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Chapter 4

Chapter 4. AC Networks. 4.1 Introduction. Alternating Waveforms Sinusoidal AC Voltage Square Wave Triangular Waveform Function Generators are devices that can create all of these waveforms. These devices will be used in lab.

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Chapter 4

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  1. Chapter 4 AC Networks

  2. 4.1 Introduction • Alternating Waveforms • Sinusoidal AC Voltage • Square Wave • Triangular Waveform • Function Generators are devices that can create all of these waveforms. • These devices will be used in lab. • Sketch the waveforms and circuit symbols for the waveforms above.

  3. 4.2 Sinusoidal AC Waveform • AC Generators (a.k.a. alternators) are fueled by: • water power • oil • gas • nuclear fusion • wind • solar power

  4. Power to a shaft turns a rotor inside of a stator. • Rotor - rotating magnetic • Stator - stationary coil of wire

  5. Simple Harmonic Motion (SHM) • Position versus time for simple harmonic motion is shown as sine or a cosine

  6. Uniform Circular Motion Video

  7. 90°, p/2 radians y 180°, p radians q 0°, 0 radians x 270°, 3p/2 radians

  8. Wave Description • Amplitude (Vpeak) - maximum displacement from the zero line • Period (T) - the time required for one cycle • Frequency (f) - the number of cycles that appear in a time span • 1 cycle per second = 1 Hertz • T=1/f

  9. Example Problems • Sketch the following waveform: v=Vpeaksinq • Example 4.2: • Determine the period and frequency of the waveform in Figure 4.7. • (5 cycles in 4ms for v versus t) • Example 4.3: • How long will it take a sine wave with a frequency of 0.2 kHz to complete 10 cycles?

  10. Example Problems • Example 4.4: • For v=20sinq, at what angle will the voltage rise to 5 Volts?

  11. Angular Velocity (w) - rotation rate of a vector (or generator shaft) • w is measure in radians per second. • New Waveform Equation: v=Vpsinwt

  12. Sinusoidal Voltage Relationships • v = Vpeaksin q • v = Vpeaksin wt • v = Vpeaksin 2pt/T • v = Vpeaksin 2pft • Which form you use depends on what you know

  13. Example Problems • Example 4.5: • Given i=2010-3 sin 400t, find the time at which i rises to 10 mA.

  14. Phase Relationship • f = phase angle • For initial intersections before 0 v=Vpsin(wt+f) • For initial intersections after 0 v=Vpsin(wt-f)

  15. Resistive Circuit • Only for a resistive circuit can current and voltage be in phase. • If the current and voltage go through zero at different times, they are said to be out of phase

  16. Current and Voltage out of Phase • View as two vectors rotating around the circle • Must mean they have the same w V w f I

  17. Sinusoidal Plot

  18. Example Problems • Example 4.6: • (a) Write the sinusoidal expression for each waveform appearing in Fig. 4.12. • (Current Amplitude = 5mA, f i = +60°, w = 400 rad/sec) • (Voltage Amplitude = 100 V, fv = -30°, w = 400 rad/sec) • (b) What is the phase relationship between the two waveforms?

  19. Example Problems • Example 4.7: • Determine the phase relationship between the following two waveforms: v=8.6 sin(300t+80°) i=0.12 sin(300t+10°)

  20. Graphs of Ex. 4.7

  21. 4.4 Effective (RMS) Value • These mean the same thing for AC circuits: • “equivalent DC voltage” • “effective voltage” • “rms voltage” • RMS = root mean square • Draw the “relating effective values” plots.

  22. 4.5 Average Value • For a sine wave, the average value is zero. • For other waves….

  23. Example Problems • Example 4.9: • Write the sinusoidal expression for a voltage having an rms value of 40 mV, a frequency of 500 Hz and and initial phase shift of +40. • Vp = 1.414(Vrms) = 1.414(40 mV) = 56.56 mV • w = 2pf = (2p)(0.5 kHz) = 3.142 X 103 rad/s • v = 56.56 mV sin(3142t + 40°)

  24. Example Problems • Example 4.10: Calculate the average value of the waveform below over one full cycle.

  25. What are the following for a wall outlet? • f = 60 Hz • Veff = Vrms = 120 V • Vpeak = Vrms = 170 V • Vave = 0 V • Ieff = Irms = limited by breaker • Ipeak = Irms • Iave = 0 A • f • Veff = Vrms • Vpeak • Vave • Ieff = Irms • Ipeak • Iave

  26. Effective Values • Average values for I or V are zero • If we use IV to measure power with a dc meter, we will get zero • Yet the resistor gets hot – power is dissipated • I and V are not always zero – you can get shocked from ac just like dc • We need a new definition for power

  27. Use Effective (rms) Values • PR = VRIR = IR2R = VR2/R • Look just like the dc forms of power • But we use rms values instead

  28. 4.7 The R, L, and C Elements • Ohm’s Law for Peak Values Resistors: Vpeak=IpeakR Capacitors: Vpeak=IpeakXC Inductors: Vpeak=IpeakXL

  29. Inductive Reactance • Recall the form of inductance • Since I for ac is constantly changing, there is a constant opposition to the flow of current • Inductive Reactance XL = wL = 2pfL • Measured in ohms • Energy is stored in the coil, not dissipated like in the resistor.

  30. Example • An inductor of 400 mH is connected across a 120 V, 60 Hz ac source. Find XL and the current through the coil • XL = 2pfL = 2p(60 Hz)(0.4 H) XL = 151 W • I = V/XL = 120 v/151 W I = 0.8 A

  31. XL vs. f graph for ideal inductor • XL = 2pfL • Should give a straight line graph of slope 2pL • Higher the L the greater the slope XL f

  32. Current Highest positive v Zero here Most negative here V and I for an Inductor • From we see that voltage is greatest when the change of current is greatest.

  33. Current Voltage Voltage and Current (Coil) Voltage leads current by 90°

  34. Example Problems • Example 4.13: • The current iL through a 10-mH inductor is 5 sin(200t +30). Find the voltage vL across the inductor.

  35. Capacitive Reactance • The effect of a capacitor is to prevent changes in voltage • Keep the potential difference from building up in the circuit • Since the build up of potential is minimized, the flow of current is reduced. • Capacitors act like a resistance in an ac circuit.

  36. Current Voltage Capacitive Reactance • V is changing in ac circuits • But Q = VC, so the charge changes also • Since , current is greatest when change in voltage is greatest. I leads V by 90°

  37. CIVIL • CIVIL is a memory aid. • For a capacitor C, current I leads the voltage V by 90 degrees. • For an inductor L, voltage V leads the current I by 90 degrees.

  38. Capacitive Reactance Xc also measured in ohms What is the capacitive reactance of a 50 mf capacitor when an alternating current of 60 Hz is applied?

  39. Form of Reactance (Fig. 4.28) C=1mF C=2mF

  40. Example Problems • Example 4.14: • The voltage across a 2-mF capacitor is 4mV (rms) at a phase angle of -60. If the applied frequency is 100kHz, find the sinusoidal expression for the current iC.

  41. 3.6 Phasors and Complex Numbers • A vector is a quantity has both magnitude and direction. • A scalar quantity has only a magnitude. • Examples: • Scalar: 50mph • Vector: 50mph North

  42. A phasor is a complex number used to represent a sine wave’s amplitude and phase. • A complex number can be written as C = A + Bj where C is a complex number, A and B are real numbers and

  43. Imaginary C=A+Bj q Real What is the magnitude of C? What is the direction of C?

  44. XL R XC Phasor Diagrams • We make the real axis the resistance • The imaginary axis is reactance • Inductive reactance is plotted on the +y axis • Capacitive reactance is plotted on the –y axis • The vector addition of R and X is Z X = XL - XC

  45. XL X R R XC X Z R Phasors • Form Reactance First X = XL – Xc • Form Impedance • Calculate phase angle  

  46. Example Series R-L Circuit • A coil has a resistance of 2.4 W and a inductance of 5.8 mH. If it is connected to a 120-V, 60 Hz ac source, find the reactance and the current

  47. XL X R R Z f Solution XL = 2pfL = 2p(60 Hz)(5.8 X 10-3 H) XL = 2.19 W

  48. Solution (cont’d) • I = V/Z • I = 120 V/3.25 W = 36.9 A ================================ • Notice that if f = 120 Hz, then • XL = 2p(120 Hz)(0.0058 H) = 4.37 W • Z = 4.99 Wf = 61.2° • I = 24 A

  49. Example Series R-C Circuit • What is the current flow through a circuit with 120 V, 60 Hz source, an 80 mf capacitor, and a 24 W resistor?

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