Average, RMS, Form Factor, and Peak Factor (PDF)

 Average Value Algebraic sum of all the values divided by the total number of values. Same concept is for a waveform applicable that varies with time. Thus, the average value over full cycle is ZERO The average value of current in a A. 𝐼𝑚 2 sinusoidal...

 Average Value Algebraic sum of all the values divided by the total number of values. Same concept is for a waveform applicable that varies with time. Thus, the average value over full cycle is ZERO The average value of current in a A. 𝐼𝑚 2 sinusoidal signal over full cycle is: B. 𝐼𝑚 2 C. 1 D. 0 However, an average value can be defined for the half-cycle (positive or negative) for a sinusoidal signal.  Average Half Wave Rectifier ValueFull Wave Rectifier  Average Half Wave Rectifier Value Full Wave Rectifier Based on the previous results, we can say that the average value of half wave rectifier is than full wave rectifier. A. Double B. Half C. Same D. None of these  RMS (Effective) Value The r.m.s value is defined in terms of heating effect. The r.m.s. value of an alternating current is given by that steady (d.c.) current which when flowing through a given circuit for a given time produces the same heat as produced by the alternating current when flowing through the same circuit for the same time. RMS (Effective) Value  RMS Value Half Wave Full Wave Rectifier Rectifier  RMS Value Half Wave Full Wave Rectifier Rectifier  Form Factor and = Peak Factor Factor, 𝐾 Form 𝑉𝑎𝑣 𝑉 𝑟𝑚𝑠 𝑔 𝑉 Factor, 𝐾 Peak Factor or Crest 𝑚 𝑉 𝑟𝑚 = 𝑠 For a pure sinusoidal waveform the Form Factor will always be equal to: 1 2 A. B. 0.637 C. 1.11 D. 1.414  Importance of Form Factor and Peak Fact  Actually some of our meters are designed to measure the RMS or values but that is of pure sinusoidal waveforms, if there comes any distortion in the waveform, the meter won't give the correct RMS value. For meter the waveform is still a sinusoidal but it doesn't detect the distortion that's why we use form factor to get accurate value of RMS by just multiplying form factor with the average value of that distorted waveform. It is helpful in finding the RMS values of waveforms other than pure sinusoidal.  Similarly, Some loads, such as switching power supplies or lamp ballasts, have current waveforms that are not sinusoidal. They draw a high current for a short period of time, and their crest factors, therefore, can be quite a bit higher than 1.414.  current is found by dividing the area enclosed by the half cycle by the length of the base of the half cycle. a)RMS current b)Average current c)Instantaneous current d)Total current In a sinusoidal wave, average current is always rms current. a)Greater than b)Less than c)Equal to d)Not related Which of the following is not ac waveform? a)sinusoidal b)square c)constant d)triangular Peak value divided by the rms value gives us? a)Peak factor b)Crest factor c)Both peak and crest factor d)Neither peak nor crest factor  Calculate the crest factor if the peak value of current is 10A and the rms value is 2A. a)5 b)10 c)5A d)10A Find the average value of current when the current that are equidistant are 4A, 5A and 6A. a)5A b)6A c)15A d)10A  Impedan ce Impedance is a complex number, with the same units as resistance, for which the SI unit is the ohm (Ω). Its symbol is usually Z, and it may be represented by writing its magnitude and phase in the polar form |Z|∠θ. However, cartesian complex number representation is often more powerful for circuit analysis purposes. Z=R+jX Impedance Representation  Basic Formulae of Z 1. Impedance Z = R or X or X (if only one is present) L C 2. Impedance in series only Z = √(R2 + X2) (if both R and one type of X are present) 3. Impedance in series only Z = √(R2 + (|XL - XC|)2) (if R, XL, and XC are all present) 4. Impedance in any circuit = R + jX (j is the imaginary number √(-1)) 5. Resistance R = ΔV / I 6. Inductive reactance XL = 2πƒL = ωL 7. Capacative reactance XC = 1 / 2πƒC = 1 / ωC CIRCUIT WITH RESISTANCE AND INDUCTANCE IN SERIES CIRCUIT WITH RESISTANCE AND CAPACITANCE IN SERIES Series RC circuit Series RC circuit Series RC circuit SERIES R-L-C CIRCUIT Basic Formulae Series RLC circuit  A coil has a resistance of 30Ω and an inductance of 0.5H. If the current flowing through the coil is 4amps. What will be the rms value of the supply voltage if its frequency is 50Hz.  A capacitor which has an internal resistance of 10Ω and a capacitance value of 100uF is connected to a supply voltage given as V(t) = 100 sin (314t). Calculate the peak current flowing into the capacitor. Problem on XL and XC calculation  Find the value of the instantaneous voltage if the resistance is 2 ohm and the instantaneous current in the circuit is 5A. a)5V b)2V c)10V d)2.5V Problem on Series RL circuit Problem on Series RL circuit  A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit. a)2.2A b)4.2A c)6.2A d)8.2A  A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuits current, power factor and draw the voltage phasor diagram. Problem on Series RLC circuit Problem on series RLC circuit A 10-Ω resistor, 10-mH inductor, and 10-µF capacitor are connected in series with a 10-kHz voltage source. The rms current through the circuit is 0.20 A. Find the rms voltage drop across each of the 3 elements Power and power factor in AC circuits TRUE POWER: The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts (symbolized by the capital letter P, as always) REACTIVE POWER: We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps- Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q. APPARENT POWER: The combination of reactive power and true power is called apparent power, and it is the product of a circuit’s voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S. POWER IN AC CIRCUITS True Power, Reactive Power and Apparent Power with Resistive and Inductive Load Current and Voltages in Resistive, Inductive and Capacitive Circuits Phasor Diagram for Purely Resistive, Capacitive and Inductive Circuits RESONANCE IN R-L-C SERIES CIRCUIT Let us consider as R-L-C series circuit We know that the impedance in R-L-C series circuit is | Z | R2  X  X C 2 L 1 Where XL =2πfL & XC 2πfL = Such a circuit shown in figure is connected to an a.c. source of constant supply voltage V but having variable frequency. The frequency can be varied from zero, increasing and approaching infinity.  Since XL and Xc are functions of frequency, at a particular frequency of the applied voltage, XL and Xc will became equal in magnitude. Since XL = Xc XL - Xc =0 = 2 +0= The circuit, when XL = Xc and hence = , is said to be in resonance. In a series circuit current I remains the same throughout we can write,  𝐿 = i.e. 𝐿 So, at resonance = 𝐶 will cancel out each the supply voltage other. 2 = 𝐿 + ( − )2 2 = 𝑅 =.. The entire supply voltage will drop across the resistor R  Resonant frequency At resonance X L =X 1 2πc = 2π ( is the resonant frequency) 𝑟 1 2= (2π)2L 1 = 2π L Where L is the inductance in henry, C is the capacitance in farad and the resonant frequency in Hz Under resonance condition the net reactance is zero. Hence the impedance of the circuit. X  0o rX  X  0 Z R  X  R 2 2 L C This is the minimum possible value of impedance. Hence, circuit current is maximum for the given value of R and its value is given by Im  V  Z  R The circuit behaves like a pure resistive circuit because net reactance is zero. So, the current is in phase with applied voltage.obviously, the power  factor of the circuit is unity under resonance condition. as current is maximum it produces large voltage drop across L and C. V Z Voltage across the inductance at resonance is given by VL  Im X L  Imr L  I m 2fr L  I m 2 1 LI L I L2 m m 2 LC LC LC L I m C At resonance, the current the current flowing in the circuit is equal to V V LR L V  V L R CR2 C Similarly voltage across capacitance at resonance is given by VC  I m X C 1 1 I  I m Cr m 2f Cr 1 I m 1  Im LC  I m 2 C LC C C2 2 LC I m L V L Thus voltage drop across L and C are equal and many times the C R Hence voltage magnification occurs at the applied voltage. C condition.so series resonance condition is often refers resonance to as voltage resonance. Q-FACTOR IN R-L-C SERIESCIRCUIT Q-FACTOR: In case of R-L-C series circuit Q-Factor is defined as the voltage magnification of the circuit at resonance. Current at resonance is given by I m  VR V  I m R And voltage across inductance or capacitor is given by = Im X C Im XL OR Voltage magnification = voltage across L or C /applied voltage  VL OR  VC  V L  VC  V V  Im X L Im X C OR Im R Im R  X L OR  X C R R Thus Q-factor  X LROR  X C R 1  r R L OR  rCR 1  2frRL OR  2fr CR  2L OR  1 2 LCR 1 2 CR 2 LC 1  1 L2 OR  C2 R LC R LC  1 L Q  factor  1 R L  2fr L  1 RC C R 2frCR Problem on Resonance Frequency A 10-Ω resistor, 10-mH inductor, and 10-µF capacitor are connected in series with a 10-kHz voltage source. The rms current through the circuit is 0.20 A. Find the rms voltage drop across each of the 3 elements What is not a frequency for ac current? a)50 Hz b)55 Hz c)0Hz d)60 Hz  What is a Single-Phase Power? Single-phase power simultaneously changes the supply voltage of an AC power by a system. More often, single-phase power is known as “residential voltage,” since it is that most homes use. In the distribution of power, a single-phase uses the phase and neutral wires. Phase wire carries the current load, while the neutral wire provides a path where the current returns. It creates a single sine wave (low voltage). The common voltage for a single- phase power starts at 230V. Also, its frequency approximates to 50Hz. Single-phase motors require extra circuits to work since a single-phase supply connecting to an AC motor doesn’t generate a rotating magnetic field. The power output of a single-phase supply is not constant, meaning its voltage supply rises and falls. What is a Three-Phase Power? Three-phase power provides three alternating currents, with three separate electric services. Each leg of alternating current reaches a maximum voltage, only separated by 1/3 of the time in a full cycle. In other words, the power output of a three-phase power remains to be constant, and it never drops into zero. In a three-phase power supply, it requires four wires, namely one neutral wire and three-conductor wires. These three conductor wires are 120-degree distant from each other. Also, each AC Power Signal is 120 degree out of phase with each other. Moreover, there are two types of circuit configurations in a three-phase power supply, such as the Delta and the Star. The Delta Configuration requires no neutral wire and only all high voltage systems use it; while, Star Configuration requires a neutral wire and a ground wire. Basic Three-Phase Circuit ECE 441 30 What is the voltage across the inductor when the source voltage is 200V and the Q factor is 10? a)100V b)20V c) 2000V d) 0V Three-Phase Power ECE 441 32  Definitio 4 wires ns – 3 “active” phases, A, B, C – 1 “ground”, or “neutral” Color Code – Phase A Red – Phase B Black – Phase C Blue – NeutralWhite or Gray ECE 441 33  THREE PHASE SYSTEM BASICS Line voltage VL= voltage between lines Phase voltage Vph= voltage between a line and neutral Star connection Derivation between line voltage and phase voltage Delta Connection Derivation of Deltaconnection Applying Kirchhoff’s Law at junction 1, The Incoming currents are equal to outgoing currents.  A complex number can be represented in one of three ways: Z=x+» Rectangular jy Form ∠Φ Z = A » Polar Form Z=Ae » Exponential jΦ Form Converting Polar Form into Rectangular Form, Converting Rectangular Form into Polar Form