16. 16 Numerator Determinant N3, and i3 | 1 -1 0| N(i3) = N 3 =| 100 010| | 0200 -20| N 3 = 1 | 0 10| -100|-1 0| | 200-20|| 200 -20| N 3 =-2,000 - 100 * (20 )=-4,000 i3=-4,000 / -110,000 i3 = 0.0363636 A = 36.36 mA

17. 17 Solve Problem 1 Using NoVoMo

18. 18 Solve Problem 1 by Node Voltage Method Ignoring the current or voltage directions from the substitution method, we use the Node Voltage Method at node n1, currents flowing toward reference node n2 We generate 1 equation with unknown V300, voltage at the 300 Ω resistor, yielding i3 Once known, we can compute the voltages at R1 and R2, and thus compute the currents i1 and i2, using Ohm’s law

19. 19 Solve Problem 1 by Node Voltage Method

20. 20 Solve Problem 1 by Node Voltage Method 3 currents flowing from n1 toward reference node n2: V300/300 + (V300-10)/100 + (V300-20)/200 = 0 V300 + 3*V300 + V300*2/3=30 + 3*20/2 V300*( 1 + 3 + 2/3 )=60 Students Compute V300

21. 21 Solve Problem 1 by Node Voltage Method 3 currents flowing from n1 toward reference node n2: V300/300 + (V300-10)/100 + (V300-20)/200 = 0 V300 + 3*V300 + V300 * 3/2=30 + 3*20/2 V300*( 1 + 3 + 3/2 )=60 V300=60 * 2 / 11 V300=10.9090 V Students Compute i3

22. 22 Solve Problem 1 by Node Voltage Method 3 currents flowing from n1 toward reference node n2: V300/300 + (V300-10)/100 + (V300-20)/200 = 0 V300 + 3*V300 + V300 * 3/2=30 + 3*20/2 V300*( 1 + 3 + 3/2 )=60 V300=60 * 2 / 11 V300=10.9090 V i3= V300 / 300 i3=36.363 mA

23. 23 Solve Problem 1 by Node Voltage Method V(R1)=v1 - V300 V(R1)=10 - 10.9090=-0.9090 V i1=V(R1) / R1 i1 = -0.9090 / 100 i1=-9.09 mA Students Compute i2

24. 24 Solve Problem 1 by Node Voltage Method V(R1)=v1 - V300 V(R1)=10 - 10.9090=-0.9090 V i1=V(R1) / R1 i1 = -0.9090 / 100 i1=-9.09 mA From this follows i2 using KCL: i2=i1 - i3 i2=-9.0909 – 36.3636 i2=-45.45 mA

25. 25 Solve Problem 1 Using MeCuMo

26. 26 Solve Problem 1 by Mesh Current Method The mesh current is fictitious, one such current associated with its own individual mesh Fictitious in the sense as if it were uniquely tied to a mesh; yet depending on the branch of the mesh, mesh currents from other parts flow though that very mesh as well Kirchhoff’s current law is trivially satisfied, but mesh currents are not everywhere measurable with an Ampere meter: not measurable, when currents from other meshes super-impose In Sample Problem 1 we have 2 meshes, with mesh currents indicated as i a and i b But we must track that, R3 for example, has both flowing though it in opposing directions

27. 27 Solve Problem 1 by Mesh Current Method

28. 28 Solve Problem 1 by Mesh Current Method KVL for mesh with ia yields: (1)R1*ia + R3*(ia-ib)= v1 KVL for mesh with ib yields: (2)R3*(ib-ia) + R2*ib= -v2 Students Compute (1) for ib Then substitute ib in (2)

29. 29 Solve Problem 1 by Mesh Current Method KVL for mesh with ia yields: (1)R1*ia + R3*(ia–ib)= v1 KVL for mesh with ib yields: (2)R3*(ib-ia) + R2*ib= -v2 From (1) follows: (1)ib= ( R1*ia + R3*ia - v1 ) / R3 Substitute ib in (2): (2)-v2= ib*(R2+R3) - R3*ia -v2= ia*(R1+R3)*(R2+R3)/R3 - v1*(R2+R3)/R3 - R3*ia

30. 30 Solve Problem 1 by Mesh Current Method v1*(R2+R3)/R3 - v2 = ia*( (R1+R3)*(R2+R3)/R3 – R3) -20 + 10*5/3= ia*(400*500/300 – 300) ia= -10 / 1100 ia = -0.00909 A = -9.09 mA Since ia = i1: i1 = -9.09 mA

31. 31 Solve Problem 1 by Mesh Current Method Recall (1): (1)R1*ia + R3*(ia–ib)= v1 R3*ib=ia*(R1+R3) - v1 ib=ia*(R1+R3)/R3 - v1/R3 ib= -10*400/(1,100*300) - 10/300 ib= -0.04545 A = -45.45 mA since i2 = ib: i2 = -45.45 mA

32. 32 Conclusion Problem 1 via Mesh Current Since i3 = i1 - i2,  i3 = -9.09 mA - -45.45 mA it follows: i3 = 36.36 mA We see consistency across 3 different approaches to circuit analysis

33. 33 Problem 2

34. 34 Sample Problem 2 We’ll analyze another, similar circuit, named Sample Problem 2 With 2 constant voltage sources of 3 V and 4 V Plus 3 resistors at 100, 200, and 300 Ohm Again we compute 3 branch currents i 1, i 2, and i 3 Using 3 methods: First we use substitution, applying Kirchhoff’s Laws Then we use the Node Voltage Method Thirdly the Mesh Current Method Any of these methods may use Cramer’s Rule

35. 35 Circuit for Sample Problem 2

36. 36 Sample Problem 2: Three Equations KCL states: (1)i1=i2+i3 KVL in the upper mesh labeled ia yields: (2)i1*100 + i2*200-3=0 KVL in the lower mesh, labeled ib yields: (3)-i2*200+ i3*300+ 4 + 3=0

37. 37 Solve Problem 2 by Substitution -200*i2+ (i1-i2)*300=-7// (1)in(3) -500*i2+300*i1=-7// (3’) 100*i1+200*i2= 3// (2)*3 300*i1+600*i2= 9// (2’) (3’)-(2’) -500*i2-600*i2=-7 -9=-16 i2*1,100=16 i2=16 / 1,100 i2 = 14.54 mA

38. 38 Solve Problem 2 by Substitution i1*100 + i2*200=3 i1*100= 3-200*(16/1,100) i1*100=100/1,100 i1=1 / 1,100 i1 = 0.91 mA i3=i1 - i2 i3=-15 / 1,100 i3 = -13.63 mA

39. 39 Solve Problem 2 Via KCL, KVL Using Cramer’s Rule

40. 40 Solve Problem 2 Using Cramer’s Rule i1=i2+i3 i1*100 + i2*200-3=0 -i2*200+ i3*300+4 +3=0 Normalized: i1 - i2 - i3=0 100*i1+ 200*i2+0= 3 0- 200*i2+ 300*i3= -7

41. 41 Cramer’s Characteristic Determinant Normalize i1, i2, i3 positions |-1 1 1||0| D=| 100 200 0|,R =|3| | 0-200 300|| -7| | 1 -1 1| S=|-1 1-1| | 1 -1 1|

42. 42 Cramer’s Characteristic Determinant Δ= -1| 200 0| -100| 1 1| + 0 | 200 -300||-200 300| Δ= -60,000 – 50,000=-110,000 Δ= -110 k

43. 43 Numerator Determinant N1, and i1 | 011| N(i1) = N1 =| 32000| | -7 -200 300| N1 = 0 - 3| 11| -7| 1 1| |-200 300||200 0| Students Compute N1, Given Δ = -110 k

44. 44 Numerator Determinant N1, and i1 | 011| N(i1) = N1 =| 3 2000| | -7 -200 300| N1 = 0 - 3| 11| -7| 11 | |-200 300||200 0| N1 = -3*(300+200) -7*(-200)= N1 =-1,500 + 1,400 N1 = -10 Now Students Compute i1

45. 45 Numerator Determinant N1, and i1 | 011| N(i1) = N1 =| 3 2000| | -7 -200 300| N1 = 0 - 3| 11| -7| 11 | |-200 300||200 0| N1=-3*(300+200) -7*(-200)= N1=-1,500 + 1,400 N1=-100 i1=-100 / -110,000 i1= 0.000909 A i1 =0.91 mA

46. 46 Numerator Determinant N2, and i2 | -101| N(i2) = N2 =|10030| | 0 -7300| N2 =-1| 30| -100|0 1| + 0 | -7 300|| -7 300| N2 = -(900)- 100* (7)=-1,600 i2 = -1,600 / -110,000 i2 = 14.54 mA With i3 = i1 - i2 it follows: i3 = -13.63 mA

47. 47 Solve Problem 2 Using NoVoMo

48. 48 Solve Problem 2 by Node Voltage Method

49. 49 Solve Problem 2 by Node Voltage Method There are 2 essential nodes, n1 and n2 One will be selected as reference node: pick n2 Compute 3 currents from n1 to n2, express as function of v200 Students compose single KCL equation For node n1, using single unknown v200

50. 50 Solve Problem 2 by Node Voltage Method Use KCL to compute 3 current from n1 toward reference node n2: V200/200 + (V200-3)/100 + (V200-3-4)/300=0 Students compute v200, and i2

51. 51 Solve Problem 2 by Node Voltage Method Use KCL to compute 3 current from n1 toward reference node n2: V200/200 + (V200-3)/100 + (V200-3-4)/300=0 V200*(3/2 + 3 + 1 )=9 + 7 V200*11/2=16 V200=2.9090 V i2=V200 / 200 i2= 14.54 mA

52. 52 Solve Problem 2 by Node Voltage Method KVL in the lower mesh, with V300 being the voltage drop across the 300 resistor, yields: V300=-7+ V200=-7 + 2.9090=-4.091 V i.e. i3= V300/300=-0.013637 mA i3 = -13.63 mA i1 = i2 + i3 = 14.54 - 13.63 i1 = 0.91 mA

53. 53 Solve Problem 2 Using MeCuMo

54. 54 Solve Problem 2 by Mesh Current Method Again we analyze 2 meshes, with fictitious currents ia and ib Circuit is repeated below for convenience

55. 55 Mesh Current In Sample Problem 2

56. 56 Solve Problem 2 by Mesh Current Method KVL for mesh with ia yields: (1) 100*ia + 200*( ia-ib )=3 (1) 300*ia - 200*ib=3 (1) ib=(300*ia-3)/200 KVL for mesh with ib yields: (2)300*ib+200*( ib – ia )=-7 (2)500*ib-200*ia=-7

57. 57 Solve Problem 2 by Mesh Current Method Substitute ib from (1) in (2): 500*(300*ia - 3)/200 - 200*ia=-7 ia = 1/1,100= 0.91 mA i1 = ia, hence: i1 = 0.91 mA ib = 3*ia/2-3/200 = 3/(1,100 * 2) - 3/200 ib =-13.63 mA i3 = ib, hence i3 = -13.63 mA

58. 58 Solve Problem 2 by Mesh Current Method With i2 = i1 - i3, it follows: i2 = 14.54 mA

59. 59 Which Method is easiest? It seems the Node Voltage Method is simplest for these problems With the smallest number of equations Mesh Current method has smaller number of equations than pure KCL and KVL Small number of equations yields less chances for sign confusion But for a large number of unknowns Cramer’s Rule is THE methodical way to compute