# 1 ECE 221 Electric Circuit Analysis I Chapter 10 Circuit Analysis 4 Ways Herbert G. Mayer, PSU Status 11/23/2014 For use at Changchun University of Technology.

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1 ECE 221 Electric Circuit Analysis I Chapter 10 Circuit Analysis 4 Ways Herbert G. Mayer, PSU Status 11/23/2014 For use at Changchun University of Technology CCUT

2 Syllabus Goal Goal Sample Problem1 Sample Problem1 Solve by Substitution KCL Solve by Substitution KCL Solve by Cramer’s Rule Solve by Cramer’s Rule Solve by Node-Voltage Method Solve by Node-Voltage Method Solve by Mesh-Current Method Solve by Mesh-Current Method Conclusion Problem1 Conclusion Problem1 Same for Problem2 Same for Problem2

3 Goal We’ll analyze simple circuits, named Sample Problem1 and Sample Problem2 Both with 2 constant voltage sources, 3 resistors Computing 3 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 Cramer’s Rule, with these 3 equations, normalizing them to determinant form Then, we use the Node-Voltage Method Finally we compute fictitious currents i a and i b, using the Mesh-Current Method

4 Problem 1

5 Circuit for Sample Problem1

6 Sample Problem1: 3 Equations KCL at node n1 states: (1)i1=i2+i3 KVL in the left mesh labeled ia yields: (2)R1*i1 + R3*i3 - V1=0 KVL in the right mesh, labeled ib: (3)R2*i2 + V2 - R3*i3=0 i.e.i3 = (R2*i2)/R3 + V2/R3

7 Solve Problem1 by 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

8 Solve Problem1 by 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

9 Solve Problem1 by Cramer’s Rule Using 3 Equations from Substitution 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

10 Cramer’s Characteristic Determinant Normalize i1, i2, i3 positions in matrix |1 -1 -1||0| Δ:|R1 0 R3|,R =| V1| |0 R2-R3||-V2| |1 -1 -1| Δ # =|100 0300| |0200 -300| | 1 -1 1| S=|-1 1-1| | 1 -1 1|

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

12 Numerator Determinant N1, and i1 |0 -1 -1| N(i1):| 10 0300| |-20200 -300| N 1 = 0 -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

13 Numerator Determinant N2, and i2 |1 0 -1| N(i2):| 100 10300| | 0-20 -300| N 2 = 1| 10300| -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

14 Numerator Determinant N3, and i3 | 1-1 0| N(i3):| 100 0 10| | 0200-20| N 3 = 1 | 0 1| -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

15 Solve Problem1 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, generating i3 Once known, we can compute the voltages at R1 and R2, and thus compute the currents i1 and i2, using Ohm’s law

16 Problem1 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 V300=60 * 2 / 11 V300=10.9090 V i3= V300 / 300 i3=36.36a mA

17 Problem1 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

18 Solve Problem1 by Mesh-Current Method The mesh current is fictitious, one each associated with each individual mesh Fictitious in the sense that we act 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 directly measurable with an Ampere meter, when currents from other meshes super- impose In the Sample Problem 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

19 Sample Problem1: Two Mesh Currents

20 Solve Problem1 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

21 Solve Problem1 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

22 Solve Problem1 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

23 Conclusion Problem1 via Mesh-Current Since i3 = i1 - i2,  i3 = -9.09 mA - -45.45 mA it follows: i3 = 36.36 mA We see consistency across 4 different methods of circuit analysis

24 Problem 2

25 Sample Problem2 We’ll analyze a similar circuit, named Sample Problem2 With 2 constant voltage sources of 3 V and 4 V Plus 3 resistors at 100, 200, and 300 Ohm But now the elements are connected differently Again we compute 3 branch currents i 1, i 2, and i 3 Using 4 methods: substitution method, using Kirchhoff’s Laws Then we use Cramer’s Rule Thirdly we use the Node-Voltage Method Finally the Mesh-Current Method

26 Circuit for Sample Problem2

27 Sample Problem2: 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

28 Solve Problem2 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

29 Solve Problem2 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

30 Solve Problem2 by 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

31 Cramer’s Characteristic Determinant Normalize i1, i2, i3 positions |-1 1 1||0| D # =| 100 200 0|,R =|3| | 0-200300|| -7| | 1 -1 1| S=|-1 1-1| | 1 -1 1|

32 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

33 Numerator Determinant N1, and i1 | 011| N(i1):| 32000| | -7 -200 300| N1 = 0 - 3| 11| -7| 1 1| |-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

34 Numerator Determinant N2, and i2 |-1 0 1| N(i2):| 100 3 0| | 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 From this follows i3 = i1-i2 i3 = -13.63 mA

35 Node-Voltage With Sample Problem2

36 Solve Problem2 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

37 Solve Problem2 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

38 Solve Problem2 by Mesh-Current Method Again we analyze 2 meshes, with fictitious currents ia and ib The circuit is repeated here for convenience

39 Mesh-Current With Sample Problem2

40 Solve Problem2 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

41 Solve Problem2 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

42 Solve Problem2 by Mesh-Current Method With i2 = i1 – i3, it follows: i2 = 14.54 mA

43 Which Method is easiest? It seems the Mesh-Current Method is simplest for these problems With the smallest number of equations And less chances for students’ sign confusion, as we follow the same sign (or: direction) through the whole mesh But for a large number of unknowns Cramer’s Rule is the only way to compute accurately

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