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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH CHAPTER 1 A L T E R N A T IN G VO L TA G E S A N D C URRENT (We e k 2 & 3 )
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.1) UNDERSTAND ALTERNATING CURRENT (AC)1.1.1 Differentiate between AC and DC• Direct Current (DC), which is electricity flowing in a constant direction, and/or possessing a voltage with constant polarity.• Electricity made by a battery.• DC circuit and waveform:
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH• AC is short for alternating current.• This means that the direction of current flowing in a circuit is constantly being reversed back and forth.• This is done with any type of AC current/voltage source.• AC circuit and waveform:
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.1.2 AC is used in preference to DC AC is commonly used because: - More voltage can be generated than DC - AC voltage can be increased and decreased with the help of static machine called transformers. - AC transmission and distribution is more economical as line material can be saved by transmitting power at higher voltage - AC can be converted to DC easily and required but DC cannot be converted to DC easily and will not be economical.1.1.3 List the sources of alternating current AC Source of AC: i) Rotating electrical machine (AC generator) - Rotating conductor - Rotating flux magnetic ii) Electronic Oscillator Circuit (electrical signal generator)
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.2 ) GENERATION OF ALTERNATING CURRENT1.2.1 Faraday’s and Lenz’s Law involved in generating a.c currentFARADAY LAW• The amount of voltage induced in a coil is directly proportional to the rate of change of the magnetic field with respect to the coil.• The amount of voltage induced in a coil is directly proportional to the number of turns of wire in the coil (N).
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAHLENZ’S LAW• Lenz’s Law is used to find the direction of induced e.m.f and hence current in a conductor or coil.• Lenz’s Law is stated as follows: The direction of the induced current is such as to oppose the change causing it.
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM1.2.2 AC A D Z A M S H A H by a simple alternating current M U ’ waveforms produced generator (one loop in 2-pole magnet)
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH• Coil voltage versus angular position.
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAHSinusoidal voltages are produced by ac generators and electronic oscillators.When a conductor rotates in a constant magnetic field, asinusoidal wave is generated. C C N B B DD S A A A B C D M o tio n o f c o n d u c t o r C o n d u c to r When the loop is movingis moving When the conductor perpendicular to the parallel with the lines of flux, no lines of flux, the maximum voltage is induced. is induced. voltage
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM1.2.3 ’Develop M S H A H of a sinusoidal waveform, M U A D Z A an equation e = Em sin (ωt + θ) • From emf formula: e = - d(Nφ) / dt volt = - N dφ / dt = - N d(φm cos ωt) /dt = -Nφm ω (-sin ωt ) • So, formula become: e = Nφm ω sin ωt = Nφm ω sin θ (1) • When the coil has turned through 900 i.e. when θ = 900, then sin θ = 1, hence e has maximum values, say Em . • Therefore, from Eq. we get Em = Nφm ω (2)
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• Substituting eq (1) and (2), we get e = Em sin θ = Em sin ωt• Similarly, the equation of induced alternating current is i = Im sin ωt.• Since ω = 2πf , where f is the frequency of rotation of the coil, the above equations of the voltage and current can be written as e = Em sin 2πf t and i = Im sin 2πf t.• Instantaneous value : e = Em sin ωt e = Em sin (ωt + θ) where, Em = maximum coil voltage (peak voltage)• Xn ωt = instantaneous angular position of the coil ω is angular velocity = 2πf θ = phase shift (angle of lag or lead)
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.3) UNDERSTANDING A SINUSOIDAL VOLTAGE AND CURRENT VALUES
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH 1.3.1 Define frequency, period, peak value or amplitude and their relations Cycle A complete consist of ½ cycle positive and ½ cycle negative. Period (T) The time interval for one complete cycle of a periodic waveform. T = 1/f Frequency (f) A cycle within one second. Unit used is hertz (Hz) f = 1/TPeak value (Vp) / Value of voltage or current at the positive or negative Amplitude maximum (peaks) with respect to zero.
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.3.2 Determine the various voltage and current values of a sine wave Instantaneous value The value at different points along the curve. Peak-to-peak value Voltage or current from the positive peak to the negative peak. Vpp = 2Vp or Ipp = 2Ip rms value Term rms stands for root mean square. RMS value referred to as the effective value and actually a measure of the heating effect of sine wave. Vrms = 0.707Vp or Irms = 0.707Ip Average value Defined over a half cycle rather than over a full cycle. Vavg = 0.637Vp or Iavg = 0.637Ip
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH Form Factor RMS over average value. Form Factor = RMS value Average value = 0.707Vp 0.637Vp = 1.11 Peak Factor Peak value over RMS value Form Factor = Peak value RMS value = Vp . 0.707Vp = 1.414
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.3.3 Calculate mean value, rms value and peak factor for a given waveform. 20 V 15 V The peak-to-peak 10 V voltage is 40 V. Vrms 0 V 0 VPP 25 3 7 .5 5 0 .0 t (µs) The rms voltage is -1 0 V 14.1 V -1 5 V -2 0 V
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH 20 V 15 V 10 V The average value for Vavg the sinusoidal voltage is 0 V t (µs) 0 25 3 7 .5 5 0 .0 12.7 V. -1 0 V -1 5 V -2 0 V
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH Calculate Vp, Vpp, Vrms, Vavg, frequency and peak factor. Vp = 30V Vpp = 60V Vrms = 21.21V Vave = 19.11V f = 125 Hz Peak factor = 1.414
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH 1.4) UNDERSTAND ANGULAR MEASUREMENT OF A SINE WAVE1.4.1) Show how to measure a sine wave in terms of angles• Sinusoidal voltage can be produced electromagnetically by rotating electromechanical machines. As the rotor of the ac generator goes through a full 360 of rotation, the resulting 0 output is one full cycle of a sine wave
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.4.2 Define radian• A radian (rad) is the angle formed when the distance along the circumference of a circle is equal to the radius of the circle.• One radian is equivalent to 57.30 . 0
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.4.3 Convert radians to degree Because there are 2π radians in one complete revolution and 360o 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 = degree X π 180 To find the number of degrees, given the radians: Degree = rad X 180 π
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.4.4 Determine the phase angle of a sine waveThe phase angle of a sine wave is an angularmeasurement that specifies the position of that sinewave relative to a reference. e = Em sin ( ωt ±θ ) where θ = Phase angle
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMPhase differenceS H A H the angular displacement between different U ’ A D Z A M refers to waveforms of the same frequency. e = Em sin ωt i = im sin ωt e = Em sin ωt i = im sin (ωt + θ) e = Em sin ωt i = im sin (ωt – θ)
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAHExample of a wave that lags the reference …and the equation has a negative phase R e fe re n c e shift 40 30 P e a k v o lta g e 20 v = 30 V sin (θ − 45o) V o lta g e ( V ) 1 0 0 0° 45° 90° 1 35° 1 80° 2 2 5° 270° 31 5° 360° 405° -2 0 -3 0 Notice that a lagging sine wave is below the axis at 0o -40 A n g le (° )
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAHExample of a wave that leads the reference Notice that a leading sine wave R e fe re n c e is above the axis at 0o 40 30 P e a k vo lta g e 20 v = 30 V sin (θ + 45o) V o lta g e ( V ) 1 0 -45 ° 0 0° 45° 90° 1 35° 1 80° 225° 2 7 0° 31 5° 360° -1 0 -2 0 …and the equation -3 0 has a positive phase shift -4 0 A n g le (° )
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM1.5) U ’ A D Z A M S H A HPHASOR TO REPRESENT A SINE M UNDERSTAND AWAVE 1.5.1 Define phasor The sine wave can be represented as the projection of a vector rotating at a constant rate. This rotating vector is called a phasor. Phasors allow ac calculations to use basic trigonometry. h y p o te n u s e o p p o s ite s id e rig h t θ a n g le o p p o s ite s id e a d ja c e n t s id e s in θ = h y p o te n u s e
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.5.2 Explain how phasors are related to the sine wave formula 90 180 0 0 90 180 360
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH 1.5.3 Draw a phasor diagramThe position of a phasor at any instant can be expressed as apositive angle, measured counterclockwise from 0° or as a negativeangle equal to θ − 360°. positive angle of θ negative angle of θ − 360° phasor
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAHA plot of the example in the previous slide (peak at 25 V) is shown.The instantaneous voltage at 50o is 19.2 V as previously calculated. 9 0° Vp Vp= 25 V v = V p s in = 1 9 .2 V = 50° 0° 50° V p
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH1.5.4 Identify angular velocity When a phasor rotates through 360° or 2π radians, one complete cycle is traced out. The velocity of rotation is called the angular velocity (ω). ω = 2πf (Note that this angular velocity is expressed in radians per second.)
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH• Sometimes voltages and currents are expressed in terms of cos ωt rather than sin ωt. For sines or cosines with an angle, the following formulas apply:To illustrate, consider cos(ωt + 30°). From Equation above, cos(ωt + 30°) = sin(ωt + 30° + 90°) = sin(ωt + 120°)
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH• Figure below illustrates this relationship graphically. The red phasor in as was shown • Figure (a) generates cos ωt • Figure (b) generates sin ωt, and the green phasor generates a waveform that leads it by 120°
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1.6) UNDERSTAND THE BASIC CIRCUIT LAWS OF RESISTIVE AC CIRCUITS 1.6.1 Apply Ohms Law to resistive circuits with AC sources • In general, Ohms law for resistive AC circuits is given by: V = IR Where, V = voltage I = current R = resistance
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAH• When an AC circuit consists of a voltage source and a resistor, the current is in phase with the voltage, meaning that each quantity rises and declines in step.
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAHThis expression is also accurate for the maximum values of the potential andcurrent. Where; ν =Vm sin ωt R ν =iR therefore Vm sin ω =iR t Vm i= sin ωt R i become maximum when sin ω =1 t V ∴ i= m or Vm =iR ν =Vm sin ωt R Vavg Vmax I avg = or I max = R R V peak −to − peak I peak −to − peak = R
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ET201: ELECTRICAL CIRCUIT P O L IT E K N IK S U L T A N A B D U L H A L IM MU’ ADZAM S HAHA series circuit consists of two resistors (R1 = 5 Ω and R2 = 15 Ω) and analternating voltage source of 120 volts. What is Iavg?Find the current and voltage drop at all the resistors in the circuit shown below:
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1.6.2 Apply Kirchhoffs Voltage Law and Current Law to resistive circuits with AC sourcesKirchhoff’s Law can be divided into 2:• Kirchhoff’s Current LawAt any point in an electrical circuit where change density isnot changing in time, the sum of current flowing towardsthat point is equal to the sum of currents flowing away fromthat point. or
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• Kirchhoff’s Voltage Law The algebraic sum of various potential drops across an electrical circuit is equal to the electromotive force acting on the circuit Vs = V1 + V2 + V3
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH Kirchhoff Current Law Calculate value current that flows to each branch use Kirchhoff’s Law
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ET201: ELECTRICAL CIRCUITP O L IT E K N IK S U L T A N A B D U L H A L IMMU’ ADZAM S HAH Kirchhoff Voltage Law Determine Vx
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1.6.3 Determine power in resistive AC circuits• This power, measured in watts, is the power associated with the total resistance in the circuit• To calculate true power, the voltage and current associated with the resistance must be used.
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1.7) USE AN OSCILLOSCOPE TO MEASURE WAVEFORM1.7.1 Identify common oscilloscope controls Oscilloscopes are commonly used to observe the exact wave shape of an electrical signal. Type of electronic test instrument that allows observation of constantly varying signal voltages
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1.7.2 Use an oscilloscope to measure the amplitude, period and frequency of a waveform
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