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Thevenin’s theorem, Norton’s theorem and Superposition theorem for AC Circuits

Thevenin’s theorem, Norton’s theorem and Superposition theorem for AC Circuits. 3/2/09. Problem10-60(b). 1. Find TEC at terminals c-d. TEC: Thevenin Equivalent Circuit. Problem10-60(b). (i) To obtain VTH. + Voc -. V oc =V TH. Problem10-60(b). Nodal equations:. Problem10-60(b).

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Thevenin’s theorem, Norton’s theorem and Superposition theorem for AC Circuits

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

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

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

  4. Problem10-60(b) Nodal equations:

  5. Problem10-60(b)

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

  7. Problem10-60(b) In Mathcad:

  8. + - Problem10-60(b) (iii) Using TEC to find Vo Let ZL = 10+j10 + Vo - ZTH VTH ZL Let’s assume,

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

  10. IN V1 Problem10-60(b) (i) To obtain IN

  11. Problem10-60(b) (i) To obtain IN

  12. Problem10-60(b) (ii) To obtain ZN ZN = ZTH ZN = ZTH= 2.667 – j4

  13. Vo = (ZN//ZL)IN Problem10-60(b) (iii) Using NEC to find Vo Let’s assume, ZL = 10+j10 + Vo - IN ZN ZL

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

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

  16. Problem10-60(b) Nodal Equations

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

  18. Nodal Equations

  19. 1 W 4 W + vo - 2H 0.1F 5V 2sin 5t 10cos 2t This circuit operates at the different frequencies: • w = 0 ; for the DC Voltage source • w = 2 rad/s ; for the AC voltage source • w = 5 rad/s ; for the AC current source Find vo

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

  21. 1W 4 W + vo1 - 5 V jwL = 0 ; Short-circuited w = 0 ; 1/jwC = infinity ; Open-circuited (i) 5V (w = 0) acting alone

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

  23. j4 W 1 W + vo2 - 4 W -j5 W 100 In time domain: V02(t)=2.498cos(2t-30.79) (ii) 10V (w = 2 rad/s) acting alone

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

  25. I1 1 W 4 W + vo3 - J10 W 20 -j2 W In time domain: V03(t)=2.328sin(2t+10) (ii) 2A (w = 5 rad/s) acting alone

  26. Total response of the circuit: v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10) Note that we can only add the individual responses in the time domain, not in the phasor.

  27. Total response of the circuit: v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10) Note that we can only add the individual responses in the time domain, not in the phasor.

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

  29. Convert time-domain quantities to phasor quantities: w = 4 rad/s

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

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

  32. Convert time-domain quantities to phasor quantities: w = 4 rad/s We use the sine function as the reference for the phasor.

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

  34. Find vo 4 W 1/12 F 2H + vo - 6 W 1 W 2sin (4t+30) 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.

  35. Convert time-domain quantities to phasor quantities. Use the cosine function as a reference. w = 4 rad/s

  36. OR Convert time-domain quantities to phasor quantities. Use the sine function as a reference. w = 4 rad/s

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