1. CHAPTER 2: DC Circuit Analysis and AC Circuit Analysis Motivation Sinusoids’ features Phasors Phasor relationships for circuit elements Impedance and admittance Kirchhoff’s laws in the frequency domain Impedance combinations SINUSOIDAL STEADY-STATE ANALYSIS – SINUSOIDAL AND PHASOR

2. Continued… Basic Approach Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent Circuits

3. Instantaneous and Average Power Maximum Average Power Transfer Effective or RMS Value Apparent Power and Power Factor Complex Power Conservation of AC Power Power Factor Correction Power Measurement Continued… AC POWER ANALYSIS

4. MOTIVATION v s (t) = 10V How can we apply what we have learned before to determine i(t) and v(t)?

5. A sinusoid is a signal that has the form of the sine or cosine function. A general expression for the sinusoid, where Vm = the amplitude of the sinusoid ω = the angular frequency in radians/s Ф = the phase Sinusoids

6. Continued… A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n. Only two sinusoidal values with the same frequency can be compared by their amplitude and phase difference. If phase difference is zero, they are in phase; if phase difference is not zero, they are out of phase.

7. Example 1 Given a sinusoid,, calculate its amplitude, phase, angular frequency, period, and frequency. Solution: Amplitude = 5, phase = –60 o, angular frequency = 4 rad/s, Period = 0.5 s, frequency = 2 Hz. Continued…

8. Example 2 Find the phase angle between and, does i 1 lead or lag i 2 ? Solution: Since sin(ωt+90 o ) = cos ωt therefore, i 1 leads i 2 155 o. Continued…

9. PHASORS A phasor is a complex number that represents the amplitude and phase of a sinusoid. It can be represented in one of the following three forms: a.Rectangular b.Polar c.Exponential where

10. Continued… Example 3 Evaluate the following complex numbers: a. b. Solution: a. –15.5 + j13.67 b. 8.293 + j2.2

11. Continued… Mathematic operation of complex number: Addition Subtraction Multiplication Division Reciprocal Square root Complex conjugate Euler’s identity

12. Continued… Transform a sinusoid to and from the time domain to the phasor domain: (time domain) (phasor domain) Amplitude and phase difference are two principal concerns in the study of voltage and current sinusoids. Phasor will be defined from the cosine function in all our proceeding study. If a voltage or current expression is in the form of a sine, it will be changed to a cosine by subtracting from the phase.

13. Example 4 Transform the following sinusoids to phasors: i = 6cos(50t – 40 o ) A v = –4sin(30t + 50 o ) V Solution: a. I A b. Since –sin(A) = cos(A+90 o ); v(t) = 4cos (30t+50 o +90 o ) = 4cos(30t+140 o ) V Transform to phasor => V VContinued…

14. Example 5: Transform the sinusoids corresponding to phasors: a. b. Solution : a) v(t) = 10cos(t + 210 o ) V b) Since i(t) = 13cos(t + 22.62 o ) AContinued…

15. The differences between v(t) and V: v(t) is instantaneous or time-domain representation V is the frequency or phasor-domain representation. v(t) is time dependent, V is not. v(t) is always real with no complex term, V is generally complex. Note: Phasor analysis applies only when frequency is constant; when it is applied to two or more sinusoid signals only if they have the same frequency. Continued…

16. Relationship between differential, integral operation in phasor listed as follow: Continued…

17. Example 6 Use phasor approach, determine the current i(t) in a circuit described by the integro-differential equation. Answer: i(t) = 4.642cos(2t + 143.2 o ) A Continued…

18. However, the derivation may sometimes be very tedious. In-class exercise for Unit 6a, we can derive the differential equations for the following circuit in order to solve for vo(t) in phase domain Vo. Continued…

19. Phasors Relationship for circuit elements Resistor:Inductor:Capacitor:

20. Continued… Summary of voltage-current relationship Element Time domainFrequency domain R L C

21. Example 7 If voltage v(t) = 6cos(100t – 30 o ) is applied to a 50 μF capacitor, calculate the current, i(t), through the capacitor. Answer : i(t) = 30 cos(100t + 60 o ) mA Continued…

22. IMPEDANCE AND ADMITTANCE The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms Ω. where R = Re, Z is the resistance and X = Im, Z is the reactance. Positive X is for L and negative X is for C. The admittance Y is the reciprocal of impedance, measured in siemens (S).

23. Continued… Impedances and admittances of passive elements ElementImpedanceAdmittance R L C

24. Continued…

25. Example 8 Refer to Figure below, determine v(t) and i(t). Answers: i(t) = 1.118cos(10t – 26.56 o ) A; v(t) = 2.236cos(10t + 63.43 o ) V Continued…

26. Kirchhoff’s Law in the Frequency Domain Both KVL and KCL are hold in the phasor domain or more commonly called frequency domain. Moreover, the variables to be handled are phasors, which are complex numbers. All the mathematical operations involved are now in complex domain.

27. IMPEDANCE COMBINATIONS The following principles used for DC circuit analysis all apply to AC circuit. For example: –voltage division –current division –circuit reduction –impedance equivalence

28. Continued… Example 9 Determine the input impedance of the circuit in figure below at ω =10 rad/s. Answer: Z in = 32.38 – j73.76

29. Continued… Basic Approach Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent Circuits

30. BASIC APPROACH 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 Steps to Analyze AC Circuits:

31. NODAL ANALYSIS 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  ) V v 2 (t) = 33.02 sin(2t + 57.12  ) V

32. MESH ANALYSIS Example 2 Find I o in the following figure using mesh analysis. Answer: I o = 1.194  65.44  A

33. SUPERPOSITION THEOREM 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.

34. Continued… 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

35. SOURCE TRANSFORMATION

36. Continued… Example 4 Find Io in the circuit of figure below using the concept of source transformation. I o = 3.288  99.46  A

37. THEVENIN AND NORTON EQUIVALENT CIRCUITS Thevenin transform Norton transform

38. Continued… Z th =12.4 – j3.2  V TH = 18.97  -51.57  V Example 5 Find the Thevenin equivalent at terminals a–b of the circuit below.

39. Instantaneous and Average Power Maximum Average Power Transfer Effective or RMS Value Apparent Power and Power Factor Complex Power Conservation of AC Power Power Factor Correction Power Measurement Continued… AC POWER ANALYSIS

40. Instantaneous and Average Power Sinusoidal power at 2  t Constant power The instantaneously power, p(t) p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.

41. Continued… 1.P is not time dependent. 2.When θ v = θ i, it is a purely resistive load case. 3.When θ v – θ i = ±90o, it is a purely reactive load case. 4.P = 0 means that the circuit absorbs no average power. The average power, P, is the average of the instantaneous power over one period.

42. Continued… Example 1 Calculate the instantaneous power and average power absorbed by a passive linear network if: Answer:

43. Continued… Answer: 927.2W Example 2 A current flows through an impedance. Find the average power delivered to the impedance.

44. MAXIMUM AVERAGE POWER TRANSFER The maximum average power can be transferred to the load if X L = –X TH and R L = R TH If the load is purely real, then

45. Continued… Example 3 For the circuit shown below, find the load impedance Z L that absorbs the maximum average power. Calculate that maximum average power. Answer: 3.415 – j0.7317, 1.429W

46. EFFECTIVE OR RMS VALUE Hence, I eff is equal to: The rms value is a constant itself which depending on the shape of the function i(t). The total power dissipated by R is given by: The effective of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

47. Continued… The average power can be written in terms of the rms values: The rms value of a sinusoid i(t) = I m cos(  t) is given by: Note: If you express amplitude of a phasor source(s) in rms, then all the answer as a result of this phasor source(s) must also be in rms value.

48. APPARENT POWER AND POWER FACTOR Apparent Power, S, is the product of the r.m.s. values of voltage and current. It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts. Power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance. Apparent Power, SPower Factor, pf

49. Continued… Purely resistive load (R) θ v – θ i = 0, Pf = 1 P/S = 1, all power are consumed Purely reactive load (L or C) θ v – θ i = ± 90 o, pf = 0 P = 0, no real power consumption Resistive and reactive load (R and L/C) θ v – θ i > 0 θ v – θ i < 0 Lagging - inductive load Leading - capacitive load

50. COMPLEX POWER Complex power S is the product of the voltage and the complex conjugate of the current:

51. Continued… P: is the average power in watts delivered to a load and it is the only useful power. Q: is the reactive power exchange between the source and the reactive part of the load. It is measured in VAR. Q = 0 for resistive loads (unity pf). Q < 0 for capacitive loads (leading pf). Q > 0 for inductive loads (lagging pf). S = P + j Q

52. Continued… Apparent Power, S = |S| = Vrms*Irms = Real power, P = Re(S) = S cos(θ v – θ i ) Reactive Power, Q = Im(S) = S sin(θ v – θ i ) Power factor, pf = P/S = cos(θ v – θ i ) S = P + j Q

53. Continued… Impedance Triangle Power Factor Power Triangle

54. CONSERVATION OF AC POWER The complex real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. For parallel connection: The same results can be obtained for a series connection.

55. POWER FACTOR CORRECTION Power factor correction is the process of increasing the power factor without altering the voltage or current to the original load. Power factor correction is necessary for economic reason.

56. Continued… Q 1 = S 1 sin θ 1 = P tan θ 1 Q 2 = P tan θ 2 Q c = Q 1 – Q 2 = P (tan θ 1 - tan θ 2 ) = ωCV 2rms P = S 1 cos θ 1

57. POWER MEASUREMENT Ifand The basic structure Equivalent Circuit with load The wattmeter is the instrument for measuring the average power.