Chapter 10

1. Chapter 10 Sinusoidal steady-state analysis SJTU

2. Steps to analyze ac circuit • Transform the circuit to the phasor or frequency domain • Solve the problem using circuit techniques(nodal analysis, mesh analysis, superposition,etc) • Transform the resulting phasor to the time domain SJTU

3. Nodal analysis Fig. 8-28: An example node SJTU

4. Mesh analysis planar circuits: Circuits that can be drawn on a flat surface with no crossovers the sum of voltages around mesh A is Fig. 8-29: An example mesh SJTU

5. EXAMPLE 8-21 Use node analysis to find the current IX in Fig. 8-31. SOLUTION: or Fig. 8-31 SJTU

6. SJTU

7. EXAMPLE 8-24 The circuit in Fig. 8-32 is an equivalent circuit of an ac induction motor. The current IS is called the stator current, IR the rotor current, and IM the magnetizing current. Use the mesh-current method to solve for the branch currents IS, IR and IM. SJTU

8. SJTU

9. EXAMPLE 8-25 Use the mesh-current method to solve for output voltage V2 and input impedance ZIN of the circuit below. SOLUTION: SJTU

10. SJTU

11. Example Frequency domain equivalent of the circuit SJTU

12. Example SJTU

13. Find Vo/Vi, Zi See F page417 SJTU

14. where X is the input phasor, Y is the output phasor, and K is the proportionality constant. Circuit Theorems with Phasors PROPORTIONALITY The proportionality property states that phasor output responses are proportional to the input phasor SJTU

15. EXAMPLE 8-13 Use the unit output method to find the input impedance, current I1, output voltage VC, and current I3 of the circuit in Fig. 8-20 for Vs= 10∠0° • Assume a unit output voltage            . • By Ohm's law,                       . • By KVL, • By Ohm's law, • By KCL, • By KCL, SOLUTION: SJTU

16. Given K and ZIN, we can now calculate the required responses for an input SJTU

17. SUPERPOSITION • Two cases: • With same frequency sources. • With different frequency sources EXAMPLE 8-14 Use superposition to find the steady - state voltage vR (t) in Fig. 8 - 21 for R=20 , L1 = 2mH, L2 = 6mH, C = 20 F, V s1= 100cos 5000t V , and Vs2=120cos (5000t +30 )V. SJTU

18. SOLUTION: Fig. 8-22 SJTU

19. SJTU

20. EXAMPLE 8-15 Fig. 8-23 Use superposition to find the steady-state current i(t) in Fig. 8-23 for R=10k , L=200mH, vS1=24cos20000t V, and vS2=8cos(60000t+30 ° ). SOLUTION: With source no. 2 off and no.1 on SJTU

21. With source no.1 off and no.2 on The two input sources operate at different frequencies, so that phasors responses I1 and I2 cannot be added to obtain the overall response. In this case the overall response is obtained by adding the corresponding time-domain functions. SJTU

22. More examples See F page403 SJTU

23. THEVENIN AND NORTON EQUIVALENT CIRCUITS The thevenin and Norton circuits are equivalent to each other, so their circuit parameters are related as follows: SJTU

24. Source transformation SJTU

25. + EXAMPLE 8-17 Both sources in Fig. 8-25(a) operate at a frequency of =5000 rad/s. Find the steady-state voltage vR(t) using source transformations. SOLUTION: SJTU

26. EXAMPLE 8-18 Use Thevenin's theorem to find the current Ix in the bridge circuit shown in Fig. 8-26. Fig. 8-26 SJTU

27. SOLUTION: SJTU

28. SJTU

29. SJTU