1. 1 Circuit Theory Chapter 9B Sinusoidal Steady-State : Node Voltage, Mesh Current, Thevenin Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 Sinusoidal Steady-State Analysis Chapter 10 10.1 Basic Approach 10.2 Nodal Analysis 10.3 Mesh Analysis 10.4 Superposition Theorem 10.5 Source Transformation 10.6 Thevenin and Norton Equivalent Circuits

3. 3 Steps to Analyze AC Circuits: 1.Transform the circuit to the phasor or frequency domain. 2.Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). 3.Transform the resulting phasor to the time domain. Time to Freq Solve variables in Freq Freq to Time 10.1 Basic Approach (1)

4. as circuits get more "complex"... 4

5. ...we'll start writing matlab scripts to solve them 5

6. Complex Polar Helper Functions download into your working folder ang.m: helps convert angle to rect Here's how to use it: --> 5*ang(45) ans = 3.5355 + 3.5355i 6

7. magPhs.m – prints in polar --> magPhs(40 + 30j) 50.000 < 36.87 this just PRINTS in polar. Do not use in calculations. Only at end to get polar 7

8. Find v(t) using NV is=10cos(wt) A, vs = 100sin(wt) V 8

9. 9 % NVdemo.m % solution to first NV demo in lecture 9b % time domain values w = 50000; C = 9e-6; L=100e-6; R1=5; R2=20; % phasor domain Is=10*ang(0); Vs=100*ang(-90); ZC = -j/(w*C); ZL = j*w*L; V = (Is + Vs/20)/(1/5 + 1/ZC + 1/ZL +1/20) magPhs(V); % print out V in polar form

10. 10 10.2 Nodal Analysis (1) Example 1 Using nodal analysis, find v 1 and v 2 in the circuit of figure below. v 1 (t) = 11.32 sin(2t + 60.01  ) Vv 2 (t) = 33.02 sin(2t + 57.12  ) V Answer:

11. 11 % NVdemo2.m % % time domain w = 2; C = 0.2; L = 2; % phasor domain Is = 10*ang(-90); ZC = -j/(w*C); ZL = j*w*L; % Node eq at V1: % -Is + V1/2 + (V1-V2)/ZC % Node eq at V2: % (V2-V1)/ZC + V2/ZL + (V2-3*Vx)/4 % Vx = V1 % Rearranging % V1( 0.5 + 1/ZC) - V2/ZC = Is % V1( -1/ZC -3/4) + V2(1/ZC + 1/ZL +1/4) = 0 % A * Vunknown = B % 1/R * V = I -> 1/Z *V = I -> Y*V = I % matrix of admittances between nodes: Y = [.5 + 1/ZC, -1/ZC ; -1/ZC - 3/4, 1/ZC + 1/ZL + 1/4 ] I = [ Is ; 0] V = inv(Y)*I % V = Y\I % solution vector of node voltages magPhs(V) % Shows V1 and V2 in polar form

12. Output from NVdemo2 Y = 0.5000 + 0.4000i 0.0000 - 0.4000i -0.7500 - 0.4000i 0.2500 + 0.1500i I = 0.0000 - 10.0000i 0 V = 9.8113 - 5.6604i (rectangular form) 27.7358 - 17.9245i 11.327 < -29.98 (polar form) 33.024 < -32.87

13. 13 Find the indicated currents expressed as cosine functions. Use the node voltage analysis method first. i 1 (t) = cos(5t - 30°) A; i 2 (t) = 0.658cos(5t + 14.6°) A

14. 14

15. Mesh Analysis 15 Use mesh current analysis to find the phasor voltages V 1 and V 2. V 1 = 22.9 ∠ 110° V; V 2 = 19.6 ∠ -101° V

16. 16 % I*Z = V in phasor domain Z = [ -2j, 5+3j ; -1,1] V = [0; 10*ang(0) ] I = inv(Z)*V magPhs(I(1)) magPhs(I(2)) V1 = -2j*I(1) magPhs(V1) V2 = 5*I(2) magPhs(V2) Solution to previous in Freemat

17. 17 Z = -0.0000 - 2.0000i 5.0000 + 3.0000i -1.0000 + 0.0000i 1.0000 + 0.0000i V = 0 10.0000 + 0.0000i I = -10.7692 - 3.8462i -0.7692 - 3.8462i 11.435 < -160.35 3.922 < -101.31 V1 = -7.6923 + 21.5385i (rectangular) 22.871 < 109.65 (polar) V2 = -3.8462 - 19.2308i (rectangular) 19.612 < -101.31 (polar) Output from previous slide

18. 18 10.3 Mesh Analysis (1) Example 2 Find I o in the following figure using mesh analysis. Answer: I o = 1.194  65.44  A

19. Find the indicated mesh currents. 19 I 1 = 3.88 ∠ -153° A; I 2 = 3.48 ∠ 176° A

20. 20

21. 21

22. 22 10.4 Superposition Theorem (1) When a circuit has sources operating at different frequencies, The separate phasor circuit for each frequency must be solved independently, and The total response is the sum of time- domain responses of all the individual phasor circuits.

23. 23 10.4 Superposition Theorem (2) Example 3 Calculate v o in the circuit of figure shown below using the superposition theorem. V o = 4.631 sin(5t – 81.12  ) + 1.051 cos(10t – 86.24  ) V

24. 24 10.5 Source Transformation (1)

25. 25 10.5 Source Transformation (2) Example 4 Find Io in the circuit of figure below using the concept of source transformation. I o = 3.288  99.46  A

28. 28 10.6 Thevenin and Norton Equivalent Circuits (1) Thevenin transform Norton transform

29. Thevenin: 29 Find the Thevenin equivalent circuit at the terminals F-G. Express all complex values in your solution in both rectangular and polar form. V TH = 250 - j200 V = 320 ∠ -38.7° V; Z TH = 100 + j20 Ω = 102 ∠ 11.3° Ω

31. 31 10.6 Thevenin and Norton Equivalent Circuits (2) Example 5 Find the Thevenin equivalent at terminals a–b of the circuit below. Z th =12.4 – j3.2  V TH = 18.97  -51.57  V

32. 32

33. 33 1) Use nodal analysis to determine which impedance element has the lowest voltage magnitude across its terminals. 3.3 kΩ resistor (voltage magnitude is 2.07 V, where other voltage magnitudes are 25.6 V and 13.7 V)

34. 34 Find the indicated mesh currents. I 1 = 3.88 ∠ -153° A; I 2 = 3.48 ∠ 176° A 2)

35. 35 3) Apply repeated source transformations to reduce this to an equivalent circuit at the terminals G-H. The simplified circuit will consist of a voltage source in series with two series-connected passive elements. Voltage source = 1140cos(60t + 68.4°); 200 Ω resistor; 74.1 μF capacitor

36. 36 4) Find the Norton equivalent circuit at the terminals Q-R. Express all complex values in your answer in both rectangular and polar form. I N = 0.769 - j1.15 A = 1.39 ∠ -56.3° A; Z N = 0.523 - j1.27 W = 1.38 ∠ -67.7° W

37. 37

38. 38