Sinusoidal Steady State Analysis (AC Analysis) Part II

Transcription

1 Sinusoidal Steady State Analysis (AC Analysis) Part II Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University

2 OUTLINE Previously on ELCN102 AC Power Analysis Instantaneous Power Average Power Complex Power Root Mean Square Maximum Average Power Transfer 2

3 Previously on ELCN102 Methods of Solution of AC Circuits To solve a AC circuit you can use one or more of the following methods: Simplification Method Loop Analysis Method Node Analysis Method Superposition Method Thevenin equivalent circuit Norton equivalent circuit

4 Previously on ELCN102 Simplification Method In step by step simplification we can use: Source transformation Combination of active elements Combination of series and parallel elements Star-delta & delta-star transformation

5 Previously on ELCN102 Source Transformation A voltage source V AC with a series impedance Z can be transformed into a current source I AC = V AC /Z and a parallel impedance Z A current source I AC with a parallel impedance Z can be transformed into a voltage source V AC = I AC Z and a series impedance Z

6 Previously on ELCN102 Loop Analysis Method The Loop Analysis Method (Mesh Method) uses KVL to generate a set of simultaneous equations. 1) Convert the independent current sources into equivalent voltage sources 2) Identify the number of independent loop (L) on the circuit 3) Label a loop current on each loop. 4) Write an expression for the KVL around each loop. 5) Solve the simultaneous equations to get the loop currents.

7 Previously on ELCN102 Matrix Form Z 11 Z 12 Z 1N Z 21 Z 22 Z 2N Z N1 Z N2 Z NN I 1 I 2 I N = V 1 V 2 V N Z ii = impedance in loop i Z ij = Common impedance between loops i and j = Z ji V i = voltage sources in loop i V is +ve if it supplies current in the direction of the loop current

8 Previously on ELCN102 Node Analysis Method The Node Analysis Method (Nodal Analysis) uses KCL to generate a set of simultaneous equations. 1) Convert independent voltage sources into equivalent current sources. 2) Identify the number of non simple nodes (N) of the circuit. 3) Write an expression for the KCL at each N 1 Node (exclude the ground node). 4) Solve the resultant simultaneous equations to get the voltages. 8

9 Previously on ELCN102 Matrix Form Y 11 Y 12 Y 1N Y 21 Y 22 Y 2N Y N1 Y N2 Y NN V 1 V 2 V N = I 1 I 2 I N Y ii = admittance of node i Y ij = common admittance between node i and j = Y ji I i = current sources at node i I is +ve if it supply current into the node

10 Previously on ELCN102 Superposition Theorem For a linear circuit containing multiple independent sources, the voltage across (or current through) any of its elements is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone o V 5 0 o A Total I a I b I = I a + I b

11 Previously on ELCN102 Thevenin s Theorem A linear two-terminal circuit, can be replaced by an equivalent circuit consisting of a voltage source V th in series with a impedance Z th.

12 Previously on ELCN102 Thevenin s Theorem Steps 1) Identify the load impedance and introduce two nodes a and b 2) Remove the load impedance between node a and b 3) Calculate the open circuit voltage between nodes a and b. This voltage is V th of the Thevenin equivalent circuit. 4) Set all the independent sources to zero (voltage sources are SC and current sources are OC) and calculate the impedance seen between nodes a and b. This impedance is Z th of the Thevenin equivalent circuit.

13 Previously on ELCN102 Norton s Theorem A linear two-terminal circuit can be replaced by equivalent circuit consisting of a current source I N in parallel with a impedance Z N

14 Previously on ELCN102 Norton s Theorem Steps 1) Identify the load impedance and introduce two nodes a and b 2) Remove the load impedance between node a and b and set all the independent sources to zero (voltage sources are SC and current sources are OC) and calculate the impedance seen between nodes a and b. This resistance is Z N of the Norton equivalent circuit. 3) Replace the load impedance with a short circuit and calculate the short circuit current between nodes a and b. This current is I N of the Norton equivalent circuit.

15 Previously on ELCN102 Thevenin and Norton equivalent circuits Thevenin equivalent circuit must be equivalent to Norton equivalent circuit Z N = Z th, V th = I N Z N, I N = V th Z Z th = V th th I N

16 AC Power Analysis Instantaneous Power Instantaneous power p t is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it p t = v t i t v t = V m cos ωt + φ v, i t = I m cos ωt + φ i p t = V m cos ωt + φ v I m cos ωt + φ i = V m I m cos ωt + φ v cos ωt + φ

17 AC Power Analysis Instantaneous Power Instantaneous power p t is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it p t = V m I m cos ωt + φ v cos ωt + φ i Hint: cos A cos B = 1 2 cos A + B + cos A B p t = V m I m 2 cos 2ωt + φ v + φ i + cos φ v φ i

18 AC Power Analysis Instantaneous Power p t = 1 2 V mi m cos 2ωt + φ v + φ i + cos φ v φ i

19 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = 1 T න 0 T p t dt = 1 T න 0 T V m I m 2 cos 2ωt + φ v + φ i + cos φ v φ i dt = V mi m 2T sin 2ωt + φ v + φ i 2ω T + t cos φ v φ i 0

20 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = V mi m 2T sin 2ωt + φ v + φ i 2ω T + t cos φ v φ i 0 sin 2ωt + φ v + φ i sin 2ωt + φ v + φ i t=0 t=t sin φ v + φ i sin 2ωT + φ v + φ i ω = 2π T sin 2 2π + φ v + φ i sin φ v + φ i

21 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = V mi m 2T T cos φ v φ i = 1 2 V mi m cos φ v φ i

22 Average Power AC Power Analysis Average Power (P av ) is the average of p t over one period P av = 1 2 V mi m cos φ v φ i = A cos φ v φ i cos φ v φ i is called the Power factor. A = 1 2 V mi m is referred to as the Apparent power. Hint: cos x is even function cos φ v φ i = cos φ i φ v

23 Average Power AC Power Analysis For an impedance Z P av = 1 2 V mi m cos φ v φ i V = V m φ v, I = I m φ i Z = V I = V m I m φ v φ i = Z Z P av = 1 2 Z I m I m cos φ v φ i = I m 2 2 Z cos Z

24 Average Power AC Power Analysis Z = Z Z = R + jx R = Z cos Z, X = Z sin Z P av = I m 2 2 Z cos Z = I m 2 R 2 For impedance Z, the average power is given by P av = I m 2 R 2 R = Re Z is the resistive part of Z

25 Average Power AC Power Analysis For φ v φ i = 0 P av = 1 2 V mi m cos φ v φ i Z is pure resistive and the average power absorbed by the impedance will be maximum. For φ v φ i = ±90 o Z is pure reactive and the average power absorbed by the impedance will be zero.

26 Example (1) AC Power Analysis For a linear circuit, v t = 120cos(377t + 45 o ), and i(t) = 10cos(377t 10 o ). Calculate the average power.

27 Example (2) AC Power Analysis For the shown circuit, calculate the average power delivered by the source and the average power absorbed by the circuit.

28 Example (3) AC Power Analysis Verify the power balance for the shown circuit

29 AC Power Analysis Root Mean Square Root Mean Square (rms) or Effective value of a varying signal is defined as the value of DC signal that would produce the same power dissipation in a resistive load DC Signal Sinusoidal Signal 2 P av = I eff R 2 I eff = I m 2 2 I eff = I m 2 P av = I m 2 2 R The eff value of a sinusoidal signal = Maximum of the sinusoidal 2

30 AC Power Analysis Root Mean Square Root Mean Square (rms) or Effective value of a varying signal is defined as the value of DC signal that would produce the same power dissipation in a resistive load DC Signal 2 P av = I eff R Any periodic signal P av = 1 T න 0 T i t v t dt I eff = 1 T න 0 T i 2 t dt = R T න 0 T i 2 t dt

31 AC Power Analysis Root Mean Square For any periodic function x t, the eff value is given by X eff = 1 T න 0 T x 2 t dt The effective value is the (square) root of the mean (or average) of the square of the periodic signal. Thus, the effective value is often known as the root-mean-square value, or rms value for short. X eff = X rms

32 AC Power Analysis Root Mean Square For any periodic function x t, the eff value is given by X rms = X eff = 1 T න 0 T x 2 t dt For a voltage signal v t = V m cos ωt v 2 t = V 2 m cos 2 ωt 1 Tv T න 2 t dt = V m 2 0 T න 0 T cos 2 ωt dt

33 AC Power Analysis Root Mean Square V m 2 Tcos 2 ωt dt = V m 2 T න 0 2T න 0 T 1 + cos 2ωt dt Hint: cos 2 θ = cos 2θ = V m 2 2T t + sin 2ωt 2ω T = V m V rms = 1 T න 0 T v 2 t dt = V m 2

34 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = I m 2 2 R = I m 2 I m 2 R = I2 rms R = V rms 2 R

35 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf A = V rms I rms is called the Apparent power because it similar to the average power of the DC circuit. Apparent power is measured in volt-amperes (VA) to distinguish it from the average power, which is measured in watts.

36 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf pf = cos φ v φ i is called the power factor. It is factor by which the apparent power must be multiplied to obtain the average power. For a purely resistive load, pf = 1. For a purely reactive load, pf = 0.

37 AC Power Analysis Root Mean Square For a resistor the average power is given by P av = 1 2 V mi m cos φ v φ i = V rms I rms cos φ v φ i = A pf pf is said to be leading if the current leads the voltage, which implies a capacitive load. pf is said to be lagging if the current lags the voltage, which implies a inductive load.

38 Example (4) AC Power Analysis Determine the rms value of the current waveform shown. If the current is passed through a 2Ω resistor, find the average power absorbed by the resistor.

39 Complex Power AC Power Analysis Complex power is important in power analysis because it contains all the information about the power absorbed by a given load. The complex power S absorbed by an AC load is the product of the voltage and the complex conjugate of the current S = 1 V I 2 V = V m φ v, I = I m φ i S = 1 2 V m φ v I m φ i

40 Complex Power For an impedance Z AC Power Analysis S = 1 2 V mi m φ v φ i = V m I m 2 2 φ v φ i = V rms I rms φ v φ i Z = V I = V m φ v I m φ i = V m φ v φ i I m = V rms φ v φ i I rms 2 S = Z I rms I rms = I rms Z = V 2 rms Z = I2 rms R + jx = I2 rms R + ji2 rms X

41 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq P = Re S is called the Active Power and its unit is Watt. Q = Im S is called the Reactive Power and its unit is VAR. A = S = V rms I rms is the Apparent Power and its unit is VA. φ v φ i is the Power Factor Angle.

42 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq The real power P is the average power in watts delivered to a load. P is the actual power dissipated by the load. The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load.

43 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq For resistive load, φ v φ i = 0 Q = 0. For capacitive load (leading pf), φ v φ i < 0 Q < 0. For inductive load (lagging pf), φ v φ i > 0 Q > 0.

44 Complex Power AC Power Analysis S = V rms I rms φ v φ i = V rms I rms cos φ v φ i + jv rms I rms sin φ v φ i 2 2 = I rms R + ji rms X = P + jq S, P, and Q construct the power triangle.

45 Example (5) AC Power Analysis The voltage across a load is v(t) = 60cos(ωt 10 o ) V and the current through the element in the direction of the voltage drop is i(t) = 1.5cos(ωt + 50) A. Find: (a) The complex and apparent powers. (b) The real and reactive powers. (c) The power factor and the load impedance.

46 Maximum Average Power Transfer Definition For maximum average power transfer, the load impedance Z L must be equal to the complex conjugate of the Thevenin/Norton impedance Z Th. For maximum power P ZL Z L = Z th = Z N

47 Maximum Average Power Transfer Definition Z th = R th + jx th Z L = Z th = R th jx th Z T = Z th + Z L = 2R th I = V th Z T = V th 2R th V th = V m 0 o I = V m 2R th 0 o P ZL = I m 2 2 R th = V m 2R th 2 Rth 2

48 Maximum Average Power Transfer Definition P ZL = V m 2 R th 2 4R th 2 P ZL = V m 2 8R th V rms = V m 2 V 2 rms = V m 2 2 P ZL = V 2 rms 4R th P ZL = I m 2 2 R th = V m 2R th 2 Rth 2

49 Maximum Average Power Transfer Example (6) For the circuit shown: a) Determine the load impedance Z L that maximizes the average power drawn from the circuit. b) What is the maximum average power?