DC Pandey Electricity And Magnetism

Understanding Physics JEE Main & Advanced ELECTRICITY AND MAGNETISM



Understanding Physics JEE Main & Advanced ELECTRICITY AND MAGNETISM DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722] ARIHANT PRAKASHAN (Series), MEERUT

Understanding Physics JEE Main & Advanced ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved © SARITA PANDEY No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only. ADMINISTRATIVE & PRODUCTION OFFICES Regd. Office 'Ramchhaya' 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550; Fax: 011- 23280316 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tel: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648 SALES & SUPPORT OFFICES Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune. ISBN 978-93-13190-57-8 Published by ARIHANT PUBLICATIONS (I) LTD. For further information about the books published by Arihant, log on to www.arihantbooks.com or e-mail at [email protected] /arihantpub /@arihantpub Arihant Publications /arihantpub

Understanding Physics JEE Main & Advanced PREFACE The overwhelming response to the previous editions of this book gives me an immense feeling of satisfaction and I take this an opportunity to thank all the teachers and the whole student community who have found this book really beneficial. In the present scenario of ever-changing syllabus and the test pattern of JEE Main & Advanced, the NEW EDITION of this book is an effort to cater all the difficulties being faced by the students during their preparation of JEE Main & Advanced. The exercises in this book have been divided into two sections viz., JEE Main & Advanced. Almost all types and levels of questions are included in this book. My aim is to present the students a fully comprehensive textbook which will help and guide them for all types of examinations. An attempt has been made to remove all the printing errors that had crept in the previous editions. I am extremely thankful to (Dr.) Mrs. Sarita Pandey, Mr. Anoop Dhyani, Nisar Ahmad for their endless efforts during the project. Comments and criticism from readers will be highly appreciated and incorporated in the subsequent editions. DC Pandey

Understanding Physics JEE Main & Advanced CONTENTS 23. CURRENT ELECTRICITY 1-108 23.1 Introduction 23.8 The Battery and the Electromotive 23.2 Electric Current Force 23.3 Electric Currents in Conductors 23.4 Drift Velocity and Relaxation Time 23.9 Direct Current Circuits, Kirchhoff ’s 23.5 Resistance of a Wire Laws 23.6 Temperature Dependence of 23.10 Heating Effects of Current Resistance 23.11 Grouping of Cells 23.7 Ohm’s Law 23.12 Electrical Measuring Instruments 23.13 Colour Codes for Resistors 24. ELECTROSTATICS 109-231 24.1 Introduction 24.10 Equipotential Surfaces 24.2 Electric Charge 24.11 Electric Dipole 24.3 Conductor and Insulators 24.12 Gauss’s Law 24.4 Charging of a Body 24.13 Properties of a Conductor 24.5 Coulomb’s Law 24.14 Electric Field and Potential Due To 24.6 Electric Field 24.7 Electric Potential Energy Charged Spherical Shell or Solid 24.8 Electric Potential Conducting Sphere 24.9 Relation Between Electric Field and 24.15 Electric Field and Potential Due to a Solid Sphere of Charge Potential 25. CAPACITORS 233-333 25.1 Capacitance 25.6 Two Laws in Capacitors 25.2 Energy Stored in a Charged Capacitor 25.7 Energy Density 25.3 Capacitors 25.8 C-R Circuits 25.4 Mechanical Force on a Charged 25.9 Methods of Finding Equivalent Conductor Resistance and Capacitance 25.5 Capacitors in Series and Parallel

Understanding Physics JEE Main & Advanced 26. MAGNETICS 335-454 26.1 Introduction Carrying Wires 26.2 Magnetic Force on a Moving 26.11 Magnetic Poles and Bar Magnets 26.12 Earth’s Magnetism Charge(Fm) 26.13 Vibration Magnetometer 26.3 Path of a Charged Particle in Uniform 26.14 Magnetic Induction and Magnetic Magnetic Field Materials 26.4 Magnetic Force on a Current 26.15 Some Important Terms used in Carrying Conductor Magnetism 26.5 Magnetic Dipole 26.16 Properties of Magnetic Materials 26.6 Magnetic Dipole in Uniform 26.17 Explanation of Paramagnetism, Magnetic Field Diamagnetism and Ferromagnetism 26.7 Biot Savart Law 26.18 Moving Coil Galvanometer 26.8 Applications of Biot Savart Law 26.9 Ampere's Circuital Law 26.10 Force Between Parallel Current 27. ELECTROMAGNETIC INDUCTION 455-560 27.1 Introduction 27.6 Self-Inductance and Inductors 27.2 Magnetic Field Lines and 27.7 Mutual Inductance 27.8 Growth and Decay of Current in an L- Magnetic Flux 27.3 Faraday’s Law R Circuit 27.4 Lenz’s Law 27.9 Oscillations in L-C Circuit 27.5 Motional Electromotive Force 27.10 Induced Electric Field 28. ALTERNATING CURRENT 561-607 28.1 Introduction 28.5 Series L-R Circuit 28.2 Alternating Currents and Phasors 28.6 Series C-R Circuit 28.3 Current and Potential Relations 28.7 Series L-C-R Circuit 28.4 Phasor Algebra 28.8 Power in an AC Circuit Hints & Solutions 609-731 1-36 JEE Main & Advanced Previous Years' Questions (2018-13)

Understanding Physics JEE Main & Advanced SYLLABUS JEE Main ELECTROSTATICS Electric charges Conservation of charge, Coulomb’s law-forces between two point charges, forces between multiple charges; Superposition principle and continuous charge distribution. Electric field Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long uniformly charged straight wire, Uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its calculation for a point charge, Electric dipole and system of charges; Equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, Dielectrics and electric polarization, Capacitor, Combination of Capacitors in series and in parallel, Capacitance of a parallel plate capacitor with and without dielectric medium between the plates, Energy stored in a capacitor. CURRRENT ELECTRICITY Electric current, Drift velocity, Ohm’s law, Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and non-ohmic conductors, Electrical energy and power, Electrical resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. Electric cell and its Internal resistance, Potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff ’s laws and their applications. Wheatstone bridge, Meter bridge. Potentiometer – principle and its applications.

Understanding Physics JEE Main & Advanced MAGNETIC EFFECTS OF CURRENT AND MAGNETISM Biot-Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields. Cyclotron. Force on a current-carrying conductor in a uniform magnetic field. Force between two parallel current-carrying conductors-definition of ampere. Torque experienced by a current loop in uniform magnetic field; Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. Bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferro- magnetic substances. Magnetic susceptibility and permeability, Hysteresis, Electromagnets and permanent magnets. ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS Electromagnetic induction; Faraday’s law, Induced emf and current; Lenz’s law, Eddy currents. Self and mutual inductance. Alternating currents, Peak and rms value of alternating current/voltage; Reactance and impedance; LCR series circuit, Resonance; Quality factor, power in AC circuits, Wattless current. AC generator and transformer.

Understanding Physics JEE Main & Advanced JEE Advanced GENERAL Verification of Ohm’s law using voltmeter and ammeter. Specific resistance of the material of a wire using meter bridge and post office box. ELECTRICITY AND MAGNETISM Coulomb’s law, Electric field and potential, Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field, Electric field lines, Flux of electric field, Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, Uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Capacitance, Parallel plate capacitor with and without dielectrics, Capacitors in series and parallel, Energy stored in a capacitor. Electric current, Ohm’s law, Series and parallel arrangements of resistances and cells, Kirchhoff ’s laws and simple applications, Heating effect of current. Biot-Savart’s law and Ampere’s law, Magnetic field near a current-carrying straight wire, Along the axis of a circular coil and inside a long straight solenoid, Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop, Effect of a uniform magnetic field on a current loop, Moving coil galvanometer, Voltmeter, Ammeter and their conversions. ELECTROMAGNETIC INDUCTION Faraday’s law, Lenz’s law, Self and mutual inductance, RC, LR and LC circuits with DC and AC sources.

Current Electricity Chapter Contents 23.1 Introduction 23.2 Electric current 23.3 Electric currents in conductors 23.4 Drift velocity and Relaxation time 23.5 Resistance of a wire 23.6 Temperature dependence of resistance 23.7 Ohm's law 23.8 The battery and the electromotive force 23.9 Direct current circuits, Kirchhoff's laws 23.10 Heating effects of current 23.11 Grouping of cells 23.12 Electrical measuring instruments 23.13 Colour codes for resistors

2 — Electricity and Magnetism 23.1 Introduction An electrical circuit consists of some active and passive elements. The active elements such as a battery or a cell, supply electric energy to the circuit. On the contrary, passive elements consume or store the electric energy. The basic passive elements are resistor, capacitor and inductor. A resistor opposes the flow of current through it and if some current is passed by maintaining a potential difference across it, some energy is dissipated in the form of heat. A capacitor is a device which stores energy in the form of electric potential energy. It opposes the variations in voltage. An inductor opposes the variations in current. It does not oppose the steady current through it. Fundamentally, electric circuits are a means for conveying energy from one place to another. As charged particles move within a circuit, electric potential energy is transferred from a source (such as a battery or a cell) to a device in which that energy is either stored or converted to another form, like sound in a stereo system or heat and light in a toaster or light bulb. Electric circuits are useful because they allow energy to be transported without any moving parts (other than the moving charged particles themselves). In this chapter, we will study the basic properties of electric currents. We’ll study the properties of batteries and how they cause current and energy transfer in a circuit. In this analysis, we will use the concepts of current, potential difference, resistance and electromotive force. 23.2 Electric Current Flow of charge is called electric current. The direction of electric current is in the direction of flow of positive charge or in the opposite direction of flow of negative charge. Current is defined quantitatively in terms of the rate at which net charge passes through a cross-section area of the conductor. Thus, I = dq or i = dq dt dt We can have the following two concepts of current, as in the case of velocity, instantaneous current and average current. Instantaneous current = dq = current at any point of time and dt Average current = q t Hence-forth unless otherwise referred to, current would signify instantaneous current. By convention, the direction of the current is assumed to be that in which positive charge moves. In the SI system, the unit of current is ampere (A). 1 A =1 C/s Household currents are of the order of few amperes. Flow of Charge If current is passing through a wire then it implies that a charge is flowing through that wire. Further, i = dq ⇒ dq = idt …(i) dt

Chapter 23 Current Electricity — 3 Now, three cases are possible : Case 1 If given current is constant, then from Eq. (i) we can see that flow of charge can be obtained directly by multiplying that constant current with the given time interval. Or, ∆q = i × ∆t Case 2 If given current is a function of time, then charge flow can be obtained by integration. Or, tf ∆q = ∫ i dt ti Case 3 If current versus time is given, then flow of charge can be obtained by the area under the graph. ∆q = area under i -t graph Extra Points to Remember ˜ The current is the same for all cross-sections of a conductor of non-uniform cross-section. Similar to the water flow, charge flows faster where the conductor is smaller in cross-section and slower where the conductor is larger in cross-section, so that charge rate remains unchanged. ˜ Electric current is very similar to water current, consider a water tank kept at some height and a pipe is connected to the water tank. The rate of flow of water through the pipe depends on the height of the tank. As the level of water in the tank falls, the rate of flow of water through the pipe also gets reduced. Just as the flow of water depends on the height of the tank or the level of water in the tank, the flow of current through a wire depends on the potential difference between the end points of the wire. As the potential difference is changed, the current will change. For example, during the discharging of a capacitor potential difference and hence, the current in the circuit decreases with time. To maintain a constant current in a circuit a constant potential difference will have to be maintained and for this a battery is used which maintains a constant potential difference in a circuit. ˜ Though conventionally a direction is associated with current (opposite to i1 the motion of electrons), it is not a vector as the direction merely represents the sense of charge flow and not a true direction. Further, current does not θ i = i1+ i2 obey the law of parallelogram of vectors, i.e. if two currents i1 and i 2 reach a point we always have i = i1 + i 2 whatever be the angle between i1 and i 2. i2 ˜ According to its magnitude and direction, current is usually divided into two Fig. 23.1 types : (i) Direct current (DC) If the magnitude and direction of current does not vary with time, it is said to be direct current (DC). Cell, battery or DC dynamo are its sources. (ii) Alternating current (AC) If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC). AC dynamo is the source of it. ˜ If a charge q revolves in a circle with frequency f, the equivalent current, i =qf ˜ In a conductor, normally current flow or charge flow is due to flow of free electrons. ˜ Charge is quantised. The quantum of charge is e. The charge on any body will be some integral multiple of e, i.e. q = ± ne where, n = 1, 2, 3K

4 — Electricity and Magnetism V Example 23.1 In a given time of 10 s, 40 electrons pass from right to left. In the same interval of time 40 protons also pass from left to right. Is the average current zero? If not, then find the value of average current. Solution No, the average current is not zero. Direction of current is the direction of motion of positive charge or in the opposite direction of motion of negative charge. So, both currents are from left to right and both currents will be added. ∴ I av = I electron + I proton = q1 + q2 t1 t2 = 40 e + 40 e (q = ne) 10 10 = 8e = 8 × 1.6 × 10−19 A = 1.28 × 10−18 A Ans. V Example 23.2 A constant current of 4 A passes through a wire for 8 s. Find total charge flowing through that wire in the given time interval. Solution Since, i = constant ∴ ∆q = i × ∆t =4×8 = 32 C V Example 23.3 A wire carries a current of 2.0 A. What is the charge that has flowed through its cross-section in 1.0 s ? How many electrons does this correspond to? Solution Q i = q t ∴ q = it = (2.0 A) (1.0 s ) = 2.0 C Ans. q = ne ∴ n = q = 2.0 e 1.6 × 10–19 = 1.25 × 1019 Ans. V Example 23.4 The current in a wire varies with time according to the relation i = (3.0 A) + (2.0 A/ s) t (a) How many coulombs of charge pass a cross-section of the wire in the time interval between t = 0 and t = 4.0 s? (b) What constant current would transport the same charge in the same time interval?

Chapter 23 Current Electricity — 5 Solution (a) i = dq ∴ dt ∴ =∫ ∫q 4 (b) i = q = 28 = 7 A dq t4 idt 00 4 q = ∫ 0 (3 + 2t ) dt = [3t + t 2 ] 4 = [12 + 16] 0 = 28 C Ans. Ans. V Example 23.5 Current passing through a wire decreases linearly from 10 A to 0 in 4 s. Find total charge flowing through the wire in the given time interval. Solution Current versus time graph is as shown in figure. i (A) Area under this graph will give us net charge flow. 10 Hence, ∆q = Area = 1 × base × height 4 t (s) 2 Fig. 23.2 = 1 × 4 × 10 2 = 20 C Ans. INTRODUCTORY EXERCISE 23.1 1. How many electrons per second pass through a section of wire carrying a current of 0.7 A? 2. A current of 3.6 A flows through an automobile headlight. How many coulombs of charge flow through the headlight in 3.0 h? 3. A current of 7.5 A is maintained in wire for 45 s. In this time, (a) how much charge and (b) how many electrons flow through the wire? 4. In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 × 10−11 m with a speed of 2.2 × 106 m /s. Determine its frequency f and the current I in the orbit. 5. The current through a wire depends on time as, i = (10 + 4t ) Here, i is in ampere and t in seconds. Find the charge crossed through a section in time interval between t = 0 to t = 10 s. 6. In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current? If yes, in what direction?

6 — Electricity and Magnetism 23.3 Electric Currents in Conductors Conductors are those materials which can conduct electricity. Conductors can be broadly classified into two groups : (i) Solid conductors (ii) Electrolyte conductors Normally in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. In solid conductors (notably metals), some of the electrons (called free electrons) are free to move within the bulk materials. In these conductors, current flow takes place due to these free electrons. Positive ions in these conductors are almost fixed. They do not move. So, they do not contribute in the current. In electrolyte solutions however, both positive and negative ions can move. In our following discussions, we will focus only on solid conductors so that the current is carried by the negatively charged free electrons in the background of fixed positive ions. Theory of Current Flow through Solid Conductors At room temperature, the free electrons in a conductor move randomly with speeds of the order of 105 m /s.Since, the motion of the electrons is random, there is no net charge flow in any direction. For any imaginary plane passing through the conductor, the number of electrons crossing the plane in one direction is equal to the number crossing it in the other direction. Therefore, net current is zero from any section. –– – Fig. 23.3 When a constant potential difference V is applied between the ends of the conductor as shown in Fig. 23.4, an electric field E is produced inside the conductor. The conduction electrons within the conductor are then subjected to a force – eEand move overall in the direction of increasing potential. i – vd – – E –– – – – V V +– +– Fig. 23.4 However, this force does not cause the electrons to move faster and faster. Instead, a conduction electron accelerates through a very small distance (about 5 ×10−8 m) and then collides with fixed ions or atoms of the conductor. Each collision transfers some of the electron’s kinetic energy to the ions (or atoms). Because of the collision, electron moves slowly along the conductor or we can say that it acquires a drift velocity vd in the direction opposite to E (in addition to its random motion.) The drift motion of free electrons produce an electric current in the opposite direction of this motion or in the direction of electric field (from higher potential to lower potential). It is interesting to note

Chapter 23 Current Electricity — 7 that the magnitude of the drift velocity is of the order of 10–4 m /s or about 109 times smaller than the average speed of the electrons of their random (or thermal) motion. The above discussion can be summarized as follows : 1. Free electrons inside a solid conductor can have two motions : (i) random or thermal motion (speed of the order of 105 m /s) (ii) drift motion (speed of the order of 10–4 m /s) 2. Net current due to random (or thermal motion) is zero from any section, whereas net current due to drift motion is non-zero. 3. In the absence of any electric field (or a potential difference across the conductor) free electrons have only random motion. Hence, net current from any section is zero. 4. In the presence of an electric field (or a potential difference across the conductor) free electrons have both motions (random and drift). Therefore, current is non-zero due to drift motion. 5. Drift motion of free electrons is opposite to the electric field. Therefore, direction of current is in the direction of electric field from higher potential to lower potential. V Example 23.6 Electric field inside a conductor is always zero. Is this statement true or false? Solution False. Under electrostatic conditions when there is no charge flow (or no current) in the conductor, electric field is zero. If current is non-zero, then electric field is also non-zero. Because the drift motion (of free electrons) which produces a net current starts only due to electric force on them. INTRODUCTORY EXERCISE 23.2 1. All points of a conductor are always at same potential. Is this statement true or false? 23.4 Drift Velocity and Relaxation Time As discussed before, in the presence of electric field, the free electrons experience an electric force of magnitude. F = qE or eE (as q = e ) This will produce an acceleration of magnitude, (m = mass of electron) a = F = eE mm Direction of force (and acceleration) is opposite to the direction of electric field. After accelerating to some distance an electron will suffer collisions with the heavy fixed ions. The collisions of the electrons do not occur at regular intervals but at random times. Relaxation time τ is the average time between two successive collisions. Its value is of the order of 10−14 second. After every collision, let us assume that drift motion velocity of electron becomes zero. Then, it accelerates for a time interval τ, then again it collides and its drift motion velocity becomes zero and so on.

8 — Electricity and Magnetism If vd is the average constant velocity (called drift velocity) in the direction of drift motion, then relation between vd and τ is given by vd = eEτ m Extra Points to Remember ˜ In some standard books of Indian authors, the relation is given as vd = eEτ 2m Initially, I was also convinced with this expression. But later on after consulting many more literatures in this. I found that vd = eEτ is correct. But at this stage it is very difficult for me to give its correct proof. m Because the correct proof requires a knowledge of high level of physics which is difficult to understand for a class XII student. Current and Drift Velocity Consider a cylindrical conductor of cross-sectional area A in which an electric field E exists. Drift velocity of free electrons is vd and n is number of free electrons per unit volume (called free electron density). i E vd vd ∆t Fig. 23.5 Consider a length vd ∆t of the conductor. The volume of this portion is Avd ∆t. Number of free electrons in this volume = (free electron density ) × (volume) = (n) (Avd ∆t) = nAvd ∆t All these electrons cross the area A in time ∆t. Thus, the charge crossing this area in time ∆t is ∴ ∆q = (nAvd ∆t) (e) or i = ∆q = neAvd ∆t or i = neAvd Thus, this is the relation between current and drift velocity.

Chapter 23 Current Electricity — 9 Current Density Current per unit area (taken normal to the current), i / A is called current density and is denoted by j. The SI units of current density are A/ m 2. Current density is a vector quantity j directed along E. j= i A But i = neAvd , therefore j = nevd Extra Points to Remember ˜ Drift velocity of electrons in a conductor is of the order of 10−4 m/s, then question arises in everybody’s mind that why a bulb glows instantly when switched on? Reason is : when we close the circuit, electric field is set up in the entire closed circuit instantly (with the speed of light). Due to this electric field, the free electrons instantly get drift velocity in the entire circuit and a current is established in the circuit instantly. The current so set up does not wait for the electrons to flow from one end of the conductor to the other end. ˜ If a current i is flowing through a wire of non-uniform cross-section, 12 then current will remain constant at all cross-sections. But drift velocity and current density are inversely proportional to the area of i i cross-section. This is because i = neAvd or vd = i or vd ∝ 1 neA A Fig. 23.6 Further, j = i or j ∝ 1 AA So, in the figure i1 = i 2 = i but, (vd )2 > (vd )1 and j 2 > j1 because A 2 < A1 Note Later we can also prove that electric field at 2 is also more than electric field at 1 . V Example 23.7 An electron beam has an aperture of 1.0 mm2 . A total of 6.0 × 1016 electrons go through any perpendicular cross-section per second. Find (a) the current and (b) the current density in the beam. Solution (a) The current is given by i = q = ne tt Substituting the values we have, Ans. i = (6.0 × 1016 ) (1.6 × 10−19 ) 1 = 9.6 × 10−3 A (b) The current density is j = i = 9.6 × 10−3 A 10−6 = 9.6 × 103 A/ m2 Ans.

10 — Electricity and Magnetism V Example 23.8 Calculate the drift speed of the electrons when 1 A of current exists in a copper wire of cross-section 2 mm2 . The number of free electrons in 1 cm3 of copper is 8.5 × 1022 . Solution n = free electron density, (Given) = 8.5 × 1022 per cm3 = (8.5 × 1022 ) (106 ) per m3 = 8.5 × 1028 per m3 From i = neAvd , we get vd = i neA Substituting the values in SI units we have, vd = (8.5 × 1028 1 )(1.6 × 10−19 )(2 × 10−6 ) = 3.6 × 10−5 m/s Answer INTRODUCTORY EXERCISE 23.3 1. When a wire carries a current of 1.20 A, the drift velocity is 1.20 × 10−4m /s. What is the drift velocity when the current is 6.00 A? 2. Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1cm2 and length 10 km. Free electron density of copper is 8.5 × 1028/ m3. How long will it take the electric charge to travel from one end of the conductor to the other? 23.5 Resistance of a Wire Resistance of a wire is always required between two points or two surfaces (say P and Q). l PQ A Fig. 23.7 Here, l = length of wire, A = area of cross-section Now, R ∝l …(i) …(ii) and R ∝ 1 A …(iii) Combining Eqs. (i) and (ii), we get R =ρ l A

Chapter 23 Current Electricity — 11 Here, ρ is called resistivity of the material of the wire. This depends on number of free electrons present in the material. With increase in number of free electrons the value of ρ decreases. Note (i) ρ = 1 , where σ = conductivity. σ (ii) SI units of resistivity are Ω-m (ohm-metre). (iii) SI units of conductivity are (Ω-m)−1. (iv) In Eq. (iii), l is that dimension of conductor which is parallel to P and Q and A is that cross-sectional area, which is perpendicular to P and Q. V Example 23.9 Two copper wires of the same length have got different diameters, (a) which wire has greater resistance? (b) greater specific resistance? Solution (a) For a given wire, R = ρ l , i.e. R ∝ 1 AA So, the thinner wire will have greater resistance. (b) Specific resistance (ρ) is a material property. It does not depend on l or A. So, both the wires will have same specific resistance. V Example 23.10 A wire has a resistance R. What will be its resistance if it is stretched to double its length? Solution Let V be the volume of wire, then V = Al ∴ A=V l Substituting this in R = ρ l , we have R = ρ l2 A V So, for given volume and material (i.e.V and ρ are constants) R ∝l2 When l is doubled, resistance will become four times, or the new resistance will be 4R. V Example 23.11 The dimensions of a conductor of specific resistance ρ are shown below. Find the resistance of the conductor across AB, CD and EF. A D b a Ec F CB Fig. 23.8

12 — Electricity and Magnetism Solution R = ρ l A Resistance across AB, CD and EF in tabular form is shown below. Table 23.1 I A B AB c a×b ρc CD b a×c ab EF a b×c ρb ac ρa bc V Example 23.12 A copper wire is stretched to make it 0.1% longer. What is the percentage change in its resistance? (JEE 1978) Solution R = ρ l = ρl ( V = volume of wire) A V/l = ρl2 V ∴ R ∝l2 ( ρ and V = constant) For small percentage change % change R = 2 (% change in l ) = 2 (0.1%) = 0.2% Since R ∝ l2 , with increase in the value of l, resistance will also increase. INTRODUCTORY EXERCISE 23.4 1. In household wiring, copper wire 2.05 mm in diameter is often used. Find the resistance of a 35.0 m long wire. Specific resistance of copper is 1.72 × 10−8 Ω-m. 2. The product of resistivity and conductivity of a conductor is constant. Is this statement true or false? 3. You need to produce a set of cylindrical copper wires 3.50 m long that will have a resistance of 0.125 Ω each. What will be the mass of each of these wires? Specific resistance of copper = 1.72 × 10–8 Ω-m, density of copper = 8.9 × 103 kg/m3. 4. Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is (JEE 2010) (a) directly proportional to L (b) directly proportional to t t (c) independent of L L Fig. 23.9 (d) independent of t

Chapter 23 Current Electricity — 13 23.6 Temperature Dependence of Resistance If we increase the temperature of any material, the following two effects can be observed : (i) Numbers of free electrons increase. Due to this effect conductivity of the material increases. So, resistivity or resistance decreases. (ii) The ions of the material vibrate with greater amplitude and the collision between electrons and ions become more frequent. Due to this effect resistivity or resistance of the material increases. In Conductors There are already a large number of free electrons. So, with increase in temperature effect-(i) is not so dominant as effect-(ii). Hence, resistivity or resistance of conductors increase with increase in temperature. Over a small temperature range (upto 100°C), the resistivity of a metal (or conductors) can be represented approximately by the equation, ρ(T ) = ρ0 [1 + α (T – T0 )] …(i) where, ρ0 is the resistivity at a reference temperature T0 (often taken as 0°C or 20°C) and ρ (T ) is the resistivity at temperature T, which may be higher or lower than T0. The factor α is called the temperature coefficient of resistivity. The resistance of a given conductor depends on its length and area of cross-section besides the resistivity. As temperature changes, the length and area also change. But these changes are quite small and the factor l/ A may be treated as constant. Then, R ∝ ρ and hence, R (T ) = R0 [1 + α (T – T0 )] …(ii) In this equation, R (T ) is the resistance at temperature T and R0 is the resistance at temperature T0, often taken to be 0°C or 20°C. The temperature coefficient of resistance α is the same constant that appears in Eq. (i), if the dimensions l and A in equation R = ρ l do not change with temperature. A In Semiconductors At room temperature, numbers of free electrons in semiconductors (like silicon, germanium etc.) are very less. So, with increase in temperature, effect-(i) is very dominant. Hence, resistivity or resistance of semiconductors decreases with increase in temperature or we can say that temperature coefficient of resistivity α for semiconductors is negative. V Example 23.13 The resistance of a thin silver wire is 1.0 Ω at 20°C. The wire is placed in a liquid bath and its resistance rises to 1.2 Ω. What is the temperature of the bath? α for silver is 3.8 × 10–3 /°C. Solution R(T ) = R0 [1 + α (T − T0 )] Here, R(T ) = 1.2 Ω, R0 = 1.0 Ω, α = 3.8 × 10–3 / ° C and T0 = 20° C Substituting the values, we have 1.2 = 1.0 [1 + 3.8 × 10–3 (T – 20)] or 3.8 × 10–3 (T – 20) = 0.2 Solving this, we get T = 72.6° C Ans.

14 — Electricity and Magnetism V Example 23.14 Read the following statements carefully (JEE 1993) Y : The resistivity of semiconductor decreases with increase of temperature. Z : In a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature. Select the correct statement (s) from the following (a) Y is true but Z is false (b) Y is false but Z is true (c) Both Y and Z are true (d) Y is true and Z is the correct reason for Y Solution Resistivity of conductors increases with increase in temperature because rate of collisions between free electrons and ions increase with increase of temperature. However, the resistivity of semiconductors decreases with increase in temperature, because more and more covalent bonds are broken at higher temperatures and free electrons increase with increase in temperature. Therefore, the correct option is (c). V Example 23.15 An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it. Its resistance at room temperature (27.0 oC) is found to be 75.3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature o range involved, is 1.70 × 10 −4 −1. C Solution Given, T0 = 27° C and R0 = 75.3 Ω At temperature T, RT = VT R = V  iT i = 230 = 85.82 Ω 2.68 Using the equation, RT = R0 [1 + α (T − T0 )] We have 85.82 = 75.3 [1 + (1.70 × 10−4 )(T − 27)] Solving this equation, we get T ≈ 850 oC Ans. Thus, the steady temperature of the nichrome element is 850 oC . INTRODUCTORY EXERCISE 23.5 1. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of (JEE 1988) (a) each of them increases (b) each of them decreases (c) copper increases and germanium decreases (d) copper decreases and germanium increases 2. The resistance of a copper wire and an iron wire at 20°C are 4.1 Ω and 3.9 Ω, respectively. Neglecting any thermal expansion, find the temperature at which resistances of both are equal. α Cu = 4.0 × 10−3 K−1 and αFe = 5.0 × 10−3 K−1.

Chapter 23 Current Electricity — 15 23.7 Ohm’s Law The equation,V = iR is not Ohm’s law. It is a mathematical relation between current passing through a resistance, value of resistance R and the potential difference V across it. According to Ohm’s law, there are some of the materials (like metals or V conductors) or some circuits for which, current passing through them is proportional to the potential difference applied across them or tan θ = slope =R i ∝V or V ∝ i θ i ⇒ V = iR or V = R = constant Ohmic i Fig. 23.10 or V - i graph for such materials and circuits is a straight line passing through origin. Slope of this graph is called its resistance. The materials or circuits which follow this law are called ohmic. The materials or circuits which do not follow this law are called non-ohmic.V - i graph for non-ohmic circuits is not a straight line passing through origin. V or R is not constant and i is not proportional toV . i V V2 Q i V1 P i1 i2 Non-ohmic Fig. 23.11 Note Equation V = iR is applicable for even non-ohmic circuits also. For example, V1 = R1 = resistance at P. i1 V2 = R2 = resistance at Q, but i2 R1 ≠ R2 V Example 23.16 The current-voltage graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. The temperature T2 is greater than T1. Is this statement true or false? (JEE 1985) I T1 T2 V Fig. 23.12

16 — Electricity and Magnetism Solution I = Slope of given graph = 1 VR or R= 1 ⇒ (Slope)T2 < (Slope)T1 ⇒ 1> 1 Slope (Slope)T2 (Slope)T1 Resistance of a metallic wire increases with increase in temperature. or RT2 > RT1 or T2 > T1 Therefore, the statement is true. 23.8 The Battery and the Electromotive Force Before studying the electromotive force (emf) of a cell let us take an example of Overhead tank a pump which is more easy to understand. Suppose we want to recycle water between a overhead tank and a ground water tank. Water flows from overhead tank to ground water tank by itself (by gravity). No external agent is required Pump for this purpose. But to raise the water from ground water tank to overhead tank a pump is required or some external work has to be done. In an electric circuit, a battery or a cell plays the same role as the pump played in the above example. Ground water tank Suppose a resistance (R) is connected across the terminals of a battery. Fig. 23.13 A potential difference is developed across its ends. Current (or HL positive charge) flows from higher potential to lower potential across the resistance by itself. But inside the battery, work has to be done to R bring the positive charge from lower potential to higher potential. The influence that makes current flow from lower to higher potential (inside the battery) is called electromotive force (abbreviated emf). If HL W work is done by the battery in taking a charge q from negative Fig. 23.14 terminal to positive terminal, then work done by the battery per unit charge is called emf (E) of the battery. Thus, E =W q The name electromotive force is misleading in the sense that emf is not a force it is work done per unit charge. The SI unit of emf is J/C or V ( 1 V =1 J /C ). 23.9 Direct Current Circuits, Kirchhoff’s Laws Single current in a simple circuit (single loop) can be found by the relation, 6V 10V 6V E i ii 3Ω 2Ω R (a) (b) (c) Fig. 23.15

Chapter 23 Current Electricity — 17 i = net emf = E net net resistance R net For example : In Fig. (a) Net emf is 6 V and net resistance is 3 Ω. Therefore, In Fig. (b) i=6 =2A 3 Net emf = (10 – 6) V = 4 V and Net resistance = 2 Ω Therefore, i= 4 =2A 2 In Fig. (c) We have n cells each of emf E. Of these polarity of m cells (where n > 2m) is reversed. Then, net emf in the circuit is (n – 2m) E and resistance of the circuit is R. Therefore, i = (n – 2m) E R Resistors in Series and in Parallel In series : A A i i R1 V1 R V B V⇒ R2 V2 B Fig. 23.16 Figure represents a circuit consisting of a source of emf and two resistors connected in series. We are interested in finding the resistance R of the network lying between A and B. That is, what single equivalent resistor R would have the same resistance as the two resistors linked together. Because there is only one path for electric current to follow, i must have the same value everywhere in the circuit. The potential difference between A and B is V. This potential difference must somehow be divided into two parts V1 and V2 as shown, ∴ V = V1 + V2 = iR1 + iR2 or V = i (R1 + R2 ) …(i) Let R be the equivalent resistance between A and B, then V = iR …(ii) From Eqs. (i) and (ii), R = R1 + R2 for resistors in series This result can be readily extended to a network consisting of n resistors in series. ∴ R = R1 + R2 + KK + Rn

18 — Electricity and Magnetism In parallel : i i i1 i2 ⇒V V R R1 R2 Fig. 23.17 In Fig. 23.17, the two resistors are connected in parallel. The voltage drop across each resistor is equal to the source voltage V. The current i, however, divides into two branches, which carry currents i1 and i2 . i = i1 + i2 …(iii) If R be the equivalent resistance, then i=V , i1 = V and i2 = V R R1 R2 Substituting in Eq. (iii), we get 1 = 1 + 1 for resistors in parallel R R1 R2 This result can also be extended to a network consisting of n resistors in parallel. The result is 1 = 1 + 1 +…K + 1 R R1 R2 Rn V Example 23.17 Compute the equivalent resistance of the network shown in figure and find the current i drawn from the battery. 18 V ii 6Ω 4Ω 3Ω Fig. 23.18 Solution The 6 Ω and 3 Ω resistances are in parallel. Their equivalent resistance is 18 V i 4Ω 2Ω Fig. 23.19

Chapter 23 Current Electricity — 19 1 = 1 + 1 or R = 2 Ω R 63 Now, this 2 Ω and 4 Ω resistances are in series and their equivalent resistance is 4 + 2 = 6 Ω. Therefore, equivalent resistance of the network = 6 Ω. Ans. i 18V 6Ω Fig. 23.20 Current drawn from the battery is i = net emf = 18 Ans. net resistance 6 = 3A Kirchhoff’s Laws Many electric circuits cannot be reduced to simple series-parallel combinations. For example, two circuits that cannot be so broken down are shown in Fig. 23.21. R1 E1 A B D A B C R2 E2 R1 R2 R3 R4 D C E2 E3 R3 E E1 F E F (a) I G R5 H (b) Fig. 23.21 However, it is always possible to analyze such circuits by applying two rules, devised by Kirchhoff in 1845 and 1846 when he was still a student. First there are two terms that we will use often. Junction A junction in a circuit is a point where three or more conductors meet. Junctions are also called nodes or branch points. For example, in Fig. (a) points D and C are junctions. Similarly, in Fig. (b) points B and F are junctions. Loop A loop is any closed conducting path. For example, in Fig. (a) ABCDA, DCEFD and ABEFA are loops. Similarly, in Fig. (b), CBFEC, BDGFB are loops.

20 — Electricity and Magnetism Kirchhoff’s rules consist of the following two statements : Kirchhoff’s Junction Rule The algebraic sum of the currents into any junction is zero. That is, Σ i=0 junction This law can also be written as, “the sum of all the currents directed towards a i1 i2 point in a circuit is equal to the sum of all the currents directed away from that i3 point.” i4 Thus, in figure i1 + i2 = i3 + i4 Fig. 23.22 The junction rule is based on conservation of electric charge. No charge can accumulate at a junction, so the total charge entering the junction per unit time must equal to charge leaving per unit time. Charge per unit time is current, so if we consider the currents entering to be positive and those leaving to be negative, the algebraic sum of currents into a junction must be zero. Kirchhoff’s Loop Rule The algebraic sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero. That is, Σ ∆V = 0 closed loop Kirchhoff’s second rule is based on the fact that the A E B A E B electrostatic field is conservative in nature. This result states that there is no net change in electric potential Path Path around a closed path. Kirchhoff’s second rule applies only ∆V = VB – VA = +E ∆V = VB – VA = –E for circuits in which an electric potential is defined at each Fig. 23.23 point. This criterion may not be satisfied if changing electromagnetic fields are present. In applying the loop rule, we need sign conventions. First assume a direction for the current in each branch of the circuit. Then starting at any point in the circuit, we imagine, travelling around a loop, adding emf’s and iR terms as we come to them. When we travel through a source in the direction from – to +, the emf is considered to be positive, when we travel from + to –, the emf is considered to be negative. When we travel through a resistor in the same direction as the assumed current, the iR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the iR term is positive because this represents a rise of potential. R i R i A B B A Path Path ∆V = VB – VA = –iR ∆V = VB – VA = +iR Fig. 23.24

Chapter 23 Current Electricity — 21 Note It is advised to write H (for higher potential) and L (for lower potential) across all the batteries and resistances of the loop under consideration while using the loop law. Then write – while moving from H to L and + for L to H. Across a battery write H on positive terminal and L on negative terminal. Across a resistance keep in mind the fact that current always flows from higher potential (H) to lower potential (L). For example, in the loop shown in figure we have marked H and L across all batteries and resistances. Now let us apply the second law in the loop ADCBA. E1 R1 C B L HLH L E2 i H LH i A D R2 Fig. 23.25 The equation will be + i R2 − E2 + i R1 + E1 = 0 V Example 23.18 Find currents in different branches of the electric circuit shown in figure. A 4Ω B 2Ω C 2V 4V 6V F D 2Ω E 4Ω Fig. 23.26 HOW TO PROCEED In this problem there are three wires EFAB, BE and BCDE. Therefore, we have three unknown currents i1 , i2 and i3 . So, we require three equations. One equation will be obtained by applying Kirchhoff’s junction law (either at B or at E) and the remaining two equations, we get from the second law (loop law). We can make three loops ABEFA, ACDFA and BCDEB. But we have to choose any two of them. Initially, we can choose any arbitrary directions of i1 , i2 and i3 . Solution Applying Kirchhoff’s first law (junction law) at junction B, A H 4Ω L B H 2Ω L C i1 i3 H 1 2 H 2V L i2 L 6V L 4V H i1 L H i3 LH E 4Ω F D 2Ω Fig. 23.27 i1 = i2 + i3 …(i) Applying Kirchhoff’s second law in loop 1 ( ABEFA ), …(ii) – 4 i1 + 4 – 2 i1 + 2 = 0

22 — Electricity and Magnetism Applying Kirchhoff’s second law in loop 2 (BCDEB ), – 2i3 – 6 – 4 i3 – 4 = 0 …(iii) Solving Eqs. (i), (ii) and (iii), we get Ans. i1 = 1 A i2 = 8 A 3 i3 = – 5 A 3 Here, negative sign of i3 implies that current i3 is in opposite direction of what we have assumed. V Example 23.19 In example 23.18, find the potential difference between points F and C. HOW TO PROCEED To find the potential difference between any two points of a circuit you have to reach from one point to the other via any path of the circuit. It is advisable to choose a path in which we come across the least number of resistors preferably a path which has no resistance. Solution Let us reach from F to C via A and B, VF + 2 – 4 i1 – 2 i3 = VC ∴ VF – VC = 4 i1 + 2 i3 – 2 Substituting, i1 = 1 A and i3 = – 5 A, we get 3 VF – VC =– 4 volt Ans. 3 Here, negative sign implies that VF < VC . Internal Resistance (r ) and Potential Difference (V ) across the Terminals of a Battery The potential difference across a real source in a circuit is not equal to the emf of the cell. The reason is that charge moving through the electrolyte of the cell encounters resistance. We call this the internal resistance of the source, denoted by r.If this resistance behaves according to Ohm’s law r is constant and independent of the current i.As the current moves through r, it experiences an associated drop in potential equal to ir. Thus, when a current is drawn through a source, the potential difference between the terminals of the source is V = E – ir This can also be shown as below. Er iB A Fig. 23.28 VA – E + ir =VB or VA – VB = E – ir

Chapter 23 Current Electricity — 23 The following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a battery), then V = E + ir (ii) V = E, if the current through the cell is zero. (iii) V = 0, if the cell is short circuited. This is because current in the circuit i= E Short r circuited or E = ir Er ∴ E – ir = 0 Fig. 23.29 or V = 0 Thus, we can summarise it as follows : Er V = E – ir or V < E i Er V = E + ir or V > E i Er V =E if i = 0 i = E V =0 if short circuited r 4Ω Er 2A + r 12 V 2Ω 12 V ir = 4V 8 V E = 12V iR = 8V O Fig. 23.30 Potential rise and fall in a circuit.

24 — Electricity and Magnetism Extra Points to Remember ˜ In figure (a) : There are eight wires and hence, will have eight currents or A 1 B eight unknowns. The eight wires are AB, BC, CE, EA, AD, BD, CD and ED. 4 2 Number of independent loops are four. Therefore, from the second law we C can make only four equations. Total number of junctions are five (A, B, C, D E D B 3 and E). But by using the first law, we can make only four equations (one (a) less). So, the total number of equations are eight. A 1 In figure (b) : Number of wires are six (AB, BC, CDA, BE, AE and CE). 3E 2 Number of independent loops are three so, three equations can be obtained from the second law. Number of junctions are four (A, B, C and E) so, we can D C make only three (one less) equations from the first law. But total number of (b) equations are again six. Fig. 23.31 ˜ Short circuiting : Two points in an electric circuit directly connected by a R1 conducting wire are called short circuited. Under such condition both points R2 E are at same potential. Fig. 23.32 For example, resistance R1 in the adjoining circuit is short circuited, i.e. potential difference across it is zero. Hence, no current will flow through R1 and the current through R2 is therefore, E/R2. ˜ Earthing : If some point of a circuit is earthed, then its potential is taken to 6V E be zero. D 2Ω B For example, in the adjoining figure, 3V C VA = VB = 0 VF = VC = VD = – 3 V FA 4Ω VE = – 9 V ∴ VB – VE = 9 V Fig. 23.33 or current through 2 Ω resistance is (from B to E) VB – VE or 9 A 22 Similarly, VA – VF = 3 V and the current through 4 Ω resistance is VA – VF or 3 A (from A to F) 44 ˜ For a current flow through a resistance there must be a potential difference A across it but between any two points of a circuit the potential difference may be zero. 1V 2Ω 1V For example, in the circuit, 2Ω i net emf = 3 V and net resistance = 6 Ω ∴ current in the circuit, i = 3 = 1 A B C 62 2Ω 1V VA – VB VA + 1– 2 × 1 = VB or VA – VB = 0 Fig. 23.34 2 or by symmetry, we can say that VA = VB = VC So, the potential difference across any two vertices of the triangle is zero, while the current in the circuit is non-zero.

Chapter 23 Current Electricity — 25 ˜ Distribution of current in parallel connections : When more than i1 R one resistances are connected in parallel, the potential difference i i2 2R across them is equal and the current is distributed among them in 3R inverse ratio of their resistance as i3 i=V R Fig. 23.35 or i ∝ 1 for same value of V R e.g. in the figure, i1 : i2 : i3 = 1 : 1 : 1 = 6: 3:2 R 2R 3R ∴ i1 =  6  i = 6i  3+  11  6 + 2  i2 =  3  i = 3 i  3  11  6 + + 2  and i3 =  2  i = 2 i  3  11  6 + + 2  Note In case of only two resistances, i1 = R2 i2 R1 ˜ Distribution of potential in series connections : When more R 2R 3R V1 V2 V3 than one resistances are connected in series, the current through V them is same and the potential is distributed in the direct ratio of i Fig. 23.36 their resistance as V = iR or V ∝ R for same value of i. For example in the figure, V1 : V2 : V3 = R : 2R : 3R = 1: 2 : 3 ∴ V1 =  1  V = V  2+  6  1 + 3  V2 =  2  V = V  2 3 3  1 + + and V3 =  3  V = V  2+ 3 2  1 + V Example 23.20 In the circuit shown in figure, E1 r1 E2 r2 R Fig. 23.37 E1 = 10 V , E2 = 4 V , r1 = r2 = 1 Ω and R = 2 Ω. Find the potential difference across battery 1 and battery 2.

26 — Electricity and Magnetism Solution Net emf of the circuit = E1 – E2 = 6 V E1 r1 E2 r2 Total resistance of the circuit = R + r1 + r2 = 4 Ω ∴ Current in the circuit, i = net emf = 6 = 1.5 A i total resistance 4 V1 V2 Now, V1 = E1 – ir1 = 10 – (1.5) (1) R = 8.5 V Fig. 23.38 Ans. and V2 = E2 + ir2 = 4 + (1.5) (1) Ans. = 5.5 V INTRODUCTORY EXERCISE 23.6 1. Find the current through 2 Ω and 4 Ω resistance. 2Ω 10V 4Ω 2. In the circuit shown in figure, find the potentials of A, B, C and D and C Fig. 23.39 B the current through 1 Ω and 2 Ω resistance. 1Ω 2V 5V 10 V D A 2Ω Fig. 23.40 3. For what value of E the potential of A is equal to 15V 1Ω A E 2 Ω B the potential of B? 5Ω Fig. 23.41 4. Ten cells each of emf 1 V and internal resistance 1 Ω are connected in series. In this arrangement, polarity of two cells is reversed and the system is connected to an external resistance of 2 Ω. Find the current in the circuit. 5. In the circuit shown in figure, R1 = R2 = R3 = 10 Ω. Find the R1 R3 10 V currents through R1 and R2. R2 10 V Fig. 23.42

Chapter 23 Current Electricity — 27 23.10 Heating Effects of Current An electric current through a resistor increases its thermal energy. Also, there are other situations in which an electric current can produce or absorb thermal energy. Power Supplied or Power Absorbed by a Battery When charges are transported across a source of emf, their potential energy changes. If a net charge ∆ q moves through a potential difference E in a time ∆ t, the change in electric potential energy of the charge is E ∆ q. Thus, the source of emf does work, ∆W = E ∆q Dividing both sides by ∆t, then taking the limit as ∆t → 0, we find dW = E dq dt dt By definition, dq = i, the current through the battery and dW = P, the power output of (or input to) the dt dt battery. Hence, P = Ei The quantity P represents the rate at which energy is transferred from a discharging battery or to a charging battery. In Fig. 23.43, energy is transferred from the source at a rate Ei Ei Fig. 23.43 In Fig. 23.44, energy is transferred to the source at a rate Ei Ei Fig. 23.44 Power dissipated across a resistance Now, let’s consider the power dissipated in a conducting element. Suppose it has a resistance R and the potential difference between its ends is V. In moving from higher to lower potential, a positive charge ∆ q loses energy ∆U =V ∆ q. This electric energy is absorbed by the conductor through collisions between its atomic lattice and the charge carriers, causing its temperature to rise. This effect is commonly called Joule heating. Since, power is the rate at which energy is transferred, we have, P = ∆U =V ⋅ ∆q =V i ∆ t ∆t ∴ P =V i which with the help of equation V = iR can also be written in the forms, P = i2R or V2 P= R

28 — Electricity and Magnetism Power is always dissipated in a resistance. With this rate, the heat V i produced in the resistor in time t is HL H = Pt or H =Vit = i2Rt = V2 R t Fig. 23.45 R Joule heating occurs whenever a current passes through an element that has resistance. To prevent the overheating of delicate electronic components, many electric devices like video cassette recorders, televisions and computer monitors have fans in their chassis to allow some of the heat produced to escape. Extra Points to Remember ˜ We have seen above that power may be supplied or consumed by a battery. It depends on the direction of current. Ei Fig. 23.46 In the above direction of current power is supplied by the battery (= Ei ) Ei Fig. 23.47 In the opposite direction of current shown in Fig. 23.47, power is consumed by the battery. This normally happens during charging of a battery. ˜ A resistance always consumes power. It does not depend on the direction of current. i i or Fig. 23.48 In both cases shown in figure, power is only consumed and this power consumed is given by the formula. P = i 2R = V2 = Vi R In the above equations V and i are the values across a resistance in which we wish to find the power consumed. ˜ In any electrical circuit, law of conservation of energy is followed. Net power supplied by all batteries of the circuit = net power consumed by all resistors in the circuit. V Example 23.21 In the circuit shown in figure, find 10V 4V i 3Ω Fig. 23.49 (a) the power supplied by 10 V battery (b) the power consumed by 4 V battery and (c) the power dissipated in 3 Ω resistance.

Chapter 23 Current Electricity — 29 Solution Net emf of the circuit = (10 – 4 ) V = 6 V Ans. ∴ Current in the circuit Ans. i = net emf = 6 = 2 A Ans. total resistance 3 (a) Power supplied by 10 V battery = Ei = (10) (2) = 20 W (b) Power consumed by 4 V battery = Ei = (4 ) (2) = 8 W (c) Power consumed by 3 Ω resistance = i2 R = (2)2 (3) = 12 W Note Here, we can see that total power supplied by 10 V battery (i.e. 20 W) = power consumed by 4 V battery and 3 Ω resistance. Which proves that conservation of energy holds good in electric circuits also. V Example 23.22 In the circuit shown in figure, find the heat developed across each resistance in 2 s. 6Ω 3Ω 3Ω 5Ω 20 V Fig. 23.50 Solution The 6 Ω and 3 Ω resistances are in parallel. So, their combined resistance is 1 = 1+ 1= 1 R 632 or R = 2 Ω The equivalent simple circuit can be drawn as shown. 3Ω 2Ω V i 5Ω 20 V Fig. 23.51 Current in the circuit, i = net emf = 20 = 2 A total resistance 3 + 2 + 5 V = iR = (2) (2) = 4 V i.e. Potential difference across 6 Ω and 3 Ω resistances are 4 V. Now, H 3 Ω (which is connected in series) = i2 Rt = (2)2 (3) (2) = 24 J H6Ω =V2 t = (4)2 (2) = 16 J R 6 3

30 — Electricity and Magnetism H3Ω (which is connected in parallel) =V2 t = (4)2 (2) = 32 J R 3 3 and H 5 Ω = i 2 Rt = (2)2 (5) (2) = 40 J Ans. INTRODUCTORY EXERCISE 23.7 1. In the circuit shown in figure, a 12 V battery with unknown internal resistance r is connected to another battery with unknown emf E and internal resistance 1 Ω and to a resistance of 3 Ω carrying a current of 2 A. The current through the rechargeable battery is 1 A in the direction shown. Find the unknown current i, internal resistance r and the emf E. 12 V r i E 1Ω 1A 3Ω 2A Fig. 23.52 2. In the above example, find the power delivered by the 12 V battery and the power dissipated in 3 Ω resistor. 23.11 Grouping of Cells Cells are usually grouped in the following three ways : Series Grouping Suppose n cells each of emf E and internal resistance r are connected in series as shown in figure. Er Er Er i R Fig. 23.53 Then, Net emf = nE Total resistance = nr + R ∴ net emf or i = nE Current in the circuit, i = nr + R total resistance Note If polarity of m cells is reversed, then equivalent emf = (n – 2m) E, while total resistance is still nr + R ∴ i = (n – 2m) E nr + R

Chapter 23 Current Electricity — 31 Parallel Grouping Here, three cases are possible. Case 1 When E and r of each cell has same value and positive terminals of Er Er all cells are connected at one junction while negative at the other. Er r i i In this situation, the net emf is E. The net internal resistance is as n R Fig. 23.54 n resistances each of r are in parallel. Net external resistance is R. Therefore, total resistance is  r + R  and so the current in the circuit will be,  n  Net emf or i = E i= R + r/n Total resistance Note A comparison of series and parallel grouping reveals that to get maximum current, cells must be connected in series if effective internal resistance is lesser than external and in parallel if effective internal resistance is greater than external. Case 2 If E and r of each cell are different but still the positive terminals of all cells are connected at one junction while negative at the other. A E1 r1 F i1 E2 r2 E i2 B i3 E3 r3 ii R D C Fig. 23.55 Applying Kirchhoff’s second law in loop ABCDEFA, E1 – iR – i1r1 = 0 or i1 =– i R + E1 …(i) r1 r1 …(ii) Similarly, we can write i2 =– i R + E2 r2 r2 … … …… Adding all above equations, we have (i1 + i2 +… + in ) = – iR Σ 1 + Σ E  r   r  But i1 + i2 +… + in = i ∴ i = – iR Σ 1 + Σ  E   r   r 

32 — Electricity and Magnetism ∴ i = Σ (E / r) = Σ (E / r)/ Σ (1/ r) = Eeq 1 + R Σ (1/ r) {1/ Σ (1/ r)} + R Req where, E eq = Σ (E /r) 1 Σ (1/ r) and Req = R + Σ (1/ r) From the above expression, we can see that i = E if n cells of same emf E and internal R + r/n resistance r are connected in parallel. This is because, Σ (E / r) = nE / r and Σ (1/ r) = n/ r ∴ i = nE / r 1 + nR / r Multiplying the numerator and denominator by r / n, we have i= E R + r/n Exercise In parallel grouping (Case 2) prove that, Eeq = E if E1 = E2 = K = E and r1 = r2 = K = r Case 3 This is the most general case of parallel grouping in which E and r of different cells are different and the positive terminals of few cells are connected to the negative terminals of the others as shown figure. E1 r1 i1 i2 E2 r2 i3 E3 r3 i i R Fig. 23.56 Kirchhoff’s second law in different loops gives the following equations : E1 – iR – i1r1 = 0 or i1 = E1 – iR …(i) r1 r1 …(ii) …(iii) – E2 – iR – i2r2 = 0 or i2 = – E2 – iR r2 r2 Similarly, i3 = E3 – iR r3 r3 Adding Eqs. (i), (ii) and (iii), we get i1 + i2 + i3 = (E1 / r1 ) – (E2 / r2 ) + (E3 / r3 ) – iR (1/ r1 + 1/ r2 + 1/ r3 ) or i [1 + R (1/ r1 +1/ r2 +1/ r3 )] = (E1/ r1 ) – (E2/ r2 ) + (E3/ r3 )

Chapter 23 Current Electricity — 33 ∴ i = (E1 / r1 ) – (E2 / r2 ) + (E3 / r3 ) 1 + R (1/ r1 +1/ r2 +1/ r3 ) or i = (E1 / r1 – E2 / r2 + E3 / r3 ) / (1/ r1 + 1/ r2 + 1/ r3 ) R +1 / (1/ r1 +1/ r2 +1/ r3 ) Mixed Grouping The situation is shown in figure. Er ii R Fig. 23.57 There are n identical cells in a row and number of rows are m. Emf of each cell is E and internal resistance is r. Treating each row as a single cell of emf nE and internal resistance nr, we have Net emf = nE nr Total internal resistance = m Total external resistance = R ∴ Current through the external resistance R is i = nE + nr R m This expression after some rearrangements can also be written as i= mnE ( mR – nr ) 2 + 2 mnrR If total number of cells are given then mn is fixed. E, r and R are also given, we have liberty to arrange the given number of cells in different rows. Then in the above expression the numerator nmE and in the denominator 2 mnrR all are fixed. Only the square term in the denominator is variable. Therefore, i is maximum when, mR = nr or R = nr m or total external resistance = total internal resistance Thus, we can say that the current and hence power transferred to the load is maximum when load resistance is equal to internal resistance. This is known as maximum power transfer theorem.

34 — Electricity and Magnetism Extra Points to Remember ˜ Regarding maximum current in the circuit or maximum power consumed by the external resistance R, there are three special cases. One we have discussed above, where we have to arrange the cells in such a manner that current and power in the circuit should be maximum. And this happens when we arrange the cells in such a manner that total internal resistance comes out to be equal to total external resistance. Rest two cases are discussed below. E, r iR Fig. 23.58 In the figure shown, i= E R+ r  E  2  R+ PR = i 2R =  r  R  (i) Now if r is variable. E and R are fixed, then i and PR both are maximum when r = 0. (ii) If R is variable. E and r are fixed. Then, current in the circuit is maximum when R = 0. But PR will be maximum, when R = r or external resistance = internal resistance V Example 23.23 Find the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in figure. 10V 2Ω 6V 1Ω 4V 2Ω Fig. 23.59 Solution The given combination consists of two batteries in parallel and resultant of these two in series with the third one. For parallel combination we can apply, E1 – E2 10 – 4 r1 + r2 2 2 E eq = 1 1 = 1 1 = 3V r1 r2 + 22 Further, 1 = 1 + 1 = 1+ 1=1 req r1 r2 2 2 ∴ req = 1 Ω

Chapter 23 Current Electricity — 35 Now this is in series with the third one, i.e. 6V 3V 1Ω 1Ω Fig. 23.60 The equivalent emf of these two is (6 – 3) Vor 3 V and the internal resistance will be (1+ 1) or 2Ω E = 3V r = 2Ω Fig. 23.61 V Example 23.24 In the circuit shown in r1 + − figure E1 = 3V, E2 = 2V, E3 = 1 V and E1 R = r1 = r2 = r3 = 1 Ω. r2 + − (JEE 1981) R B A (a) Find the potential difference between the E2 points A and B and the currents through each r3 + − branch. E3 (b) If r2 is short-circuited and the point A is Fig. 23.62 connected to point B, find the currents through E1 , E2 , E3 and the resistor R. Solution (a) Equivalent emf of three batteries would be E eq = Σ(E / r) = (3/ 1+ 2/ 1+ 1/ 1) = 2V Σ(1/ r) (1/ 1+ 1/ 1+ 1/ 1) Further r1 , r2 and r3 each are of 1 Ω. Therefore, internal resistance of the equivalent battery will be 1 Ω as all three are in parallel. 3 AR 2V 1 Ω B 3 Fig. 23.63 The equivalent circuit is therefore shown in the figure. Since, no current is taken from the battery. VAB = 2 V (FromV = E − i r) Further, V A − VB = E1 − i1 r1 ∴ i1 = VB −VA + E1 = − 2+ 3 = 1A r1 1 Similarly, i2 = VB −VA + E2 = − 2+ 2= 0 r2 1 and i3 = VB −VA + E3 = − 2+ 1= − 1A r3 1

36 — Electricity and Magnetism (b) r2 is short circuited means resistance of this 1Ω 3V branch becomes zero. Making a closed circuit i1 with a battery and resistance R. Applying R =1Ω i2 2V Kirchhoff’s second law in three loops so formed. i3 1 Ω 3 − i1 − (i1 + i2 + i3 ) = 0 …(i) (i1+i2+i3) 1V 2 − (i1 + i2 + i3 ) = 0 …(ii) 1 − i3 − (i1 + i2 + i3 ) = 0 …(iii) Fig. 23.64 From Eq. (ii) i1 + i2 + i3 = 2 A ∴ Substituting in Eq. (i), we get i1 = 1A Substituting in Eq. (iii), we get i3 = − 1A ∴ i2 = 2 A INTRODUCTORY EXERCISE 23.8 1. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistances r1 and r2respectively, with polarities as shown in figure r2 V2 +– AB r1 V1 Fig. 23.65 2. Find the net emf of the three batteries shown in figure. 2V 1Ω 4V 0.5 Ω 1Ω 6V Fig. 23.66 3. Find the equivalent emf and internal resistance of the arrangement shown in figure. 10V 1Ω 4V 2Ω 6V 2Ω Fig. 23.67

Chapter 23 Current Electricity — 37 23.12 Electrical Measuring Instruments So far we have studied about current, resistance, potential difference and emf. Now, in this article we will study how these are measured. The basic measuring instrument is galvanometer, whose pointer shows a deflection when current passes through it. A galvanometer can easily be converted into an ammeter for measuring current, into a voltmeter for measuring potential difference. For accurate measurement of potential difference or emf, a potentiometer is more preferred. Resistances are accurately measured by using post office box or meter bridge which are based on the principle of “Wheatstone bridge”. All these are discussed here one by one in brief. Galvanometer Many common devices including car instrument panels, battery chargers measure potential difference, current or resistance using d’Arsonval Galvanometer. It consists of a pivoted coil placed in the magnetic field of a permanent magnet. Attached to the coil is a spring. In the equilibrium position, with no current in the coil, the pointer is at zero and spring is relaxed. When there is a current in the coil, the magnetic field exerts a torque on the coil that is proportional to current. As the coil turns, the spring exerts a restoring torque that is proportional to the angular displacement. Thus, the angular deflection of the coil and pointer is directly proportional to the coil current and the device can be calibrated to measure current. The maximum deflection, typically 90° to 120° is called full scale deflection. The essential electrical characteristics of the galvanometer are the current ig required for full scale deflection (of the order of 10 µA to 10 mA) and the resistance G of the coil (of the order of 10 to 1000 Ω). The galvanometer deflection is proportional to the current in the coil. If the coil obeys Ohm’s law, the current is proportional to potential difference. The corresponding potential difference for full scale deflection is V = igG Ammeter A current measuring instrument is called an ammeter. A galvanometer can be converted into an ammeter by connecting a small resistance S (called shunt) in parallel with it. Suppose we want to convert a galvanometer with full scale ig G – current ig and coil resistance G into an ammeter with full scale i b reading i. To determine the shunt resistance S needed, note S that, at full scale deflection the total current through the +a i – ig parallel combination is i, the current through the galvanometer Fig. 23.68 is ig and the current through the shunt is i – ig. The potential difference Vab (=Va – Vb ) is the same for both paths, so igG = (i – ig ) S ∴ S =  i ig  G  – ig 

38 — Electricity and Magnetism Voltmeter A voltage measuring device is called a voltmeter. It measures the potential difference between two points. A galvanometer can be converted into a voltmeter by connecting a high resistance (R) in series with it. The whole assembly called the voltmeter is connected in parallel between the points where potential difference has to be measured. For a voltmeter with full scale reading V, we need a series resistor R such that G ig R – ig + ia Circuit bi element V Fig. 23.69 V = ig (G + R ) or R = V – G ig Extra Points to Remember ˜ Conversion of galvanometer into an ammeter. S (i) A galvanometer is converted into an ammeter by connecting a i – ig G low resistance (called shunt) in parallel with galvanometer. i This assembly (called ammeter) is connected in series in the wire in which current is to be found. Resistance of an ideal ig ammeter should be zero. Fig. 23.70 In parallel, current distributes in the inverse ratio of resistance. Therefore, S = ig G i − ig ∴ S = shunt =  ig  G  i − ig  (ii) Resistance of an ammeter is given by 1 = 1 + 1 ⇒ A = GS AG S G+S

Chapter 23 Current Electricity — 39 (iii) The reading of an ammeter is always lesser than actual current in the circuit. R R A i¢ i i¢ GS i¢ EE (a) (b) Fig. 23.71 For example, in Fig. (a), actual current through R is …(i) i=E …(ii) R while the current after connecting an ammeter of resistance A  = GS  in series with R is  G + S i′ = E R+ A From Eqs. (i) and (ii), we see that i ′ < i and i ′ = i when A = 0. ig R i.e. resistance of an ideal ammeter should be zero. G ˜ Conversion of a galvanometer into a voltmeter (i) A galvanometer is converted into a voltmeter by connecting Fig. 23.72 a high resistance in series with galvanometer. The whole assembly called voltmeter is connected in parallel across the two points between which potential difference is to be found. Resistance of an ideal voltmeter should be infinite. V = Ig (G + R) ∴ R = high resistance required in series = V −G ig (ii) Resistance of a voltmeter is RV = R + G (iii) The reading of a voltmeter is always lesser than the true value. For example, if a current i is passing through a resistance r, the actual value is r r R i G ii RV V Fig. 23.73 V = ir …(i) Now, if a voltmeter of resistance RV (= G + R) is connected across the resistance r, the new value will be V′ = i × (rRV ) or V′ = ir r …(ii) r + RV 1+ RV From Eqs. (i) and (ii), we can see that, V′ < V and V′ = V if RV = ∞ Thus, resistance of an ideal voltmeter should be infinite.