Current and static electricity
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- 1. 5 Current and Static ElectricityUNIT 1 CURRENT AND STATIC ELECTRICITY Structure 1.1 Introduction Objectives 1.2 DC Circuit and Steady Current and Voltage 1.3 Basic Definitions Related to Circuit Analysis 1.4 Ohms Law and its Limitations 1.5 Resistance 1.5.1 Classification of Substances 1.5.2 Resistance Law 1.5.3 Effect of Temperature on Resistance 1.5.4 Temperature Coefficient of Resistance 1.5.5 Resistance in Series 1.5.6 Resistance in Parallel 1.5.7 Classification of Resistors 1.6 Kirchhoffs Laws 1.6.1 Kirchhoffs Voltage Law 1.6.2 Kirchhoffs Current Law 1.6.3 Application of Kirchhoffs Laws 1.7 Source Transformations 1.8 Heating Effect of Electric Current 1.9 Summary 1.10 Answers to SAQs 1.1 INTRODUCTION The development of electricity and its application has a great history. In 600 BC, Thales, a Greek Mathematician, philosopher could produce a feel of electricity. He found that when amber is rubbed against wool or fur it gets electrified and attracts small particles of straw and produces spark. Static positive and negative charges can be created by rubbing amber with a piece of wool or fur. Some of the valence electrons from the one material become free electrons and are transferred to another material. The charges on the amber and the wool tend to remain stationary, thus the name static electricity. The names like electricity, electrons and electronics originated from the word elektron which is the Greek word of amber. Above was the first experiment in the field of static electricity. After a long time in 1785, Coulomb, in France, first established the mathematical relation for such experiment. He defined the charge and force between two charges. In 1800, Galvani and Volta produced the electricity by chemical means. They made the first DC source known as battery. In 1825-26, Ohm established the relation between voltage and current which became the basis for Kirchhoffs Voltage and Current law in 1845. Most of the useful application of static electricity do not rely on static discharge which ionize the air and occurs when the electric force field between a positive charge and a negative charge becomes too strong. Rather they make use of the force of attraction between unlike charges or the force of repulsion between like charges. These forces are used to move charged particles to desired location.
2. 6 In this unit, we will learn basic definitions of the terms which are related to electricity, such as voltage, current, power and energy. Circuit element resistance will be discussed in detail. Electricity The relationship between current (I), voltage (V) and resistance (R) was discovered by a German scientist named Georg Ohm. In this honor this relationship is named Ohms law. Here, in this unit, we will discuss Ohms law with its limitations and two basic laws of the behaviour of current and voltage, which are known as Kirchhoffs Laws. These laws enable scientists to understand and, therefore, evaluate the behaviour of electrical networks. How to convert a voltage source into a current source and vice-versa will also be discussed under source transformation technique. Objectives After studying this unit, you should be able to explain the concept of DC circuit and steady voltage, current, power and energy, define Ohms law and its limitations, explain the resistance law and its dependency on temperature, find the equivalent resistance for parallel and series circuits, write the voltage and current equations using Kirchhoffs voltage and current laws, understand source transformation technique, and calculate the power loss and energy dissipated in electric circuit. 1.2 DC CIRCUIT AND STEADY CURRENT AND VOLTAGE Figure 1.1 shows a simple DC circuit which consists of DC source and passive element like resistance. Different types of DC sources are cell, battery, DC generator etc. In DC Analysis voltage and current remain constant with respect to time, while alternating current (or AC voltage) changes direction and magnitude with respect to time. Voltage and current waveforms for given simple DC circuit are shown in Figures 1.2(a) and (b). RV I Figure 1.1 : Simple DC Circuit V 0 t I 0 t V R (a) (b) Figure 1.2 : Waveforms (a) Direct Voltage; and (b) Direct Current 3. 7 Current and Static Electricity1.3 BASIC DEFINITIONS RELATED TO CIRCUIT ANALYSIS In general, we will use uppercase letters V, I, P to indicate constant voltage, current and power respectively and lower case letters v, i, p to indicate instantaneous values of time varying signals. Let us know about some basic definition such as voltage, current, power and energy which are commonly used in electrical. Voltage (or Potential) Voltage is the electric pressure that causes current to flow. Voltage is also known as electromotive force (e.m.f.), or potential difference. All these terms refer to the same thing, that is, the force that sets charges in motion. Potential difference is actually a potential energy difference that exists between two points. An electric charge possesses potential energy (energy at rest). Potential energy is capable of doing work when we provide the right conditions to convert it from its stored form into another form (potential to kinetic energy). Voltage is the work done for moving unit positive charge from one point to another in electric field. Energy ( ) Voltage ( ) Charge ( ) W V Q = Potential difference can also be defined as the work done for bringing a unit positive charge from infinity to any point in electric field. Instantaneous value of voltage can be given as joules/coulombor volts. dw v dq = Current Electric current is the moment of charged particles in a specified direction. The charged particle is often referred to as a current carrier. In a solid, such as copper wire the charged particle (current carrier) is the electron. However, in both liquids and gases, the ions are free to move about and become current carriers. In semiconductors, the charge is carried by electron and holes, the holes behaving like positive charges. In conducting materials, a large number of free electrons are available which move from one atom to other at random when a potential difference is applied between two points of the conducting material and the current starts flowing. In electricity the amount of current is specified in terms of the charge and time required to move the charge past a given point. The amount of electric current is, therefore, specified in Coulombs per second (ampere). When we think about current, keep two points in mind. First, the effect of current is almost instantaneous. Current in a wire travels at nearly the speed of light, i.e. 186,000 miles per second (3 108 meters per second). Second an individual electron moves much more slowly. It may take minutes for an electron to travel a few feet in the wire. The rate of flow of charges through any cross-section of conductor is called a current and is denoted as i. Amperes dq i dt = where i is the instantaneous value of the current (value at any particular instant). 4. 8 Hence one ampere is the current, which flows when a charge of one coulomb moves across the cross-section of a conductor in one second. Electricity The steady current I is given as Amperes Q I t = where Q is the uniform flow of the charges through the cross-section of the conductor in time t. Otherwise CoulombsQ i dt= The term static electricity comes into the picture when flow of electron is steady or the flow of electron is at constant drift velocity. Static electricity includes the study of current, potential and power related to DC circuits. Electrical Power Power is defined as the rate of doing work and it is expressed in Joules per second Joules/second W P t = When one coulomb of electrical charge moves through a potential difference of one volt in one second the work done is one Joule/sec and in electrical engineering it is expressed as Watt. So, power supplied WattsP V I= By Ohms law ( ) WattsP I R I= 2 WattsI R= or, 2 Watts V V P V R R = = For AC circuit or circuits with unidirectional source, electrical power is expressed as WaP v i tt= 2 i R= where, p, v, i all stands for instantaneous values. Energy Electrical energy is the total amount of work done and it is expressed in Joules or in Watt-Second in electrical engineering. In DC circuit, energy is dissipated in the form of heat in resistor for a time t seconds and is given by Watt-sec.W P t= V I t= 2 Watt-sec.W I Rt= or 2 Watt-sec. V W t R = For AC circuits, energy can be expressed as 5. 9 Current and Static Electricity c.Watt-seW pd t= Watt-sec.W v id t= where, p, v, i are instantaneous values. Energy expressed in terms of Kilo-Watt-hour (kWh) or units, sine Watt-sec. is a small unit, 1 unit of energy = 1 kWh Electrical energy supplied to the consumer is charged by Power Distribution Companies in terms of Standard Energy Units known as Kilo-Watt-Hours. 1.4 OHMS LAW AND ITS LIMITATIONS Ohms law is a central concept to most electrical engineering theories. In 1825-26, Ohm gave the relation between electric current and potential. This relation is known as Ohms law. According to Ohms law potential difference across a conductor is directly proportional to the current flowing through the conductor, the temperature of the conductor remains constant. i.e. V I V RI= where R is the constant of proportionality is known as resistance, V R I = (resistance in ohms) Electrical resistance is the hindrance to the flow of electrons in a given material. V and I represent constant value of current and voltage. Above relation is shown in the form of resistance curve in Figure 1.3, which is linear in nature. (a) Resistance Curve for Ohms Law (b) Circuit Represents Ohms Law Figure 1.3 Ohms law can also be applied to AC circuits volv Ri= ts where v and i are instantaneous values of voltage and current respectively. Limitations of Ohms Law (a) Ohms law is valid only if the physical conditions like temperature, pressure remain constant. 6. 10 (b) Ohms law is not applicable to non-metallic conductors. For example, for silicon carbide, the relationship is given by , where k and n are constants. This relation is not linear. n V k I= Electricity (c) Since the resistance also depends on the length and area of cross section of conductor, so for the application of Ohms law the dimensions of conductor should remain constant. Example 1.1 A lamp load of 1000 resistance is connected across the DC supply of 25 V. What is power absorbed in lamp and what amount of heat will be released in 10 sec. Solution The current taken by the lamp load 25 0.025 Amp 100 I = = 25 m Amp= Power loss 2 I R= 2 (0.025) 1000= 3 625 10 Watts = 625 mW= Heat released or energy consumed in 10 sec. 2 W I Rt= Pt= 3 625 10 10 = 2 625 10 = 6.25 joules= . SAQ 1 An electric iron operates from a 230 volt outlet and draws 8 amperes of current. At Rs. 4/ kWh, how much does it cost to operate the iron for 2 hours? 1.5 RESISTANCE It is the property of any substance due to which it opposes the flow of current through it. It has the same role in electric circuit as that of friction in mechanical system. This opposition is basically due to the molecular structure of the substance. When electrons flow through any substance then they collide with the other molecules or atoms of the substance. In each collision, some energy is dissipated in the form of heat. So we can say due to resistance some energy is wasted in the form of heat (which is given by I2 Rt). The resistance is defined as the ratio of voltage and current in any circuit and its unit is ohm (). ohm ( ) V R I = 7. 11 Current and Static Electricity Reciprocal of resistance is known as the conductance (G) and its unit is mho (). 1 mho I G R V = = . 1.5.1 Classification of Substances On the basis of their resistance, substances may be classified as good conductor, semiconductor and bad conductor. Good Conductor Materials with low resistance and high conductance are known as the good conductors of electricity. Example Metals (like copper, aluminum, silver etc.), acids and electrolytes. Semiconductor Materials which are bad conductors at low temperature and good conductors at high temperature are classified as the semiconductors. Such materials are partially conducting, but also has properties of an insulator. The amount of current conduction that can be supported can be varied by doping the material with appropriate materials, which results in the increased presence of free electrons for current flow. They have medium resistance (between good and bad conductors) at room temperature. Example Germanium, Silicon, GaAs. Bad Conductors Materials which offer very high resistance to flow of electricity are known as the bad conductors of electricity. They are normally used as the insulator in electrical machines. Example Mica, Glass, Paper, Rubber, Wood, Bakelite etc. 1.5.2 Resistance Law Resistance of any material is directly proportional to the length of material and inversely proportional to its area of cross-section. i.e. R l 1 R A l R A or l R A = where l = length in metre, A = area of cross section in metre2 , and resistivity of material (specific resistance).= It is defined as the resistance between the opposite faces of a metre cube of any material. A R l = 2 metre = ohm metre = Ohm-metre 8. 12 Reciprocal of resistivity is known as conductivity or the specific conductance. It is denoted by and its unit is /metre. Electricity 1 = . mho/metr l G A = e 1.5.3 Effect of Temperature on Resistance In ideal conditions, resistance is the constant element of the circuit. But as the current flows, heat is produced, and temperature is increased. With increase in temperature following effects are observed : (a) Resistance of the metal conductors increases with increase in temperature. The Resistance-temperature graph is a straight line and shown in Figure 1.4. Figure 1.4 : Resistance-Temperature Curve for Metals (b) Resistance of alloy also increases with increase in temperature. But this increment is relatively slow and irregular. (c) Resistance of semiconductors decreases with increase in temperature. At very low temperature, they acts like insulators but at high temperature they show the property of conductors. (d) Resistance of electrolytes and insulators (bad conductors) decreases with increase in temperature. 1.5.4 Temperature Coefficient of Resistance Let R0 is resistance at any initial temperature t0 and Rt is the resistance at higher temperature t. Then increment in resistance (Rt R0) is directly proportional to initial value of resistance R0 and increment in temperature (t t0). i.e. 0 0 0 0( )tR R R t t = where 0 is the temperature coefficient of resistance referred to temperature t0 0 0 0 0( ) tR R R t t = If initial temperature t0 = 0o C then 0 0 tR R R t = . Unit of temperature coefficient is per o C. Resistance at any temperature is defined as 0 (1 )tR R t= + (here t0 = 0o C) 9. 13 Current and Static Electricity The value of temperature coefficient of resistance () is not constant. Its value depends upon the initial temperature on which the increment of resistance is based. If the value of at t1 o C is 1 then its value at t2 o C will be 2 2 1 1 1 1 ( )t t = + The temperature coefficient of resistance is positive for metals and alloys. is negative for semiconductors, electrolytes and insulators. Example 1.2 A platinum coil has a resistance of 3.146 at 40o C and 3.767 at 100o C. Find the resistance at 0o C and the temperature-coefficient of resistance at 40o C. Solution 0 0 0 0( )tR R R t= + t 0 Take t0 = initial temperature = 0o . 0 0tR R t= + R 0 0(1 )R t= + 100 0 0(1 100 )R R= + at t = 100o C . . . (i) and 40 0 0(1 40 )R R= + at t = 40o C . . . (ii) From Eqs. (i) and (ii) o0 0 0 1 1003.767 0.00379 per C 3.146 1 40 + = = + From Eq. (i) 03.767 (1 100 0.00379)R= + 0 2.732,R = o 40 2 1 0 1 1 1 per C 1 1 304( ) (40 0) 0.00379 t t = = = + + 1.5.5 Resistance in Series The equivalent resistance of series combination of resistances is the summation of all resistances connected in series. eq 1 2 3R R R R= + + R1 R3 I V V2V1 V3 R2 Figure 1.5 : Resistances in Series In series circuit, current remains same through all the resistances while the voltage is divided in all resistances. 1 2V V V V= + + 3 31 2IR IR IR= + + 10. 14 Electricity q1 2 3 e( )I R R R I R= + + = Voltage Division Rule Since in series circuit, shown in Figure 1.5, current is given by equation eq V I R = then 1 1 1 . eq R V IR V R = = 2 2 2 . eq R V IR V R = = 3 3 3 . eq R V IR V R = = 1.5.6 Resistance in Parallel Figure 1.6 shows the resistance connected is parallel. I V R1 R3 R2 I1 I3 I2 Figure 1.6 : Parallel Resistors In parallel circuits, current is divided while the voltage remains same across all parallel branches. eq 1 2 3 1 1 1 1 R R R R = + + 2 3 1 3 1 2 1 2 3 R R R R R R R R R + + = 1 2 3 eq 1 2 2 3 3 1 R R R R R R R R R R = + + It is better to represent the circuit in conductance form. i.e. eq 1 2 3G G G G= + + Current Division Rule Current, 1 1 V I R = But, . eqV I R= eq 2 3 1 1 1 2 2 3 3 . . R R R I I I 1R R R R R R R = = + + Similarly 1 3 2 1 2 2 3 3 1 . R R I I R R R R R R = + 11. 15 Current and Static ElectricityAnd, 1 2 3 1 2 2 3 3 1 . R R I I R R R R R R = + + Voltage and Current Division Rules for Combination of Two Resistances Voltage Division Rule 1 2 1 2 1 2 1 2 . , . R R V V V V R R R = = + + R I V R1 R2 V2V1 Figure 1.7 : Series Resistances Current Division Rule 2 1 1 2 1 2 1 2 . , . R R I I I I R R R = = + + R R1 R2 I V I1 I2 Figure 1.8 : Parallel Resistances 1.5.7 Classification of Resistors On the basis of construction : (a) Carbon composition resistor (b) Wire-wound resistor on ceramic or plastic forms (example : rheostat) (c) Deposit carbon resistor (d) Metal film resistor (e) Printed, painted or earthed circuit resistance On the Basis of Value Fixed Resistors Their values remain constant. Fixed resistors are commonly made of carbon composition. R Figure 1.9 : Fixed Resistor Tapped Resistors There are tapping on the resistor so that any desired value can be achieved. 12. 16 Electricity Figure 1.10 : Tapped Resistor Variable Resistor If the value of resistor can be changed with the help of movable contact then it is known as the variable resistor (like rheostat). Figure 1.11 : Variable Resistor Example 1.3 Find the equivalent resistance for the circuit shown below. Figure for Example 1.3 Solutions After adding the series resistances, the circuit becomes : 5 4 6 15seR = + + = Now solve the parallel branch parallel 2 || 2R = 2 2 1 2 2 = = + eq se parallelR R R= + = 15 + 1 = 16 . Example 1.4 13. 17 Current and Static Electricity Find the current through each element of the given network and also find potential difference across 15 resistor. 15 5 100V 10 10 Figure for Example 1.4 Solution The circuit can be reduced in the following form : Total current i in the circuit is obtained by 100 100 4.76 Amp 5 10 6 21 V i R = = = = + + So the current through 5 and 10 resistor will be i = 4.76 Amp. Now, current through 10 resistor in parallel branch is determined by current division rule : 2 1 1 2 r i i r r = + 15 4.76 2.85 Amp 10 15 = = + Similarly, current through 15 resistor is 2 1 2 10 i i r r = + 10 4.76 1.9 Amp 10 15 = = + Potential difference across 15 resistor will be 2 15V i= 1.9 15 28.5 volts= = Example 1.5 14. 18 Find the equivalent resistance of network across the source terminals and determine the current drawn from the source. Electricity 4100V 6 2 2 24 Figure for Example 1.5 Solution After solving both the parallel branches, we get eq 4 0.67 2.9 7.067R = + + = The current drawn from the source eq 100 14.16 Amp 7.067 V I R = = = . SAQ 2 (a) A tungsten filament has a temperature of 2,050o C and a resistance of 500 when taking normal working current, calculate the resistance of the filament when it has a temperature of 25o C. Temperature coefficient at 0o C is 0.005/o C. (b) Find equivalent resistance for the following passive network. Figure for SAQ 2(b) 1.6 KIRCHHOFFS LAWS 15. 19 Current and Static Electricity In 1845, Kirchhoff, at the age of 23, published his paper regarding two basic law for solving the electrical circuits. These laws became basis for the development of the network analysis or circuit theory. 1.6.1 Kirchhoffs Voltage Law (KVL) This law is based on the law of conservation of energy. According to this, electrical energy supplied to any circuit (closed loop) is equal to the energy consumed by its passive elements. Since this law relates the voltage in a closed circuit of an electrical network it is known as Kirchhoffs Voltage Law (KVL) or Kirchhoffs Mesh Law. Kirchhoffs voltage law states that in a closed electric circuit the algebraic sum of emfs and voltage drops is zero. By convention, the emfs or voltage rises are taken to be positive and voltage drops are taken to be negative. I1 R1 R2 V1 V2 E2E1 A B C D Figure 1.12 : Closed Loop In the closed circuit ABCDA, shown in Figure 1.12, applying Kirchhoffs voltage law, we have 1 1 2 2 0E V V E = 1 1 2 2 0E IR IR E = 1 2 1 2E E IR IR = + Sum of voltage rises = Sum of voltage drop or Energy supplied = Energy consumed. 1.6.2 Kirchhoffs Current Law (KCL) Kirchhoffs current law states that algebraic sum of all the current meeting at a point or junction is zero. Since this law relates the current flowing through the circuit, it is known as Kirchhoffs Current Law (KCL). In Figure 1.13, applying KCL to the junction, we have 1 3 4 2 5 0I I I I I+ + = or, 1 3 4 2I I I I I+ + = + 5 I5 I4 I3 I2 I1 O Figure 1.13 : KCL at Junction Current entering the junction = Current leaving the junction 1.6.3 Application of Kirchhoffs Laws Application of Kirchhoffs Voltage Law 16. 20 Kirchhoffs voltage law is used to determine the value of current in the multi loop circuit. For this purpose, Maxwells loop method is used as explained by an example given below. Electricity Figure 1.14 shows the circuit having two loops and one voltage source in first loop. I1 R4 I2 V R1 R3 R2 Figure 1.14 : Maxwells Loop Method Assume, two loop currents flowing in the directions shown in Figure 1.14. Now we can write the KVL equations for each loop. KVL Equation for First Loop Voltage rise = Voltage drops 1 1 2 1 2( )V R I R I I= + 1 2 1 2( ) 2R R I R I= + . . . (i) KVL Equation in Second Loop 2 2 1 3 2 4 20 ( )R I I R I R I= + + 2 1 2 3 4 2( )R I R R R I= + + + . . . (ii) After solving equations, we can find two loop currents I1 and I2. Voltage division rule is also one of the applications of KVL. Application of Kirchhoffs Current Law Kirchhoffs current law is used to determine the voltages at different nodes or junctions of the circuit. For this purpose, Maxwells node method is used. This method is being explained with the help of an example given below. Again consider the same circuit of Figure 1.14. For Applying Maxwells Node Method Assume voltage V1 at node as shown in Figure 1.15 and one reference node with zero voltage. I1 R4 Reference Node (Voltage = 0) I3 V R1 R3V1 I2 R2 Figure 1.15 : Maxwells Node Method Assume three currents I1, I2 and I3 leaving the node V1, then apply KCL : Algebraic of currents at any node is zero. 17. 21 Current and Static Electricity 1 2 3 0I I I+ + = 1 1 1 1 2 3 4 0 V V V V R R R R + + = + 1 1 2 3 4 1 1 1 V V 1R R R R R + + = + From this equation node, voltage V1 can be obtained and then current passing through different elements can be determined. For example, current through R1 is 1 1 1 V V I R = If value of I1 comes negative, this means actual direction of current I1 will be opposite to the assumed direction as shown in Figure 1.15. Current division rule is also one of the applications of KCL. Example 1.6 Find currents I1 and I2 in the given circuits by applying KVL. Figure for Example 1.6 Solution We apply KVL for first loop : 1 110 1 1 ( )I I I= + 2 2110 2I I= . . . (i) KVL for second loop : 2 1 21 ( ) 4 0I I I + = 2 15 0I I = . . . (ii) 1 2 0.2 5 I I = = 1I 1I . . . (iii) Put the value of I2 in Eq. (i) 1 110 2 0.2 1.8I I= = 1 5.55 AmpI = From Eq. (iii), .2 1.11 AmpI = Example 1.7 Find the node voltage V and then current passing through each element by using KCL. 18. 22 Electricity Figure for Example 1.7 Solution Apply KCL at the node whose voltage is V. Algebraic summation of all the current meeting at node is zero. 10 0 1 1 4 V V V + + = 1 1 1 10 4 V + + = 2.25 V = 10 V = 4.44 volt. Now we have to find current through each element. 10 10 4.44 5.56 Amp 1 V I = = = 1 4.44 Amp 1 V I = = 2 1.11 Amp 4 V I = = . SAQ 3 (a) In the given DC circuit, determine the currents I1 and I2 by using KVL. Figure for SAQ 3(a) (b) In the following circuit, find the node voltage V and then determine the current passing through each element (use KCL). 19. 23 Current and Static Electricity Figure for SAQ 3(b) (c) Find the current passing through 2 resistor by using Maxwell loop method. 10V 5 25V 5 2 1 Figure for SAQ 3(c) (d) Find the current passing through 1 resistor by using KVL (by Maxwell loop method) Figure for SAQ 3(d) (e) Solve the above problem (SAQ 3(d)) by using KCL (apply node method). 1.7 SOURCE TRANSFORMATIONS The voltage and current sources are mutually transferrable. Any practical voltage source (or simply, a voltage source) consists of an ideal voltage source in series with an internal resistance (for ideal source, this impedance is being zero and the output becomes independent of the load current). Any practical current source is demonstrated by ideal current source in parallel with their internal resistance. Let Figures 1.16(a) and (b) demonstrates their equivalence. (a) A Practical Voltage Source (b) A Practical Current Source Figure 1.16 Assume some load resistance rL connected at terminal a-b of both voltage and current source. aL a L V I r r = + b b L b L r I I r r = + 20. 24 Two sources to become identical they should deliver same current to loadElectricity b a L b L rV I r r r r = + + However, for the current source, the terminal voltage at a-b would be I rb, a-b being open i.e. bV I r= Now, we finally get a L br r r rL+ = + i.e. a br r= So, for any practical voltage source, if ideal voltage be V and internal resistance be ra, the voltage source can be replaced by a current source I with the internal resistance in parallel to the current source. Example 1.8 Convert the following current source to equivalent voltage source. Figure for Example 1.8 Solution For equivalent voltage source V is given by V I r= 10 10A= 100 V= Series resistance for voltage source will be same as parallel resistance in case of current source (that is 10 as given). So, equivalent voltage source is Figure 1.8 HEATING EFFECT OF ELECTRIC CURRENT We know that the resistance opposes the flow of current in a conducting material. This is due to atomic structure of the material. When current flows in a conductor (flow of free electrons) then electrons collide with the immobile charges of the material. In each collision, kinetic energy of electron is converted into heat energy. This is the heating effect of electric current. Material which has the minimum value of resistance show the minimum heating effect. Power dissipated in the resistor = I 2 R watts 21. 25 Current and Static Electricity Energy dissipated in the form of heat = power time = I 2 Rt Joules. This is the expression for total heat developed in resistor of R in time t sec and this is directly proportional to square of current I flows in the conductor. Example 1.9 Find total power lost in the circuit. Also find the energy dissipated in 5 sec. Figure for Example 1.9 Solution The given circuit can be redrawn as : Current I is given by 250 10 Amp 25 I = = So, power loss = I 2 R = (10) 2 25 = 2500 watts Energy dissipated = I 2 Rt = 2500 5 = 12500 joules = 12.5 k joules. SAQ 4 Find the power loss in 10 resistor and also find energy dissipated in 10 seconds. Figure for SAQ 4 22. 26 Electricity 1.9 SUMMARY In this unit, we have discussed basic terms related to electricity like current, emf, power, energy etc. Concept of three basic laws of electrical engineering Ohms law, Kirchhoffs Voltage Law and Kirchhoffs Current Law with their applications have also been discussed. This unit provides the platform to understand the concept of complicated electrical problems which will be discussed in Unit 2. Effect of temperature on resistance and the heating effect of electric current have also been presented here. Solved examples along with SAQs related to each topic have been given for better understanding. 1.10 ANSWERS TO SAQs SAQ 1 Given V = 230 V I = 8 Amp t = 2 h Rate = Rs. 4 per kWh (Rs. 4 for one unit) Cost = ? W = Pt, P = V I P = 8 Amp 230 V = 1840 Watt Energy consumed, W = 1840 Watt 2 h = 3,680 Wh = 3.68 kWh Cost = Rs. 4 per kWh 3.68 kWh = Rs. 14.72 The cost is Rs. 14.72. SAQ 2 (a) 0 0tR R t= + R At t = temperature = 2050o C Rt = 500 500 = R0 (1 + 2050 ) . . . (i) at t = 25o C Rt = R0 (1 + 25 ) . . . (ii) Now, dividing Eqs. (ii) by (i) 1 25 500 1 2050 tR + = + But = 0.005 per o C 23. 27 Current and Static Electricity 1 25 0.005 500 50 1 2050 0.005 tR + = = + 50tR = (b) Figure for Answers to SAQ 2(b) Req in the parallel combination of 2 and 15 resistors. eq 2 15 30 2 15 17 R = = + . SAQ 3 (a) Figure for Answers to SAQ 3(a) By Maxwell loop method Apply KVL for loop 1 : 1 15 3 (I I I= + 2 ) 2 2 or, . . . (i)1 25 4 3I I= KVL for loop 2 : 2 1 20 3 ( ) 2 4I I I I= + + 1 20 3 9I I= + . . . (ii) 1 3I I= . . . (iii) Put Eq. (iii) in Eq. (i) to find value of I2 2 25 4 (3 ) 3I I= 2 212 3I I= 29I= So, 2 5 Amp 0.555 Amp 9 I = = From Eq. (iii) 1 5 5 3 A 9 3 I = = mp or, 1 1.667 AmpI = 24. 28 (b)Electricity Figure for Answers to SAQ 3(b) KCL at node : 5 0 1 1 6 V V V + + = 1 1 1 5 6 V + + = V = 3.33 volt (c) Figure for Answers to SAQ 3(c) For loop 1 1 1 1 2 125 5 2 5( ) 25 12 5I I I I I 2I= + + = . . . (i) For loop 2 2 2 1 110 5( ) 10 5 6I I I I 2I = + = + . . . (ii) Solving Eqs. (i) and (ii) 2 25 5 10 6 2.128 Amp 12 5 5 6 I = = (d) Assume loop currents I1, I2, I3 as shown in figure. Figure for Answers to SAQ 3(d) Here current passing through 1 is I1 KVL equations : Loop 1 : 1 215 6( )I I I1= + + . . . (i) Loop 2 : 1 2 2 315 6( ) 10 6( )I I I I= + + + . . . (ii) Loop 3 : 3 25 6( ) 2I I I3 = + . . . (iii) 25. 29 Current and Static Electricity After solving Eqs. (i), (ii) and (iii), we get I1 = 6.364 Amp (current passing through 1 resistor). (e) (a) (b) Apply source transformation Now, network after assuming node voltages V1, V2, V3 Here we have to use super node analysis V2 and V3 from the super node. Apply KCL at node 1 : 1 1 2 5 3 3 V V V + = 1 2 1 2 1 1 1 5 2 1 3 3 3 V V V V + = = 5 . . . (i) KCL at node 2 and 3 (super node) 3 32 1 2 5 0 3 1 6 2 V VV V V + + + = 1 2 30.333 1.333 0.666 2.5V V V + + = . . . (ii) And the voltage source is equal to V2 V3 V2 V3 = 10 V V3 = V2 10 . . . (iii) After solving Eqs. (i), (ii) and (iii), we get V1 = 10.682 V V2 = 6.3648 V V3 = 3.635 V Current through 1 , 2 6.3648 Amp 1 V I = = SAQ 4 Figure for Answers to SAQ 4 26. 30 Electricity We have to find I2 for power dissipation in 10 . Calculation for I : eq 5 10 50 25 5 5 (5 10) 15 3 R = + = + = + eq 100 12 AmpI R = = By current division rule : 2 5 1 12 4 Amp 5 10 3 I I= = = + Power loss in 10 resistor : 2 2 2 10 4 10 160 WattP I= = = Energy dissipated : .2 160 10 1600 JouleW I Rt= = =