Electrical

ELT-114 Electrical Essentials & Networks Page 198 It is possible to get smooth and variation in voltages. Disadvantages of Auto-Transformer There is possibility of high short circuit currents for short circuits on the secondary side. The full primary current will appear across the secondary causing higher voltage on secondary resulting danger of accidents. Risk factor appears as there is no electrical isolation between primary and secondary. It is economical only, if the voltage ratio is less than 2. Applications: (1) Is used where primary and secondary voltage have no large difference. (2) Is used to provide neutrals to three wire lighting systems. (3) These are used for light dimmers. (4) These are used to get two phase supply from three phase supply. (5) Used to control single and three phase locomotive devices. 6.1.9 STAR DELTA CONNECTIONS OF 3 PHASE TRANSFORMER: The primary and secondary windings of three phase transformers can be connected in several ways like star and delta and the voltage can be raised or lowered. The following are the most useful and commonly used three phase transformers. Star-star Delta-delta Star-delta Delta-star Delta Connection: The phase coils in delta connection are connected in a closed loop. In this connection instantaneous and phasor sums of the three phase voltages are zero. Therefore no current flows in the three phase coils unless a lead is connected from the source. The three phase coils must be properly phased in delta connection. The reference end of one coil should be connected to the non-reference end of another coil. The symbol for delta connection is given in figure below:

ELT-114 Electrical Essentials & Networks Page 199 Fig.6.11 delta connections Proper phasing of the source coils can be checked. As shown as figure 7.11, before the last connection is made to form delta connection ring. (b) Correct phasing 0 (a) Incorrect phasing 0 Fig.6.12 Phasing in delta connections If the coils are properly phased, the volt meter read the OV as shown in figure 6.12(b) when the phase coils are exactly equal. If the phase coils are not properly balanced, the volt meter may read some voltage as shown in figure 6.12(a). The line voltage is equal to the phase voltage in the delta system whereas the line current is greater than the phase current because one line carries current from two phases. Star Connection: This is also called Y connection. The Y connection of a three phase winding is shown in figure 6.13. In the Y connection the junction of the three phase coils is referred to as the neutral point because this point is connected to ground. Thus it is neutral with respect to the earth ground. It can also be referred to as the star point.

ELT-114 Electrical Essentials & Networks Page 200 The line current is equal to the phase current in the star system. However the line voltage of a Y connected system is larger than the phase voltages of the system. Fig.6.13 Star connections Star-Star Connection: Star- Star connection is shown in figure below. It is economical for small, high voltage transformers because the phase voltage is 1 √3 times the line voltage. (a) Y – Y Connection

ELT-114 Electrical Essentials & Networks Page 201 R Y B r y b N n (b) Star – Star Connection Fig.6.14 (a) & (b) star to star connections Delta-Delta Connection: – Connection (Phase Diagram) Fig.6.15 (a) & (b) Delta to Delta connections R Y B r y b (b) – Connection

ELT-114 Electrical Essentials & Networks Page 202 Star-Delta Connections: (a) Star – dalta phasor Diagram R Y B r y b (b) Star – delta Connection Fig.6.16 (a) & (b) Star to Delta connections

ELT-114 Electrical Essentials & Networks Page 203 Delta-Star Connection: (a) –Y phasor Diagram Fig.6.17 (a) & (b) Delta to Star connections 6.1.10 PHASE AND LINE VOLTAGE FOR STAR & DELTA CONNECTIONS: Delta Connection: In this connection, ends of each armature coil are connected to those of its two neighbored. These coils are so connected that the generated voltages are in the same direction around in the triangular form. Since these voltages are out phase with each other, they must be added vectorialy. The vector diagram is R Y B r y b delta – star Connection

ELT-114 Electrical Essentials & Networks Page 204 shown in figure (b). As the resultant is zero, thus there will be no current flowing around the triangle. EIII 1 2 3 EII EI (a) (b) Vectorial Diagram I3 –II I V31 EIII IIII I II –I1 V23 EII 30° I2 30° 30° –IIII V12 EI Fig.6.18 Delta Connections Voltages Connection of each junction between the coils are taken from three phase lined. The two voltage produced by an armature coil is called as the phase voltage (E). The voltage produced across the lines as a result of the phase voltage is called as the line voltage. In delta connections the phase voltage is same as the line voltage. Mathematically, VL = E Star Connections: In this connection, the end of one armature coil is connected with the end of other two armature coils. Three phase lines are connected to free ends of the armature coils. In star connection the line voltage is equal √3 or 1.73 times the phase voltage. Voltage between lines = Vector difference between two phase emfs. As E = EI = EII = EIII Therefore, Similarly, Hence V =E –E = 2 E Cos30° 12 I II 3 2 2E = 3EV =E –E = 3E 23 II III V =E –E = 3E 31 III I V = 3E L

ELT-114 Electrical Essentials & Networks Page 205 (a) Star Connection 1 E1 2 3 EIII E1 EII E2 E3 Neutral (b) Vector Diagram –EI I V1 2 I I I3 –EII V23 EII I V31 2 EIII 30° 30° 30° Fig.6.19Star Voltages & Currents 6.1.11 PHASE AND LINE CURRENT FOR STAR & DELTA CONNECTIONS: Delta Connections: In delta connections, the phase voltage is same as line voltage. However the line current turns out to be 1.73 times the phase current. We can generally say that in a delta connection phase current equals line current divided by 1.73. Therefore referring to figure 6.19 (b) Line current = Vector difference between the phase currents As phase currents: I = II = III = IIII I1 = II – IIII = 2I cos 300 Or Similarly, = 2 I √3 2 = √3 2 II I I =I –I = 3I 3 III II I =I –I = 3I

ELT-114 Electrical Essentials & Networks Page 206 Star connection: Referring to figure 6.20(b), suppose that the Ii, Iii and Iiii are the phase currents lagging behind their respective phase EMF by a constant angle Ø. Line current in the star connection is equal to the phase voltage. Therefore, IL = I 6.1.12 APPLICATION OF TRANSFORMER IN ELECTRONICS: i. Step down transformer: A step down transformer is one in which output voltage is less than the input voltage. The main uses of step down transformers have been given below: To step down the 14KV voltage to 440V for industry applications. To step down 440V for house hold and other uses. To isolate the high primary current from the secondary current. To isolate the primary and secondary windings in order to avoid shorts. ii. Step Up Transformer: A step up transformer is one in which output voltage is greater than the input voltage. The main uses of step up transformer are where the voltage needs to be stepped up from low value to high value. iii. Impedance Matching: For maximum transfer of power from one circuit to another, the both of two should have equal impedances. If they do not have equal impedances, a transformer with suitable turn ratio can be used to achieve this impedance match. A certain circuit working at a high voltage but low current (hence high impedance) has some times to be coupled to another circuit which requires low voltage but high current (hence low impedance). If two such circuits are coupled directly, energy transfers will not maximum. In such cases, a transformer is used as impedance matching devices because it can do the job of increasing or decreasing the voltages and currents very efficiently. Consider, S P P S V I T = = V I 2 S P P S V I T = × V I 2 S S P P P S I Z I T = × I Z I

ELT-114 Electrical Essentials & Networks Page 207 Where ZP=Impedance of primary Zs = Impedance of Secondary T = Turn ratio Suppose a circuit of output impedance 200Ω is to be coupled to a circuit of input impedance 2π. The turn ratio Ns/Np should be such that the impedances match to each other. From the formula: = 1 10 This means that the secondary turns should be one-tenth the primary turns. Often, auto transformer is also sued for impedance matching purpose. iv. Coupling: Two AC circuits are said to be coupled when they are linked in such a way that energy is transferred from one circuit to another. When there is an existence between the coils that are in separate circuits, then they are inductively coupled. Mutual inductance makes possible the transfer of energy from one circuit to the other by transformer action. It means that the alternating current established in the first or primary circuit produces magnetic flux which is linked with, and induces a voltage in the coupled or secondary circuit. This does not of course; apply to PC circuits since the flux must be changing for electromagnetic induction to occur. 6.1.13 TRANSFORMER LOSSES: The various losses associated with transformer are listed below: i. Copper Losses: Copper losses are due to the resistance of primary and secondary windings. ii. Core/Iron Losses Core/Iron Losses are of following types: 2 S P Z T = Z S P 2 Z Z = T 2 2 200 = T 2 2 T = 200 2 1 T = 100 1 T= 100

ELT-114 Electrical Essentials & Networks Page 208 (a). Magnetizing current Loss: In the case of ideal transformer, the primary inductance will offer infinite impedance and therefore no magnetizing current will flow. Practically, the magnetizing current does flow. (b). Eddy current Loss: Resistive losses are caused by eddy currents induced in the core of the transformer. (c ). Hystersis Loss: Resistive (heating) losses occure in taking the core through its magnetisation cycle. Fig.6.20 Hystersis Loss iii. Flux leakage losses: As all the flux of primary is not linked with the secondary, and vice versa. The leakage of flux has been shown in figure 6.20. The induced voltages are therefore smaller than those indicated by a coupling factor of unity. 6.1.14 HYSTERESIS LOSS & CORE LOSS: Hysteresis Loss: When iron core changes its polarity due to the passing of an A.C in every cycle, it consumes a little energy at each alternation. This loss of energy is known as the hysteresis loss. Due to the effect of hysteresis, the flux changes, loss behind the current changed producing them. Hence energy is lost due to hysteresis and appears as heat in the core. The higher the frequency of the alternating current and greater the flux density. The greater will be the hysteresis loss. As soft iron has smaller hysteresis loss than hard steel, hence the cores of the transformers are generally made of soft iron. The alloyed iron and other alloys often used for cores are stalloy, Permalloy and mumetal.

ELT-114 Electrical Essentials & Networks Page 209 Core Losses: There is always some loss of energy in the core material of a practical transformer. This loss is seen as a heating of ferrite and iron cores, but it does not occur in air cores. Part of this energy is consumed in the continuous reversal of the magnetic field due to the changing direction of the primary current, this energy loss is called hysteresis loss. The rest of the energy is caused by eddy currents induced in the core material by the changing magnetic flux. The eddy-current loss is greatly reduced by the use of laminated construction of iron cores. The thin layers of ferromagnetic material are insulated from each other to minimize the build-up of eddy currents by confining them to a small area and keep core losses to a minimum.

ELT-114 Electrical Essentials & Networks Page 210 Multiple Choice Questions Q.1 A transformer can operate from_______ d.c. (a)fixed (b)Changing (c) Positive (d) Negative Q.2 A _______ can operate from changing dc. (a) Transformer (b) opto- coupler (c)source (d) Battery Q.3 An autotransformer has only _______ winding. (a) one (b) two (c) three (d) any of above Q.4 An _______ transformer has only one winding. (a) step up (b) step down (c) tapped (d) auto Q.5 A transformer represents an example of __inductance. (a)Linear (b)Non-linear (c) Mutual (d)Self Q.6 A ____ represents an example of mutual inductance. (a)Transformer (b) Capacitor (c) Conductor (d) Insulator Q.7 Thin sheets of silicon steel used for making transformer core are called: (a) Windings (b) coils (c) laminations (d) Mutual Q.8 Unit of inductance is called _______. (a) Farad (b) Henry (c)Ampere (d)Ohm Q.9 Unit of _______ is called Henry. (a)Capacitance (b) Resistance (c) Conductance (d) Inductance Q.10 A transformer consists of _______ or more coils. (a) Infinite (b) stepped (c) two (d) Longitudinal Q.11 Transformer coils are _______ coupled. (a)Electrically (b) Magnetically(c) Horizontally (d)Vertically Q.12 A _______ transformer has more than 1 turn ratio. (a) Step-up (b) Step-down (c) Auto (d) Tapped Q.13 A _______ transformer has less than 1 turn ratio. (a)Step-up (b) Step-down (c) Auto (d) Tapped Q.14 A transformer cannot respond to _______ source. (a) Constant voltage (b) DC voltage (c) Professional (d) Resonance Q.15 _______ cannot be increased by transform. (a)Current (b) Power (c) Resistance (d) Inductance Q.16 Power cannot be increased by _______. (a) Transformer (b) Capacitor (c) Inductor (d) None of these Q.17 Transformer working depends on _______.

ELT-114 Electrical Essentials & Networks Page 211 (a) Self inductance (b) magnetic flux (c) Mutual inductance (d) Any of these Q.18 Working of _______ depends on mutual inductance. (a) Transformer (b) Capacitor (c) Resistor (d) All of these Q.19 Step-up transformer steps up _______. (a)Voltage (b) Current (c) Power (d)Capacitance Q.20 _______ transformer increases the level of voltage. (a)Step down (b)step up (c)Auto (d)power Q.21 A step up transformer always decreases the ______ (a)Voltage (b)Current (c)Power (d) Turn ratio Q.22 If primary voltage is 200v with a secondary turns 100 and primary turns 500 then the output voltage at secondary will be _______. (a)40 (b)60 (c) 80 (d)100 Q.23 The purpose of laminating is _______ (a) To decrease resistance (b) To decrease eddy current loss (c) To increase resistance (d) To increase power Q.24 A transformer can be used only for _______. (a)DC voltage (b)DC Current (c)DC power (d)AC voltage Q.25 The turn ratio to match a 50W source to 2000W load is _______. (a) 0.10 (b) 0.5 (c) 0.20 (d) 0.17 ANSWER KEY 1.(b) 2. (a) 3.(a) 4.(d) 5.(c) 6.(a) 7.(c) 8.(b) 9.(d) 10.(c) 11(b) 12.(a) 13.(b) 14.(b) 15.(b) 16.(a) 17.(c) 18.(a) 19.(a) 20.(b) 21.(b) 22.(a) 23.(b) 24.(d) 25.(b)

ELT-114 Electrical Essentials & Networks Page 212 Short Questions 1. Define transformer. 2. Define mutual induction. 3. Define self-inductance. 4. Define co-efficient of mutual induction. 5. Describe the turn ratio of transformer. 6. One coil produces a magnetic flux of 50mWb while other 20mWb. Determine K. 7. A transformer primary has 100 turns while secondary has 400 turns. Determine turn ratio. 8. Describe the construction of transformer. 9. Enlist the types of transformer. 10. Enlist core material of transformer. 11. Describe auto transformer. 12. Describe step down transformer. 13. Describe step up transformer. 14. List the transformer losses. 15. Two 250 mH inductor has mutual inductance of 250mH. Determine K. 16. The coefficient of coupling between a coil of 2H and a coil of 0.9His 0.7. Determine mutual inductance. 17. If Vp = 120V, f = 60Hz and turn ratio = 5 then find Vs. Long Questions 1. What is meant by transformer? Write down its working principle. 2. Explain the construction of transformer. 3. Write a detail note on Auto transformer. 4. Explain the losses occurs in transformers. 5. Write down the applications of transformers in electronics. 6. List the advantages and disadvantages of auto transformer. 7. Explain co-efficient of coupling and co-efficient of mutual inductance. 8. Write a detail note on three phase transformer.

ELT-114 Electrical Essentials & Networks Page 213 CHAPTER 7 Photo & Thermo-Electricity . OBJECTIVES After completion of this chapter students will be able to: Understand Photo and Thermo-electricity Describe photoelectric effects (photo emissive, photo conductive and photo voltaic effects) Name the applications of photoelectric effect Explain thermoelectric effect (Thompson & See-beck effects) Explain thermocouples Draw the diagram of incandescent lamp Explain the working of incandescent lamp Draw the diagram of fluorescent tube light Explain the working of fluorescent tube light 7.1 Understand Photo & Thermo-Electricity Definition of Photoelectricty: Electricity generated by light or affected by light. Electricity - a physical phenomenon associated with stationary or moving electrons and protons. Definition of Thermoelectricity: Electricity produced by the direct action of heat (as by the unequal heating of a circuit composed of two dissimilar metals) is called Thermoelectricity. 7.2 Photoelectric Effect: An English scientist Joseph J. Thomson discovered that when ultraviolet light falls on a metal surface, the electrons are emitted. These electrons are called as Photoelectrons and the phenomenon is known as Photo Electric Effect. This emission of electrons due to the effect of light energy is

ELT-114 Electrical Essentials & Networks Page 214 called as Photo Electric Emission. Photovoltaic effect is explained in the following diagram7.1. An evacuated glass take light is shown in figure 7.1 comprising of two electrode P & C.When light falls on the surface of metal plate P, electrons are emitted. These are photoelectrons and are attached with collector electrode C, being the +ve terminal of the battery. Due to this flow of electrons, galvanometer shows a current as a deflection in the meter. When light rays are stopped, the currents also stops. Since according to photon theory, light consist of packets of energy and each packet is termed as Photon and the energy of the photon is given by: E(photon) = hv. Where v is the frequency of the light and h is plank's constant. It should be noted that photo electrons are emitted from a metal surface if the light of certain minimum frequency falls upon it. This minimum frequency is called as Threshold Frequency. From the photovoltaic effect, following important results are found: i. Photo electrons are only emitted if the frequency of light is equal or greater than a certain frequency called as Threshold Frequency. ii. Photo electrons emission does not depend upon, intensity of light. However the no. of photo electrons depends upon the intensity of light. iii. The energy of the emitted photo electrons increase with the increase of frequency above the threshold frequency. iv. The threshold frequency depends upon the nature of the metal.

ELT-114 Electrical Essentials & Networks Page 215 Photo Conductive Effect: Some metals or substances have the property that their conductivity changes when light energy falls on the surface of these metals. This property is called as Photo Conductive Effect or photo conductivity. Such materials are called photoconductive. Examples of such materials is Cadmium Selenide, Cadmium Sulphides and Selenium etc. Such materials offer high resistance in the absence of the light so their conductivity is low and offer low resistance in the presence of light. A resistance showing photoconductive effect is called as Photo Resistor. Photo Voltaic Effect: When light energy falls on the surface of a PN junctions photons of energy are absorbed and a voltage difference is produced. This process of development of voltage difference due to the fall of light is called photo Voltaic Effect. When a photo sensitive material is placed in the light, it emits electrons and these electrons are absorbed by the other material. The material which emits electrons loses some negative charge and obtained positive charge. On the other hand, the material which gains electrons is negatively charged and obtained negative charge. Due to the development of positive and negative charges a cell is formed which is called as photovoltaic cell. Solar cell works on the same principal. The amount of produced voltage depends on the intensity of light, size of semiconductor material and nature of semiconductor material.

ELT-114 Electrical Essentials & Networks Page 216 7.3Applications of Photoelectric Effect: Some applications of photoelectric effect are as follows: i. Phototube ii. Automatic Door opening/ Closing system iii. Light Dependent Resistors iv. In Counting and Sorting System v. Fire alarm and control system vi. In photometry system vii. Solar cells (used in calculators, electronic instruments in satellites etc). 7.4 Thermoelectric Effect: Thermoelectric effect as a phenomenon by which heat energy is directly converted to electric energy. If two different metals say A & B are joined at one end and the junction is heated, some voltage is produced at the open ends or cool ends. The amount of the voltage produced depends on the difference of temperature between hot and cold ends. This action is called as Thermoelectric Effect. If a voltmeter is placed at the open ends, voltage will be measured the device formed by above arrangement is called as thermocouple. In industrial applications, the thermocouple is used to measure the temperature. When many thermocouples are joined in series, a Thermopile is formed. It is a very sensitive detector of heat rays.

ELT-114 Electrical Essentials & Networks Page 217 Thomson Effect: If a metal rod is heated unequally i.e., different parts of the rod are at different temperatures, then a potential difference is developed across the ends of the rod. As a result heat energy is absorbed or evolved when the current flows along a metal rod. EMF produced is directly proportional to the temperature difference. The phenomenon of absorption or evolution of heat energy if a current flows along a metal rod when different parts of the conductor are at different temperature is known as Thomson Effect. Materials like Copper, Silver, Zinc, Antimony heat energy is absorbed when current flows from a point at a lower temperature to a point at higher temperature. Heat energy is therefore evolved when current flows from a point at higher temperature to a point at lower temperature. Such materials/ metals have positive Thomson Effect. It is shown in figure (a) In substances like Iron, Bismuth, Cobalt, Platinum and nickel heat energy is evolved when current flows from a point at a lower temperature to a point at a higher temperature. Such metals have Negative Thomson Effect. it is shown in figure (b).

ELT-114 Electrical Essentials & Networks Page 218 Seeback Effect: When two dissimilar metals are placed in contact with each other, a potential difference appears between the two. This potential is called as contact Potential. It is found that contact potential depends upon the type of the metal and the temperature of the junction between them. If the two dissimilar are joined together so as to form a loop, the contact potentials at the junctions, being just equal and opposite, cancel each other and there is no net EMF around the loop, provided the two junctions are at the same temperature. If the junction are at different temperatures, the covalent potential will slightly different and net EMF exist which can flow a current around the loop. The EMF generated depends upon the pair of metals forming the thermocouple and the temperature between the junctions. The EMF exist in such case is called Thermal EMF or thermoelectric EMF. This effect was discovered by Seeback Effect.

ELT-114 Electrical Essentials & Networks Page 219 7.4 Thermocouples: A thermocouple is an electrical device consisting of two dissimilar electrical conductors forming an electrical junction. A thermocouple produces a temperature- dependent voltage as a result of Seeback effect, and this voltage can be interpreted to measure temperature. Thermocouples are widely used as temperature sensors. Thermocouples are widely used in science and industry. Applications include temperature measurement for kilns, gas turbine exhaust, diesel engines, and other industrial processes. Thermocouples are also used in homes, offices and businesses as the temperature sensors in thermostats, and also as flame sensors in safety devices for gas-powered appliances. 7.6 Incandescent Lamp: Incandescent lamp was invented by Thomas A. Edison in 1879. He used a wire made of carbon called filament. He heated the filament at temperature approximately 1350°C by passing a current through it until it become incandescent. The melting point of carbon is about 3500°C, so there is no danger to burn up.

ELT-114 Electrical Essentials & Networks Page 220 The disadvantage of carbon filament is that a layer of carbon vapor was deposited on the inner surface of the bulb. Moreover, the carbon filament was quite brittle and easily broken by vibration. An American scientist William Coolidge in 1910 discovered a process in which tungsten could be made extremely ductile and thus be drawn in to a fine thin filaments. Tungsten filament does not evaporate as readily as does carbon. Also it can be heated to a greater temperature than a carbon filament i.e. about 2100°C. Hence tungsten filaments are used in incandescent lamps today instead of carbon. In order to reduce the evaporation another American scientist Irving Langmuir introduced a inert gas, such as nitrogen or argon in to the bulb. By increasing the pressure, rate of evaporation can be reduced. A filament is constructed just like a helical tight coil of fine wire as shown in Figure below and thus can be concentrated in to a small space. This results in greater heat and more intense light produced by tungsten filament lamp. The coiled tungsten filament is supported by two wires that are sealed in the base of the glass bulb. These wires are used to flow the current to the filament. One of the wire is connect to the metal shell of the base and other attached to the bottom contact. When the lamp is screwed in to a socket, the shell and the bottom contact are connected to the source 7.7 Fluorescent Tube: Fluorescent tube light was introduced in 1938. The construction is shown in Figure below, in which, a tungsten filament is sealed in to each end of a long glass tube whose inner surface is coated with a phosphors. This glass tube

ELT-114 Electrical Essentials & Networks Page 221 may be obtained in sizes ranging from 61cm to 12cm. The air is pumped out inside the tube and in its place, argon gas and a few drops of mercury are filled. There are two tungsten filament at each side of the fluorescent tube called electrode. When current is passed through the filaments and the heat produced vaporizes the mercury, filling the tube with mercury vapors. The argon gas ionizes and the mercury vapors, produced ultraviolet rays. These rays collide with the phosphors coating, causing it to glow and produce visible light. As the fast moving electrons and ions collide, the force of impact may be greater enough to damage the coils. Therefore, guards are used over the heating coils. There are two pins attached to filament at each side of the tube. These pins are then connected to holder are called tube holder. The thermal switch and its capacitor are enclosed in a plastic or metal cylinder called starter. Most of fluorescent tubes used today are of the instant start type. They do not require starters. Fluorescent tubes are made in various form such as straight, circles and semicircle tubes. They operate at about 300°C and produce a yellowish white light. The average life of fluorescent tube is about five times longer than that of incandescent bulb.

ELT-114 Electrical Essentials & Networks Page 222 MCQs Q.1 When ultraviolet light falls on a _______surface, electrons are emitted. (a) metal (b) non-metal (c) wooden (d) ceramic Q.2 Emission of electrons due to effect of light is called as _______ emission. (a) thermionic (b) secondary (c) photoelectric (d) electric Q.3 Photoconductive materials offer _____ resistance in the absence of light. (a) low (b)high (c) very low (d) none of these Q.4 In photo-sensitive material the amount of produced voltage is depends on. (a) intensity of light (b) size of semiconductor (c) nature of semiconductor (d) All of these Q.5 Thermocouple is used for measuring. (a) resistance (b) temperature (c) velocity (d) flow Q.6 Thomas A. Edison invented incandescent lamp in. (a) 1879 (b) 1897 (c) 1987 (d)1978 Q.7 _______ filament used in incandescent lamp. (a) carbon (b) copper (c) silver (d) tungsten Q.8 Fluorescent tube operate on_______ working temperature. (a) 300℃ (b) 400℃ (c) 500℃ (d) 600℃ ANSWER KEY 1.(a) 2. (c) 3.(b) 4.(d) 5.(b) 6.(a) 7.(d) 8.(c)

ELT-114 Electrical Essentials & Networks Page 223 Short Questions 1. Define Photoelectricty. 2. Define Thermoelectricity. 3. State Photoelectric Effect. 4. State Photo Conductive Effect 5. State Photo Voltaic Effect. 6. Enlist Applications of Photoelectric Effect. 7. Define Thomson Effect. 8. What is Seeback Effect. 9. Describe working of thermocouple. 10. Enlist Applications of thermocouple. Long Questions 1. Explain photoelectric effects. 2. Explain thermoelectric effect. 3. Explain the working of incandescent lamp. 4. Explain the working of fluorescent tube light.

ELT-114 Electrical Essentials & Networks Page 224 CHAPTER 08: NETWORK THEOREM OBJECTIVES After completion of this chapter students will be able to: 1. Learn and solve the problems related to Superposition theorem for complex circuits. 2. Learn Thevenin's Theorem circuit’s Simplification. 3. Solve the problems base on the Thevenin's theorem 4. Learn Norton theorem and current source concept 5. Solve the problems based on the Norton's Theorem. 6. Understand Star and Delta transformation 7. Solve the problems based on Star and Delta transformation. 8.1.1 SUPERPOSITION THEOREM: Superposition Theorem Superposition theorem states that "The voltage or current present in a component comprising of more than one voltage or current sources is equal to the algebraic sum of the voltages or currents which exist independently as if a single source present in the circuit." Procedure of calculations based on the superposition theorem: 1. If the circuit comprise of a more than one voltage or current sources, consider a single voltage or current source at a time and replace other voltage sources as short circuited and current source as open circuited. 2. Then find the current or voltage at the particular branch. 3. Then consider the next source in the circuit and repeat the second step. 4. The total voltage or current is found by adding the voltages or currents found in step 2 & 3. 8.1.2 SOLVED PROBLEMS ON SUPERPOSITION THEOREM: Problem.1

ELT-114 Electrical Essentials & Networks Page 225 Find the voltage drop and the current across the 6Ω resistor shown in the circuit below using superposition theorem. + – + – 2 12V 24V 4 6 Fig.3.1circuit of problem#1 Solution We have to find Suppose first V2 is short circuited so + – 2 12V SC 4 6 Fig.3.1 (B) circuit of problem#1 Find V6Ω using series parallel simplification + – 2 12V (6||4) Fig.3.1(C) circuit of problem#1 + – 2 12V Fig.3.1 (D) circuit of problem#1 Since 6Ω and 4Ω are in parallel so the voltage will be same across them and will be same as across their resultant So, Now short circuit V1 and keep V2 present in the circuit. Let R1= 2 ,R2 6 , R3 4 , V1 12volts, V2 24volts V ?, I ? 6 6 6 4 24 2.4 6 4 10 2.4 2.4 V 12 6.54volts 2.4 2 V V V 6.54v 6 4 2.4

ELT-114 Electrical Essentials & Networks Page 226 + – 2 24V 4 SC 6 Fig.3.1 (E) circuit of problem#1 Now find V6Ω: 2Ω and 6Ω are now in parallel and its resultant is Rt=6x2/(6+2)=12/8=1.5Ω + – 24V 1.5 4 Fig.3.1 (F) circuit of problem#1 Using voltage division rule V1.5 = 1.5 1.5+4 ×24=6.55 volts V1.5Ω = V2Ω = V6Ω = 6.55 volts Apply Superposition Theorem: V6Ω = V6 (V2 short circuited) + (V1 short circuited) = 6.54 + 6.55 = 13.09 Volts Problem.2 (Self Test Problem) Find the voltage and current across 10Ω resistor using the superposition theorem for the circuit shown below. + – R1 15V 5 15 10 + – R3 R2 30 Fig.3.2 circuit of problem#2

ELT-114 Electrical Essentials & Networks Page 227 Problem.3 In the circuit shown below, find VP using superposition theorem. – + 600 400 VP 15V 3V – + Fig.3.3 circuit of problem#3 Solution: Now find VP if V2 is short circuited. – + 600 400 VP 15V Fig.3.3 (B) circuit of problem#3 = 4 10 x 15 = 6volts So VP = VP (V1 is short circuited) +VP (V2 is short circuited) = 1.8 V+ 6 V=7.8 Volts Problem.4 (Self Test) Find the voltage at point P shown in circuit below using superposition theorem. – + 30K 60K P 24V 9V + – Fig.3.4 circuit of problem#4 Let V 15V, V 3volts 1 2 R 600 , R 400 1 2 P 400 V 15 400 600 400 15 1000

ELT-114 Electrical Essentials & Networks Page 228 8.1.3 THEVENIN’S THEOREM: Thevenin's theorem states that "Any two terminal linear complex networks containing a number of resistances and sources (voltage or current) can be replaced by a single voltage source called Thevenin’s voltage source Vth and in series with a resistance which is called as Thevenin’s resistance Rth." Complex Linear Network A B Vth + – Rt h A B Fig.3.5circuit for Thevenin’s Theorem Method of finding Thevenin's Equivalent: 1. Remove the load across which Thevenin equivalent is required. 2. Keeping load terminal open find Rth. In order to find Rth, open circuit voltage sources and use basic rules of series parallel simplifications. 3. Now keeping the load terminals open find the open circuit voltage which is called as Vth. 4. Replace the entire circuit with Vth in series with Rth. 5. Connect the removed load resistance and find the current or voltage. 8.1.4 SOLVED PROBLEMS ON THEVENIN’S THEOREM: Problem.1 Solve the circuit shown below using Thevenin’s theorem to find the current across the 20Ω resistance. + – 5 24V 15 R2 R1 R3 10 R4 20 Fig.3.6 circuit for problem#1 Solution: Current is to be found across 20Ω so 20 is load and remove the load from the circuit.

ELT-114 Electrical Essentials & Networks Page 229 + – 5 24V 15 A 10 B Fig.3.6 (B) circuit of problem#1 Now find Rth so find Rth short circuit 24V supply and find the resistance across the points A&B. 5 15 A 10 B Rth Fig.3.6(C) circuit for problem#1 Now find VAB=Vth=? Keep load removed from the circuit and find the voltage VAB keeping voltage source present in the circuit. + – 5 24V 15 A 10 B Fig.3.6 (D) circuit for problem#1 Since points A B are open so the VAB will be the same as voltage across 10Ω resistor. So th 5 10 50 R 5 10 15 15 15 5 10 15 R 3.33 15 18.33 th 10 10 V 24 10 5 10 24 16volts 15 20 16 16 I 0.4115 Amp. 18.33 20 38.33

ELT-114 Electrical Essentials & Networks Page 230 Problem.2 (Self Test Problem) Determine the Thevenin’s equivalent circuit for the circuit shown below. + – 1K 15V 3K A 2K B Fig.3.7 circuit forproblem#2 Problem.3 (Self Test) Using Thevenin's Theorem to find the current flowing in 6Ω in the circuit shown below. + – 12 6 3 84V 21 + – Fig.3.8 circuit forproblem#3 8.1.5 NORTON’S THEOREM: Norton's theorem states that two terminal linear complex network can be replaced with current source called Norton current (IN) and with a parallel resistance which is called as Norton's resistance RN. It is shown in figure below. Complex Linear Network A B Vth IN RN A B Fig.3.9 circuit of Norton’s theorem Method of finding Norton's Equivalent: 1. First of all locate the load across which current or voltage is to be found and remove the load from the circuit. 2. Find Norton's resistance RN. To find RN, short circuit voltage source and open circuit any current source present in the circuit. Find resistance as seen from the load terminals.

ELT-114 Electrical Essentials & Networks Page 231 3. Now find the Norton's current. To find Norton’s current keep the voltage source or current source in the circuit and short circuit the load terminals. Find the current passing from the terminals A & B. 4. Draw the Norton's equivalent circuit by connecting RN and IN in parallel and connect the load across the points AB. And proceed for unknown current using simple rules for series parallel simplification. 8.1.6 SOLVE PROBLEMS OF NORTON’S THEOREM: Problem.1 Calculate the current in the 2Ω resistor using Norton's theorem in the circuit shown below. + – 3 36V 15 6 2 Fig.3.10 circuit for problem#1 Solution: Since current is to be found in 2Ω resistor so our load is 2 Ω and say RL=2 Ω. Find the RN (Norton resistance). Remove the load from the circuit and short circuit voltage source. 3 6 A B Fig.3.10 (B) circuit for problem#1 Now find the resistance at terminal A & B so = 6 ⫽ 3 = 6 × 3 6 + 3 = 18 9 = 2 Ω Now find IN (Norton's current) to find IN short circuit the A + B terminals and keep the voltage source present in the circuit. + – 3 36V 6 + – 36V 3 IN Fig.3.10 (C) circuit of problem#1

ELT-114 Electrical Essentials & Networks Page 232 Now the Norton's equivalent is I = 12Amp N RN 2 R = 2 L Fig.3.10 (D) circuit for problem#1 Connect the load resistance back into the circuit so I = 12Amp N RN 2 R = 2 L Fig.3.10 (E) circuit for problem#1 Problem.2 (Self-Test) Find the Norton's equivalent of the circuit shown in figure3.11. + – 4 48V 12 A 6 B Fig.3.11 circuit for problem#2 Problem.3 Using Norton's theorems find the current flowing through the 12Ω resistor shown in figure N V 36 I 12Amp R 3 2 2 I 12 6Amp 2 2

ELT-114 Electrical Essentials & Networks Page 233 Fig.3.12 circuit for problem#3 Solution: RL = 12 Ω First of all we will find RN. To find the value of RN, by short circuitingV1 and V2 4 2 A B S.C S.C Fig.3.12 (B) circuit for problem#3 So find the Norton's current IN, short at the RL, so 12V I 1 I 2 6V + – + – 4 V 2 Fig.3.12(C) circuit for problem#2 So, 8.1.7 TRANSFORMATION OF STAR TO DELTA AND DELTA TO STAR NETWORK: In some cases, it is very difficult to solve the complex network by using Kirchhoff's laws due to a large no. of the simultaneous equations that AB N 4 2 8 R R 4 2 1.33 4 2 6 12 1.33 I 6 12 1.33 I 0.6 Amps 12 + – 4 V = 12V 1 2 R2 R1 R3 12 6V = V2 + –

ELT-114 Electrical Essentials & Networks Page 234 need determinants to have their solutions. In such cases in the analysis of the networks, it is often helpful to convert a ∆ to Y or vice versa. In some cases even it may be impossible to solve the circuit without such conversion. Consider the following Y and ∆ networks. Fig.3.13 Star delta connections 8.1.8 PROBLEMS BASED ON STAR , DELTA TRANSFORMATION: Problem.1 For the delta circuit shown below find the Y equivalent circuit. RA =4 a b c RC=10 RB=6 Fig.3.14 circuit for problem#1 Solution: Suppose the equivalent Y circuit as shown below. R2 R3 R1 a b c Fig.3.14(B) circuit for problem#1

ELT-114 Electrical Essentials & Networks Page 235 R2=2 a b c R3=1.2 R1=3 Fig.3.14(c) circuit for problem#1 Problem.2 Solve the following bridge circuit using ∆ to Y or Y to ∆ conversions and find the current supplied by source. + – P3 P2 P4 P1 6 2 4 I T 30V 6 3 Fig.3.15 circuit for problem#2 Solution: Converting the lower delta network to star R2 R3 R1 RA RC RB 2 P2 P1 P4 6 4 Fig.3.15 (B) circuit for problem#2 B C 1 A B C R R 6 10 60 R 3 R R R 4 6 10 20 C A 2 A B C R R 10 4 40 R 2 R R R 4 6 10 20 A B 3 A B C R R 4 6 24 R 1.2 R R R 4 6 10 20 B C 1 A B C R R 6 4 24 R 2 R R R 2 4 6 12 C A 2 A B C R R 6 2 12 R 1 R R R 2 4 6 12 A B 3 A B C R R 2 4 8 2 R R R R 2 4 6 12 3

ELT-114 Electrical Essentials & Networks Page 236 P1 3 6 R = 6 C R = 4 B P2 P3 P4 R 2 A R2 R3 R1 Fig.3.15 (C) circuit for problem#2 P3 3 6 R = 1 2 R = 2/3 3 R = 2 1 P4 Simplifying the above circuit Fig.3.15 (D) circuit for problem#2 P3 2.5 2 P4 P3 4.5 P4 Fig.3.15 (E) circuit for problem#2 T T V 30 I 6.67 Amps R 4.5

ELT-114 Electrical Essentials & Networks Page 237 Multiple Choice Questions Q.1 An ideal constant voltage source has _______ resistance. (a) 5Ω (b) 10Ω (c) Infinite (d) zero Q.2 An ideal constant current source has _______ resistance. (a) Infinite (b) zero (c) Small (d) very large Q.3 In superposition theorem, all the voltage sources except ___ are removed. (a) One (b) Two (c) Three (d) Any of above Q.4 In superposition, all the voltage sources except one are removed leaving behind their _______ resistance, if any. (a) Extrinsic (b) Intrinsic (c) External (d) Internal Q.5 Rth is _______ (a) Thevenin resistance (b) Thevenin rating (c) Thevenin resistances (d) Thevenin resisting Q.6 While calculating Rth _______ voltage sources are removed. (a) All (b) None (c) One (d) Two Q.7 While calculating Rth, all voltage sources are removed but not their _______ resistance (a) Internal (b) External (c) Load (d) All of above Q.8 _______ equivalent of a circuit consists of a constant current source and a resistance in parallel. (a) Superposition’s (b) Thevenin’s (c) Norton's (d) Any of above Q.9 Norton’s equivalent of a circuit consists of a constant current source with a _______ resistance. (a) Series (b) Parallel (c) Infinite (d) External Q.10 Thevenin resistance Rth is found _______. (a) Between any two open terminals (b) By short circuiting the given two terminals (c) By removing voltage and internal resistance (d) Same as for Vth Q.11 While thevenizing a circuit Vth equals _______.

ELT-114 Electrical Essentials & Networks Page 238 (a)Short circuit terminal voltage (b) Open circuit terminal voltage (c)EMF of the battery (d) none of above Q.12 Rth is the same as _______ (a) RN (b) Vth Open Circuit Voltage (c) Infinite resistance (d) internal resistance Q.13 RL in a circuit is _______. (a) Linear resistor (b) Load resistor (c) Light resistor (d) Liquid resistance Q.14 There are _______ resistances in delta connections. (a)One (b) Two (c)Three (d) Four Q.15 If the resistances are same, both star and delta arrangements will be __. (a) Equal (b) Reciprocal (c) Un-equal (d) Opposite Q.16 The term Vth is called _______. (a) Thevenin's voltage (b) Thevenin’s velocity (c) Theoretical voltage (d) none of above Q.17 Network theorems are used to _______ the complex circuits. (a) Usage (b) construct (c) Analyze (d) derail Q.18 Constant voltage source has very _______ internal impedance. (a)High (b) Low (c) Random (d) fixed Q.19 Voltage source provides _______. (a)Current (b) Voltage (c) Both a and b (d) None of above Q.20 Current source provides _______ (a)Voltage (b) impedance (c) Current (d) a & c ANSWER KEY 1.(d) 2. (a) 3.(a) 4.(d) 5.(a) 6.(a) 7.(a) 8.(c) 9.(b) 10.(d) 11.(b) 12.(a) 13.(b) 14.(c) 15.(a) 16.(a) 17.(c) 18.(b) 19.(b) 20.(c)

ELT-114 Electrical Essentials & Networks Page 239 Short Questions 11. State the superposition theorem. 12. State Thevenin’s Theorem. 13. Define Vth. 14. Define Rth. 15. State the Norton's theorem. 16. Describe Norton's equivalent Current (IN). 17. Find RN for the circuit of figure 3.12 18. Describe conversion from ∆ to Y. 19. Describe the application of ∆ to Y conversion. 20. Draw the star and delta diagrams Long Questions 5. Explain Super position theorem. 6. Explain how you can find Thevenin equivalent circuit with examples. 7. Explain how you can find Norton’s equivalent circuit with examples 8. What do you mean star delta transformation. , explain. 9. Compare Thevenin Theorem with Norton’s Theorem.

ELT-114 Electrical Essentials & Networks Page 240 CHAPTER 09 RESONANCE . OBJECTIVES After completion of this chapter students will be able to: 1. Introduction to Resonance 2. Series Resonance & its Characteristics 3. Series RLC Impedance 4. Parallel Resonance & Characteristics 5. Comparison of Series & Parallel Resonance 6. Q of a Circuit, Selectivity 7. Application of a Resonant Circuit 9.1 RESONANCE: The resonance effect occurs when XL becomes equal to XC in RLC circuits. The main application of resonance is in RF circuits for tuning an AC signal to desired frequency. All examples of tuning in radio and television, receivers, transmitters and electronic equipment in general are application of resonance. At particular frequency, in a circuit comprising of XL and XC, the inductive reactance is equal to the capacitive reactance i.e XL = XC then this case of equal and opposite reactance is called resonance, and the circuit is called the Resonant Circuit. The frequency at which the XL = XC is called the resonant frequency (ƒr). Generally, we can say that large values of L and C provide relatively a low resonant frequency and small values of L and C provides large resonant frequency. The most common application of resonance in RF circuit is called Tuning. In this use, the LC circuit provides maximum voltage output at the resonant frequency compared with the amount of output at any other frequency either below or above resonance. This is illustrated in figure 8.1, where the LC circuit resonates at 1000KHZ. The result is maximum output at 1000 KHZ, compared with lower or higher frequencies. There are almost unlimited uses for resonance in AC circuits.

ELT-114 Electrical Essentials & Networks Page 241 500kHz 750kHz 1000kHz 1250kHz 1500kHz Max. output Resonant at 1000 kHz LC Circuits = 1000kHz fr =2 LC 1 Fig.9.1 Resonant circuit 9.2 RELATION BETWEEN RESONANT FREQUENCY , INDUCTANCE AND CAPACITANCE: We know that with the increase of frequency, the inductive reactance is increased. Therefore, the frequency and inductive reactance are directly proportional to each other. In equation form we can write this relation as: XL=2πfL ----------------- (i) Conversely, the capacitive reactance is decreased with the increase of frequency. Thus the frequency and the capacitive reactance are inversely proportional to each other. In equation form we can write this relation as: XC = 1 2 ----------------- (ii) We know that when the resonance of an ac circuit occurs then the inductive and capacitive reactance is equal, i.e. XL=XC ------------------------ (iii) 2 = 1 2 ()() = 1 (2) 2 () 2 = 1 (2)2 = 1 2√ Numerical Problems Example 8.1 What is the resonant frequency if 500mH inductance connected in series with a 2000 nF capacitor? Solution: As

ELT-114 Electrical Essentials & Networks Page 242 So 9.3 & 9.4 SERIES RESONANT CIRCUIT: A series resonant circuit is shown in figure below: Fig.9.2 Series Resonant Circuit In this resonant circuit, inductive and capacitive reactance are equal. The net reactance of such a circuit is zero and the impedance of the circuit is equal to the resistance(R) of the circuit. XL = XC So Z = R I = Current, is in phase with the applied voltage and power factor of such a circuit is unity. Maximum current flows through series resonant circuit. With the change of value in capacitance the resonance state is achieved. The large value of current in series resonant circuit is controlled by the resistor R. A very large voltage drop across the L and C is appeared which are equal and opposite of each other, cancel out the effect of each other. So it is also called Voltage Resonance. r 1 f = 2 LC r –3 –9 1 f = 6.28 500×10 ×2000×10 6 1 6.28 10 3 1 6.28 10 2 2 Z = R + X – X L C

ELT-114 Electrical Essentials & Networks Page 243 9.5 CHARACTERISTICS OF SERIES RESONANT CIRCUIT: Impedance at resonance is equal to resistance R when XL = XC. For a series RLC circuit. As we know that for lower frequencies the value of inductive reactance XL will be smaller than capacitive reactance XC. However at higher frequencies the value of capacitive reactance will be lower than the inductive reactance. If we draw a graph between the frequency and an impedance Z, then we obtain a curve as shown in figure 8.3. The figure shows that the impedance gradually increases as the frequency separation from resonance is decreased. The main characteristic of series resonance is that, the current is maximum at resonance because the impedance is minimum and decreases both sides if the frequency separation from resonance increases. Fig.9.3Impedance maximum on Resonance point The relation between current and frequency is shown in figure below: 2 2 Z = R + (X – X ) L C

ELT-114 Electrical Essentials & Networks Page 244 Fig 9.4The relation between current and frequency in series resonance The main characteristics of the series resonance circuit are listed as follows: (1) The inductive reactance is small below the resonant frequency while the capacitive reactance has high values that limit the amount of current. (2) Capacitive reactance is small above the resonant frequency. However the inductive reactance is having high values that limit the amount of current. (3) Inductive reactance is equal to the capacitive reactance at the resonant frequency and they cancel out to allow maximum current. (4) There is minimum impedance offered by the circuit i.e Zmin=R at resonance. (5) The voltage drops across inductor and capacitor are maximum and equal in magnitude but they cancel out each other because they are 180° out of phase with each other. (6) The resonant frequency is given by the following formula. 9.6 CURRENT , VOLTAGE & IMPEDANCE OF SERIES RESONANT CIRCUIT: Series RLC Impedance: As we know that below fr, XC>XL. Therefore the circuit acts as capacitive. However at resonance XC = XL, the circuit is purely resistive. At frequency above the resonant XL> XC, so the circuit is inductive. At Z=R the impedance magnitude is minimum and increases in value above and below the resonant point. The figure below shows the graph between the frequency and impedance. At zero frequency XL is zero while XC and Z are infinitely large, because the capacitor is behaving like an open and inductor is behaving like a short circuit. As the frequency increases XC decreases and XL increases. Since XC is larger than XL at frequencies below ƒr, Z decreases along with XC. At ƒr, XC=XL and Z=R because the impedance of simple RLC circuit is: But at resonance XL=XC 1 0.16 fr = = 2 LC LC 2 2 Z = R + (X – X ) L C

ELT-114 Electrical Essentials & Networks Page 245 Z = R So and voltage drop = VR = I × R. Current and voltage in a series RLC circuit: The current is maximum in the series resonant circuit at resonant frequency. Because the impedance increases, above and below resonant frequency, hence the current decreases. So I = V/R = Maximum current The voltage of resistor, follows the current and is maximum at resonance and 0 at f = 0. On the other hand, the voltage is maximum at resonant frequency but drops off above and below ƒr. The voltages across L and C at resonance are exactly equal in magnitude but 180° out of phase, so they cancel. Thus the total voltage across the L and C is zero. Fig.9.5 Current and voltage in a series RLC circuit 9.7 & 9.8 PARALLEL RESONANT CIRCUIT: A parallel resonant circuit is shown in figure below: Fig.9.6Parallel Resonant Circuit: 2 2 Z = R +(0) V I at resonance R

ELT-114 Electrical Essentials & Networks Page 246 It consists a capacitor in parallel with a coil of negligibly small resistance, connected across an ac voltage source V having frequency ƒ. In such a case the coil draw lagging current while the capacitor laws of leading current. When both the inductive reactance and capacitive reactance are equal at resonant frequency, they cancel out the affect of each other because though they are equal in magnitude but opposite in direction. Hence reactive branch currents are equal and opposite at resonance. They cancel each other to produce minimum current in the main lines. Since the line current is minimum, the impedance is maximum. 9.9 CHARACTERISTICS OF PARALLEL RESONANT CIRCUIT: The characteristics of parallel resonant circuit are exactly opposite to that of the series resonant circuit. If a graph is drawn between the changing line current with the changing frequency, then a curve as shown in figure 8.7 is obtained. As from the figure the line current is zero or minimum at resonance frequency and increases from both sides of resonance with the increase or decrease the frequency from resonant frequency. Therefore, in parallel LC circuit, the circuit impedance is infinite and line current is zero or at minimum level at resonance. Fig 9.7 Current at resonance in parallel resonant circuit If the graph is drawn between impedance and frequency then a curve as shown in figure below is obtained. Figure 8.6 Parallel resonant circuitR L V,f C IL IC I Figure 8.6 Parallel resonant circuitR L V,f C IL IC I

ELT-114 Electrical Essentials & Networks Page 247 Fig 9.8 Impedance at resonance in parallel resonance circuit. The figure shows that there is infinite impedance at resonance and reduced impedance after resonance frequency. The main characteristics of parallel resonance circuit have been listed below: (1) The line current is minimum at the resonant frequency. (2) It offers maximum impedance at the resonant frequency. (3) If the coil resistance is considered negligible then there is no DC voltage drop across it. (4) Circuit impendence is Zmax = (5) To find the current of this circuit (6) The resonant frequency can be found by ƒr = 1 2√ ƒr = 0.16 √ T max V = V VRC I = Z L L RC