1. 1 Thevenin’s theorem, Norton’s theorem and Superposition theorem for AC Circuits 3/2/09

2. 2 Problem10-60(b) 1. Find TEC at terminals c-d TEC: Thevenin Equivalent Circuit

3. 3 Problem10-60(b) (i) To obtain VTH + Voc - V oc =V TH

4. 4 Problem10-60(b) Nodal equations:

5. 5 Problem10-60(b)

6. 6 Problem10-60(b) (i) To obtain Z TH Z TH Z TH =[(10//j5)+4]//-j4

7. 7 Problem10-60(b) In Mathcad:

8. 8 Problem10-60(b) (iii) Using TEC to find V o + - ZLZL Z TH V TH +Vo-+Vo- Z L = 10+j10Let Let’s assume,

9. 9 Problem10-60(b) 2. Find NEC at terminals c-d NEC: Norton Equivalent Circuit

10. 10 Problem10-60(b) (i) To obtain I N ININ V1V1

11. 11 Problem10-60(b) (i) To obtain I N

12. 12 Problem10-60(b) (ii) To obtain Z N Z N = Z TH Z N = Z TH = 2.667 – j4

13. 13 Problem10-60(b) (iii) Using NEC to find V o ZLZL ZNZN ININ +Vo-+Vo- Z L = 10+j10 Let’s assume, VoVo = (Z N //Z L )I N

14. 14 Problem10-60(b) 3. Find V o using superposition theorem

15. 15 Problem10-60(b) (i) 4A acting alone

16. 16 Problem10-60(b) Nodal Equations

17. 17 Problem10-60(b) (i) 20V acting alone

18. 18 Nodal Equations

19. 19 Find v o 2H 10cos 2t + v o - 1  4  0.1F5V 2sin 5t This circuit operates at the different frequencies:  = 0 ; for the DC Voltage source  = 2 rad/s ; for the AC voltage source  = 5 rad/s ; for the AC current source

20. 20 We must use superposition theorem Different frequencies problem reduces to single-frequency problem. Due to 5V Due to 2sin 5t A Due to 10cos 2t V

21. 21 (i) 5V (  = 0) acting alone + v o1 - 11 4  5 V  = 0 ; j  L = 0 ; Short-circuited 1/j  C = infinity ; Open-circuited

22. 22 (ii) 10V (  = 2 rad/s) acting alone 2H 0.1F Convert time-domain quantities to phasor quantities:

23. 23 (ii) 10V (  = 2 rad/s) acting alone + v o2 - 4  1  j4  -j5  10  0  In time domain: V 02 (t)=2.498cos(2t-30.79  )

24. 24 (iii) 2A (  = 5 rad/s) acting alone 2H 0.1F Convert time-domain quantities to phasor quantities

25. 25 (ii) 2A (  = 5 rad/s) acting alone In time domain: V 03 (t)=2.328sin(2t+10  ) J10  1  4  -j2    + v o3 -

26. 26 Total response of the circuit: Note that we can only add the individual responses in the time domain, not in the phasor. v 0 (t)= -1 + 2.498cos(-30.79  )+2.328sin(5t+10  )

27. 27 Total response of the circuit: Note that we can only add the individual responses in the time domain, not in the phasor. v 0 (t)= -1 + 2.498cos(-30.79  )+2.328sin(5t+10  )

28. 28 Find v o 2H 2cos (4t+30  ) +vo-+vo- 6  1/12 F 6 cos 4t Case #1: This circuit operates at a single frequency. The sources are represented by a cosine function. 1  4 

29. 29 2H 1/12 F Convert time-domain quantities to phasor quantities:  = 4 rad/s

30. 30 A phasor circuit 2  30  +vo-+vo- 6  -j3  1  j8  4  6060

31. 31 Find v o 2H 2sin (4t+30  ) +vo-+vo- 6  1/12 F 6 sin 4t Case #2: This circuit operates at a single frequency. The sources are represented by a sine function. 1  4 

32. 32 2H 1/12 F Convert time-domain quantities to phasor quantities:  = 4 rad/s We use the sine function as the reference for the phasor.

33. 33 A phasor circuit 2  30  +vo-+vo- 6  -j3  1  j8  4  6060

34. 34 Find v o 2H 2sin (4t+30  ) +vo-+vo- 6  1/12 F 6 cos 4t Case #3: This circuit operates at a single frequency. One source is represented by a sine function; Another source is represented by cosine function. 1  4 

35. 35 2H 1/12 F Convert time-domain quantities to phasor quantities. Use the cosine function as a reference.  = 4 rad/s

36. 36 2H 1/12 F Convert time-domain quantities to phasor quantities. Use the sine function as a reference.  = 4 rad/s OR