ELL 100 - Introduction to Electrical Engineering LECTURE 13: SINGLE PHASE AC CIRCUITS 1 OUTLINE Introduction Basics of sinusoid and phasors Average, RMS value of an AC quantity Steady state AC response for a pure resistive circuit Exercise/Numerical Analysis 2 INTRODUCTION Why alternating current (AC) power system became popular? While direct current (DC) electricity flows in one direction through a wire, AC electricity alternates its direction in a back-and-forth motion AC electricity is created by an AC electric generator, which determines the frequency The major advantage of AC electricity is that the voltage can be readily changed, thus making it more suitable for long-distance transmission than DC electricity 3 INTRODUCTION AC can also be employed using capacitors and inductors in electronic circuitry, allowing for a wide range of applications Various AC Waveforms 4 SINGLE PHASE APPLIANCES Single Phase Transformer 5 SINGLE PHASE APPLIANCES Single Phase Meter 6 SINGLE PHASE APPLIANCES Single Phase Power Supply with Single Phase Grid Connected System Battery 7 SINGLE PHASE APPLIANCES Single Phase Fan Induction Motor 8 APPLICATIONS Phase-Shifter Circuits 9 APPLICATIONS 10 BASICS OF SINUSOID A sinusoid is a signal that has the form of the sine or cosine function A sinusoidal current is usually referred to as alternating current (AC) Such a current reverses at regular time intervals and has alternately positive and negative values as shown, Fig. Sinusoidal Variation 11 BASICS OF SINUSOID A sinusoidal signal is easy to generate and transmit It is the dominant form of signal in the communications and electric power industries Through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids Circuits driven by sinusoidal current or voltage sources are called AC circuits 12 BASICS OF SINUSOID We experience sinusoidal variation in, Fig. Motion of a pendulum Fig. Vibration of a string 13 BASICS OF SINUSOID Fig. Ripples on the ocean surface 14 BASICS OF SINUSOID The terms which we most commonly use: Waveform: The variation of a quantity such as voltage or current shown on a graph to a base of time is a waveform. Cycle: Each repetition of a variable quantity, recurring at equal intervals, is termed a cycle. Period: The duration of one cycle is termed its period. Fig. Cycles and Periods 15 BASICS OF SINUSOID Instantaneous value: The magnitude of a waveform at any instant in time. They are denoted by lower-case symbols such as e, v and i. Peak value: The maximum instantaneous value measured from its zero value is known as its peak value. Peak-to-peak value: The maximum variation between the maximum positive instantaneous value and the maximum negative instantaneous value is the peak-to-peak value. Fig. Peak Values 16 BASICS OF SINUSOID Peak amplitude: The maximum instantaneous value measured from the mean value of a waveform is the peak amplitude. Frequency: The number of cycles that occur in 1 second is termed the frequency of that quantity. Frequency is measured in hertz (Hz) Fig. Effect on waveforms by varying frequency 17 BASICS OF SINUSOID Sinusoids, play an important role in the analysis of periodic signals The derivative and integral of a sinusoid are themselves sinusoids. For these reasons, the sinusoid is an extremely important function in circuit analysis Consider the sinusoidal voltage, Where, V = the amplitude of the sinusoid v( t ) Vm sin t m ω = the angular frequency in radians/s ωt = the argument of the sinusoid 18 BASICS OF SINUSOID It is evident that the sinusoid repeats itself every T seconds thus, T is called the period of the sinusoid 2 From the figures, we observe that ωT = 2π T Fig. Vm sin ωt as a function of ωt Fig. Vm sin ωt as a function of t 19 BASICS OF SINUSOID The fact that v(t) repeats itself every T seconds is shown by replacing t by t + T, T v( t T ) V sin ( t T ) V sin ( t ) mm2 Vmmsin( t 2 ) V sin t v ( t ) Hence, v()() t T v t Thus, v has the same value at t + T as it does at t and v(t) is said to be periodic. A periodic function is one that satisfies f (t ) = f (t + nT ), for all t and for all integers n 20 BASICS OF SINUSOID The period T of the periodic function is the time of one complete cycle or the number of seconds per cycle The reciprocal of this quantity is the number of cycles per second, known as the cyclic frequency (f) of the sinusoid. Thus, 1 f T Therefore, 2 f While ω is in radians per second (rad/s), f is in hertz (Hz) 21 BASICS OF SINUSOID A more general expression for the sinusoid is, v( t ) Vm sin( t ) where, (ωt+ϕ) is the argument and ϕ is the phase Both argument and phase can be in radians or degrees Let us examine the two sinusoids, Fig. Two sinusoids with different phases v12( t ) Vmm sin( t ), v ( t ) V sin( t ) 22 BASICS OF SINUSOID The starting point of v2 in Fig. occurs first in time Therefore, v2 leads v1 by ϕ or v1 lags v2 by ϕ If ϕ ≠ 0, v1 and v2 are out of phase Moreover, if ϕ = 0 then v1 and v2 are said to be in phase as they reach their minima and maxima at exactly the same time Fig. Two sinusoids with different phases 23 BASICS OF SINUSOID A sinusoid can be expressed in either sine or cosine form When comparing two sinusoids, it is convenient to express both as either sine or cosine with positive amplitudes using, sin(ABABAB ) sin cos cos sin cos(ABABAB ) cos cos sin sin With these identities, it is easy to show that, sin(t 180) sin t ,cos( t 180) cos t sin(t 90) cos t ,cos( t 90) sin t Using these, sinusoid can be transformed from sine form to cosine form or vice versa. 24 BASICS OF SINUSOID A graphical approach can also be used to relate or compare sinusoids as an alternative to using the trigonometric identities, The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine Angles are measured positively counter Fig. A graphical means of relating cosine and sine, clockwise from the horizontal, as usual in cos(ωt − 90°) = sin ωt polar coordinates 25 BASICS OF SINUSOID Similarly, by adding 180° to the argument of sin ωt gives −sin ωt, or sin(ωt + 180°) = −sin ωt as shown below, + sin ωt Fig. A graphical means of relating cosine, sin(ωt + 180°) = −sin ωt 26 BASICS OF SINUSOID The graphical technique can also be used to add two sinusoids of the same frequency when one is in sine form and the other is in cosine form To add A cosωt and B sinωt as shown, Here, A is the magnitude of cosωt while B is the magnitude of sinωt Acos t B sin t C cos( t ) B where, CAB2 2, tan 1 A Fig. Adding A cosωt and B sinωt 27 SOLVED NUMERICALS ON SINUSOIDS 28 Numerical 1: Determine frequency in Hz, angular frequency in rad/s, and the amplitude of the harmonic voltage signals as shown. Solution: All four signals in Fig. have the same amplitude of 0.6 V, the same frequency of f = 1 kHz, and the same angular frequency of ω = 2πf = 6283.1 rad/s. 29 Numerical 2: Determine the frequency, amplitude, and phase of the harmonic voltage signal as shown versus the base cosine signal. Solution: The amplitude by inspection is observed to be Vm = 1.0V. The frequency is determined by observing the entire interval from 0 to 3 ms containing three full periods; hence T = 1 ms, and f = 1/T = 1000 Hz =1 kHz. 30 For the phase determination, we note that the first maximum occurs later in time than for the base cosine, which already peaks at t = 0. Therefore, the phase must be negative, that is, φ < 0. The absolute value of the phase is, T 2 T which gives |φ| = π/3 for ΔT = T/6 Alternatively, the same result can be obtained by observing that the -1 cosine function is equal to 0.5 Vm at t = 0, so that φ = -cos (0.5) = -π/3. 31 Numerical 3: Find the amplitude, phase, period, and frequency of the sinusoid v(t) = 12 cos(50t + 10°) V. Solution: The amplitude is Vm = 12 V The phase is ϕ = 10° The angular frequency is ω = 50 rad/s The period T =2π/ω =2π/50= 0.1257 s The frequency is f = 1/T= 7.958 Hz Numerical 4: Given the sinusoid 45 cos(5πt + 36°), calculate its amplitude, phase, angular frequency, period, and frequency. Solution: 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz 32 ° Numerical 5: Calculate the phase angle between v1 = −10 cos(ωt + 50 ) ° and v2 = 12 sin(ωt − 10 ). State which sinusoid is leading. Solution: METHOD 1- In order to compare v1 and v2, we must express them in the same form. If we express them in cosine form with positive ° ° ° amplitudes, v1 = −10 cos(ωt + 50 ) = 10 cos(ωt + 50 − 180 ) ° ° v1 = 10 cos(ωt − 130 ) or v1 = 10 cos(ωt + 230 ) ° ° ° v2 = 12 sin(ωt − 10 ) = 12 cos(ωt − 10 − 90 ) ° v2 = 12 cos(ωt − 100 ) = 12 cos(ωt − 100° + 360°) = 12 cos(ωt + 260°) ° It can be deduced that the phase difference between v1 and v2 is 30 ° Thus, it can be observed v2 leads v1 by 30 33 METHOD 2- Alternatively, we may express v1 in sine form: ° ° ° v1 = −10 cos(ωt + 50 ) = 10 sin(ωt + 50 − 90 ) = 10 sin(ωt − 40°) = 10 sin(ωt − 10° − 30°) ° ° But v2 = 12 sin(ωt − 10 ).