1 Bakiss Hiyana binti Abu Bakar JKE, POLISAS
1. Explain AC circuit concept and their analysis using AC circuit law. 2. Apply the knowledge of AC circuit in solving problem related to AC electrical circuit. 2
CHAPTER CONTENT Use oscilloscope to measure waveforms Understand alternating current Understand the generation of an alternating current Understand the basic circuits laws of resistive AC circuits ALTERNATING VOLTAGE AND CURRENT Understand a sinusoidal voltage and current values Understand a phasor to represent a sine wave Understand angular measuremen t of a sine wave
1.1.1 DIFFERENTIATE BETWEEN DIRECT CURRENT AND ALTERNATING CURRENT DC The flow of electrical charge is only in one direction AC The movement of electrical charge periodically reverses directions. The output voltage will remain essentially constant over time AC source of electrical power charges constantly in amplitude & regularly changes polarity
1.1.2 EXPLAIN WHY AC IS USED IN PREFERENCE TO DC DC CRITERIA AC When a large amount of electrical energy is required, it is much difficult to generate DC (Expensive) Difficult to convert voltage. DC AC : complex, expensive & less efficient. DC does get used in some local commercial applications COST CONVERT VOLTAGE When a large amount of electrical energy is required, it is much economical & easier to generate & transmit AC (Cheaper) Easy to change AC voltage to a higher @ low voltage using transformer. Easy to convert to DC, so can be used to operate various types of DC circuits @ equipment USAGE AC is the form in which electrical power is delivered to business & residences. AC may also converted into electromagnetic waves (radio waves) which can radiate @ travel through space. (wireless). Use extensively in electronic to carry information from 1 point to another. 5
1.1.3 LIST THE SOURCES OF ALTERNATING CURRENT Dry cell battery Solar cell Car battery DC AC Alternating current generator Generating plant Wind power station **WHERE IS AC USED? - In any application where a large quantities of power are needed. 6
1.2.1 EXPLAIN FARADAY S & LENZ S LAW INVOLVED IN GENERATING AC CURRENT - Faraday s Law: Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be induced in the coil. - Lenz s Law: There is an induced current in a closed conducting loop if and only if the magnetic flux through the loop is changing. The direction of the induced current is such that the induced magnetic field always opposes the change in the flux. 7
» In accordance with Faraday's Law of electromagnetic induction, an alternating current (AC) is generated when: 1. a magnet field is rotated within a wire coil or 2. a wire coil is rotated around a magnet field. 8
Generation of a sine wave:» Sinusoidal voltages are produced by ac generators and electronic oscillators.» 2 way to generate AC current: (a) Conductor rotates in a constant magnetic field, a sinusoidal wave is generated N B C D S A B C D A When the conductor is moving parallel with the lines of flux, no voltage is induced. Motion of conductor Conduc tor 9
(b) Conductor remain constant whole the magnetic field moved. - A bar magnet passes through a coil When magnet s S-pole is leaving the coil, induced I flows in such a direction as to produce a N-pole to oppose the leaving of magnet. The induced I become zero. I is about to change direction. When magnet s N-pole is moving into coil, induced I flows in such a direction as to produce a N-pole to oppose the approaching of magnet. Induced Voltage: the voltage produced within the conductor. The voltage induced in a conductor is directly proportional to the rate at which the conductor cuts the magnetic lines of forces. 10
The speed of conductor movement The faster the conductor moves, the greater the induced voltage coz it cut more lines of force in a given period of time, voltage increase. The strength of magnetic field Stronger magnetic field will result in more lines of force, induced voltage will higher. The length of the conductor in the field The longer the conductor, the greater the induced voltage coz longer conductor cut more line of force as it moves through the field. The angle at which the conductor cuts the field. If the conductor moves at a right angle with respect to the field, maximum amount of voltage is induced. 11
1.2.2 DRAW AC WAVEFORMS PRODUCED BY A SIMPLE ALTERNATING CURRENT GENERATOR ( 1 LOOP 2 POLE MAGNET )» Generators convert rotational energy to electrical energy. When a conductor is in a magnetic field and either the field or the conductor moves, an emf (voltage) is induced in the conductor. This effect is called electromagnetic induction.» A loop of wire rotating in a magnetic field produces a voltage which constantly changes in amplitude and direction.» The waveform produced is called a sine wave and is a graphical picture of alternating current (ac). One complete revolution (360 ) of the conductor produces one cycle of ac.» The cycle is composed of two alternations: a positive alternation and a negative alternation. One cycle of ac in one second is equal to 1 hertz (1 Hz). 12
AC GENERATOR: - The AC generator has slip rings that pick up the induces voltage through a complete relation cycle. - The induced voltage is related to the number of lines flux cut. - When the loop in moving parallel with the lines of flux, no voltage is induced. - When the loop in moving perpendicular to the lines of flux, the maximum voltage is induced. N S brushes arm ature slip rings 13
14
15
Basic Single Coil AC Generator 16
MULTI POLE AC GENERATOR: - By increasing the number of poles, the number of cycle per revolution can be increased. 17
1.2.3 DEVELOP AN EQUATION OF A SINUSOIDAL WAVEFORM, e = Em sin ( ωt ± θ ) e = Em sin ( ωt ± θ )» Em = is the peak voltage @ current. (unit: volt @ A).» ω = is the angular frequency (unit: radians per second; rads) The angular frequency is related to the physical frequency, (unit = hertz), which represents the number of cycles per second, by the equation.» t = is the time (unit: second).» θ = the phase, specifies where in its cycle the oscillation begin at t= 0. 18
Basic trigonometry: 19
Sinusoidal equation: θ 20
» An elementary four pole generator with a six turn rotor coil generates the following voltage wave e = 24.2 sin 36t» Determine the frequency. So; ω = 2πf f = ω/2π = 36 / 2π = 5.7296Hz 21
1.3.1 DEFINE FREQUENCY, PERIOD, PEAK VALUE OR AMPLITUDE AND THEIR RELATIONS. FREQUENCY: - Frequency ( f ) is the number of cycles that a sine wave completes in one second. - Frequency is measured in hertz (Hz). - The more cycles completed in 1 sec, the higher the frequency. - Relationship between frequency (f) & period (T) is f = 1/T 1.0 s 22 If 3 cycles of a wave occur in one second, the frequency is 3.0 Hz
- The period and frequency are reciprocals of each other. 1 f T AND 1 T f - Thus, if you know one, you can easily find the other. If the period is 50 ms, the frequency is 0.02 MHz = 20 khz. 23
PERIOD: - The time required for a sine wave to complete 1 full cycle is called a period (T). - A cycle consists of 1 complete +ve and 1 complete ve alternation. - The period of sine wave can be measured between any 2 corresponding points on the waveform. AMPLITUDE@ PEAK VALUE (Vp/Ip): - The amplitude is the maximum value of a voltage or current. - The amplitude of a sine wave is only measured from the center to the maximum point. 20 V The amplitude (A) of this sine wave is 20 V The period is 50.0 ms 15 V 10 V 0 V A 0 25 37.5 50.0 t ( ms) -10 V -15 V -20 V T 24
PEAK TO PEAK VALUE: - The voltage @ current from the +ve and ve peak. - The peak to peak values are represented as Vpp @ Ipp. - Where: Vpp = 2Vp @ Ipp = 2Ip 20 V 15 V 10 V 0 V 0 25 37.5 50.0 t ( ms) -10 V -15 V -20 V V PP 25
1.3.2 DATERMINE THE VARIOUS VOLTAGE AND CURRENT VALUES OF A SINE WAVE a. INSTANTANEOUS VALUE AT ANY POINT: - The instantaneous values of a sine wave voltage @ current are different at any different point along the curve, having +ve and ve value. - Represent as: v @ I b. RMS VALUE: - The rms ( root mean square ) value @ effective value of a sinusoidal voltage is equal to the dc voltage that produces the same amount of heat in a resistance as does the sinusoidal voltage. - V rms = 0.707 Vp - I rms = 0.707 Ip NOTE: 0.707 = 1_ 2 26
c. AVERAGE VALUE: - By definition, the average value is as 0.637 times the peak value - The average value is the total area under the half cycle curve divided by the distance in radians of the curve along the horizontal axis. - Vavg = 0.637 Vp @ 2/π Vp - Iavg = 0.637 Ip @ 2/π Ip The peak voltage of this waveform is 20 V. 20 V 15 V The rms voltage is 10 V V P V avg V rms 14.1 V. 0 V 0 25 37.5 50.0 t ( ms) The average value for the sinusoidal voltage is 12.7 V. -10 V -15 V -20 V 27
d. FORM FACTOR: - Rms value _ = 1.11 Average value e. PEAK FACTOR: - Peak value @ maximum value = 1.414 0.707 peak value rms value 28
1.3.3 CALCULATE MEAN VALUE, RMS VALUE AND PEAK FACTOR FOR A GIVEN WAVEFORM: EXAMPLE: Mean/ average value = = 0.637 Vp = 0.637. 20V = 12.74V Rms value = = 0.707 Vp = 0.707. 20V Peak factor = = Vp / rms value = 20V / 14.14 = 1.414 = 14.14 V 29
FORMULA UNIT Frequency 1 Hz f T Period 1 Sec T f Amplitude Vp @ Ip Volt @ A Peak to Peak value 2 x Vp @ 2 x Ip Volt @ A RMS value 0.707 x Vp @ 0.707 x Ip Volt @ A Average value 0.637 x Vp @ 0.637 x Ip Volt @ A Form Factor RMS value = 1.11 Average value Peak Factor Peak value = 1.414 RMS value - - 30
» An alternating voltage is given by v = 282.8 sin 314t v. Find; a) The r.m.s voltage b) The frequency c) The instantaneous value of voltage when t = 4ms Solution: a) Vrms = 0.707 x Vp = 0.707 x 282.8V = 200V b) ω = 314 rad/s = 2πf f = 314 / 2π = 50Hz Note: 1.256 radians. Convert radians degree 1.256 x 180 / π = 71.96 a) v = 282.8 sin ( 314 x 4ms ) = 282.8 sin 1.256 = 282.8 sin 71.96 = 268.9V 31
» An alternating voltage is given by v = 310 sin 100πt + 30. Determine; a) The amplitude b) The root mean square voltage c) The average voltage d) The instantaneous value of voltage when t = 5ms e) The time when the voltage first reach maximum value Solution: a) Amplitude = Vp = 310V b) Vrms = 0.707 x Vp = 0.707 x 310V = 219.17 V c) Vavg = 0.637 x Vp = 0.637 x 310V = 197.47 V d) V = 310 sin [ 100 π x 5ms ] + 30 = 310 sin [ 1.571 + 30 ] = 310 sin [ 90 + 30 ] = 268.47 V e) 310 = 310 sin [100πt + 30 ] 310 / 310 = sin [100πt + 30 ] 1 = sin [100πt + 30 ] sin -1 1 = 100πt + 30 90-30 = 100πt Convert radian degree 1.571 x 180 / π = 90 Convert degree radian 60 x π / 180 = 1.047 rad 60 = 100πt 1.047 rad = 100πt 32 so; t = 1.047 rad / 100π = 3.33ms
1.4.1 SHOW HOW TO MEASURE A SINE WAVE IN TERMS OF ANGLES» Angular measurements can be made in degrees ( o ) or radians. 33
1.4.1 SHOW HOW TO MEASURE A SINE WAVE IN TERMS OF ANGLES R R As angle A increases, the values of the trigonometric functions of A undergo a periodic cycle from 0, to a maximum of 1, down to a minimum of -1, and back to 0. There are several ways to express the measure of the angle A. One way is in degrees, where 360 degrees defines a complete circle. Another way to measure angles is in a unit called the radian, where 2π radians defines a complete circle. 34
35
1.4.2 DEFINE RADIAN - The Radian, (rad) is defined mathematically as a quadrant of a circle where the distance subtended on the circumference equals the radius (r) of the circle. - There are 360 o or 2p radians in one complete revolution. - Since the circumference of a circle is equal to 2π x radius, so 1 radian = 360 o /2π = 57.3 o. - Radian = the standard unit of angular measurement. 36
1.4.3 CONVERT RADIANS TO DEGREE - Because there are 2p radians in one complete revolution and 360 o in a revolution, the conversion between radians and degrees is easy to write. - To find the number of radians, given the number of degrees: rad 2 rad p 360 degrees - To find the number of degrees, given the radians: 360 deg 2 p rad rad 37
1.4.4 DETERMINE THE PHASE ANGLE OF A SINE WAVE Phase shift: - The phase of a sine wave is an angular measurement that specifies the position of a sine wave relative to a reference. - To show that a sine wave is shifted to the left or right of this reference, a term is added to the equation given previously. e Em sin t Where, θ = phase shift 38
Voltage (V) Example of a wave that lags the reference: 40 30 20 10 0 Referenc e Peak voltage v = 30 V sin (wt - 45 o ) 0 45 90 135 180 225 270 315 and the equation has a negative phase shift 360 405-20 -30-40 Notice that a lagging sine wave is below the axis at 0 o Angle ( ) 39
Voltage (V) Example of a wave that leads the reference: 40 30 20 10 Referenc e Notice that a leading sine wave is above the axis at 0 o Peak voltage v = 30 V sin (wt + 45 o ) -45 0-10 -20-30 -40 0 45 90 135 180 225 270 315 and the equation has a positive phase shift 360 Angle ( ) 40
1.5.1 DEFINE PHASOR - A phasor is a straight line drawn in such a way that its length is related to the amplitude of the sine wave represented, and its angular position relative to other phasors is related to the phase difference between the quantities. - Phase denotes the particular point in the cycle of a waveform, measured as an angle in degrees. 41
1.5.2 EXPLAIN HOW PHASORS ARE RELATED TO THE SINE WAVE FORMULA - The sine wave can be represented as the projection of a vector rotating at a constant rate. This rotating vector is called a phasor. - The phasor represented by the arrow is rotating in an anticlockwise direction about the centre origin point, describing the sine wave as it rotates. - Phasors allow AC calculations to use basic trigonometry. The sine function in trigonometry is the ratio of the opposite side of a right triangle to the adjacent side. 90 V p v = V p sin = 19.2 V V p = 25 V = 50 0 50 V p 42
1.5.3 DRAW A PHASOR DIAGRAM Phasor Diagram a. At any point in time, the length of the red dotted line represents the instantaneous value of the wave. b. The length of the phasor represents the amplitude of the wave. c. The angle of the phasor gives the phase of the waveform. d. Increments in phasor angle in the circular diagram are equivalent to time or angle increments along the horizontal axis of the waveform diagram. e. So with this addition of angular information, the phasor gives a relatively simple way to show the complex relationships that exist between sine waves in an ac circuit. 43
» The position of a phasor at any instant can be expressed as a positive angle, measured counterclockwise from 0 or as a negative angle equal to - 360. positive angle of phasor negative angle of - 360 44
1.5.4 DISCUSS ANGULAR VELOCITY» When a phasor rotates through 360 or 2p radians, one complete cycle (since 1 revolution = 360 )» In 1 second, phasor will rotate through f revolutions @ through f x 360» In calculation, it is more common to use angular unit RADIAN (rad) where 360 = 2π rads.» The phasor therefore rotate through 2π f radians per second.» The velocity of rotation is called the angular velocity (w). w = 2pf (Note that this angular velocity is expressed in radians per second.) 45
1.6.1 APPLY OHM S LAW TO RESISTIVE CIRCUITS WITH AC SOURCES - The voltage V across a resistor is proportional to the current I travelling through it. - This is true at all times: V = RI. ohm law 46
1.6.1 APPLY OHM S LAW TO RESISTIVE CIRCUITS WITH AC SOURCES - The voltage V across a resistor is proportional to the current I travelling through it. - This is true at all times: V = RI. ohm law 47
1.6.2 APPLY KIRCHHOFF S VOLTAGE LAW AND CURRENT LAW TO RESISTIVE CIRCUITS WITH AC SOURCES» Kirchhoff's Voltage and Current Laws apply to all AC circuits as well as DC circuits. Kirchhoff's Current Law: - The sum of current into a junction equals the sum of current out of the junction. - i 2 + i 3 = i 1 + i 4 - The sum of all currents at a node must equal to zero. 48
1.6.2 APPLY KIRCHHOFF S VOLTAGE LAW AND CURRENT LAW TO RESISTIVE CIRCUITS WITH AC SOURCES Kirchhoff's Voltage Law: - The algebraic sum of the voltage (potential) differences in any loop must equal zero. - Example: V1 + V2 Vs = 0 49
1.6.3 DETERMINE POWER IN RESISTIVE AC CIRCUITS - In a direct current circuit the power is equal to the voltage times the current, or P = E X I. - The TRUE POWER depends upon the phase angle between the current and voltage. - True power of a circuit is the power actually used in the circuit. - Measured in watts. 50
1.6.3 DETERMINE POWER IN RESISTIVE AC CIRCUITS - Note that the waveform for power is always positive, never negative for this resistive circuit. - This means that power is always being dissipated by the resistive load, and never returned to the source as it is with reactive loads. 51
Example: In this example, the current to the load would be 2 amps. The power dissipated at the load would be 240 watts. Because this load is purely resistive (no reactance), the current is in phase with the voltage, and calculations look similar to that in an equivalent DC circuit. 52
Calculate the current and power consumed in a single phase 240V AC circuit by a heating element which has an impedance of 60 Ohms. Also draw the corresponding phasor diagram.» The Active power consumed by the AC resistance is calculated as:» The corresponding phasor diagram is given as: 53
» A sinusoidal voltage supply defined as: V(t) = 100 x cos(ωt + 30 o ) is connected to a pure resistance of 50 Ohms. Determine its impedance and the value of the current flowing through the circuit. Draw the corresponding phasor diagram. Converting this voltage from the time-domain expression into the phasor-domain expression gives us: Applying Ohms Law gives us: The corresponding phasor diagram will be 54
1. In North America, the frequency of ac utility voltage is 60 Hz. The period is A. 8.3 ms B. 16.7 ms C. 60 ms D. 60 s 55
2. The amplitude of a sine wave is measured.. A. at the maximum point B. between the minimum and maximum points C. at the midpoint D. anywhere on the wave 56
3. Which property of a sine wave does the length of a phasor represent? A. Frequency B. Phase C. Amplitude D. Instantaneous value 57
4. In the equation v = V p sin ωt ±, the letter v stands for the A. peak value B. average value C. rms value D. instantaneous value 58
Voltage (V) Referenc e 40 30 Peak voltage 20 10 V 1 0 0 45 90 135 180 225 270 315 360 405-20 -30-40 Angle ( ) 5. Give the suitable sinusoidal equation for waveform V1 above A. V 1 = 30 sin ωt B. V 1 = 30 sin ωt + 45 C. V 1 = 30 sin ωt - 45 D. V 1 = 30 sin ωt ± 45 59
6. The number of radians in 90 o is A. p/2 B. p C. 2p/3 D. 2p 60
7. For the waveform shown, the same power would be delivered to a load with a dc voltage of A. 21.2 V 60 V B. 37.8 V 45 V 30 V C. 42.4 V D. 60.0 V 0 V 0 25 37.5 50.0 t ( ms) -30 V -45 V -60 V 61
8. A control on the oscilloscope that is used to set the desired number of cycles of a wave on the display is A. volts per division control B. time per division control C. trigger level control D. horizontal position control 62
9. Convert the angle of 3π/5 radian to degree unit. A. 180 B. 118 C. 108 D. 110 63
10. Alternating current changes in.. A. Direction only B. Value only C. Both value and direction D. Frequency and value but not direction 64
1. B 2. A 3. C 4. D 5. C 6. A 7. C 8. B 9. C 10. C 65