Electric Circuits 2.pptx

HOUSE RULES 1. Attendance comprises 5% of each term grade. It is the duty of the student to come to class on the specified class schedule. 2. The name of the students are called at the end of each period. 3. If a student is absent during quiz or exam, a written explanation signed by guardian or parents noted by the Dean should be presented in order to have a special quiz or exam.

4. N0 special quiz will be given after the result of the quizzes has already been distributed to the students. This will prevent the students to know the solution of the original quiz. 5. Assignment should be submitted on time, late assignments will be strictly not accepted except if the student is absent from the time the assignment should be submitted. 6. Assignment comprises 5% of grade in each term.

7. There will be no surprise quiz so as to make the student well prepared during the date of the quiz. 8. Students who are absent during laboratory activity will have their own lab activity. Laboratory final report comprises 40% of the trimestral grade. 9. Permits should be presented during examination days. Failure to do so will automatically disqualify the student to have their exam. In this case, the student will pay to the cashier a certain amount for special examination.

10. Silence is a must during lecture time. There is allotted time for argument, questions and interactions. 11. Remove your hats, sunglasses, earphones and other paraphernalia that is not needed inside the classroom. If you do not obey the rules the door is always open for you. 12. Observe courtesy when asking questions. It might come back to you ten times fold. 13. Write and specify your answer clearly as possible to avoid questionable corrections.

14. Class monitor will be designated every meeting. Monitoring means erasing the board, picking up pieces of papers, turning on and off the lights and airconditioning unit, borrowing of overhead projector, etc. there will be two monitoring student per session. 15. Portfolio should be submitted at the end of each trimester. Therefore do not throw away your test papers and assignments. They are legal documents in case there is a question in your grades.

TEXTBOOK AND REFERENCES Electrical Circuit Theory and Technology, Bird, John, 5th ed, Routledge, 2014 Alexander, Charles (2013), fundamentals of electric circuits, McGraw Hill Nahvi, mahmood (2011), Schaum’s outline of electric circuit 5th ed, McGraw Hill Boylestad, Introductory Circuit Analysis, 10th ed, 2003, Prentice Hall Hayt, Engineering Circuit Analysis, 6th ed, 2002, McGraw Hill

COURSE OUTLINE 1. 2. 3.

4. 5. 6.

7. 8.

REVIEW OF NETWORK THEOREMS DC AND AC MEASURING INSTRUMENTS AC ANALYSIS IMPEDANCE AND ADMITTANCE AC NETWORK ANALYSIS POWER AND POWER FACTOR CALCULATIONS RESONANT AND TUNED CIRCUITS TRANSIENT RESPONSE

LESSON NO. 1 REVIEW OF NETWORK THEOREMS At the end of the lesson, the student will be able to 1. Recall the types of electric circuit network theorems 2. Apply network theorems in a given electric circuit

Thevenin’s and Norton’s Thevenin’s theorem states that a linear network terminating on any two nodes and containing any number of sources can be replaced by a single ideal voltage source in series with an internal resistance.

The Thevenin equivalent circuit provides an equivalence at the terminals only-the internal construction and characteristics of the original network and thevenin equivalent circuit are usually quite different. It was formulated by Leon Charles Thevenin, French telegraph engineer. it is similar to the study of Hermann von Helmholtz(eye theory of vision) which was applied to animal physiology and not to communication or generator systems.

EXAMPLE NO. 1 1. Find the equivalent Thevenin of the circuit to the left of a-b and find the current thru RL = 6Ω, 16Ω and 36Ω

2. For the given sets of circuits, if 5Ω is connected across point a-b, find the current flowing using Thevenin equivalent circuit.

3. Find the current and voltage across the 3Ω resistor.

Norton Theorem states that any two terminal linear bilateral DC network can be replaced by an equivalent circuit consisting of a current source and a parallel resistor as shown. It was formulated by Edward L. Norton of bell Lab.

EXAMPLE NO. 2 1. Find the Norton equivalent circuit fo the given circuit then find the current and voltage across 5Ω at point ab.

2. Find the current and voltage across 3Ω resistor.

MAXIMUM POWER TRANSFER In many practical situations, a circuit is designed to provide power to a load. There are applications in areas such as communications where it is desirable to maximize the power delivered to a load. We now address the problem of delivering the maximum power to a load when given a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (RL = RTH ). To prove the maximum power transfer theorem, we differentiate p with respect to RL and set the result equal to zero.

EXAMPLE NO. 3 Find the value of RL for maximum power transfer

SOURCE TRANSFORMATION Source transformation is another tool for simplifying circuits. Basic to these tools is the concept of equivalence. We recall that an equivalent circuit is one whose v-i characteristics are identical with the original circuit. A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.

EXAMPLE NO. 4 Using source transformation, find the current and voltage across 8Ω resistor.

2. Find the voltage Vs if Is = 0.25A

ASSIGNMENT NO. 1 1. Using Thevenin Equivalent circuit, find the voltage and current in the 3Ω resistor.

2. If an 8Ω resistor is connected across point a-b, find its current and voltage using Norton’s equivalent circuit.

3. Use source transformation to find the Vo

4. Using maximum power transfer theorem, find the power being delivered to R.

LESSON NO. 2 MEASURING INSTRUMENTS At the end of the lesson, the student would be able to: 1. State the operations of a DC and AC instruments 2. Know how to properly measure voltage, current and resistive values of a circuit.

What is measurement? -essentially an act or result of comparison between the quantity (whose magnitude is unknown) and predetermined or predefined standard. -measurand is the physical quantity to be measured. -comparison of two quantities expressed in numerical values.

Basic requirements for a meaningful measurements: 1. The standard used for comparison purposes must be accurately defined and should be commonly accepted. 2. The apparatus used and the method adopted must be probable (verifiable)

Functions and characteristics of instruments: 1. Indicate 2. Record 3. Control A particular instrument may serve any one or all three of these functions simultaneously. General purpose electrical and electronic test instruments primarily provides indicating and recording functions. The instrument used in industrial process situation frequently provides control functions. The entire system may be called a control or automated system.

Evolution of instruments: 1. Mechanical 2. Electrical 3. Electronics Mechanical instruments are very reliable for static and stable conditions, however, they are unable to respond rapidly to measurements of transient and dynamic condition.

Electrical instruments are faster than mechanical system indicating the output rapidly but uses the mechanical movement of the meters. The average response is from 0.5 to 26 secs. Electronics instruments is much faster because it uses semiconductor devices and weak signals can be detected.

Classification of Instrument 1. Absolute instrument – this instrument gives the original magnitude of the quantity. 2. Secondary instrument – this instrument is calibrated with comparison with absolute instrument which are already been calibrated.

3. Deflection type instrument 4. Null type instrument

standard is defined as International standard – they are defined by international agreement. They are maintained at the international bureau of weight and measure in Paris and are periodically evaluated and checked by absolute measurements in terms of fundamental units of Physics. Primary standard – are maintained at the national standards laboratories in different countries. The NBS in Washington is responsible for maintaining the primary standard in North America. Their primary function is to calibrate secondary standard of the area.

Secondary standard – are the basic reference standard used by measurement and calibration laboratories in the industry to which they belong. Each industrial laboratory is completely responsible for its own secondary standards. Each laboratory periodically sends its secondary standards to the national standards lab for calibration. Working standard – are the principal tools of a measurements laboratory, they are used to check and calibrate the instrument used in the lab or to make comparison measurements in industrial application.

Error in measurement: - Limiting errors: they are guarantor errors - Instrumental errors: these arises from inherent short comings of the instrument, misuse of the instrument and loading of the instrument. - Environmental error: due to conditions external to the instrument. - Observational error: due to parallax error - Residual error: due to the change or fluctuation in measurement.

Limiting errors: Most manufacturer of measuring instruments state that an instrument is accurate with a certain percentage of full scale reading. For example, the manufacturer of a certain voltmeter may specify the instrument to be accurate within +/-2% with full scale deflection. This specification is called the limiting error and means that a full scale reading is guaranteed to be within the limits of 2% of a perfectly accurate reading. However, with readings that are less than full scale, the limiting error will increase. Therefore, it is important to obtain measurements as close as possible to full scale.

Units of measurements Measurement units are the reference quantities that have been selected as a basis for comparison for most measurable quantities. Quantity Base Unit Abbreviation Length meter m Mass kilogram Kg Time second s Electric current ampere A Temperature kelvin K Luminous intensity candela cd Amount of substance mole mol

DC Meters The history of the basic meter movement used in DC measurements can be traced to Hans Oersted’s discovery of the relationship between current and magnetism. Various types of devices that made use of Oersted’s discovery were used. In 1881, Jacques d’Arsonval developed and patented the moving coil galvanometer. The same basic construction is widely used in meter movements today.

This basic moving coil system, generally referred to as a d’Arsonval meter movement or a permanent magnet moving coil (PMMC) meter movement is shown in the next slide. The moving coil mechanism is generally set in a jewel and pivot suspension system to reduce friction.

The d’Arsonval operates on the basic principle of the DC motor. The figure shows a horseshoe shaped permanent magnet with soft iron pole pieces attached to it. Between the north-south pole pieces is a cylindrical shaped soft iron core about which a coil of fine wire is wound. This fine wire is wound on a very light metal frame and mounted in a jewel setting so that it can rotate freely. A pointer attached to the mounting coil deflects up scale as the moving coil rotates.

Current from a circuit in which measurements are being made with the meter passes through the windings of the moving coil. Current through the coil causes it to behave as an electromagnet with its own north and south poles. The poles of the electromagnet interact with the poles of the permanent magnet causing the coil to rotate. The pointer deflects up scale whenever current flows in the proper direction in the coil. For this reason, all DC meter movements show polarity markings.

It should be noted that the d’Arsonval meter movement is a current responding device. Regardless of the units for which the scale is calibrated, the moving coil responds to the amount of current through its windings

The DC Ammeter The basic DC Ammeter circuit is shown in the figure below.

Where Rsh = resistance of the shunt Rm = internal resistance of the meter movement Ish = current through the shunt Im = full scale deflection current of the meter movement I = full scale deflection current for the ammeter

the voltage across the meter movement is Vm = ImRm Since the shunt resistor is in parallel with the meter movement, the voltage drop across the shunt is equal to the voltage drop across the meter movement. That is, Vsh = Vm

The current through the shunt is equal to the total current minus the current through the meter movement; Ish = I - Im

Therefore the shunt allows us to determine the shunt resistance as

Rsh

Vsh = -----Ish

ImRm = -----------I - Im

The purpose of designing the shunt circuit is to allow us to measure a current I that is some number n times larger that Im. The number n is called a multiplying factor and relates total current and meter current as I = nIm Substituting this in place of I in the previous equation yields

Rsh

Rm = ----------n - 1

EXAMPLE NO. 5 Calculate the value of the shunt resistance required to connect a 1 mA meter movement with a 90Ω internal resistance into a 0 to 5mA ammeter 2. A 100μA meter movement with an internal resistance of 100Ω is used in a 0 to 100mA ammeter. Find the value of the required shunt resistance. 1.

The Ayrton Shunt It is also known as the universal shunt. One advantage of the Ayrton shunt is that it eliminates the possibility of the meter movement being in the circuit without any shunt resistance. Another advantage is that it may be used with a wide range of meter movements. Its circuit is shown in the next slide.

EXAMPLE NO. 6 Design an ayrton shunt to provide an ammeter with the current ranges. A basic meter resistance of 1000Ω and the full scale deflection current of 1mA. Rm=1000

Rsh Rc +

Ra

Rb 1A

100mA

10mA

The DC Voltmeter The basic d’Arsonval meter movement can be converted to a DC voltmeter by connecting a multiplier Rs in series with the meter movement. +

Rs

Im

Rm

The purpose of the multiplier is to extend the voltage range of the meter and to limit current through the d’Arsonval meter movement to a maximum full scale deflection current. To find the value of the multiplier resistor, we first determine the sensitivity S of the meter movement. It is found by taking the reciprocal of the full scale deflection current S = 1/Ifs ohms per volt

The units of sensitivity express the value of the multiplier resistance for the 1V range. To calculate the value of the multiplier for voltage ranges greater than 1V, Rs = S x range – internal resistance

EXAMPLE NO. 7 Calculate the sensitivity of 100uA meter movement which is to be used as a DC voltmeter. 2. Calculate the value of the multiplier resistance on the 50V range of a DC voltmeter that used 500uA meter movement with an internal resistance of 1KΩ 1.

3. Calculate the value of the multiplier resistances for the multiple range DC voltmeter shown.

4. Convert a basic D’Arsonval meter movement with an internal resistance of 100Ω and a full scale deflection current of 1 mA into a multirange DC voltmeter with ranges from 0-15v and 0-50v.

Voltmeter Loading Effects When using a voltmeter to measure the voltage across a circuit component, the voltmeter circuit itself is in parallel with the circuit component. Since the parallel combination of the two resistors is less than either resistor alone, the resistance seen by the source is less with the voltmeter connected than without; therefore the voltage across the component is less whenever the voltmeter is connected.

The decrease in voltage may be negligible or it may be appreciable, depending on the sensitivity of the voltmeter being used. This effect is called voltmeter loading and the resulting result is loading error.

EXAMPLE NO. 8 Two different voltmeters are used to measure the voltage across RB in the circuit. The meter specifications are Meter A: S = 1KΩ/V, Rm = 0.2KΩ, range = 10V Meter B: S = 20KΩ/V, Rm = 1.5KΩ, range = 10V Calculate: 1. Voltage across RB without any meter connected across it. 2. Voltage across RB when each meter is connected 3. Error in voltmeter reading.

Ra = 25K E = 30V Rb = 5K

Ammeter Insertion Effect Another source of error is that inserting an ammeter in a circuit to obtain a current reading. All ammeters contain some internal resistance that may range from a low value of current meters capable of measuring in the ampere range to an appreciable value of 1KΩ or greater for microammeters. Inserting an ammeter always increases the resistance of the circuit and therefore always reduces the current in the circuit. The insertion error is 1 – Im/Ie

EXAMPLE NO. 9 A current meter has an internal resistance of 78Ω is used to measure the current through resistor Rc in the figure. Determine the percent of error of the reading due to ammeter loading. Ra=1K

E=3V

Rb=1K

Rc=2K

The Ohmmeter The basic d’Arsonval meter movement may also be used in conjunction with a battery and a resistor to construct a simple ohmmeter circuit such as shown in the figure.

The current through the meter movement can be determined by Ohm’s Law Ifs = E / (Rz + Rm) And the value of the current at the circuit is E I = -------------------Rz + Rm + Rx

The current I is less than the full scale current Ifs because of the additional resistance Rx. The ratio of the current I to the full scale is I Rz + Rm ----- = -----------------Ifs Rz + Rm + Rx And let p = I/Ifs which is very useful when marking off the scale on the meter face of the ohmmeter to indicate the value of the resistor being measured.

EXAMPLE NO. 10 1. An ohmmeter uses 1.5V battery and a basic 50uA movement in an internal resistance of 1KΩ Calculate the value of Rz required and the value of Rx that would cause half scale deflection in the circuit.

2. A 1mA full scale deflection current meter movement is to be used in an ohmmeter circuit. the meter movement has an internal resistance Rm of 100Ω and a 3V battery will be used in the circuit. Mark off the face of the meter for reading resistances.

3. Determine the current thru the meter Im when a 20Ω resistor between terminals X and Y is measured on the RX1 range. Show that this same current flows thru the meter movement when a 200Ω resistor is measured on the RX10 range Show that the same current flows when a 2KΩ resistor is measured on the RX100 range.

ASSIGNMENT NO. 2 Calculate the voltage drop developed across a D’Arsonval meter movement having an internal resistance of 850Ω and a full scale deflection current of 100μA. 2. Which meter has a greater sensitivity: meter A having of 0 to 10v and a multiplier resistor of 18KΩ or meter B with a range of 0 to 300V and a 298KΩ multiplier resistor? They have both an internal resistance of 2KΩ. 1.

3. Find the current through meter A and B in the circuit

4. Calculate the value of resistors R1 thru R5 in the circuit.

5. Calculate the value of the resistors R1 to R3 of the circuit

LESSON NO. 3 AC AND PHASOR CONCEPTS Lesson Objectives: at the end of the lesson, the student would be able to: 1. Write the general form of a sinusoidal waveform 2. Obtain the average and rms values of a waveform 3. Identify the phase relationship of multiple waveforms

SINUSOIDAL ALTERNATING WAVEFORM Definition of terms: Waveform – the path traced by a quantity, such as the voltage in the figure as a function of time, position, degrees, radians, temperature and so on. Instantaneous value – the magnitude of a waveform at any instant of time. Sinusoid – a signal that has the form of sine or cosine function

Peak amplitude – the maximum value of a waveform as measured from its average or mean value. Peak value – the maximum instantaneous value of a function as measured from the zero volt level Peak to peak value – the full voltage between positive and negative peaks of the waveform. It is the sum of the magnitude of the positive and negative peaks.

Periodic waveform – a waveform that continually repeats itself after the same time interval. Period – the time interval between successive repetitions of a periodic waveform. Cycle – the portion of a waveform contained in one period of time. Frequency – the number of cycle that occur in 1 sec.

The general format for sinusoidal waveform is i(t) = Im sin ωt or v(t) = Vm = sin ωt Where i(t) and v(t) = instantaneous current or voltage Im and Vm = maximum current or voltage(peak value of the waveform) ω = angular velocity = 2πf = 2π/T T = period f = frequency

The sinusoidal waveform is the only alternating waveform whose shape is unaffected by the response characteristics of R, l, and C elements. It means that if the voltage across or current thru a resistor, a coil or capacitor is sinusoidal in nature, the resulting voltage or current is also sinusoidal in characteristics.

The frequency is the number of cycle that occur in 1 sec. its unit is Hertz from German physicist Heinrich Rudolph Hertz.

CCIR FREQUENCY BANDS

Allocations: For television : 2 VHF and 1 UHF for low band and high band respectively. Channel 1 is used for military purposes (44MHz 50MHz)

Low band VHF

High band VHF

UHF channels: channel 14-83 470Mhz – 890 Mhz 3G SMART 850MHz – 2100MHz (LTE-long term evolution) 3G GLOBE 2100MHz 4G GLOBE 1800MHz 4G SMART 2100MHz

Phasor relations: Recall: cos ωt = sin (ωt + 90) sin ωt = cos (ωt – 90) cos (-ωt) = cos ωt sin(- ωt) = - sin ωt - sin ωt = sin (ωt +/- 180) - cos ωt = sin (ωt – 90) = sin (ωt + 270) etc Acos(ωt) + Bsin(ωt) = C cos(ωt - θ) where C = √A2 + B2 θ = tan-1 B/A

EXAMPLE NO.10 1. Determine the angle at which the magnitude of the sinusoidal function v(t) = 10 sin377t is 4V. What will be the time for which this will be attained?

2. Add 4cos(100t) – 3sin(100t) 3. Find the value of A, B, C and θ if 44 cos (120πt – 30) + 27 sin (120πt + 45) = A cos 120πt + B sin 120πt = C cos (120πt + θ)

What is the phase relationship between the sinusoidal waveform for each of the following sets? 1. v(t) = 10 sin (ωt + 30) i(t) = 5 sin (ωt + 70) 2. v1(t) = 3 sin (ωt + 60) v2(t) = 4 sin (ωt – 20) 3. i(t) = -2 cos (ωt – 60) v(t) = 3 sin (ωt – 150)

4. i(t) = 10 cos (30t + 10) v(t) = 3 sin (30t – 10) 5. i(t) = -sin (2t + 30) v(t) = -sin (2t – 45)

Average and effective values The average value of a sinusoidal waveform is, Vave = 1/T ∫ v(t) dt Effective value (RMS) – is the amount of steady (DC) current which when flowing through a given circuit for a given time produces the same heat as the alternating current when flowing in the same circuit for the same time. Vrms = √1/T ∫ v(t)2 dt Form factor FF = Vrms/Vave

EXAMPLE NO.11 Compute for the average and rms values of the given waveforms

ASSIGNMENT NO.3 1. If v(t) = 40V at α = 300 and t = 1ms, determine the mathematical expression for the sinusoidal voltage. 2. A voltage expression is given as v1 (t) = 4 sin (20t +50)V. Find its phase relationship with a. i(t) = 6 sin (20t + 40) b. v2(t) = 10 cos (20t - 40) c. i(t) = - 3 sin (20t + 30) d. i(t) = -3 cos (20t + 90)

3. Find the angle for which i1(t) = 20 sin(ωt + 30) lags i2(t). a. i2(t) = -10 sin (ωt – 30) b. i2(t) = 5 cos (ωt + 45) c. i2(t) = -4 cos (ωt + 60)

4. Find the average and rms value of the waveform

PHASOR A phasor is a complex number that represents the amplitude and phase angle of a sinusoid. Phasor is a radius vector that has a constant magnitude at a fixed angle from the positive real axis and that represents a sinusoidal voltage or current in the vector domain. Phasor diagram is a snapshot of the phasors that represent a number of sinusoidal waveform at t = 0.

In the next slide, we can see the sum of two waveforms v1 and v2. Note that in the figure v2 passes through the horizontal axis at t = 0, requiring that the radius vector be on the horizontal axis to ensure a vertical projection of zero volts at t = 0. its length in (a) is equal to the peak value of the sinusoid as required by the radius vector. The other sinusoid has passed thru 900 of its rotation by the time t = 0 is reached and therefore has its maximum vertical projection as shown in (a)

EXAMPLE NO. 12 Convert the following to phasor domain 1. √2(50) sinωt 2. 69.9 sin(ωt + 70) 3. 45 cos(ωt) 4. 100 cos(100t -60) Convert to sinusoidal expression at 60Hz 1. I = 20∟30 2. V = 115∟-60

Find the input voltage of the circuit with v1 = 10 sin(377t + 60) and v2 = 20sin (377t – 45) at 60Hz.

Determine the current i2(t) if the total current is 120sin(ωt + 60)mA and the current i1(t) = 80sin(ωt ).

Find the sinusoids represented by a. I = -3 + j4 b. V = j8ej-0.349

RESPONSE OF PURELY RESISTIVE ELEMENT For a purely resistive element, the voltage across and the current flowing through it is always in phase and their peak value is related by Ohm’s Law.

RESPONSE OF PURELY INDUCTIVE ELEMENT For an inductor, vL leads iL by 90 or iL lags vL by 90. inductive reactance is the opposition to the flow of current.

RESPONSE OF PURELY CAPACITIVE ELEMENT For a capacitor, ic leads vc by 90, or vc lags ic by 90. Capacitive reactance is the opposition to the flow of charge.

EXAMPLE NO. 13 The voltage across the resistor is indicated to be v(t) = 100 sin 377t V. find the sinusoidal expression for the current if the resistor is 10Ω. Sketch the curve for v(t) and i(t). 2. Repeat no. 1 if the voltage is v(t) = 25 sin (377t + 60) V. 1.

3. The current through a 10Ω resistor is i = 40sin(377t + 30). Find the sinusoidal expression or the voltage. 4. The current through a 0.1H coil is i = 10 sin (377t) A. find the sinusoidal expression for the voltage across the coil. 5. Repeat no. 4 if the current is i = 7sin(377t – 60)

6. The voltage across a 1uF capacitor is v = 20sin400t. Find the sinusoidal expression of the current. 7. Repeat no. 6 if v = 10sin(400t + 30)

8. For the given pairs of voltage and current, determine whether the element involved is a capacitor, inductor or a resistor and determine the value of the element. a. v(t) = 100 sin (2t +45) i(t) = 20 sin (2t + 45) b. v(t) = 1000 sin (377t + 10) i(t) = 25 sin (377t – 80) c. v(t) = 500 sin (100t + 30) i(t) = 100 sin (100t + 120)

DC high and low frequency effects on L and C For DC circuits, f = 0 therefore the reactance of a coil is XL = 0 (shorted), at high frequency XL = ∞(open) Same as true with the capacitor. For applied frequency, the resistance of the resistor remains constant as frequency increases, while for the inductive reactance increases linearly and the capacitive reactance decreases nonlinearly.

9. At what frequency will the reactance of a 200mH inductor match the resistance level of a 5kΩ resistor? 10. At what frequency will an inductor of 5mH have the same reactance as a capacitor of 0.2uF

ASSIGNMENT NO. 4 1. What is the inductive reactance of a 2H coil for DC and for 120Hz and 2000Hz. 2. Determine the frequency at which a 10H inductance and 0.33uF capacitance same with the reactance of 243Ω

3. Determine the frequency at which a 1uF and 10mH will have the same reactance. 4. For the network shown, find the current in each element and the total current of delivered by the source in both phasor and sinusoidal functions.

4mH

e(t) = 141.42sin(10000t + 60)V

2uF

LESSON NO. 4 POWER, IMPEDANCE AND ADMITTANCE At the end of the lesson the student would be able to 1. Calculate the power of an AC circuits 2. Differentiate impedance from admittance 3. Calculate the total admittance and impedance of a circuit 4. Apply ohm’s law in an Ac circuits

POWER AND POWER FACTOR For any load in a sinusoidal AC network, the voltage across the load and the current through the load will vary in a sinusoidal nature. Let v(t) = Vm sin (ωt + θv) and i(t) = Im sin (ωt + θi) Then the power is defined by p = v(t)i(t) RECALL: sinAsinB = [cos(A-B) – cos(A+B)]/2

The magnitude of average power delivered is independent of whether v leads I or I leads v. therefore V m Im P = ----------- cos θ 2 Where P = average power in watts θ = |θv - θi| The effective or rms power is P = Veff Ieff cos θ

EXAMPLE NO. 14 1. Find the average power dissipated and the power factor in a network whose inputs are stated below: a. i(t) = 5 sin (ωt + 40) v(t) = 10 sin (ωt + 40) b. i(t) = 20 sin (ωt + 40) v(t) = 20 sin (ωt + 70) c. i(t) = - 3 sin (ωt + 60) v(t) = 15 cos (ωt – 30)

In the equation P = Vm Im /2 cos θ, the factor that has significant control over the delivery of power level is the cos θ. No matter how large the voltage or current, if cos θ = 1, the power delivered is maximum. Since it has such control, the expression was given the name power factor and is defined by Pf = cos θ

For purely resistive load, the phase angle between v(t) and i(t) is 0o and the power factor is 1. the power delivered is therefore maximum. For purely inductor or capacitive load, the power delivered is the minimum value of zero watts, even though the current has the same peak value. For situations where the load is the combination of resistive and reactive load, the power factor will vary between 0 and 1.

The terms leading and lagging are often written in conjunction with the power factor. They are defined by the current through the load, if the current leads the voltage across the load, the load has a leading power factor and so on. In other words, capacitive networks have leading power factors and inductive networks have lagging power factor.

EXAMPLE NO. 15 Determine the power factor of the following and indicate whether lagging or leading or neither. 1. v(t) = 50 sin (377t – 20) i(t) = 2 sin (377t + 40) 2. v(t) = 120 sin (100t + 80) i(t) = 5 sin (100t + 30) 3. Veff = 20V Ieff = 5A , P = 100W

What is an impedance? it is the complex ratio of the voltage to the current in an alternating current (AC) circuit. Impedance extends the concept of resistance to AC circuits, and possesses both magnitude and phase.

The impedance of a resistive element is ZR = R∟0 The impedance of an inductive element is ZL = XL∟90 The impedance of a capacitive element is ZC = XC∟-90 Their impedance diagrams are:

EXAMPLE NO. 16 1. Draw the impedance diagram of the circuit given below and find the total impedance

2. Determine the input impedance to the series network in the figure. Draw the impedance diagram.

3. Determine the current flowing in the circuit for the network below. And find the voltage of each element. Draw the phasor diagram and find the total power and power factor.

4. Find the following: a. Total impedance and current b. The voltage in each element c. The total power, power in each element and total power factor d. Impedance and phasor diagram

5. Three elements are connected in series, 6Ω resistor, 10Ω inductor and 4Ω capacitor. If the current flowing in the circuit is 2.5sin(400t + 70)A, a. Find the voltage in each element b. Find the power in each element c. Draw the phasor and the impedance diagram d. The total voltage

ADMITTANCES In AC circuits, we define admittance (Y) as being equal to 1/Z. the unit of measure for admittance as defined by the SI system is siemens, which has a symbol of S. it is the measure of how well an AC circuit will admit or allow current to flow in the circuit. The larger its value, the heavier the amount of current flow for the same applied potential.

The reciprocal of reactance (1/X) is called susceptance and is measure of how susceptible an element is to the passage of current thru it. Susceptance is also measured in siemens and is represented by B = 1/X For inductive susceptance BL = 1/XL For capacitive susceptance BC = 1/XC

The admittance diagram

EXAMPLE NO. 17 1. Find the total impedance and admittance of the of the given circuits.

2. If a 100∟45 v is impressed at the input of the circuit in number 1. find the voltage and current in each branches.

3. Find the admittance of each parallel branch and determine the total admittance and impedance.

4. If a 100∟60v is impressed at the input of the circuit in problem 3. find the voltage, current, power in each branch.

5. An ac circuit has an impressed voltage of 80∟-60v to the input and produces a current of 20∟-45A. Find the series and parallel element of the circuit

ASSIGNMENT NO. 5 1. Find the voltage in phasor and sinusoidal form for each element below

2. Find the current in each element below. f = 60hz

3. Find the type of series and parallel element that must be closed in the container for the indicated voltage and current at the terminal.

4. For the circuit shown find a. Total impedance and current(phasor and sinusoidal) b. The voltage in each element(phasor and sinusoidal) c. The average power and the power in each element and total power factor.(leading or lagging?) d. The phasor and impedance diagram

5. For the given circuit find a. The total admittance b. The total voltage and current in each element(phasor and sinusoidal) c. The average power, the power in each element and the total power factor d. The admitttance and phasor diagram.

LESSON NO. 5 AC CIRCUIT ANALYSIS Lesson objectives: at the end of the lesson, the student would be able to 1. Apply network theorems to determine the current, voltage and power of a given AC circuits

EXAMPLE NO. 18 Using the circuit below find the mesh current and node voltages

BY Thevenin and Norton Theorem find the voltage across and current flowing thru the 2Ω capacitive reactance.

ASSIGNMENT NO. 6 Find the mesh current and node voltages

LESSON NO. 6 POWER FACTOR CORRECTION Lesson Objectives: at the end of the lesson, the student would be able to 1. Relate the DC power to AC power 2. Apply power factor correction in AC circuits

In the previous lesson, we had studied the average power of a given AC circuits. In this lesson we now examine the total power equation in a slightly different form and there will be additional power of two types: apparent and reactive power. For any system, the power delivered to a load at any instant is defined by P = VI

In this case, since V and I are sinusoidal quantities, V = Vmsin(ωt + θ) And I = Imsin(ωt) The chosen V and I include all possibilities because, if the load is purely resistive, θ = 0, purely inductive or capacitive θ = 90 or – 90, respectively.

For a network that is primarily inductive, θ is positive (V leads I) and for a network is that primarily capacitive θ is negative (I leads V). In a resistive circuit, the total power delivered will be dissipated in the form of heat. The average or real power is P = VI = VmIm/2 = I2R = V2/R in watts The energy dissipated by the resistor over a one full cycle of the applied voltage is W = Pt or VI/f1 in joules

Apparent power – it is the power delivered to a load without consideration of the effects of a power factor angle of the load. It is determined solely by the product of the terminal voltage and the current of the load. S = VI in volt-ampere or VA Since V = IZ and I = V/Z S = I2Z or V2/Z

The average power to the load is P = VI cosθ But S = VI Therefore P = Scosθ And the power factor pf = cosθ = P/S

In general, power equipment is rated in volt-ampere or in KVA not in watts. By knowing the volt-ampere rating and the rated voltage of a device, we can readily determine the maximum current rating. For example, a device rated at 10KVA at 200V has a maximum current rating of 50A when operated under the rated conditions. The volt-ampere rating of a piece of equipment is equal to the wattage rating only when the pf is 1.

Reactive power – it is the power associated with reactive elements that provides a measure of the energy associated with setting up the magnetic and electric field of inductive and capacitive elements respectively. Q = VI sinθ in volt-ampere reactive, VAR Where θ is the phase angle between V and I

For the inductor QL = VI in VAR = I2XL = V2/XL The power factor pf = 0 The energy stored by the inductor during the positive portion of the cycle is W = LI2 or VI/ω1 in joules Where I = rms value of current

Or the capacitor, QC = VI = I2XC = V2/XC The energy stored by the capacitor during the positive portion of the cycle W = CV2 or VI/ω1

The power triangle The three quantities average power, apparent power and reactive power can be related in the vector domain by S=P+Q With P = P∟0 QL = QL∟90 or QC = QC∟-90 Therefore S = P + j(QL – QC)

Since the reactive power and the average power are always angled 90 to each other, the three powers are related by Pythagorean Theorem. S2 = P2 + Q2 It is particularly interesting that the equation S = VI* Will provide the vector form of the apparent power of a system.

EXAMPLE NO. 19 1. For the R-L circuit given below, find the apparent power using two methods

2. For the R-C circuit find the apparent power in two Methods

3. Find the total number of watts, volt-ampere reactive, volt-ampere and power factor of the given network. Sketch the power triangle.

4. For the system shown, find a. The average power, apparent power and reactive power and power factor b. Find the total number of watts, volt-ampere reactive and volt-ampere and power factor of the system. c. Find the source current I.

5. A 5 hp motor with a 0.6 lagging power factor and an efficiency of 90% is connected to a 208V, 60hz supply. a. Establish the power triangle for the load b. Determine the power factor capacitor that must be placed in parallel with the load to raise the power factor to unity. c. Determine the change in supply current from the uncompensated to the compensated system

ASSIGNMENT NO. 7 1. For the system shown, find a. The total real power, apparent power and reactive power b. The power factor c. Construct the power triangle d. The total current

2. The loading of a factory on a 1000v, 60hz system includes: 20kw heating load (unity power factor) Induction motor with 10kw input power 0.7 lagging power factor 5kw lighting with 0.85 lagging power factor a. Establish the power triangle for the total loading on the supply b. Determine the power factor capacitor required to raise the power factor to 0.98 and unity. c. Determine the change in supply current from the uncompensated to the compensated system at unity power factor.

3. A small industrial plant has a 10kw heating load with unity power factor and a 20KVA inductive load due to a bank of induction motors with power factor 0.7 lagging. If the supply is 1200V at 60Hz, determine the capacitive element required to raise the power factor to 95%. And compare the levels of current drawn from the power supply.

LESSON NO. 7 RESONANCE At the end of the lesson, the student would be able to 1. Differentiate series resonance to parallel resonance 2. Calculate the maximum and minimum frequency for series and parallel circuits 3. Verify the importance of quality factor

The resonant circuit is a combination of R, L and C element having a frequency response characteristics similar to the curve below

Note that the response is a maximum frequency fr, decreasing to the right and left of this frequency, in other words, for a particular range of frequencies the response will be near or equal to the maximum. The frequencies to the far left or right have very low voltage or current levels and for all practical purposes, have little effect on the system’s response. The radio or TV receiver has a response curve for each broadcast station as in the figure. When the receiver is set or tuned to a particular station it is set on or near the frequency fr

Stations at frequencies to the far right or left of this resonant frequency are not carried through with significant power to affect the program of interest. The tuning process (setting the dial to fr) is describe as a result of the terminology tuned circuits. When the response is at or near the maximum, the circuit is said to be in state of resonance.

Series Resonance A resonant circuit whether series or parallel should have an inductive and a capacitive element. A resistive element will always be present due to the internal resistance of the source Rs and any added resistance to control the shape of the resonance curve.

ZT = R + jXL – jXc At resonance XL = Xc Therefore ZTs = R Therefore the frequency at resonance fs = 1/2π√LC The current thru the circuit at resonance is I = V∟0 / R∟0 and since XL = XC then VL = VC at resonance.

The quality factor (Q) of a series resonant circuit is the ratio of the reactive power of either the inductor or capacitor to the average power of the resistor at resonance. Qs = reactive power/average power = I2 XL/ I2 R = XL / R In terms of resonant frequency Qs = 1/R [√L/C]

The inductive reactance versus the frequency

Capacitive reactance versus the frequency

Series resonance versus frequency

Impedance versus frequency

Series circuit current at resonance

Phase angle of a series resonance

Bandwidth of a series resonance circuit

Bandwidth of a series RLC circuit

1. At resonant frequency XL= XC the ω of the circuit will be 2. The current at resonant frequency at ωr ZT is minimum I is maximum Imax = Vmax/ZT 3. At low cut-off frequency PHPF = 0.5Pmax I = 0.707Imax Z = √2 x R X = -R (capacitive) Therefore ωL = -R/2L + √(R/2L)2 + 1/LC

4. Upper cut-off frequency X = R (inductive) ωH = R/2L + √ (R/2L)2 + 1/LC 5. Bandwidth BW = fr/Q = R/2πL 6. Quality factor Q = ωrL/R = 1/ωrCR = 1/R x √L/C

EXAMPLE NO. 20 1. A series resonance network consisting of a resistor of 30Ω, a capacitor of 2uF and an inductor of 20mH is connected across a sinusoidal supply voltage which has a constant output of 9 volts at all frequencies. Calculate, the resonant frequency, the current at resonance, the voltage across the inductor and capacitor at resonance, the quality factor and the bandwidth of the circuit.

2. A series resonance circuit with R = 2Ω , XL = 10Ω and XC = 10Ω has an input voltage of 12sin(377t + 30). At resonant frequency of 5000hz, Find a. The total current b. The voltage in each elements c. The quality factor of the circuit d. the bandwidth e. The maximum and the half power frequency

3. The bandwidth of a series resonant circuit is 500hz. a. If the resonant frequency is 4500hz, what is the quality factor? b. If R = 18Ω, what is the value of the XL at resonant? c. Find the inductance and capacitance of the circuit.

 For resonance to occur in any circuit it must have at

least one inductor and one capacitor.  Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor.  Resonance occurs when XL = XC and the imaginary part of the transfer function is zero.  At resonance the impedance of the circuit is equal to the resistance value as Z = R.

 At low frequencies the series circuit is capacitive

as: XC > XL, this gives the circuit a leading power factor.  At high frequencies the series circuit is inductive as: XL > XC, this gives the circuit a lagging power factor.  The high value of current at resonance produces very high values of voltage across the inductor and capacitor.  Series resonance circuits are useful for constructing highly frequency selective filters. However, its high current and very high component voltage values can cause damage to the circuit.

 The most prominent feature of the frequency response

of a resonant circuit is a sharp resonant peak in its amplitude characteristics.  Because impedance is minimum and current is maximum, series resonance circuits are also called Acceptor Circuits.

Parallel Resonance

A parallel circuit containing a resistance, R, an inductance, L and a capacitance, C will produce a parallel resonance (also called anti-resonance) circuit when the resultant current through the parallel combination is in phase with the supply voltage. At resonance there will be a large circulating current between the inductor and the capacitor due to the energy of the oscillations, then parallel circuits produce current resonance.

A parallel resonant circuit stores the circuit energy in the magnetic field of the inductor and the electric field of the capacitor. This energy is constantly being transferred back and forth between the inductor and the capacitor which results in zero current and energy being drawn from the supply.

This is because the corresponding instantaneous values of IL and IC will always be equal and opposite and therefore the current drawn from the supply is the vector addition of these two currents and the current flowing in IR

In the solution of AC parallel resonance circuits we know that the supply voltage is common for all branches, so this can be taken as our reference vector. Each parallel branch must be treated separately as with series circuits so that the total supply current taken by the parallel circuit is the vector addition of the individual branch currents. Then there are two methods available to us in the analysis of parallel resonance circuits. We can calculate the current in each branch and then add together or calculate the admittance of each branch to find the total current.

At resonance the parallel circuit produces the same equation as for the series resonance circuit. Therefore, it makes no difference if the inductor or capacitor are connected in parallel or series. Also at resonance the parallel LC tank circuit acts like an open circuit with the circuit current being determined by the resistor, R only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and Z = R

At resonance, the impedance of the parallel circuit is at its maximum value and equal to the resistance of the circuit creating a circuit condition of high resistance and low current. Also at resonance, as the impedance of the circuit is now that of resistance only, the total circuit current, I will be “in-phase” with the supply voltage, VS

We can change the circuit’s frequency response by changing the value of this resistance. Changing the value of R affects the amount of current that flows through the circuit at resonance, if both L and C remain constant. Then the impedance of the circuit at resonance Z = RMAX is called the “dynamic impedance” of the circuit

Impedance in a parallel circuit

Susceptance at resonance

we can see that at the resonant frequency point were it crosses the horizontal axis the total circuit susceptance is zero. Below the resonant frequency point, the inductive susceptance dominates the circuit producing a “lagging” power factor, whereas above the resonant frequency point the capacitive susceptance dominates producing a “leading” power factor.

that the impedance of a parallel circuit at resonance is equivalent to the value of the resistance and this value must, therefore represent the maximum dynamic impedance (Zd) of the circuit as shown.

Current in Parallel Resonance

At resonance, currents IL and IC are equal and cancelling giving a net reactive current equal to zero. Then at resonance the above equation becomes

Parallel Current Circuit at Resonance

The quality factor for parallel resonance circuit is given as Q = R/2πfL = 2πfRC = R √C/L Note that the Q-factor of a parallel resonance circuit is the inverse of the expression for the Q-factor of the series circuit. Also in series resonance circuits the Qfactor gives the voltage magnification of the circuit, whereas in a parallel circuit it gives the current magnification.

EXAMPLE NO. 21 1. A parallel resonance network consisting of a resistor of 60Ω, a capacitor of 120uF and an inductor of 200mH is connected across a sinusoidal supply voltage which has a constant output of 100 volts at all frequencies. Calculate, the resonant frequency, the quality factor and the bandwidth of the circuit, the circuit current at resonance and current magnification.

Resonant Frequency using Impure Inductor fr = 1/2π √1/LC – [RS/L]2

Where: L is the inductance of the coil, C is the parallel capacitance and RS is the DC resistive value of the coil.

ASSIGNMENT NO. 8 1. A series circuit consisting of a 20Ω resistor, an 88.3uF capacitor and a variable inductor is connected to a 120V, 60hz source. a. Calculate the inductance, current and the voltage drop across the inductor. b. Calculate the value of the inductance, the current and the voltage drop across the inductor when the latter has its maximum value

2. A series circuit consisting of a 15Ω resistor, 0.06H inductor, and a variable capacitor connected to a 120v, 60hz source. a. Calculate the value of C, the total current and the voltage drop across the capacitor for the condition at resonance. b. Determine the value of C, the total current and the voltage drop across the capacitor when the latter has its maximum value.

3. Given the circuit below

For the condition of resonance calculate the two values of L and the two values of line current