Et201 chapter1 ac voltage

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Et201 chapter1 ac voltage

  1. 1. TOPIC 1 : ALTERNATING VOLTAGE AND CURRENTA) DC and AC • Direct current (DC) is current that flows in one direction only. DC voltage has a fixed polarity. o For example, +12V a DC represents 12 volts in the positive direction, or -5V DC represents 5 volts in the negative direction. • A DC voltage source is a voltage source that produces direct current. o Examples: Batteries , dc power supplies (such as the power supply built into the trainer that you use in lab) DC generators , fuel cells and solar cells are DC voltage sources • Alternating current (AC) is current whose direction periodically reverses. o An AC voltage source is a voltage source that produces alternating current. AC voltage switches polarity back and forth. o The direction alternates between 50 and 60 times per second, depending on the electrical system of the country. o Examples: Electrical outlets in the walls of your home provide alternating current. The trainer that you use in lab also contains an AC voltage source called a function generator.Waveform • In DC circuits, current and voltage remain constant as time passes. Voltage (v) • V Time (μsec) t V Time(μs) • But in AC circuits the voltage and current change as time passes. • Graph of a current or voltage versus time is called a waveform. Below are several examples of AC voltage waveforms. because the voltage changes polarity Voltage (v) Time(μs)shs/ppd/dis2010 1/15
  2. 2. Voltage (v) Time (μs) Voltage(mv) Time(ms) Advantages of AC over DC: • The alternating current is the current which can travel with a large distances without being a large loss in energy while the direct current cannot travel through the long distances without any loss. DC power degrades as it moves away from its generating source; the further away, the less power. • AC voltages can be readily transformed to higher or lower voltage levels, by a transformer With direct current it is not possible to use a transformer to change voltage. • greater reliability and efficiency • lower cost of manufacture • Tuning circuits : AC electricity also allows for the use of a capacitor and inductor within an electrical or electronic circuit. A combination of a capacitor, inductor and resistor is used as a tuner in radios and televisions. Without those devices, tuning to different stations would be very difficult.B) Faradays Law• Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be "induced" in the coil.• No matter how the change is produced, the voltage will be generated.• The change could be produced by : o changing the magnetic field strength, o moving a magnet toward or away from the coil, o moving the coil into or out of the magnetic field, o rotating the coil relative to the magnet• In the Electromagnetic Induction, when a single wire conductor moves through a permanent magnetic field thereby cutting its lines of flux, an EMF is induced in it.• However, if the conductor moves in parallel with the magnetic field in the case of points A and B, no lines of flux are cut and no EMF is induced into the conductor,• but if the conductor moves at right angles to the magnetic field as in the case of points C and D, the maximum amount of magnetic flux is cut producing the maximum amount of induced EMF.• Also, as the conductor cuts the magnetic field at different angles between points A and C, 0 and 90o the amount of induced EMF will lie somewhere between this zero and maximum value.• Then the amount of emf induced within a conductor depends on: o the angle between the conductor and o the magnetic flux as well asshs/ppd/dis2010 2/15
  3. 3. o the strength of the magnetic field. N C B A D SBasic Single Coil AC Generator• As the coil rotates anticlockwise around the central axis which is perpendicular to the magnetic field, the wire loop cuts the lines of force set up between the north and south poles at different angles as the loop rotates.• The amount of induced EMF in the loop at any instant of time is proportional to the angle of rotation of the wire loop.• As the loop rotates, electrons in the wire loop flow in one direction around the loop. When the wire loop moves across the magnetic lines of force in the opposite direction, the electrons in the wire loop flow in the opposite direction. Then the direction of the electron movement determines the polarity of the induced voltage.• When the loop or coil rotates one complete revolution, or 360o, one full sinusoidal waveform is produced with one cycle of the waveform being produced for each revolution of the coil. As the coil rotates within the magnetic field, the electrical connections are made to the coil by means of carbon brushes and slip-rings which are used to transfer the electrical current induced in the coil.• The amount of EMF induced into a coil cutting the magnetic lines of force is determined by the following three factors. o Speed - the speed at which the coil rotates inside the magnetic field. o Strength - the strength of the magnetic field o Length - the length of the coil or conductor passing through the magnetic field• The frequency of a supply is the number of times a cycle appears in one second and that frequency is measured in Hertz. As one cycle of induced emf is produced each full revolution of the coil through a magnetic field comprising of a north and south pole as shown above,.• So by increasing the speed of rotation of the coil the frequency will also be increased.• Therefore, frequency is proportional to the speed of rotation, ( ƒ Ν ) where Ν = r.p.m.shs/ppd/dis2010 3/15
  4. 4. Displacement of a Coil within a Magnetic Field 1 cycle +Vm emax einst θ=ωt ω -Vm Instantaneous Value .C) Instantaneous Voltage• The instantaneous values of a sinusoidal waveform = Maximum value x sin θ Vi = Vmax x sin θ o Where, Vmax is the maximum voltage induced in the coil o and θ = ωt, is the angle of coil rotation.• The instantaneous value of the waveform and also its direction will vary according to the position of the coil within the magnetic field.• The waveform studied most frequently in electrical circuit theory is the sine wave.• a sine wave must pass through the origin (the point where the x-axis crosses the y-axis).Sinusoidal• The more general term sinusoid is used to describe any waveform that has the same shape as a sine wave but that may be shifted to the right or to the left along the x-axis. It does not pass through the origin:shs/ppd/dis2010 4/15
  5. 5. Periodic Waveform• A periodic waveform is a waveform whose values are repeated at regular intervals.• All of the waveforms shown above are periodic waveforms.Waveform Parameters• Important parameters associated with periodic waveforms include: o Period o Frequency o Peak Value o Peak-to-Peak Value o RMS Value (also called effective value) o Average Value• Each of these terms is explained below.• The plot of a periodic waveform shows a regularly repeating pattern of values, each of which is called a cycle. Vp Vp-p I cyclePeriod• The period of a waveform is the time required for completing one full cycle.• The symbol for period is T.• Period is measured in units of seconds, abbreviated s.• Example: The sine wave shown above has a period T of 50 ms.Frequency• The frequency of a waveform is the number of cycles that is completed one second• The symbol for frequency is f• Frequency is measured in units of cycles per second, or Hertz, abbreviated Hz. f=1/T and T=1/fPeak Value• Peak voltage is the voltage measured from the baseline of an ac waveform to its maximum, or peak, level.• Unit: Volts peak (Vp)shs/ppd/dis2010 5/15
  6. 6. • Symbol: Vp• The peak value of a waveform is also called amplitude,Peak-to-Peak Value• Peak-to-peak voltage is the voltage measured from the maximum positive level to the maximum negative level.• Unit: Volts peak-to-peak (Vp-p)• Symbol: Vp-pRMS Value (or Effective Value)• RMS voltage is the amount of voltage that is required for producing the same amount of heat as you would get if you connected a DC source across that same resistor.• Unit: Volts (V)• Symbol: Vrms• The RMS voltage of a sinusoidal waveform is equal to 0.707 times its peak value. Vrms = 0.707 Vp• A multimeter set to AC mode measures rms valuesAverage Value• The average value of a waveform is the average of its values over a time period.• Any waveform that is symmetrical about the time axis has zero average value over a complete cycle.• Sometimes, though, its useful to refer to the waveforms average value over a half cycle.• sine waves average value over a half ccyle is equal to 0.636 times its peak value.• The average value of a waveform is also called its DC value. Vave = 0.637 VpInstantaneous Value• The instantaneous value of an ac waveform is its value at a specific instant of time. You can use the mathematical expression for a waveform to find the waveforms instantaneous values at specific times.• Example: The instantaneous value of 10 V sin(377t) at time 3 s is equal to 266 mV, v = 10 x sin(377 x 3) = 266 mV• Since ω is in rad/s, your calculator must be in Radians mode when you do this calculation.Form Factor and Crest Factor• Form Factor and Crest Factor can be used to give information about the actual shape of the AC waveform.• Form Factor is the ratio between RMS value and the average value is given as. Form Factor = R.M.S value = 0.707 x Vmax = 1.11 Average Value 0.637 x vmax• Crest Factor is the ratio between the R.M.S. value and the Peak value of the waveform and is given as.shs/ppd/dis2010 6/15
  7. 7. Crest Factor (Peak Factor) = Peak Value = Vmax = 1.414 R.M.S Value 0.707 x VmaxExample1. A sinusoidal alternating current of 6 amps is flowing through a resistance of 40Ω. Calculate the average voltage and the peak voltage of the supply.Answer: The R.M.S. Voltage value is calculated as: Vrms = I x R = 6 x 40 =240 V. The Average Voltage value is calculated as: Form Factor = Vrms Vaverage Vavg = Vrms = 240 = 216.2V Form Factor 1.11 The Peak Voltage value is calculated as: Vp = Vrms x 1.414 = 240 x 1.414 = 339.4 volt 2. Two series resistors are connected to an ac source. If there are 7.5 V rms across one resistor and 4.2 V rms across the other, the peak source voltage is 3. If the rms voltage drop across a 15 k resistor is 16 V, the peak current through the resistor is 4. One sine wave has a positive-going zero crossing at 15° and another sine wave has a positive-going zero crossing at 55°. The phase angle between the two waveforms is 5. What is the angular frequency of a waveform whose period is 6.77 µs? : shs/ppd/dis2010 7/15
  8. 8. D) Phase of a Sine WaveVoltage (mV) Time (ms) • The pictures of sinusoidal waveforms shown above had voltage on the vertical axis and time on the horizontal axis. • Another way of plotting a sine wave is voltage on the vertical axis, and degrees of the rotors rotation on the horizontal axis. One complete cycle of the sine wave corresponds to 360°. Voltage (Volts) Phase (degrees) v = 300 V sin(θ) • The quantity on the horizontal axis is called the phase of the sine wave. This sine wave has a voltage of 0 V when its phase is 0°, and a voltage of 300 V when its phase is 90°, Radians The Radian, (rad) is defined as a quadrant of a circle where the distance subtended on the circumference equals the radius (r) of the circle. 2∏ rad = 360° 1 rad = 57.5° shs/ppd/dis2010 8/15
  9. 9. • Often we measure angles in degrees, but the radian (rad) is another unit for measuring angles. and we should be able to convert from degrees to radians and radian to degree. • A full circle is equal to 360°, and its also equal to 2∏ radians. And since ∏ is approximately equal to 3.14, this means that 360° is approximately equal to 6.28 radians. As an equation: 360° = 2∏ rad ≈ 6.28 rad 180° = ∏ rad ≈ 3.14 rad 90° = ∏/2 rad ≈ 1.57 rad • So we can say that the sine wave pictured above has a voltage of 300 V when its phase is ∏/2 rad, and a voltage of 0 V when its phase is ∏ rad.Converting Between Degrees and Radians • To convert any angle from radians to degrees: Degrees = 180º x radians ∏ • to convert from degrees to radians : Radians = ∏ x degrees 180ºPhase ShiftTwo or more sinusoids that have the same frequency, one of which is shifted to the right or theleft of the other one . produces an angular shift or Phase Difference between the two sinusoidalwaveforms. • Any sine wave that does not pass through zero at t = 0 has a phase shift. • The waveform shifted to the left is said to lead the other waveform. • the waveform shifted to the right is said to lag the other waveform.Two Sinusoidal Waveforms - "in-phase" v = Vm sin θ i = Im sin θshs/ppd/dis2010 9/15
  10. 10. Phase Difference of a Sinusoidal Waveform v = Vm sin θ where Vm = Vpeak i = Im sin (θ - Φ) • the equation for the current waveform above: i = Im sin (θ -30 º) t T • we can also determine the phase shift between 2 waveform above . if T is the period of the waveforms, and t is the time interval between corresponding points on the two waveforms, then the phase shift Φ is given by the equation: Φ = t × 360° TE) Phasors • A phasor is a vector that represents an AC electrical quantity that has both magnitude ("peak amplitude") and direction ("phase") which is "frozen" at some point in time. o The phasors length OP represents the voltages or currents peak value. o The phasors angle Φ represents the voltages or currents phase. P Φ 0 • Phasors is use to represent the relationship between two or more waveforms with the same frequency. • For example, consider the following diagram below, which shows two phasors labeled v1 and v2. o Phasor v1 is drawn at an angle of 0°, and it has a length of 10 units. o Phasor v2 is drawn at an angle of 45°, and it has a length of 5 units o v2 leads v1 by a phase shift of 45°.shs/ppd/dis2010 10/15
  11. 11. 5 • In terms of the equations for sinusoidal waveforms the above diagram would then be a representation of the equations v1 = 10 V sin ωt where θ =ωt v2 = 5 V sin (ωt + 45°)Angular Frequency • When a phasor is rotating about the origin, the waveforms frequency determines the speed of the phasors rotation. • If the waveforms frequency is f, then the phasor will rotate with an angular speed of 2∏f. • This is the waveforms angular frequency, symbol ω . • and the unit for ω is radians per second (rad/s): ω = 2∏f rad/s • ω is the Greek letter = omega; its not a w. (Since ω is in rad/s, your calculator must be in Radians mode when you do your calculation and firstly you must convert Φ from degrees to radians first . If you dont do this, youll be mixing degrees with radians, and youll get the wrong answer). Phase Difference of a Sinusoidal Waveform +vm Voltage (v) +im Current (i) Θ=wt 0 Φ=30 -im -vm v= Vm sin ωt where θ =ωt i = Im sin (ωt - 30°) shs/ppd/dis2010 11/15
  12. 12. Phasor Diagram for the Sinusoidal Waveform above V lead i by phase shift of 30ºF) Resistance with AC source IR VAC VR RSinusoidal Waveforms VR = VM sin ωt IR = IM sin ωtPhase in Resistors • The voltage across any resistor and the current through that resistor have the same phase angle. They reach their peak values at the same instant. • The resistors voltage and current are in phase with each other.Phasor Diagram VR is in phase with IR (Phase angle = 0º)shs/ppd/dis2010 12/15
  13. 13. Ohms Law for Resistors • Ohms law can be applied to resistors in AC circuits: Vp = Ip x R Vpp = Ipp x R Vrms = Irms x R • In AC circuits the term Impedance, symbol Z is the generally used and we can say that DC resistance = AC impedance, R = Z. • Z= V/I Ω • Expression for impedance in a purely resistive circuit given as a complex number will be. Z=R+j0 ΩKVL and KCL for AC Circuits • Kirchhoffs Voltage Law (KVL) says that the sum of the voltage drops around any closed loop in a circuit equals the sum of the voltage rises around that loop. • And Kirchhoffs Current Law (KCL) says that the sum of all currents entering a point is equal to the sum of all currents leaving that point. • You can also apply KVL and KCL to AC circuits that contain just resistors, as long as youre careful to use all peak values, or all rms values, or all peak-to-peak values.Power in a Resistor • Average AC Power Current and Voltage waveforms are in phase, Average power: Pave = IRMS x ERMS POWER P • In AC circuits, the resistors voltage and current must be in rms values (effective values): P = Irms2 x R P = Vrms2 / R P = Vrms x Irms • If you use peak values or peak-to-peak values instead of rms values, youll get the wrong answer for the power.shs/ppd/dis2010 13/15
  14. 14. G) Oscilloscope • The oscilloscope is an instrument designed to display waveforms. Using it, you can measure period, frequency, peak values, peak-to-peak values, and other important quantities. • The screen has a reference grid with usually 8 vertical and 10 horizontal divisions. Each resulting square has 5 further subdivisions per axis useful to better readings. Using the Oscilloscope to Measure Voltage Vertical axis v/div = 2 V sec/div = 2ms Horizonta l axis • The oscilloscope displays a graph of voltage versus time, with voltage plotted on the vertical axis and time plotted on the horizontal axis. • To measure a waveforms peak-to-peak voltage, you count how many vertical divisions (squares) the waveform covers on the oscilloscopes screen, and then you multiply this number times the setting of the oscilloscope VOLTS-PER-DIVISION knob. • Example above : Vpp = 7.6 division x v/div = 7.6 x 2 = 15.2 Vpp Using the Oscilloscope to Measure Period and Frequency. • To measure a waveforms period, T you count how many horizontal divisions (squares), one cycle of the waveform covers on the oscilloscopes screen, and then you multiply this number times the setting of the oscilloscope SECONDS-PER-DIVISION knob. • Once you know the waveforms period, you can use the formula f = 1 / T to find its frequency. • Example above : 1cycle = 5.2 division x sec/div T = 5.2 division x 2ms = 10.4 ms f = 1/T =1/10.4ms = 96.2 HzIn conclusion, to use an analog oscilloscope, you need to adjust three basic settings toaccommodate an incoming signal: • The attenuation or amplification of the signal. Use the volts/div control to adjust the amplitude of the signal before it is applied to the vertical deflection plates. • The time base. Use the sec/div control to set the amount of time per division represented horizontally across the screen. • The triggering of the oscilloscope. Use the trigger level to stabilize a repeating signal, as well as triggering on a single event.shs/ppd/dis2010 14/15
  15. 15. Also, adjusting the focus and intensity controls enables you to create a sharp, visible display.List of controls or adjustments • ON/OFF: To on or off the oscilloscope. • BRIGHTNESS: Also known as intensity of electronic beam.Control the brightness of screen • FOCUS: To get into focus the electronic beam, adjust it to have a well defined trace. • XY: This key must remain inactive for normal use. It switch the horizontal deflection drive from internal time base to a second external signal (usually the second channel in a dual trace oscilloscope). In such configuration it is possible to see the Lissajous figures on the screen. • CAL: Emit a square wave signal available to calibrate the probes • Voltage Input Channel :This section deal in vertical axis deflection on the screen. The double trace oscilloscopes have two identical sections, one for each channel, • Volt/Division selector • Coaxial connector (BNC) for input signal where to connect the measuring probe. • Invert. This selector allows exactly to reverse the signal on vertical axis. • Time / division selector • Trigger :This fundamental section allows to select the source of trigger and also to filter the signal arriving at comparator that generate the trigger event. • CH 1 main input channel. • CH 2 secondary input channel whereas exists. • DUAL TRACE OSCILLOSCOPE :With two channels all the vertical section doubles, there are two separated BNC input connectors, two V/Div selectors and so on. • ADD stands for additive and draws only one trace summing the two input channels. Using the invert on one of them it will get the graphic difference. • ALT stands for alternate and means that, at every end of scan, the channel driving the vertical axis is exchanged. • CHOP means that the vertical deflection drive is quickly exchanged by both channels during each scan.Questions1. What is the angular frequency of a waveform whose frequency is 100 Hz?2. If a waveforms frequency increases, its period ___________.3. If a waveforms frequency increases, its period ___________.4. Another name for rms value is ________.5. What is the frequency of a sine wave that has a period of 0.155 µs?6. Suppose a 33 kΩ resistor has an rms current of 1.27 mA rms. How much power is dissipatedin the resistor?7.. A negative phase shift (lag) moves a waveform to the _____________.8. A 15 V p AC voltage source is connected across a 470-kΩ resistor. What is the resistors rms current?shs/ppd/dis2010 15/15

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