Chapter 6: Alternating Current. An alternating current is an current that reverses its direction at regular intervals.

Transcription

1 Chapter 6: Alternating Current An alternating current is an current that reverses its direction at regular intervals.

2 Overview Alternating Current Phasor Diagram Sinusoidal Waveform A.C. Through a Resistor A.C. Through a Capacitor A.C. Through a nductor RCL Circuit in Series

3 6.1 Alternating Current Define alternating current (AC) Sketch and interpret sinusoidal AC waveform Use sinusoidal voltage and current equations: sint sint Learning Objectives

4 Alternating Current An alternating current (AC) is the electrical current which varies periodically with time in direction and magnitude. The usual circuit-diagram symbol for an AC source is

5 Sinusoidal AC Waveform The output of an AC generator is sinusoidal and varies with time. or Positive half cycle or or 1 T 1 Revolution Peak value T t Negative half cycle Peak-to-peak value

6 Sinusoidal AC Waveform Equation for alternating current (): o sint Equation for alternating voltage (): o sint Phase where : angular frequency OR angular ve locity ( f ) : peak current : peak voltage

7 Phase Angle Phase denotes the particular point in the cycle of a waveform, measured as an angle in degrees/ radian

8 Revision A A t (s) A A sin t A A A A t (s) A A sint A A A t (s) A A sint

9 Terminology in AC Peak (maximum) current ( ) Definition: Magnitude of the maximum current. Peak (maximum) voltage ( ) Definition: Magnitude of the maximum voltage. Angular frequency ( ) Unit: radian per second (rad s -1 ) Equation: f or T

10 Terminology in AC Frequency ( f ) Definition: Number of complete cycle in one second. Unit: Hertz (Hz) or s -1 Period ( T ) Definition: Time taken for one complete cycle. Unit: second (s) T 1 f

11 Example 1 Figure shows the variation of voltage with time for a sinusoidal AC. Determine a) the frequency b) the phase angle at t = 15 s c) the peak-to-peak voltage and d) write the expression for the graph

12 Example 1 Solution

13 Example 1 Solution

14 Example The current in an AC circuit is given by the expression: 5mA sin 5 t Sketch a -t graph for the AC circuit. Determine the current when t = 5 s.

15 Example Solution

16 6. Root Mean Square (rms) Define root mean square (rms) current and voltage for AC source. Use the following formula, rms and rms **Equations rms and rms are valid only for a sinusoidal alternating current and voltage** Learning Objectives

17 Average Current ( ave ) Average current is defined as the average or mean value of current in a half-cycle flows of current in a certain direction. av 1 T T T av ave for one complete cycle is zero because the current flows in one direction in one-half of the cycle and in the opposite direction in the next half of the cycle. ave ( ) t ave ave π o π o

18 Root Mean Square Current ( rms ) Root mean square current ( rms ) is defined as the effective value of AC which produces the same power (mean/average power) as the steady d.c. when the current passes through the same resistor. power dc average power ac R ave ave R rms rms

19 Root Mean Square oltage ( rms ) Root mean square voltage/ p.d ( rms ) is defined as the value of the steady direct voltage which when applied across a resistor, produces the same power as the mean (average) power produced by the alternating voltage across the same resistor power dc average power ac R R ave ave rms rms

20 Example 3 A sinusoidal, 6. Hz, ac voltage is read to be 1 by an ordinary voltmeter. Determine a) the maximum value the voltage takes on during a cycle b) the equation for the voltage

21 Example 3 Solution

22 Example 4 The alternating potential difference shown above is connected across a resistor of 1 k. Calculate a) the rms current, b) the frequency, c) the mean power dissipated in the resistor.

23 Example 4 Solution

24 Example 4 Solution

25 Example 5 An AC source = sin t is connected across a resistor of 1. Calculate a) the rms current in the resistor. b) the peak current. c) the mean power.

26 Example 5 Solution

27 Example 5 Solution

28 6.3 Resistance, Reactance and mpedance Sketch and use phasor diagram and sinusoidal waveform to show the phase relationship between current and voltage for a single component circuit consisting of pure resistor, pure capacitor and pure inductor. Use phasor diagram to analyze voltage, current, and impedance of series circuit of RL, RC and RLC. Define and use capacitive reactance, inductive reactance, impedance and phase angle. Explain graphically the dependence of R, X C, X L and Z on f and relate it to resonance. Learning Objectives

29 Phasor Diagram Phasor is defined as a vector that rotates anticlockwise about its axis with constant angular velocity. A diagram containing phasor is called phasor diagram. t is used to represent a sinusoidally varying quantity such as alternating current (AC) and alternating voltage. t also being used to determine the phase angle (is defined as the phase difference between current and voltage in AC circuit).

30 Phasor Diagram nstantaneous value N P A o A ω A A o sin t O P 1 T T T t Phasor diagram (not including the circle) Sinusoidal Waveform

31 Resistance, Reactance and mpedance Key Term/Ω Resistance, R Reactance, X Capacitive reactance, X C nductive reactance, X L mpedance, Z Meaning Opposition to current flow in purely resistive circuit. Opposition to current flow resulting from inductance or capacitance in AC circuit. Opposition of a capacitor to AC. Opposition of an inductor to AC. Total opposition to AC. (Resistance and reactance combine to form impedance)

32 Pure Resistor in the AC Circuit

33 Pure Resistor in the AC Circuit The current flows in the resistor is sinωt The voltage across the resistor R at any instant is R R R Rsinωt and R R sinωt :Supply voltage

34 Pure Resistor in the AC Circuit ω 1 T T 3 T T t Phasor diagram sinωt sin ωt

35 Pure Resistor in the AC Circuit The phase difference between and is ωt ωt n pure resistor, the voltage is in phase with the current and constant with time (the current and the voltage reach their maximum values at the same time). The resistance in a pure resistor is R rms rms

36 Pure Resistor in the AC Circuit The instantaneous power The average power t P t t P R R P sin sin sin rms ave P R R P A resistor in AC circuit dissipates energy in the form of heat

37 Pure Capacitor in the AC Circuit

38 Pure Capacitor in the AC Circuit The voltage across the capacitor C C sinωt The voltage on a capacitor depends on the amount of charge you store on its plates. The charge accumulates on the plates of the capacitor is Q C C Q C sint

39 Pure Capacitor in the AC Circuit The current flows in the ac circuit is dq dt d dt C C d dt sint sint Compare with general equation cosωt C cos( ωt) and C sinωt

40 Pure Capacitor in the AC Circuit rad ω 1 T T 3 T T t Phasor diagram sin ωt sinωt

41 Pure Capacitor in the AC Circuit The phase difference between and is Δ Δ t ωt n pure capacitor, the voltage lags behind the current by / radians or the current leads the voltage by / radians.

42 Pure Capacitor in the AC Circuit The capacitive reactance in a pure capacitor is X C rms rms C X C 1 1 C fc Definition Z X C in pure capacitor circuit

43 Pure Capacitor in the AC Circuit The instantaneous power P P P P P R cost 1 1 sin t sin t R sint The average power P ave For the first half cycle where the power is positive, the capacitor is saving the power (in electric field). For the second half of the cycle where the power is negative, the power is returned to the circuit.

44 Pure nductor in the AC Circuit

45 Pure nductor in the AC Circuit The current flows in the resistor is sinωt When the current flows in the inductor, the back emf caused by the self-induction is produced and given by d εb L dt B L d dt sin ωt B L ωcosωt

46 Pure nductor in the AC Circuit At each instant the supply voltage must be equal to the back e.m.f B (voltage across the inductor) but the back e.m.f always oppose the supply voltage. Hence, the magnitude of and B : B L L ωcosωt ωsinωt sinωt Compare with general equation cosωt L

47 Pure nductor in the AC Circuit rad ω 1 T T 3 T T t Phasor diagram sinωt sinωt

48 Pure nductor in the AC Circuit The phase difference between and is Δ Δ ωt ωt n pure inductor, the voltage leads the current by / radians or the current lags behind the voltage by / radians.

49 Pure nductor in the AC Circuit The inductive reactance in a pure inductor is X L rms rms L X L L fl Definition Z X L in pure inductor circuit

50 Pure nductor in the AC Circuit The instantaneous power P P P P R R sint cost P 1 1 sin t sin t The average power P ave For the first half of the cycle where the power is positive, the inductor is saving the power. For the second half cycle where the power is negative, the power is returned to the circuit.

51 Summary 1 Circuit Relationship between and mpedance, Z ( = Z) P instantaneous P ave Pure resistor The voltage is in phase with the current Resistor, R P sin t P ave 1 P Pure capacitor The voltage lags behind the current by / radians Capacitive reactance, X C X C 1 1 C fc P P 1 sin t P ave Pure inductor The voltage leads the current by / radians nductive reactance, X L X L L fl P P 1 sin t P ave

52 Example 6 A capacitor with C = 47 pf is connected to an AC supply with r.m.s. voltage of 4 and frequency of 5 Hz. Calculate a. the capacitive reactance. b. the peak current in the circuit.

53 Example 6 Solution

54 Example 6 Solution

55 Example 7 A 4 rms supply with a frequency of 5 Hz causes an rms current of 3. A to flow through an inductor which can be taken to have zero resistance. Calculate a. the reactance of the inductor. b. the inductance of the inductor.

56 Example 7 Solution

57 Example 7 Solution

58 RC Circuit

59 RC Circuit The rms voltage across the resistor R and the capacitor C are given by: R R and C X C C R ω Phasor diagram Based on the phasor diagram, the rms supply voltage (or total voltage) of the circuit is given by: R C leads by ϕ

60 RC Circuit C C C R X R X R C X R ω X C Z R C X R Z Z and Phasor diagram ωc X C 1 and

61 RC Circuit From the phasor diagrams, tan C R or tan X C R Graph of Z against f

62 RL Circuit

63 RL Circuit The rms voltage across the resistor R and the inductor L are given by: R R and L X L L R Phasor diagram ω Based on the phasor diagram, the rms supply voltage (or total voltage) of the circuit is given by: R L leads by ϕ

64 RL Circuit L L L R X R X R L X R L X R Z Z and and ωl X L Phasor diagram ω X L Z R

65 RL Circuit From the phasor diagrams, tan L R or tan X L R Graph of Z against f

66 RCL Circuit

67 RCL Circuit The rms voltage across the resistor R and the capacitor C are given by: R R, C X C, L X L L C L R C Phasor diagram ω Based on the phasor diagram, the rms supply voltage (or total voltage) of the circuit is given by: R ( ) c L leads by ϕ

68 RCL Circuit ) ( ) ( C L C L c L R X X R X X R L > C C L X X R Z ω X L Z X C L X C X R Phasor diagram

69 RCL Circuit From the phasor diagrams, tan L R C or tan X L R X C Graph of Z against f

70 L < C? L = C?

71 Example 8 An alternating current of angular frequency of rad s 1 flows through a 1 k resistor and a.1 F capacitor which are connected in series. Calculate the rms voltage across the capacitor if the rms voltage across the resistor is.

72 Example 8 Solution

73 Resonance in AC circuit Resonance is defined as the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency of the LRC series circuit. The graph shows that: i. at low frequency, impedance Z is large because 1/ωC is large. ii. iii. at high frequency, impedance Z is high because ωl is large. at resonance frequency, impedance Z is minimum (Z = R; X C = X L ) and is maximum

74 Resonance in AC circuit At resonance frequency: Resonance frequency X 1 L C 1 f r L f C f r L X C 1 LC r

75 Resonance in AC circuit At resonance frequency: Z R X X L C Z min R Z R min rms rms rms Z 1, Z Z min, rms max

76 Example 9 Based on the RCL series circuit in figure above, the rms voltages across R, L and C are shown. a. With the aid of the phasor diagram, determine the applied voltage and the phase angle of the circuit. Calculate: b. the current flows in the circuit if the resistance of the resistor R is 6, c. the inductance and capacitance if the frequency of the AC source is 5 Hz, d. the resonant frequency.

77 Example 9 Solution

78 Example 9 Solution

79 Example 9 Solution

80 Example 9 Solution

81 Example 9 Solution

82 6.4 Power and Power Factor Apply: i. average power, P ave rms rms cos ii. instantaneous power, P i. power factor, cos Pr P a Pav rms rms in AC circuit consisting of R, RC, RL and RLC in series. Learning Objectives

83 Power and Power Factor n an AC circuit, the power is only dissipated by a resistance; none is dissipated by inductance or capacitance. Therefore, the real power (P r ) that is used or gone is equal to that dissipated from the resistor and given by the average power (P ave ) P ave rms R rms P ave R P rms r

84 Power and Power Factor For RCL circuit: From the diagrams above cos R and cos R Z

85 Power and Power Factor P P ave ave rms or rms Z cos rms cos Papparent P a Rearranging cos P ave P a

86 Example 1 A 1 F capacitor, a. H inductor and a resistor are connected in series with an alternating source given by the equation below : Calculate : a. the frequency of the source. b. the capacitive reactance and inductive reactance. c. the impedance of the circuit. d. the maximum (peak) current in the circuit. e. the phase angle. 3sin3t f. the mean power of the circuit.

87 Example 1 Solution

88 Example 1 Solution

89 Example 1 Solution

90 Example 1 Solution

91 Example 1 Solution

92 Example 11 An oscillator set for 5 Hz puts out a sinusoidal voltage of 1 effective. A 4. Ω resistor, a 1.μF capacitor, and a 5. mh inductor in series are wired across the terminals of the oscillator. a. What will an ammeter in the circuit read? b. What will a voltmeter read across each element? c. What is the real power dissipated in the circuit?

93 Example 11 Solution

94 Example 11 Solution

95 Example 11 Solution

96 Summary Alternating Current Phasor Diagram Sinusoidal Waveform A.C. Through a Resistor A.C. Through a Capacitor A.C. Through a nductor RCL Circuit in Series

97