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1. i +1A 1 2 3 4 0 T(s) -1A AC WAVEFORMS • Alternating Current (AC) is current that periodically reverses direction • If the switch changes position every second: AC Fundamentals

2. AC WAVEFORMS • The plot, or graph, of a current (or voltage) versus time is called a waveform • The magnitude is the size of current or voltage (y-axis) • Waveforms where the current changes magnitude, but not direction (all the values remain positive or negative) are referred to as pulsating DC • Such waveforms can also be regarded as the superposition (addition) of an AC waveform and a DC level • An AC voltage is one that periodically reverses polarity • The voltage across the 10Ω resistor reverses polarity as 1s intervals • An AC voltage source produces an EMF whose polarity reverses at periodic intervals • The AC waveform used the most in circuit theory is the sinusoidal waveform or sine wave AC Fundamentals

3. AC WAVEFORMS i(t) AC increasing and decreasing linearly with time, triangular waveform t Pulsating DC. Current does not reverse direction, doesn’t go negative t Sawtooth waveform t Pulsating DC. Not AC. AC superimposed on DC level t Max current in positive direction Sinusoidal AC t Min current in negative direction AC Fundamentals

4. y Positive angle x Negative angle REVIEW OF TRIG FUNCTIONS • Sine is a function of angle • Positive angles are measured in the anticlockwise direction from the positive x-axis. Negative angles are measured in a clockwise direction. • In AC circuit theory, angles greater than 180° are expressed as negative equivalent angles. • Eg. 225° = -135° (225 - 360) • Angles more negative than -180° are expressed in equivalent positive angles • Top two quadrants expressed in positive angles • Bottom two quadrants expressed in negative angles • Angle on negative x-axis is 180° • 200°=200°-360°= -160°; -250°= -250°+360°= 110° • 400°=400°-360° = 40°; 800°=800°-2(360°) = 80° • Sin(-θ) = -sin(θ) AC Fundamentals

5. 90° 90° PROJECTION OF ROTATING RADIUS • As the radius of the circle rotates anticlockwise, the angle it generates between itself and the positive x-axis varies from 0 to 360° • At any instant, the radius is the hypotenuse of a right-angled triangle, containing the angle θ • Sin θ = 0 when θ = 0 and θ = 180° • Sin (θ+90°) = cos θ • Sin θ = cos (θ - 90°) • Cos(- θ) = cos θ AC Fundamentals

6. r b θ a USEFUL TRIG RELATIONSHIPS • Sin(θ180°) = -sin θ • Cos(θ180°) = -cos θ • Sin(θ180°) = -sin(θ) • Cos(θ180°) = -cos(θ) • Tan90° = +∞ • Tan(-90°) = -∞ • Tan θ = b/a • Sin θ = b/r; b = r sin θ • Cos θ = a/r; a = r cos θ • Sin-1(b/r) = θ or arcsin(b/r) = θ • Cos-1(a/r) = θ or arcos(a/r) = θ • Tan-1(b/a) = θ or arctan(b/a) = θ AC Fundamentals

7. T = 0.4s T = 0.4s WAVEFORM PARAMETERS:PERIOD AND FREQUENCY i 0.3 0.4 0.6 t 0 0.1 0.2 0.5 • An AC waveform can be considered to exist for all time • Yet we need a reference time, t =0, where plot begins • Helps us to express waveform mathematically • Periodic AC waveform repeats at regular intervals • The time required to complete a cycle is called the periodT • Period can be measured between any two corresponding points on successive cycles • Frequency is 1/T (Hertz – Hz) • 1 Hz = 1 cycle/second AC Fundamentals

9. v(t) +3V Peak value =3V Peak-to-peak value = 6V t 0 -3V WAVEFORM PARAMETERS:PEAK & INSTANTANEOUS VALUES • Max value reached by AC waveform - peak value (pk) • Peak-to-peak (p-p) value is the difference between positive peak and negative peak values (3 - -3 = 6) • The peak value is also called amplitude • Any sin function can be expressed as • Vpand Ip are the peak values • Lowercase i and v used for AC quantities • Uppercase I and V used for DC quantities • Instantaneous value of AC waveform is the value at specific instant of time: i(t) = 3sin100t at t = 2ms? AC Fundamentals

10. PHASE RELATIONS • Adding angle  to angle θ in the sine function: sin(θ ) causes to sine waveform to shift left (+ ) or right (- ) • ωt is in radians, but  is expressed in degrees • v(t) = 5 sin (100t + 30°) V means v(t) is shifted left by 30° • Example: Find the instantaneous value at t = 0.25s of i(t) = 0.5 sin (8105t + 50°) A • Answer: 0.439 A AC Fundamentals

11. LAG AND LEAD • When two waveforms have different phase angles, the one shifted farthest to the left is said to lead the other • v1(t) = 6 sin(ωt + 50°) leads v2(t) = 0.1 sin(ωt + 20°) because v1 is shifted left by 50°, while v2 is shifted left by 20° • v1 has a phase shift 30° greater than that of v2, i.e. v1leadsv2 by 30°, or v2lagsv1 by 30° • Lead-lag terminology derived from observation of relative positions of the waveforms when plotted versus time • The waveform with greater positive phase reaches its peak first (earliest in time), i.e. it leads the other AC Fundamentals

12. 80° 30° 50° 80° LAG AND LEAD EXAMPLE ωt • v(t) is shifted left by 30°, i(t) is shifted right by 50° • v(t) lies to the left by 80°, thus v(t) leads i(t) by 80° • Equivalently i(t) lags v(t) by 80° • Notice how v(t) reaches its peak value 80° before i(t) • Phase comparisons of this type can only be made if the two waveforms have the same frequency ω • i1 = 75sin(ωt - 18°)A; i2 = 3sin(ωt - 31°)A • i1 shifted right by 18°, i2 shifted right by 31°. i1 lies 31-18=13° to left of i2. i1 leads i2 by 13° AC Fundamentals

13. AVERAGE VALUES (1) • Average value of waveform is the average of all its values over a period of time • Computing the average over time involves adding all the values that occur in a specific time interval and dividing that sum by time • This is done by computing the area of waveform over a period of time • Area above time axis is a positive area • Area below time axis is a negative area • Algebraic signs must be taken into account when computing total (net) area • Time interval is the period T • Find the average value of the following AC waveform v(t) A1 +15V A2 0.2 0.6 0.8 1.2 0 t(s) -3V T AC Fundamentals

14. i(t) 0.4A 0 0.1 0.2 0.3 0.4 0.5 t (s) AVERAGE VALUES (2) • Sine waves are symmetrical about time axis, thus average is zero: positive area cancels negative area • Must use calculus to compute average, and we take into consideration the area of a half cycle of a sinusoidal waveform, called a pulse • Using calculus it can be shown that: • Area = 2Vp/ω V-s or 2Ip/ω A-s • What is the average of the following waveform AC Fundamentals

15. AVERAGE VALUES (3) • Average value of waveform also called dc value • Knowledge of average value important in the design of high power devices such as power supplies and power amplifiers • Waveforms in circuits regarded as sine waves with dc offset • DC level (offset) is simply added to ac waveform • Equation for ac voltage with dc component Vdc is • v(t) = Vdc + Vp sin(ωt + ) • Vdc can either be positive or negative • Minimum v(t) = Vdc – Vpvolts • Maximum v(t) = Vdc + Vp volts • Find the average value of v(t), and derive mathematical expression as a function of time +12V Vdc 0 2 10 t (s) -2V AC Fundamentals

16. Average values not useful for comparing sinusoidal ac waveforms because average value is zero Use measure called effective or root mean square (rms) value Measure shows how effective the waveform produces heat in a resistance RMS measure eliminates consideration of waveform polarity The DC voltage that causes the same heating in resistance R as an ac voltage is the effective value (rms value) of the ac voltage (likewise for current) Find the effective value of the ac current i(t) = 4.2sin(5000t + 45°)A The ac voltage supplied at a terminal has a frequency of 60Hz and effective value of 120V rms. Derive the mathematical expression for the voltage at the terminal, assuming zero phase angle EFFECTIVE (RMS) VALUES (1) AC Fundamentals

17. v(t) +20V 2 4 7 9 12 0 1 5 6 10 11 t (s) -12V EFFECTIVE (RMS) VALUES (2) • Relationship between peak and rms values given in previous equations valid only for sinusoids • It is possible to compute effective value of some nonsinusoidal waveforms using root mean square method • Square the waveform, find its average, then square root the answer • [average(waveform)2] • Find the effective value of the voltage of this waveform AC Fundamentals

18. AC VOLTAGE AND CURRENT IN RESISTANCE • Ohm’s Law can be applied to an ac circuit containing a resistance to determine ac current in a resistance when ac voltage is applied across it • At every instant in time, the current in the resistor is the voltage at that instant divided by the resistance, i.e. instantaneous current is instantaneous voltage divided by the resistance • For a sinusoidal voltage • Ip= Vp/R • The voltage across and the current through the resistor have the same phase angle, they are in phase AC Fundamentals

19. AC VOLTAGE AND CURRENT IN RESISTANCE: EXAMPLES • The current in a 2.2 kΩ resistor is: • i(t) = 5sin(2π 100t + 45°) mA • Write the mathematical expression for the voltage across the capacitor • What is the effective value of the resistor voltage? • What is the instantaneous value of the resistor at t = 0.4ms? • The ac voltage across a 150Ω resistor is: • 39sin(2π  103t) V. At what value of t does the current through the resistor equal -0.26A? AC Fundamentals

20. XC XC = 1/2πfC 0 f CAPACITORS AND AC (1) • AC current through a capacitor depends not only on the voltage across it, but also on the frequency of that voltage • The property of a capacitor that causes it to resist the flow of ac current through it is called capacitive reactance, denoted by XC. Its units are also Ohms • XC= 1/ωC = 1/2πfC Ohms • XCis inversely proportional to frequency • The greater the frequency, the smaller the reactance and thus the greater the current through the capacitor • The lower the frequency, the greater the reactance • Graph of reactance XC v frequency f Capacitor is an open circuit when dc voltage is connected across it i.e. f = 0 AC Fundamentals

21. CAPACITORS AND AC (2) • Example: The ac voltage across a 0.5F capacitor is: • v(t) = 16sin(2  103t) V • What is the capacitive reactance of the capacitor? • What is the peak value of the current through it? • Example: The ac current through a 20F capacitor is i(t) = 3sin(800t) A. What is the peak voltage across the capacitor? • The peak current through a capacitor does not occur at the same instant of time that the voltage across it reaches its peak value • The current through a capacitor leads the voltage across it by 90° • vC(t) = Vpsin(ωt + )V • iC(t) = Ip sin(ωt +  + 90°)A • Example: The voltage across a 0.01F capacitor is vC(t) = 240sin(1.25104t -30°)V. Write the mathematical expression for the current through it. AC Fundamentals

22. INDUCTORS AND AC • Inductance resists or impedes the flow of ac current • Property called inductive reactanceXLwhere: • XL = ωL = 2πfL ohms; waveform: • XL directly proportional to frequency • XL decreases with frequency, approaching 0Ω as frequency approaches zero (dc) • Inductor is a short circuit when a dc voltage is connected across it • Ohm’s Law: Ip = Vp/XL amperes • Voltage across inductor leads current through it by 90° • iL(t) = Ipsin(ωt + ); vL(t) = Vpsin(ωt +  + 90°) • Example: The current through an 80mH is 0.1sin(400t-25°)A. Write the mathematical expression for the voltage XL (Ω) XL f AC Fundamentals

23. AVERAGE POWER • The power at any instant in time, instantaneous power: • p(t) = v(t)i(t) = [i(t)]2R = [v(t)]2/R • More useful measure is average power, which is the average of the instantaneous power over a period of time • Example: Find the average power dissipated through a 50Ω with a voltage of 12sin(377t)V across it. Use all the equations on the left. AC Fundamentals