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th 6 Year Physics Higher Level Kieran Mills Electricity Notes (2015/16) No part of this publication may be copied, reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from The Dublin School of Grinds. Ref: 6/phy/h/lc/Electricity notes 6-HOUR CRASH COURSES MAY & JUNE 2016 %01.! čƫ+*+)%/ƫ0!$!.ƫ¨*8*ƫ1. +' The final push for CAO points... $!ƫ 1(%*ƫ $++(ƫ +"ƫ .%* /ƫ %/ƫ .1**%*#ƫ ćġ$+1.ƫ ./$ƫ +1./!/ƫ 0ƫ 0$!ƫ !* ƫ +"ƫ 5ƫ * ƫ 0$!ƫ !#%**%*#ƫ +"ƫ 1*!ċƫ $!/!ƫ +1./!/ƫ #%2!ƫ /01 !*0/ƫ 0$!ƫ !/0ƫ ,+//%(!ƫ 2*0#!ƫ /ƫ 0$!5ƫ ,.!,.!ƫ "+.ƫ 0$!ƫ ((ġ%),+.0*0ƫ 00!ƫ 4)%*0%+*/ċƫ *!ƫ (/0ƫ +1*!ƫ +"ƫ !""+.0ƫ +1( ƫ )'!ƫ ((ƫ 0$!ƫ %""!.!*!ċ !.!ƫ%/ƫ$+3ƫ0$!/!ƫ+1./!/ƫ3%((ƫ !*!"%0ƫ5+1č • They will offer students one last opportunity to avail of expert teaching before the State Examinations • They will provide students with a final boost of confidence before exam day • They will give students an exam strategy plan to help them maximise their grade on the day +0!: At these courses our teachers will predict what questions are most likely to appear on your exam paper. These questions will be covered in detail and our teachers will provide you with model A1 answers. čƫĺāćĀƫƫ To book, call us on 01ġ442 4442 or book online at www.dublinschoolofgrinds.ie ./$ƫ+1./!s Timetable 6th Year Subject Date Time Accounting Level H Sunday 29th May 9am - 3pm Biology H Saturday 28th May 9am - 3pm Business H Sunday 29th May 2pm - 8pm Chemistry H Saturday 4th June 9am - 3pm Economics H Saturday 28th May 9am - 3pm English H Sunday 29th May 9am - 3pm English H Saturday 4th June 9am - 3pm French H Saturday 4th June 9am - 3pm Geography H Saturday 28th May 9am - 3pm Irish H Saturday 4th June 9am - 3pm Maths Paper 1 H Saturday 4th June 9am - 3pm Maths Paper 2 H Sunday 5th June 9am - 3pm Maths O Saturday 28th May 9am - 3pm Maths O Saturday 4th June 9am - 3pm Physics H Saturday 28th May 9am - 3pm Spanish H Sunday 5th June 9am - 3pm Date Time 3rd Year Subject Level Business Studies H Sunday 5th June 9am - 3pm English H Sunday 5th June 9am - 3pm French H Sunday 29th May 9am - 3pm Irish H Sunday 29th May 9am - 3pm Maths H Sunday 29th May 9am - 3pm Science H Saturday 4th June 9am - 3pm H = Higher O = Ordinary (!/!ƫ*+0!ƫ0$0ƫ((ƫ+1./!/ƫ3%((ƫ0'!ƫ,(!ƫ0ƫ +1.ƫ!.*%*#ƫ!*0.!ƫ0ƫ$!ƫ.%).5ƫ$++(ƫ%*ƫ 0(* /Čƫ0%((+.#*Čƫ+ċƫ1(%*ċ Electricity Chapter 1. Charges 1.1 Charging Materials........................................................................................... 2 1.2 The Gold Leaf Electroscope (GLE)................................................................. 4 1.3 Distribution of Charge on Conductors............................................................. 5 Chapter 2. Electric Fields 2.1 Coulomb’s Law................................................................................................ 7 2.2 Electric Fields.................................................................................................. 9 2.3 Potential........................................................................................................... 12 Numerical Problems.............................................................................................. 15 Chapter 3. Capacitance 3.1 Defining Capacitance....................................................................................... 19 3.2 Parallel Plate Capacitor.................................................................................... 19 3.3 Energy in a Charged Capacitor........................................................................ 20 Numerical Problems.............................................................................................. 23 Chapter 4. Electric Current 4.1 Sources of EMF and Electric Current.............................................................. 25 4.2 Ohm’s Law....................................................................................................... 27 4.3 Conduction in Materials................................................................................... 28 4.4 Resistance........................................................................................................ 35 4.5 Potential Divider.............................................................................................. 43 4.6 Effects of an Electric Current.......................................................................... 44 4.7 Domestic Circuits............................................................................................. 47 Numerical Problems.............................................................................................. 50 Chapter 5. Electromagnetism 5.1 Magnetism........................................................................................................ 58 5.2 Current in a magnetic field............................................................................... 60 5.3 Electromagnetic induction............................................................................... 64 5.4 Alternating current........................................................................................... 68 5.5 Mutual and self induction................................................................................ 69 Numerical Problems.............................................................................................. 73 Leaving Cert Questions (2002 - 2015)................................................................ 76 © Dublin School of Grinds Page 1 Kieran Mills & Tony Kelly Chapter 1 • Charges 1.1 Charging Materials There is an electric force between all charged particles. This force can be attractive or repulsive. Protons and electrons are examples of such charged particles whereas a neutron is not a charged particle. Protons (p) and electrons (e) each have a different type of electric charge. A proton has a positive charge whereas an electron has a negative charge. e 1 1 e Electrical repulsion e 1 +1 p Electrical attraction Charges interact with one another according to the following rule: Like charges repel; unlike charges attract. An atom consists of a central core called a nucleus made up of protons and neutrons surrounded by orbiting electrons. The nucleus is a central positive core surrounded by negative charges. Electrons + Normally atoms are electrically neutral overall which means they have the same number of electrons as protons. It is possible for atoms to lose or gain electrons. An atom losing electrons will have an overall positive charge whereas one gaining electrons has an overall negative charge. Atoms with an overall charge are called ions. Nucleus Definition: Ions are atoms with an overall charge. A positive ion has lost electrons whereas a negative ion has gained electrons. Electric charge is measured in Coulombs (C). Conductors and Insulators Materials can be divided into conductors and insulators. Metals are good conductors. They consist of atoms whose outermost electrons are not very strongly attached to the atom and they wander at random throughout the metal. These are called free electrons. + + + + Fixed positive ion + + + + + Electron In conductors negatively charged electrons are wandering at random among fixed positive ions. In insulators all the electrons are firmly attached to the atoms. They do not wander among the atoms. Definition: Conductors allow charge to flow; insulators do not. Charged Rods Materials can be made to gain or lose electrons by rubbing them with another material. This is called charging by friction. If a polythene rod is rubbed with a cloth it acquires a negative charge. If a perspex rod is rubbed with a cloth it acquires a positive charge. Both these materials are insulators. This means the charged rods can be held by hand and the charge does not flow away. © Dublin School of Grinds Page 2 Kieran Mills & Tony Kelly Polythene rods become negative when rubbed by a cloth whereas perspex rods become positive. Demonstration: Force between charges The apparatus is set up as shown with one piece of polythene suspended by a nylon thread. Each piece of polythene is charged by rubbing with a cloth. A is repelled by B showing the force of repulsion between charges. The experiment was repeated with: Perspex (+) and perspex (+): Repulsion Polythene (−) and perspex (+): Attraction Nylon thread Repulsion A -- --- --- --- --- Polythene Rod -- --- --- --- --- Polythene Rod B Charging Bodies There are two ways that a body can be charged: (1) By contact (as in the polythene and perspex rods) (2) By induction. (1) By contact: Electrons can be transferred from one material to another simply by touching. A charged body placed in contact with a neutral body will transfer charge to the neutral body. If the body is a good conductor the charge will spread to all parts of the surface because like charges repel. (2) By induction: Bring a negatively charged rod close to a hollow metal sphere. Free electrons close to the rod will be repelled across the sphere. Demonstration: Charging by Induction B + + + + A The positive charge on the perspex rod induces a negative charge on face A of the metal sphere by +++ Perspex Rod attracting electrons in the metal towards it. This leaves a net positive charge on face B. The metal sphere is earthed so that electrons move from earth to neutralise the positive charge on B. B A B + + + + A +++ Perspex Rod +++ Perspex Rod The earth wire is then broken while the perspex rod is still held in place. The rod is now removed. The negative charge spreads out all over the surface. © Dublin School of Grinds Page 3 Kieran Mills & Tony Kelly A positively charged rod induced a negative charge on the sphere. The induced charge is always opposite to the inducing charge. Static Electricity in the home and industry Dust on TV screens Electrons hit the screen to produce a picture. As a result, the screen becomes charged attracting dust to it. Static on clothes Synthetic materials become charged when rubbed. As a result, sparks jump causing a crackling sound. Clothes tend to cling to each other because of the attraction between charges. Flour mills Static electricity producing sparks can be dangerous near the fine dust produced in flour mills. Explosions of dust can occur. Fuelling aircraft Aircraft flying through the air can build up lots of static charge which could cause explosions on landing. The charge is conducted harmlessly to ground through conducting rubber in the tyres. 1.2 The Gold Leaf Electroscope (GLE) This is a sensitive instrument which is used to investigate the nature of charges. Construction A metal cap attached to a thin flexible piece of gold foil is enclosed in a metal case with a glass front. The metal case stands on an insulated base and is connected to earth. A metal rod runs down through the centre of the case from which it is insulated. Metal cap Metal Rod Metal Case Glass front Metal leaf Earth connection Operation When charge is placed on the metal cap the leaves diverge due to the repulsion of like charges. The earthed metal case makes the instrument more sensitive due to the opposite charge induced on the inside of the case. The GLE measures the potential difference between the leaves and the case. The bigger the charge, the bigger the potential difference and therefore the bigger the deflection of the gold leaf. Uses of electroscope An electroscope can do the following: • Detect charge. • Indicate the approximate size of a charge. • Test whether a charge is + or −. • Test if an object is a conductor or insulator. • Indicate the size of a potential difference. Charging the GLE by induction Bring a positively charged rod close to the cap of an uncharged GLE. Earth the cap of the GLE by touching it. Break the earth connection by taking away your finger. Remove the rod. The GLE is negatively charged. The divergence of the leaves indicates the size of charge. © Dublin School of Grinds Page 4 Kieran Mills & Tony Kelly 1.3 Distribution of Charge on Conductors There are a number of ways in which a body can be charged. These methods include: 1. Charging by induction. 2. Using a Van de Graff Generator. Metal dome 3. Using a high voltage source. + + + + + The Van De Graff Generator C + + The Van de Graff generator is a machine for producing large + quantities of charge. It consists of a rubber belt which revolves + + + about two pulleys. The upper pulley is inside a hollow metal dome + + + while the lower pulley is attached to a motor which turns it. + Belt + The belt passes close to a sharply pointed comb at B which is + connected to a high tension source. Charge is deposited from the Motor driven B roller comb onto the belt where it is carried up to the second pulley. At C the charge leaves the belt and spreads out over the metal dome. All the charge resides on the surface of the generator’s dome. This is because like charges repel, moving to positions of maximum distance from every other charge. Therefore, there is no electric field inside the dome. Applications of this effect include the electrostatic shielding used in co-axial cables to transmit TV signals without interference. The distribution of charges around a conductor depends on the shape of the conductor. Charges tend to accumulate at sharper ends. A proof plane is a device which picks up charge from a surface. Now let it touch the cap of an uncharged GLE. The size of the divergence of the leaves is an indication of the charge density on the surface of the conductor. Insulating handle Metal disc Proof Plane Demonstration: Total charge resides on the outside of a metal object 1. A metal cylinder is charged by contact with the metal sphere of the Van de Graff generator. 2. A proof plane touches the inside surface of the cylinder and is then placed in contact with the cap of an + uncharged GLE. No divergence of + the leaves takes place showing that there is no charge inside the cylinder. + + 3. If the proof plane comes in contact with the outside surface of the metal cylinder and then in contact with the cap of an uncharged GLE, the leaves do diverge showing that charge resides on the outside of a metal object. © Dublin School of Grinds Page 5 + + + + + + + + Proof plane + + + + + GLE Kieran Mills & Tony Kelly Demonstration: Charges tend to accumulate at points 1. Two pear-shaped conductors are charged by induction and then separated. + ++ ++ + + + + + A + + + + + + + + B + + + ++ ++ + ++ ++ ++ + + + + + + A + + + + + + + + B + + + ++ ++ ++ + 2. A proof plane is placed at the non-pointed end of A and then placed in contact with the cap of an uncharged GLE. There is a small divergence of the leaves showing a low density of charge in this region. 3. Another proof plane is placed at the pointed end of B and is then placed in contact with the cap of an uncharged GLE. There is a large divergence of the leaves showing a high density of charge at the pointed end. At the pointed surface of a charged conductor, the high charge density produces an intense electric field near the point. The air around the point becomes ionised and these ions are attracted to or repelled from the point depending on the sign of the charge on the point. The result is that the charge on the point becomes neutralised and an electric wind is felt as ions are repelled. Air molecules Spike + ++ + + + ++ + + + + + + Ionised air Spike + ++ + + + ++ + + + + + + Electric wind Demonstration: Point Discharge This can be demonstrated by putting a metal spike in the dome of the Van de Graff generator. If a candle is held near the point of the spike, the flame is blown by the electric wind. Lightning Large amounts of static charges build up on clouds. Lightning occurs when this static discharges between clouds or from cloud to ground. This can cause damage to buildings. Lightning Conductor As a result of lightning, many high buildings have a large metal conductor connected to a large metal plate connected to the ground. A cloud with a large voltage passing overhead induces charges on the lightning rod. This causes point discharge to occur which reduces the voltage between cloud and rod thus lessening the chance of a strike. If a strike does occur the charge flows harmlessly to earth protecting the building. © Dublin School of Grinds Page 6 Kieran Mills & Tony Kelly Chapter 2 • Electric Fields 2.1 Coulomb’s Law Electrical charges exert forces on one another. Consider two charges, q1 and q2, a distance d apart. q1 q2 d Definition: Coulomb’s Law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Note: A point charge is one in which all the charge is concentrated at a point. Stated mathematically: F ∝ q1q2 qq ⇒ F = k 1 22 2 d d Experimentally the constant k is found to depend on the nature of the medium between the charges. To take this into account a medium is said to have a permittivity, denoted by e. For practical purposes the permittivity of air is taken to be the same as a vacuum and is denoted by 1 e0, and is called the permittivity of free space. The constant k is given the value: k = 4πε Formula for Coulomb’s Law: 1 q1q2 F= 4πε d 2 F: Force between charges (N) Formulae and Tables Book: Page 61 q: Quantity of Charge (C) Coulomb’s law (Electricity) d: Distance between charges (m) e: Permittivity of the medium (F m-1) The permittivity of free space constant Fundamental physical constants (Formulae and Tables Book: page 47) Permittivity of free space ε 0 = 8.854187817 ×10−12 F m −1 ≈ 8.85 ×10−12 F m −1 Note 1. Coulomb’s Law is an inverse square law. What does this mean? If the force is F at 1 m, then it is 1 4 F at 2 m, and it is 1 9 F at 3 m ...... therefore F ∝ 1 . d2 2. The permittivity of an medium e can be written as the permittivity of free space, e0, multiplied by the constant, er, the relative permittivity of the medium. ε = ε0 ×εr e: Permittivity of the medium (F m−1) e0: Permittivity of free space (F m−1) er: Relative permittivity of the medium © Dublin School of Grinds Page 7 Kieran Mills & Tony Kelly Example 1: Two point charges of +2 µC and −3 µC are 50 cm apart in air. Find the magnitude and direction of the force on the 2 µC charge. (ε0 = 8.85 × 10−12 F m−1) Solution Example 2: A negative charge of 2.5 nC and a negative charge of 4.0 nC are placed 50 mm apart in air. Find the resultant force on a positive charge of 2.0 nC placed 20 mm from the 2.5 nC negative charge and on the line between them. (ε0 = 8.85 × 10−12 F m−1) Solution © Dublin School of Grinds Page 8 Kieran Mills & Tony Kelly 2.2 Electric Fields If charges are brought close to each other there is a force between them. This is an example of an action-at-a-distance force. In Physics we use the idea of a force field to analyse these situations. Charges are visualised as being surrounded by electric fields. + When two charges are close to one another, it is the electric fields which interact. The field is represented by lines which, by convention, show the direction a positive charge would move, if free to do so. The electric field lines are directed away from a positive charge but directed towards a negative charge. Field patterns 1. Field between a positive and negative charge. The lines of force represent the direction a positive test charge would move. A positive test charge will be repelled by the positive charge and attracted by the negative charge. + 2. Field between two positive charges. The lines of force represent the direction a positive test charge would move. A positive test charge will be repelled by each of the positive charges. There is no electric field half-way between the two positive charges. + 3. Field due to two charged plates. The field is uniform between the plates. This means the field has the same strength at all points between the plates. This is indicated by the lines which are parallel and equally spaced. + + + + + + Definitions An electric field is a region in which electric charges at rest experience a force. An electric field line represents the direction a positive charge would move if free to do so. © Dublin School of Grinds Page 9 Kieran Mills & Tony Kelly Demonstration: Experiment to plot electric field lines Two electrodes of any required shape are connected to a high voltage supply. They are placed in a container of castor oil on the surface on which is spread some semolina (tiny particles of an insulating substance). When the electric supply is turned on, the particles of semolina line up along the field lines and the field patterns can be observed. Petri Dish + Castor oil + 5 kV DC Supply Electric Field Intensity Consider a single isolated charge Q. We want an expression for the strength of the electric field E around this charge at various distances d. Put a test charge q in the field and calculate the force F on it. F d q Q Definition: The electric field strength, E, at a point is defined as the force per unit positive charge at that point. Stated mathematically: E= F q F: Force (N) q: Charge (C) E: Electric field strength (intensity) (N C−1) Formulae and Tables Book: Page 61 Electric field strength (Electricity) Electric fields are described by drawing field lines. These lines are also called flux lines. The electric field strength (intensity), E, is a vector quantity, its direction being that of the force on a positive charge at the point in question. Example 3: A charge of 2 µC experiences a force of 40 N when placed at a point in an electric field. Calculate the electric field strength at that point. Solution Use Coulomb’s law to calculate the force F between the Qq two charges Q and q: F = k 2 d F Qq Q But E = ⇒ E = k 2 = k 2 q d q d Q E=k 2 d F d q Q E: Electric field strength (intensity) (N C−1) Q: Charge (C) d: Distance (m) © Dublin School of Grinds Page 10 Kieran Mills & Tony Kelly Example 4: Calculate the electric field strength at a distance of 15 cm from a 7 µC charge placed in a vacuum. (ε0 = 8.85 × 10−12 F m−1) Solution Example 5: What is the electric field strength half-way between a charge of +4 µC and a charge of +2 µC, if the distance between the charges is 20 cm? What is the force on a charge of 5 µC placed at this point? (ε0 = 8.85 × 10−12 F m−1) Solution Electric Fields Precipitators This is a device that removes dust and other small particles from dirty air. It charges the particles and then attracts them to metal plates of the opposite charge. Air purifiers and smoke removers in bars work on this principle. Xerography (The photocopier) A drum is charged electrostatically. An image of the document to be copied is focused on the drum. Effect of electric fields on integrated circuits Technicians working on integrated circuits (IC’s) must make sure they are earthed so that any static on their body does not flow through the delicate IC’s causing damage. They are usually connected to a wire connected to their wrist with a wrist band. © Dublin School of Grinds Page 11 Kieran Mills & Tony Kelly 2.3 Potential You can get information about the electric field around a charge by calculating the electric field strength E at any point. An alternative method is to consider the potential V. Consider a positive charge Q. Work must be done to carry a charge +q from point B to point A. If work has to be done to move a charge from one point to another then the points are said to be at different potentials. There is a potential difference between them which is denoted by VAB (potential difference p.d. between A and B). On the other hand, a charge +q placed at A will gain energy in moving from point A to point B. The electric field does a certain amount of work in moving +q from A to B. +Q +q A +Q B +q A B Definition: The potential difference V between two points is the work done in moving a unit charge from one point to the other. Stated mathematically: V: Potential difference between two points (Volts, V) W W: Work done in moving the charge between the points (J) V= q q: Charge being moved (C) Formulae and Tables Book: Page 61 Potential difference (Electricity) Definition: The potential difference between two points is 1 Volt if 1 Joule of work is done when a unit charge is brought from one point to the other. Absolute Potentials When a charge flows from A to B there is a potential difference between A and B. If the moving charge is positive then A is at a higher potential than B. Therefore a positive charge moves from points of higher to lower potential if free to do so. A negative charge will do the opposite. In general it is not necesary to know the absolute potential at A and B, just the difference between them. However, it is sometimes convenient to know the absolute potential at a point. To do this a zero of potential must be established. Once again consider a positive charge Q. At an infinite distance from Q (essentially a very large distance) the strength of the electric field E would be zero. Points at this distance are at zero potential. The absolute potential at A is defined as the work done in bringing a +q charge from infinity to A against the field. Definition: The absolute potential at a point is defined as the work done in bringing a positive charge from infinity to that point. In practical applications the potential of the earth is taken to be zero. This is because the earth is so large that it is unaffected by the flow of charges to and from it. © Dublin School of Grinds Page 12 Kieran Mills & Tony Kelly Example 6: The potential difference between two points is 100 kV. Find the work done when a charge of 3 µC is moved from one point to another. Solution Electron (Charge and mass) Fundamental physical constants (Formulae and Tables Book: page 46) −19 −19 Electronic charge e = 1.60217653 ×10 C ≈ 1.6 ×10 C Electron mass me = 9.1093826 ×10−31 kg ≈ 9.1×10−31 kg Example 7: The potential difference between two oppositely charged parallel plates 1 cm apart is 400 V. Find: (a) the force acting on a charge of 1 C when placed between the plates, (b) the electric field intensity between the plates, (c) the force acting on an electron put between the plates, (d) the potential energy lost by an electron if it moves from the negative to the positive plate, (e) the kinetic energy of an electron on arriving at the positive plate if its speed when released at the negative plate was zero, (f) the speed of the electron on arriving at the positive plate. (Constant: Charge on electron e = 1 ⋅ 6 ×10 Solution © Dublin School of Grinds −19 1 cm + + + + + 1C V = 400 V C; Mass of electron me = 9 ⋅1×10−31 kg ) Page 13 Kieran Mills & Tony Kelly Answers to Examples Example 1: 0.216 N towards the -3 mC charge Example 2: 3.25 ×10−5 N towards the -2.5 nC charge Example 3: 2 ×107 N C−1 Example 4: 2.8 ×106 N C−1 Example 5: 1.8 ×106 N C−1 , 9 N Example 6: 0.3 J Example 7: (a) 4 ×104 N (b) 4 ×104 N C−1 (c) 6.4 ×10−15 N (d) 6.4 ×10−17 J (e) 6.4 ×10−17 J (f) 1.2 ×107 m s −1 © Dublin School of Grinds Page 14 Kieran Mills & Tony Kelly Numerical Problems Constants: Permittivity of free space ε 0 = 8.85 ×10−12 F m −1 Charge on electron e = 1 ⋅ 6 ×10−19 C Mass of electron me = 9 ⋅1×10−31 kg Coulomb’s Law 1. Calculate the force that a charge of +4 C exerts on a charge of +5 C placed 4 m from it. 2. Two point charges of +3 µC and −4 µC are 60 cm apart in air. Find the magnitude and direction of the force on the 3 µC charge. 3. Calculate the force on a charge of +2 µC when placed a distance of 4 m form a negative charge of −8 µC in air. Is the force one of attraction or repulsion? 4. In a vacuum, a charge of +2 µC is 50 cm from a charge of −2 µC. What is the magnitude and direction of the force on the +2 µC charge? 5. Calculate the magnitude of the force between two point charges of 4.0 nC each, if the medium between them is air and the distance between them is 10 mm. 6. Calculate the force a 2 µC charge exerts on a −2 µC charge when 0.01 m apart in a vacuum. 7. Point charges of 40 µC and 20 µC are separated by a 3 mm thick sheet of plastic. If each charge exerts a force of 1 × 105 N on the other, calculate the permittivity of the plastic. 8. In a medium of permittivity 7 × 10−10 Fm−1 a charge of +2 µC is 60 cm from a charge of −4 µC. What is the magnitude and direction of the force on the +2 µC charge? 9. The force of repulsion between two identically charged particles is 0.4 N. If the charges are 5 cm apart in air, what is the size of the charge on each? 10. Three charges are arranged in air as shown. Find the magnitude and direction of the force on the 3 µC charge. 1 C 30 cm +3 C 50 cm 3 C 11. A negative charge of 2.5 nC and a positive charge of 6.0 nC are placed 60 mm apart in air. Find the resultant force on a positive charge of 3.0 nC placed 20 mm from the negative charge and on the line between them. 12. The three charges shown are collinear. A +3 µC charge is placed at A and a −4 µC charge is placed at B. What charge must be placed at C so that there is no net force on the charge at B? A © Dublin School of Grinds 12 cm B Page 15 4 cm C Kieran Mills & Tony Kelly 13. A, B, C are three points on a straight line with |AB| = 20 cm and |BC| = 15 cm, where B is between A and C. Charges of +4 µC and +5 µC are placed at A and B respectively. What charge must be placed at C so that there is zero resultant force on the charge at B? Electric Field Intensity 14. A charge of 3 µC experiences a force of 30 N when placed at a point in an electric field. Calculate the electric field strength at that point. 15. A charge of 4 µC experiences a force of 16 N when placed at a point in an electric field. Calculate the electric field strength at that point. 16. What is the force on a charge of 2 µC when placed in an electric field of strength 2 × 103 N C−1? 17. A charged particle experiences a force of 6 × 10−6 N when placed in a field of strength 3 × 103 N C−1. What is the charge on the particle? 18. Find the magnitude and direction of the electric field strength at a distance of 3 m from a charge of +9 µC. 19. Find the magnitude and direction of the electric field strength at a distance of (a) 0.1 mm, (b) 1 mm, (c) 10 cm from a negative charge of 30 µC. 20. What is the electric field strength half way between a charge of +4 µC and a charge of +2 µC, if the distance between the charges is 40 cm? What is the force on a charge of 6 µC placed at this point? 21. What is the electric field strength half-way between a charge of +2 µC and a charge of −8 µC, if the distance between the charges is 60 cm? What is the force on a charge of 3 µC placed at this point? 22. At what point between a charge of +8 µC and a charge of +2 µC is the electric field strength zero if the charges are 1 m apart? 23. Two charges are placed on the x-axis, a +3 µC charge at x = 60 cm and a −6 µC charge at x = 0 cm. Find the electric field at (a) x = 20 cm and (b) x = 100 cm. Potential 24. The potential difference between two points is 16 V. Find the work done in transferring a charge of 4 C between points. 25. The work done in bringing a charge of 2 C from one point to another is 5 J. What is the potential difference between the points? 26. The work done in bringing a charge of 4 C from one point to another is 12 J. What is the potential difference between the two points? 27. The work done in bringing a charge of 4 µC from one point to another is 4 × 10−5 J. What is the potential difference between the two points? © Dublin School of Grinds Page 16 Kieran Mills & Tony Kelly 28. Calculate the work done in transferring a charge of 2 C between two points when the potential difference between the points is 10 V? 29. The potential difference between two points is 100 kV. Find the work done when a charge of 2 µC is moved from one point to another. 30. An electron of charge 1.6 × 10−19 C loses 6.4 × 10−16 J of energy as it moves from one point to another in an electric field. What is the potential difference between the two points? 31. Calculate the work done when a charge of 8 µC moves between two points if the potential difference between the points is 12 V. 32. 9.6 × 10−16 J of work is done in moving an electron between two points in an electric field. What is the potential difference between the two points? 33. Find the work done when an electron passes through: (a) 1 volt, (b) 500 volts. 34. Two oppositely charged plates are 2 cm apart. There is a uniform electric field of strength 3 × 103 N C−1 between them. (a) What is the force on a +1 C charge placed between the plates? (b) Find the work done in bringing a charge of 1 C from one plate to the other. (c) What is the potential difference between the plates? 35. The potential difference between two points is 5000 V. An electron is released at one of the points and moves towards the other under the action of the field. Find its speed when it arrives at the second point. 36. The potential difference between two oppositely charged parallel plates 10 cm apart is 100 V. Find: (a) the force acting on a charge of 1 C when placed between the plates, (b) the electric field intensity between the plates, (c) the force acting on an electron put between the plates, (d) the potential energy lost by an electron if it moves from the negative to the positive plate, (e) the kinetic energy of an electron on arriving at the positive plate if its speed when released at the negative plate was zero, (f) the speed of the electron on arriving at the positive plate. © Dublin School of Grinds Page 17 Kieran Mills & Tony Kelly Answers Coulomb’s Law 1. 1.125 ×1010 N 2. 0.3 N towards the other charge Potential 24. 64 J 25. 2.5 V 3. 9 ×10−3 N, attraction 4. 0.144 N towards the −2 µ C charge 26. 3 V 5. 1.44 ×10−3 N away from each other 6. 360 N 28. 20 J 29. 0.2 J 7. 7.07 ×10−11 F m −1 30. 4000 V 27. 10 V −3 8. 2.53 × 10 N towards the −4 µ C charge 9. 3.33 × 10−7 C 10. 0.024 N towards the −3 µ C charge 11. 0.024 N towards the −2 ⋅ 5 nC charge 12. 3.3 ×10−7 C 13. 2.25 ×10−6 C Electric Field Intensity 14. 1×107 N C−1 31. 9.6 ×10−5 J 32. 6000 V 33. (a) 1.6 × 10−19 J, (b) 8 × 10−17 J 34. (a) 3 × 103 N, (b) 60 J, (c) 60 V 35. 4.2 × 107 m s−1 15. 4 ×106 N C−1 36. (a) 1000 N (b) 1000 N C−1 (c) 1.6 × 10−16 N (d) 1.6 × 10−17 J 16. 4 ×10−3 N 17. 2 ×10 −9 (e) 1.6 × 10−17 J (f) 5.93 × 106 m s−1 C 18. 9 ×103 N C−1 away from the charge 19. (a) 2.7 ×1013 N C−1 (b) 2.7 ×1011 N C−1 (c) 2.7 ×107 N C−1 5 −1 20. 4.5 ×10 N C , 2.7 N 6 −1 21. 1×10 N C , 3 N 22. 2 3 m from the 8 µC charge 23.(a) 1.52 ×106 N C−1 (b) 1.15 ×105 N C−1 © Dublin School of Grinds Page 18 Kieran Mills & Tony Kelly Chapter 3 • Capacitance 3.1 Defining Capacitance The capacitance of a body is its ability to store charge. Definition: Capacitance is the ratio of charge to potential. Q C= V Q: Quantity of charge stored (C) V: Potential (voltage) (V) C: Capacitance (Farads (F)) Formulae and Tables Book: Page 62 Capacitance (Electricity) Definition: The capacitance of a body is 1 Farad (F) if the addition of a charge of 1 C raises its potential by 1V. Example 1: A charge of 5 µC is placed on a conductor of capacitance 2 pF. Find the increase in potential of the conductor. Solution 3.2 The Parallel-Plate Capacitor d Capacitors can have all sorts of shapes and sizes. An example of a capacitor is two parallel plates separated by an insulator called a dielectric. The dielectric can be air having the permittivity of free space or another material like paper or mica. It can be shown that the capacitance of a parallel plate capacitor is given by: C= εA d C: Capacitance (F) e : Permittivity of the dielectric (F m−1) A: Common area of overlap of the plates (m2) d: Distance between the plates (m) Dielectric Formulae and Tables Book: Page 62 Parallel-plate capacitor (Electricity) Example 2: A parallel plate capacitor has a distance of 1 mm between the plates, each of which has an area of 25 cm2. It has a mica dielectric. Find the charge on either of the plates when the potential difference between the plates is 500 V. (Relative permittivity of mica = 7.) Constant: Permittivity of free space ε0 = 8.9 × 10−12 F m−1 Solution © Dublin School of Grinds Page 19 Kieran Mills & Tony Kelly Note: The relative permittivity of a material is the permittivity of the material compared to the permittivity of free space. In the previous example, mica has a relative permittivity of 7 which means its permittivity is 7 times that of the permittivity of free space. Demonstration: Factors affecting capacitance of a parallel-plate capacitor d A positive charge Q is put on one of the plates. Its potential is measured using a GLE. The divergence of the leaf is a measure of this potential. The capacitance C is affected by three factors: 1. The common area, A, of the plates: C∝A If the plates are moved so that their common area of overlap is reduced the leaf divergence increases, i.e. the potential increases and so the capacitance is reduced. 2. The distance, d, that the plates are apart: C ∝ Dielectric 1 d If the plates are moved further apart the potential rises and so the capacitance falls. 3. The permittivity of the material, e , between the plates: C ∝ ε A dielectric causes the leaf divergence to fall indicating an increase in capacitance 3.3 Energy in a Charged Capacitor A capacitor stores electrical energy. The energy W stored in the capacitor is the amount of work done by the battery in charging the capacitor. W = 12 CV 2 W: energy stored in a capacitor (J) C: Capacitance (F) V: Potential Difference or voltage (V) Formulae and Tables Book: Page 62 Energy stored in capacitor (Electricity) Example 3: A capacitor stores a charge of 5 µC and has a potential difference of 20 V across it. What energy does it store? Solution Example 4: A capacitor has a capacitance of 3.2 µF. What charge is on it if the energy stored is 40 mJ? Solution © Dublin School of Grinds Page 20 Kieran Mills & Tony Kelly Demonstration: A Capacitor stores charge A capacitor is a device for storing charge. This can be demonstrated as follows: When switch A is closed the capacitor charges up. + A C + Open switch A and then close switch B. B The bulb flashes as the capacitor discharges through it. Demonstration: AC and DC through Capacitors D.C. In a direct current (DC) circuit no current flows once the capacitor is fully charged. C In the Alternating Current (AC) circuit the capacitor first of all charges in one direction and then the other. As a result current constantly flows. Therefore the bulb lights. Off A.C. On C A capacitor allows A.C. to flow but blocks steady D.C. Common uses of capacitors Tuning radios A variable capacitor is used to change to a particular station on a radio. Flash guns In cameras, a capacitor is charged slowly from a battery and discharged quickly through a bulb producing a flash. Smoothing Capacitors smooth out variations in direct current. Filtering Capacitors allow certain frequencies of alternating current to pass and block others. This is called filtering. © Dublin School of Grinds Page 21 Kieran Mills & Tony Kelly Example 5: A thundercloud has a horizontal lower surface of area 30 km2 which is 800 m above the earth. Treating this system as a parallel-plate capacitor find: (a) the capacitance of the cloud, (b) the energy stored if the potential difference between the cloud and the ground is 2 × 105 V. (ε0 = 8.85 × 10-12 F m-1) Solution 800 m Ground Answers to Examples Example 1: 2.5 ×106 V Example 2: 8 ×10−8 C Example 3: 5 ×10−5 J Example 4: 5.06 ×10−4 C Example 5: (a) 3.32 ×10−7 F (b) 6.64 ×103 J © Dublin School of Grinds Page 22 Kieran Mills & Tony Kelly Numerical Problems Constants: Charge on electron = 1.6 × 10−19 C Mass on electron = 9 × 10−31 kg Permittivity of free space ε0 = 8.85 × 10−12 F m−1 1. A conductor has a potential of 6 V when a charge of 6 µC is placed on it. What is its capacitance? 2. The capacitance on a body is 4 pF. Calculate the charge on it when its potential is 300 V. 3. The capacitance of a sphere is 20 pF. If its potential is 5000 V, find the charge on it. 4. A capacitor has a capacitance of 50 mF. What is the potential difference between its plates if it stores a charge of 1.2 µC? 5. A charge of 4 µC is placed on a conductor of capacitance 3 pF. Find the increase in potential of the conductor. 6. A capacitor has a capacitance of 50 µF. What is the charge on one of its plates if the potential difference between them is 100V? 7. The area of overlap of the plates of an air spaced capacitor is 20 cm2. The distance between the plates is 1 mm. (a) Find the capacitance of the capacitor. (b) If the space between the plates is now filled with mica of relative permittivity 7 calculate the capacitance of the capacitor. 8. Find the distance between the plates of an air-spaced capacitor of 2 pF if the area of one side of one of the plates is 100 cm2. 9. Find the capacitance of a parallel plate air capacitor if the common area of the plates is 0.5 m2 and the distance between them is 1 mm. Find also the charge stored in the capacitor if the potential difference between the plates is 10 V. 10. A parallel plate air-spaced capacitor is to have a capacitance of 1 F. If the distance between the plates is 1 mm, find the area of one of the plates. 11. Find the capacitance of a parallel plate air-spaced capacitor if the area of one of the plates is 100 cm2 and the distance between the plates is 2 mm. Find the capacitance if the space between the plates is filled with perspex which has a relative permittivity of 2.6. 12. A parallel plate capacitor has a distance of 1.2 mm between the plates, each of which has an area of 30 cm2. It has a mica dielectric. Find the charge on either of the plates when the potential difference between the plates is 400 V. (Relative permittivity of mica = 7.) © Dublin School of Grinds Page 23 Kieran Mills & Tony Kelly 13. A capacitor consists of two sheets of Aluminium, each 5 m long by 3 cm wide, separated by an insulating material of permittivity 3 × 10−11 Fm−1 and thickness 25 µm. Calculate the capacitance of the capacitor and the charge on the plates when the potential difference between them is 20 V. 14. A parallel plate capacitor has plates of area 0.01 m2 and carries a charge of 12 mC. Its capacitance is 470 µF and the relative permittivity of the dielectric is 1200. Calculate (a) the potential difference between the plates, (b) and their distance apart. 15. Calculate the energy stored in a capacitor of 500 µF when the potential difference between the plates is 15 V. 16. A capacitor stores a charge of 7 µC and has a potential difference of 30 V across it. What energy does it store? 17. A capacitor has a capacitance of 2.4 µF. What charge is on it if the energy stored is 23 mJ? Answers 1. 1×10−6 F −9 2. 1 ⋅ 2 ×10 C 3. 1×10−7 C 4. 2 ⋅ 4 ×10−5 V 5. 1.33× 106 V 6. 5 × 10−3 C 7. (a) 1 ⋅ 77 ×10−11 F, (b) 1.24 × 10−10 F 8. 4.43 cm 10. 1.13 × 108 m2 11. 4.43 × 10−11 F, 1.15 × 10−10 F 12. 6.2 × 10−8 C −7 13. 1 ⋅ 8 ×10 F, 3 ⋅ 6 ×10−6 C 14. (a) 25.5 V, (b) 2 ⋅ 3 ×10−7 m 15.5.625 × 10−2 J 16. 3.15 × 10−3 J 17. 3.32 × 10−4 C 9. 4.43 nF, 44.3 nC © Dublin School of Grinds Page 24 Kieran Mills & Tony Kelly Chapter 4 • Electric Current 4.1 Sources of EMF and Electric Current A metal is a good electrical conductor. In a metal there are free electrons wandering around in no particular direction. They wander in between fixed metal ions. Positive Terminal + + + + + + + + Negative Terminal A battery acts as an electron pump. It forces the electrons to move in a particular direction. This constitutes an electric current. The battery gives energy to the electrons. Note: The flow of charge is shown as moving from the positive to the negative terminal of the battery. This is conventional current. The real current flows in the opposite direction.) + + + + + + + + Fixed positive ion + Electron The amount of energy that a battery gives to a certain quantity of electrons is measured in Volts (V). For example, a 3 V battery will give 3 Joules (J) of energy to a certain quantity of electrons (This number of electrons is 6 × 1018.) The amount of charge on this quantity of electrons is assigned a value of 1 Coulomb (C). 1 C is the charge on 6 × 1018 electrons. 1 C = 1.6 ×10−19 C. The charge on 1 electron is 18 6 ×10 Therefore, one electron has a charge of 1.6×10−19 C. You can now say that a 3 V battery gives 3 J of energy to each Coulomb of charge. This is called the EMF of a battery. Definition: The electromotive force (EMF) is the amount of electrical energy in Joules that the battery gives to each Coulomb of charge passing through it. As the charge moves around the circuit it loses energy in the various devices. The charge does work. Assume that the bulb causes 1 C of charge to lose 2.5 J of energy. The charge has to also pass through the battery - it loses the other 0.5 J of energy here. The Coulomb of charge arriving at the bulb has 3 J of energy (point A). Leaving the bulb (point B) it has only 0.5 J of energy. Therefore 2.5 J of energy has been converted to other forms. We say that there is a potential difference (p.d.) or voltage between ‘A’ and ‘B’. A 2.5J B 1C EMF=3V + 0.5J - Battery Definition: The potential difference V between two points is the work done in moving a unit charge from one point to the other. Stated mathematically: V: Potential difference between two points (Volts, V) W V= W: Work done in moving the charge between the points (J) q q: Charge being moved (C) © Dublin School of Grinds Page 25 Kieran Mills & Tony Kelly Potential difference or voltage is measured in Volts (V) by a voltmeter connected in parallel to the device. In this example V = V W 2⋅5J = 2 ⋅ 5 V. = Q 1C Voltmeter Bulb B A Defining the Volt If 1 J of of work was done or 1 J of energy was converted to other forms when 1 C of charge passed through a device then the voltage would be 1 Volt. V = V = 2.5 V Does 2.5 J of work W + W 1J = = 1V Q 1C Battery 3V Definition: The potential difference between two points is 1 Volt if 1 Joule of work is done when a unit charge is brought from one point to the other. Energy was also required to get through the battery. This is because of the internal resistance of the battery. Electric Current Current I is a measure of how long it takes a quantity of charge Q to pass a certain point. Current (I ) = I= Q t Quantity of Charge (Q) Time (t ) Q: Quantity of charge (C) I: Current (A) t: Time (s) A Ammeter + I V Current is measured in Amps (A) by an ammeter connected in series into the circuit. The unit of charge, the Coulomb, can be defined from the formula above: Q = It = 1 A × 1 s = 1 C Definition: The Coulomb is the quantity of charge transferred when a current of 1A flows for 1s. Example 1: If the heating element of an electric radiator takes a current of 5 A, what charge passes each point every minute? How many electrons pass a given point in this time? (Charge on electron e = 1 ⋅ 6 ×10−19 C ) Solution © Dublin School of Grinds Page 26 Kieran Mills & Tony Kelly Sources of EMF Mains Electricity to your home is called mains electricity. Cells An electric cell is a source of EMF. It consists of two metals called electrodes immersed in a liquid called an electrolyte. A typical example is a copper and zinc plate dipped in dilute sulphuric acid. This is a primary cell as it cannot be recharged. Almost all primary cells have electrolytes that are pastes rather than liquids. Such cells are called dry batteries. Lead-acid accumulator Some cells can be recharged. These are called secondary cells or accumulators. A car battery is a lead-acid accumulator which consists of 6 lead acid cells in a battery. Thermocouple Two wires made of different metals maintained at different temperatures provide a source of EMF. Such a device can be calibrated to act as a thermometer. 4.2 Ohm’s Law Conductors that are used because they have a resistance to electric current are called resistors. Resistors slow the current down. The more resistance there is in a circuit the less current will flow. Definition: A resistor is a conductor which has a resistance which limits the flow of current. To investigate the resistance R of a conductor, the current I flowing through it must be measured. The voltage V which gives energy to the charge must also be measured. I Using the rheostat, vary the current going through the resistor. In each case read the corresponding voltage. Draw a table of voltage and current readings. A graph of V against I yields a straight line passing through the origin showing that the voltage is directly proportional to the current, i.e. V ∝ I. Voltage V V (V) I (A) R= V I Current I Rheostat V R A V Dividing voltage by current for any pair of readings gives a constant value. V ∴ = Constant I The constant value obtained is called the V resistance R of the conductor. ∴ = R I V: Voltage or potential difference (V) I: Current (A) R: Resistance (Ω) Formulae and Tables Book: Page 61 Resistance (Electricity) Definition: The resistance of a material is the ratio of the potential difference across its ends to the current flowing through it. © Dublin School of Grinds Page 27 Kieran Mills & Tony Kelly Resistance is measured in Ohms (Ω) by an ohmmeter. The resistance of a conductor is constant as long as its temperature remains constant. An increase in temperature causes the resistance to change. Definition: Ohm’s Law states that the current flowing through a conductor is proportional to the potential difference between its ends if the temperature remains constant. Example 2: A certain quantity of charge does 4000 J of work when passing through a resistor which has a potential difference of 20 V across its ends. It takes two minutes for the charge to be transferred from one point to the other. Calculate the quantity of charge, the current and the resistance of the resistor. Solution Physics Experiment Book Experiment E3: To demonstrate the variation of the resistance of a metallic conductor with temperature. (Page 130) 4.3 Conduction in Materials Consider conduction through the following materials: [A] Metals: Metallic conductor and filament bulb. [B] Ionic Solutions: Using active and inactive electrodes. [C] Gases [D] Vacuum [E] Semiconductors In each case, the I-V characteristic (the graph of current against voltage) will be examined and the types of charge carriers used will be explained. [A] Metals Metallic Conductor In a metal, free electrons move amongst fixed positive ions. Provided the temperature remains constant, the resistance of the conductor does not change. Therefore, the voltage and the current are directly proportional. The I−V graph is a straight line passing through the origin. It obeys Ohm’s Law. Material: Metallic Conductor [Temperature remains constant] I-V Graph (Characteristic) I (A) V (V) Obeys Ohm’s Law. Charge carriers: Electrons (negative) © Dublin School of Grinds Page 28 Kieran Mills & Tony Kelly Physics Experiment Book Experiment E5[A]: To investigate the variation of current I with potential difference V for a metallic conductor. (Page 138) Filament Bulb As the voltage across a filament bulb increases, its current also increases. As the current goes up, the filament gets significantly hotter. In metals, the effect of increasing the temperature is to make the fixed positive ion cores vibrate with greater amplitudes. This means that the charge carriers find it more difficult to flow meaning the rise in the current will be less for a given rise in V voltage. From the formula R = , the resistance I of the filament bulb increases. Material: Filament Bulb [Temperature changes] I-V Graph (Characteristic) I (A) V (V) Does not obey Ohm’s Law. Charge carriers: Electrons (negative) Physics Experiment Book Experiment E5[B]: To investigate the variation of current I with potential difference V for a filament bulb. (Page 142) [B] Ionic Solutions Two metal plates are connected to a battery and an ammeter and put into a solution of distilled water. No current flows. However, when sodium chloride is dissolved in the water, a current flows. The water with the dissolved salt is an ionic solution. © Dublin School of Grinds Page 29 Anode + Cathode I + + + + + + + + + + Cathode Explanation: Sodium chloride dissolves in water into positive and negative ions. The positive ions are attracted by the cathode and the negative ions by the anode. This constitutes a flow of current which is registered on the ammeter. In distilled water there is an absence of ions and so no current flows. Therefore conduction in liquids is due to the movement of ions, both positive and negative. A Anode Terminology Metal plates: Electrodes Positive electrode: Anode Negative electrode: Cathode Conducting liquid: Electrolyte Container, electrolyte and electrodes: Voltameter + + Kieran Mills & Tony Kelly Copper Electrolysis Two copper plates are connected to a battery and put into a solution of copper sulfate. When copper sulfate is dissolved in water it breaks up into its ions: positive copper ions and negative sulfate ions. The cathode attracts the positive copper ions and they become deposited on it. The sulfate ions are attracted by the anode. They react with the copper plate to form copper sulfate which goes back into solution. As the voltage across the electrodes increases so does the current. The resistance remains constant and so the I-V graph is a straight line. The electrodes are active (they take part in the chemical reactions), the electrolyte obeys Ohm’s Law and the graph passes through the origin. Replace the copper electrodes with platinum plates which do not react with the copper sulfate solution. If the electrodes are inactive, the voltameter behaves as a cell and produces an EMF across its plates. The applied voltage must be greater than this EMF before current will flow. This I-V graph does not obey Ohm’s law as it does not pass through the origin. Note: Increasing the concentration of the solution will increase the slope of the straight line graph as the number of charge carriers has increased. Therefore, the resistance of the electrolyte decreases. Diluting the solution will have the opposite effect. Material: Ionic Solution [Using active electrodes] I-V Graph (Characteristic) I (A) s de ro t ec l ee iv ct A V (V) Obeys Ohm’s Law. Charge carriers: Positive and negative ions Material: Ionic Solution [Using inactive electrodes] I-V Graph (Characteristic) I (A) s de ro t ec l ee tiv ac In V (V) Does not obey Ohm’s Law. Charge carriers: Positive and negative ions Example 3: Inactive electrodes like Platinum are used in copper sulfate solution setting up a voltage of 0.75 V across the electrodes. A 6 V source is then connected across the electrodes giving rise to a 2.8 A current flowing between the plates. Calculate the resitance of the electrolyte. Solution Physics Experiment Book Experiment E5[C]: To investigate the variation of current I with potential difference V for copper sulfate solution with copper electrodes. (Page 145) © Dublin School of Grinds Page 30 Kieran Mills & Tony Kelly [C] Gases A gas at low pressure is put into a sealed container with two electrodes in it. It is called a discharge tube. There will always be some ions in the gas formed by background radiation. After a while these ions recombine back with their electrons. Gas at low pressure If a voltage is put across the tube, the positive ions move towards the cathode and the electrons move towards the anode. A current flows. As the voltage increases the number of ions and electrons also increases thus increasing the current. This corresponds to region OA in the graph. Material: Gases I-V Graph (Characteristic) I (A) C A Eventually the graph levels out in the region AB. Increases in voltage produce no increases in current because all the ions are now crossing the tube. As the voltage is further increased, the ions and electrons get sufficiently fast to produce further ions by collision. The current now increases with increased voltage (region BC on the graph). B O V (V) Does not obey Ohm’s Law. Charge carriers: Positive ions, electrons and a few negative ions [D] Vacuum A vacuum exists if nothing is put into the discharge tube. No electricity will flow because no charge carriers are present. If, however, the cathode is heated sufficiently, electrons in the metal will get enough energy to escape. This is called thermionic emission. These electrons will cross the tube under an applied voltage. Material: Vacuum I-V Graph (Characteristic) I (A) The current will increase as the voltage is increased until V (V) all the available electrons are used up. After this there Does not obey Ohm’s Law. is no increase in current for an increase in the voltage. Charge carriers: Electrons Hence, the graph flattens out. Neon lamps, street lamps Sodium vapour street lamps (yellow/orange) and neon lamps are examples of gas discharge tubes where conduction takes place in a gas. © Dublin School of Grinds Page 31 Kieran Mills & Tony Kelly [E] Semiconductors Definition: A semiconductor may be defined as a material whose conductivity lies between that of a good conductor and a good insulator. Silicon is an example of a semiconductor material. This is a Group IV element. Each of its outer electrons shares an electron with 4 other Si atoms to form a Si crystal. At 0 K all electrons are involved in bonding and none are available for conduction. However, as the crystal heats up (called thermal agitation) some electrons get enough energy to break free from their bonds and become conduction or free electrons. There is now a vacancy left behind into which bound electrons can move. This movement is called hole movement. Si Si Si Si Si Si Hole Free Electron Bound Electron Si Under an applied electric field both electrons (negative charge carriers) and holes (positive charge carriers) contribute to the overall current. I = Ie + Ih Free electron Hole The generation of Material: Semiconductor charge carriers in this way from the material itself is known as intrinsic conduction. As the voltage across a semiconductor increases, the current increases. As a result of the increased current it gets hotter. Therefore, more holes and electrons are produced making for an even larger current. As a result, the I−V graph gets steeper. The charge carriers are electrons (negative) and holes (positive). The graph does not obey Ohm’s law. I-V Graph (Characteristic) I (A) V (V) Does not obey Ohm’s Law. Charge carriers: Electrons and holes Semiconductors [A] Light Emitting Diodes (LEDs) Uses: 1. ON/OFF indicator lamps: e.g. calculators, stereos, etc. When a calculator is switched on the voltage of the battery is applied to the LED which emits red light. The LED is protected by a resistor. 2. Digital displays as in clocks or calculators. The seven segments are LEDs. Any of the ten numerical digits can be visually displayed by activating the correct LEDs. 3. Optical fibre transmitters. [B] Integrated Circuits (ICs) An integrated circuit (IC) is a circuit containing electronic components like transistors, diodes, resistors and capacitors on one small chip of Silicon. © Dublin School of Grinds Page 32 Kieran Mills & Tony Kelly Extrinsic Semiconductors These type of semiconductors can be manufactured by doping a material like Silicon with another material called an impurity. There are two types: Si [A] N-Type Semiconductor: Dope a Si crystal with a Group V element like Phosphorous (P). This produces free electrons. An extrinsic semiconductor in which the majority charge carriers are electrons is called an N-type semiconductor. Si Si Si Si [B] P-Type Semiconductor: Dope a Si crystal with a Group III element like Aluminium (Al). This produces free holes. An extrinsic semiconductor in which the majority charge carriers are holes is called a P-type semiconductor. P Si Si Al Si Semiconductor Diode A diode consists of a P-type and N-type material joined together (called a P-N Junction). Definition: A semiconductor junction diode allows current to flow through it in one direction only. Unbiased P-N Junction (no battery connected) • Charges diffuse across the junction • Recombination occurs (electrons meet holes killing each other off) giving rise to a depletion layer (an area of no charge carriers). • A junction voltage of ~0.6 V is formed at the depletion layer preventing the further movement of charges across the junction. Recombination N-Type Electrons P-Type Holes P-Type N-Type Depletion layer Biased P-N Junction (battery connected) 1. Forward Biased (Connect + to P and − to N) A forward biased junction can only conduct once the applied voltage exceeds the junction voltage. When the diode is in forward bias, its resistance is very small ( R ≈ 0 Ω.) © Dublin School of Grinds Page 33 P-Type Holes N-Type Electrons Kieran Mills & Tony Kelly 2. Reverse Biased (Connect + to N and − to P) As the applied voltage is increased the depletion layer widens and so practically no charge carriers can cross the barrier. Thus the current is practically zero and so the device has a very high resistance when reverse biased. P-Type N-Type Holes Electrons Characteristic Curve of S-C Diode This is the circuit to plot the characteristic curve of a diode. A potentiometer varies the voltage across the diode. IF (mA) A V V (V) .2 .4 .6 .8 1.0 IR (A) Demonstration: Current flow across a diode If a battery is connected to a diode the right way (forward bias) the bulb lights. If it is connected the wrong way (reverse bias), no current flows and the bulb is off. + + On Off Forward Bias Reverse Bias Rectification of AC Alternating current is current that flows both ways reversing direction many times per second. If a diode is put into a circuit then current is only allowed to flow one way (direct current DC). A diode rectifies AC (changes AC to DC). Physics Experiment Book Experiment E5[D]: To investigate the variation of current I with potential difference V for a semiconductor diode. (Page 149) © Dublin School of Grinds Page 34 Kieran Mills & Tony Kelly Example 4: A light emitting diode (LED) has a junction voltage of 2 V. The LED is connected in forward bias to a 12 V battery. A resistor of 25 kW is connected in series with the LED to protect it from excessive currents. Find the current flowing through the LED. Solution 12 V + I LED 25 k 4.4 Resistance l Consider a piece of wire of uniform cross-sectional area. A uniform circular cross-section means its diameter is the d same at all points. It is shown by experiment that if the length l of the wire is doubled, the resistance doubles. ∴R ∝ l 1 If the cross-sectional area A is doubled, the resistance is halved. ∴ R ∝ A l l Combining these two factors gives: R ∝ ⇒ R = (Constant) A A This constant of proportionality is called the resistivity r of the material. R: Resistance (W) Formulae and Tables Book: Page 61 l: Length (m) Resistivity (Electricity) A: Area (m2) r: Resistivity (W m) Note: The resistance of a material also depends on its temperature. R=ρ l A Definition: The resistivity of a material is numerically the resistance of a sample of unit length and unit cross-section area, at a certain temperature. Example 5: Calculate the diameter of a Copper wire 2.5 m long if its resistance is 0.1 W. −8 Constant: Resistivity of Copper r = 1.70 × 10 W m Solution Physics Experiment Book Experiment E2: To measure the resistivity of the material of a wire. (Page 123) © Dublin School of Grinds Page 35 Kieran Mills & Tony Kelly Combinations of resistors Resistors in series Resistors connected in series are placed one after another in a circuit. There is only one path for the current to flow and therefore the current is the same at all points in the circuit. The voltage across each resistor is the amount of electrical energy converted in each resistor. Therefore, the voltage across each reistor in series adds up to the main (battery) voltage. I V R1 R2 R3 V1 V2 V3 Voltages across components in series always add up to the supply voltage. The current in a series circuit is the same at all points in a circuit. The biggest resistor gets the biggest share of the voltage. Proof: Resistors in series formula V = V1 + V2 + V3 I From Ohm’s Law V = IR, where R denotes the total resistance and I the current which is the same at all points in the circuit. The voltages across each of the resistors are as follows: = V1 IR = IR2 , V3 = IR3 1 , V2 V R1 R2 R3 V1 V2 V3 ∴ IR = IR1 + IR2 + IR3 R = R1 + R2 + R3 R = R1 + R2 Formulae and Tables Book: Page 61 Resistors in series (Electricity) Example 6: Calculate (a) the current I (b) the potential difference between A and B, VAB, and the potential difference between B and C, VBC. Solution 9V A © Dublin School of Grinds Page 36 6 B 12 C Kieran Mills & Tony Kelly Resistors in parallel Resistors connected in parallel are connected side by side. The current flowing in the circuit has a choice of paths through which to flow. Therefore, the main current divides up into different currents. Each resistor is connected directly to the battery and so the same voltage is applied across each resistor. I V R1 I1 I3 R2 I2 R3 The current in a parallel circuit divides up at each arm. Every arm in a parallel network has the same voltage drop across it, even if the components are not identical. The smallest resistor gets the biggest current. Proof: Resistors in parallel formula I I = I1 + I 2 + I 3 From Ohm’s Law I = V V . R R1 As the voltage across each resistor is V then: V V V = I1 = , I2 , I3 = R1 R2 R3 V V V V ∴ = + + R R1 R2 R3 I1 I3 R2 I2 R3 1 1 1 1 = + + R R1 R2 R3 1 1 1 = + R R1 R2 Formulae and Tables Book: Page 61 Resistors in parallel (Electricity) Example 7: Calculate: (a) the current I and, (b) the current flowing in each resistor. Solution 9V I I 10 15 © Dublin School of Grinds Page 37 Kieran Mills & Tony Kelly Example 8: In the circuit shown, calculate: (a) the current flowing in the 3.125 W resistor, (b) the potential difference across the 5 W resistor, (c) the current flowing through the 5 W resistor, 6.25 V (d) the current flowing through R, (e) the current flowing through the 2 W resistor, (f) the potential difference across the 2 W resistor, (g) the potential difference across R, (h) the value of R. Solution Example 9: In the circuit shown, calculate: 10 V 3.125 5 R 12 V I (a) the total resistance and the current I, (b) the voltage between A and B, 3 (c) the current flowing through the 5 W resistor. 2 B A 5 Solution © Dublin School of Grinds 2 Page 38 Kieran Mills & Tony Kelly Example 10: The circuit shows a 75 mF capacitor connected in series with a 35 kW resistor, a 12 V battery and a switch S. When S is closed, the capacitor C starts to charge and the current flowing R at a particular instant in the circuit is 50 mA. Calculate (a) the potential difference across the resistor R and hence the potential difference across the capacitor when the current is 50 mA; (b) the charge on the capacitor at this instant; (c) the energy stored in the capacitor when it is fully charged. Solution Measuring Resistance using a Wheatstone Bridge The circuit shown is a Wheatstone bridge. Suppose the resistors are arranged in such a way that no current flows through the galvanometer. The bridge is now said to be balanced. If no current flows through the galvanometer, there is no potential difference between B and D, ie. B and D are at the same potential. This means that: VAB = VAD and VBC = VDC I S 12 V 35 k C 75 F B R1 R2 I1 I1 A I2 I2 R4 R3 From Ohm’s Law: I1 R1 = I2 R3 and I1 R2 = I2 R4 I D Dividing gives: R1 R3 = R2 R4 C G V Formulae and Tables Book: Page 61 Wheatstone bridge (Electricity) Using the formula, you can use the Wheatstone Bridge to find the resistance of an unknown resistor. This method is accurate because it does not depend on the accuracy of the galvanometer but only on its sensitivity (its ability to detect a small current). Therefore this method of measuring resistance is called the null method. © Dublin School of Grinds Page 39 Kieran Mills & Tony Kelly Example 11: Calculate the current flowing through the resistor R1 and hence calculate the potential at the point B when the potential at C is zero. What is the potential at D? What direction would a current flow through a galvanometer connected between the points B and D? What value of the resistance R4 would make the galvanometer read zero? Solution B 10 20 R1 R2 C A R4 R3 11 5 D 6V Metre Bridge A convenient way to use the Wheatstone bridge is as the metre bridge. This is a very accurate instrument for measuring the resistance of a device by comparing it with a known resistor. The wire AB is a piece of resistance wire of uniform cross-sectional area one metre long. The probe slides along this wire until the galvanometer needle points to zero, i.e. no current is flowing through it now. When this is the case the resistance wire is divided into two parts in such a way that the ratio between the two parts is the same as the ratio of the two resistors, R1 and R2. Stated mathematically: A l2 l1 R1 G B R2 R1 l1 = R2 l2 © Dublin School of Grinds Page 40 Kieran Mills & Tony Kelly Example 12: A metre bridge is balanced as shown when the probe is at the 35 cm mark. The length of wire AB is 1 metre long. Find the resistance R. Solution A0 100 B 35 15 G R Practical uses of the Wheatstone Bridge 1. Temperature Control: A Wheatstone Bridge is balanced. If one of the resistors changes then current flows through the galvanometer. A change of temperature can change the resistance. If it gets hotter the resistance will go up causing current to flow in a particular direction through the galvanometer. If it gets colder its resistance goes down causing current to flow the other way through the galvanometer. This current can be used to control a heater/cooler to restore temperature to its original value. Thermistors Thermistors are intrinsic semiconducting devices whose resistance decreases as the temperature increases. As the temperature increases, the number of electrons and holes (charge carriers) increases. Therefore, the current I which flows under an applied potential difference V also V increases. From the formula R = , the resistance R I decreases in response to an increase in temperature. Resistance R 2. Fail-Safe Device: A gas flame boiler has a pilot flame that should be continuously lighting. If the pilot flame goes out, the fuel supply should get shut off. This can be achieved by having a thermistor near the pilot flame. The thermistor is one of the resistors in a Wheatstone bridge. If the pilot flame goes out, the thermistor’s resistance changes causing the Wheatstone Bridge to be unbalanced. The unbalanced current can be used to switch off the fuel. Symbol Temperature T Thermistors are cheaply manufactured from semiconducting powders like iron oxide. Physics Experiment Book Experiment E4: To demonstrate the variation of the resistance of a thermistor with temperature. (Page 134) © Dublin School of Grinds Page 41 Kieran Mills & Tony Kelly A 400 14 V 100 200 B Photoconductive Cells Photoconductive cells (also called Light Dependent Resistors LDRs) are intrinsic semiconductor devices whose resistance decreases as the intensity of light (electromagnetic radiation) falling on them increases. As the intensity of light increases, more photons fall on the semiconductor per second and, as in the thermistor’s case, the number of electrons and holes increases. Resistance R Example 13: In the circuit diagram, the resistance of the thermistor at room temperature is 400 W. At room temperature, calculate (a) the total resistance of the circuit (b) the potential difference between A and B (c) the potential at B (d) the current flowing through the 100 W resistor (e) the temperature of the room increases changing the resistance of the thermistor to 25 W. What is the potential at B now? Solution Symbol Light Intensity One type of material used in LDRs is cadmium sulfide and their resistance can range from 10 MΩ in darkness to a few hundred Ohms in brightness. © Dublin School of Grinds Page 42 Kieran Mills & Tony Kelly Demonstration: Using LDRs and thermistors in Circuits When the thermistor is cold its resistance is very high and the bulb does not light. As the thermistor is heated the bulb gradually comes on. If the LDR is covered so that it is in darkness its resistance is very high and the bulb does not light. If it is exposed to light the bulb comes on. LDR Thermistor 4.5 Potential divider A variable potential divider consists of a resistor connected to a battery (12 V in this example). A probe slides along the resistor and allows us to tap off any voltage between 0 V and 12 V. A voltmeter reads the voltage. When the probe is half-way along the resistor, the voltmeter reads 6 V. Demonstration: Using a potentiometer A potential divider is connected to a 12 V 12 V battery. As the probe taps off a voltage from 0 V to 12 V the bulb turns on and gradually becomes fully bright. 12 V Probe V Probe 12 V bulb V Potentiometer A variable potential divider is called a potentiometer. It is used to divide a given voltage so that you can tap off any voltage between 0 V and the maximum voltage. Example 14: A potential divider is made up of a fixed resistor of 10 W and a variable resistor whose resistance can be varied from 0 W to 20 W. In the circuit shown what is the maximum and minimum potential differences between A and B? Solution 12 V 20 A © Dublin School of Grinds Page 43 10 B Kieran Mills & Tony Kelly 4.6 Effects of Electric Current An electric current has three effects on a circuit: [A] Heating Effect, [B] Magnetic Effect, [C] Chemical Effect [A] Heating Effect: The production of heat in a device depends on three factors: time (t), resistance (R) and current (I). An electric current is passed through a coil of resistance wire immersed in water so that the electrical energy converted in the wire is all transferred as heat energy to the water. The temperature rise of the water ∆q will be directly proportional to the heat energy supplied to it. The heat energy supplied to the water is the electrical energy converted in the resistor or the work done W in the resistor. Demonstration: Heating Effect of an electric Current 1. Showing how the heating effect depends on time Allow a constant current to flow. Measure the temperature rise ∆q after various times t. A graph of ∆q vs. t yields a straight line. A This proves that ∆q ∝ t. D.C. Thermometer ∴W ∝ t Calorimeter 2. Showing how the heating effect depends on the resistance of the coil Heating Coil Allow a constant current to flow through coils of different resistances for a constant time. A graph of ∆q vs. R yields a straight line. ∴W∝R 3. Showing how the heating effect depends on the current flowing Allow different currents to pass through a coil of a certain resistance for a set time. In this case we have to plot a graph of ∆θ vs. I 2 in order to obtain a straight line graph. ∴W ∝ I 2 The amount of work done W in the resistor depends on three factors: The current I flowing, the resistance R of the coil and the time t for which the current flows. This can be expressed mathematically as follows: W = I 2 Rt W: Work done (J) I: Current (A) R: Resistance (Ω) t: Time (s) Note: The above formula can be proved mathematically using W = QV and substituting Q = It and V = IR into this formula. © Dublin School of Grinds Page 44 Kieran Mills & Tony Kelly Example 15: A small heater with a resistance of 100 Ω is connected to a 220 V supply. Calculate the current and the heat energy it gives off per minute. Solution Definition: Joules Law states that the amount of electrical energy converted to internal energy in a resistor is directly proportional to the square of the current flowing through the resistor. I θ I 2 θ 1.0 1.5 2.0 2.5 I 2 3.0 3.5 4.0 Physics Experiment Book Experiment E1: To verify Joule’s heating law. (Page 117) Electrical Power Energy is produced more quickly in an electric fire than it is in a light bulb. The term power is used to describe the rate of energy conversion. An electric fire has a higher power rating than an electric bulb because it converts electrical energy into heat and light energy more quickly. Definition: The power of a device is the rate at which it converts energy from one form into another. = P W QV ItV = = = IV t t t P = IV P: Power (W) I: Current (A) V: Voltage (V) Formulae and Tables Book: Page 62 Power (instantaneous) (Electricity) Definition: 1 Watt equals an energy conversion rate of 1 Joule per second. There are two alternative expressions for power: Substituting V = IR into the above formula gives P = I 2 R. V V2 I = . On the other hand, substituting into the formula gives P = R R The last two formulae can only be applied to the conversion of electrical energy in a resistor. © Dublin School of Grinds Page 45 Kieran Mills & Tony Kelly Example 16: What is the maximum current you could safely pass through a 50 Ω, 2 W resistor? Solution [B] Magnetic Effect: The flow of electric current produces a magnetic field. Demonstration: Magnetic Effect of an electric Current When S is closed, the compass needle deflects showing the presence of a nearby magnetic field. When S is opened the compass returns to its normal orientation of pointing to magnetic north, i.e. the nearby magnetic field has disappeared. Explanation: An electric current is simply a flow of electric charges. Once the charges are in motion a magnetic field exists. The above effect is called the Oersted effect. Compass S [C] Chemical Effect: An electric current can cause a chemical reaction. Demonstration: Chemical Effect of an electric Current Sodium chloride dissolves in water into positive and negative ions. The positive ions are attracted by the cathode and the negative ions by the anode. This constitutes a flow of current which is registered on the ammeter. In distilled water there is an absence of ions and so no current flows. A Anode Cathode Effects of an Electric Current Everyday examples of the heating effect Electric heaters, cookers, hairdryers, kettles... Advantage of EHT in transmission of electrical energy Transformers can be used to step up the voltage to extra high tension (EHT) while also reducing the current. According to P = VI they continue to transmit the same power. The advantage of reducing the current is that the energy losses due to heating ( P = I 2 R ) are greatly reduced. Everyday examples of the chemical effect Electroplating, extracting metals from ores, purifying metals. © Dublin School of Grinds Page 46 Kieran Mills & Tony Kelly 4.7 Domestic Circuits Electricity is supplied to all homes at 220 V A.C. (alternating current) which is at a frequency of 50 Hz. It is supplied via two cables called the live and neutral wires. Live Wire (Colour code: Brown): The potential of this wire goes alternately negative and positive, making the current flow backwards and forward through the circuit. The potential varies from +311 V to −311 V relative to earth. This variation takes place 50 times per second. The effective voltage or root mean square value is 240 V. Neutral Wire (Colour code: Blue): The Electricity Board earths the neutral wire by connecting it to a metal plate. Earth Wire (Colour Code: Yellow and green stripes): This is a safety wire which connects the metal body of a kettle to earth and prevents it becoming live if a fault develops. If the live wire were to touch the body of the kettle, a current would flow to earth and blow the fuse. If there was no earth wire, the body of the kettle would remain live and be a source of danger. 3-Pin Plugs Plugs provide a convenient and safe method of connecting different appliances into a mains circuit. Fused plugs are normally fitted with either 3 A or 13 A fuses, though others are available. The fuse value must be greater than the current that normally flows through the appliance, but as close as possible to this value so that the fuse will blow before an overheating cable can cause a fire. Example 17: An appliance is rated 240 V, 480 W. What fuse should be connected to its plug, given they come in values of 3 A and 13 A? Solution Paying for Electricity The electrical cables entering a home pass into a meter which records the total electrical energy used by the consumers. For example, a 60 W lamp uses 60 J of energy per second. If the lamp is on for 10 s, its total energy consumption is 600 J. Therefore to work out the electrical energy used by an appliance its power rating is multiplied by the time it is on. The E.S.B. charges electricity used according to the number of kilowatts-hours used. Definition: One kilowatt-hour is the amount of electrical energy used when a 1 kW appliance is on for one hour. The kW-hr is called a unit of electricity and electricity is charged according to the number of units used. © Dublin School of Grinds Page 47 Kieran Mills & Tony Kelly Example 18: Calculate the cost of using a 4.2 kW immersion heater for four and a half hours, a 60 W bulb for 6 hours and a 2.8 kW kettle for 15 minutes if one unit costs 8 cent. Solution Main Fuse Meter Distribution (Fuse) Box L 6A 30A 16A N Live wire Neutral wire Earth wire E To cooker L N RING MAIN CIRCUIT L N E N L E Miniature Circuit Breakers (MCBs): Usually used instead of fuses in the distribution box. MCBs contain a bimetallic strip and an electromagnet. The bimetallic strip causes the switch to trip for small currents and the electromagnet does it for large currents. They operate faster than a fuse and can be reset by flicking a switch. Residual Current Devices (RCDs): They protect against electrocution when a fuse or MCB acts too slow. An RCD detects a difference in the current between the live and neutral which may arise if someone touched the live so that current flowed through them to earth. The effect of this difference is to cause the RCD to trip which disconnects the live from the circuit. Bonding: All metal pipes, taps etc.. must be connected or bonded to earth to prevent people getting electrocuted. © Dublin School of Grinds Page 48 Kieran Mills & Tony Kelly Radial Circuit: Appliances that take a large current like electric cookers have a separate live and neutral wire coming from the distribution (fuse) box. This is called a radial circuit and it has its own fuse. Ring Circuit: The live terminals of each socket are connected together in a ring so that current can flow both ways to the socket reducing the current in the wires. Light Circuits: Lights are connected in parallel so that if one light blows the other lights stay on. Answers to Examples Example 1: 1.875 ×1021 Example 2: 200 C, 1.67 A, 12 V Example 3: 1.875 W Example 4: 4 ×10−4 A Example 5: 7.4 ×10−4 m Example 6: (a) 0.5 A, (b) 3 V, 6 V Example 7: (a) 1.5 A, (b) 0.9 A, 0.6 A Example 8: (a) 2 A, (b) 3.75 V, (c) 0.75 A, (d) 1.25 A, (e) 1.25 A, (f) 2.5 V, (g) 1.25 V, (h) 1 W Example 9: (a) 3.875 W, 3.1 A, (b) 5.8 V, (c) 1.2 A Example 10: (a) 1.75 V, 10.25 V, (b) 7.7 ×10−4 C, (c) 5.4 ×10−3 J Example 11: 0.2 A, 4 V, 4.125 V, D to B, 10 W Example 12: 27.86 W Example 13: (a) 280 W, (b) 4 V, (c) 10 V, (d) 40 mA, (e) 12.73 V Example 14: Minimum: 0 V; Maximum: 8 V Example 15: 2.2 A, 29 040 J Example 16: 0.2 A Example 17: 3 A fuse (I = 2 A) Example 18: 1.60 euro © Dublin School of Grinds Page 49 Kieran Mills & Tony Kelly Numerical Problems 4. If the heating element of an electric radiator takes a current of 4 A, what charge passes each point every minute? How many electrons pass a given point in this time? Constants −19 Charge on electron e = 1.6 × 10 C −8 Copper r = 1.70 10 W m −8 Aluminium r = 2.80 × 10 W m −8 Silver r = 1.60 × 10 W m −8 Iron r = 14.0 × 10 W m −8 Platinum r = 11.0 × 10 W m 5. The current through a conductor is 4 A when the potential difference (p.d.) across it is 20 V. Calculate the resistance of the conductor. Basic Calculations 6. What p.d. will produce a current of 5 A in a 12 Ω resistor? 1. (a) Calculate the work done in bringing a charge of 4.5 mC through a potential difference of 20 V. 7. What current flows through a resistance of 100 Ω when it is connected to a 230 V supply? (b) The potential difference between two points is 6 V. How much work is done when a charge of 1000 C is transferred from one point to the other. (c) 0.8 J of energy is expended in moving a 0.5 mC charge from one point to another. Calculate the potential difference between these two points. (d) Calculate the charge transferred by a current of 3 A flowing for six minutes. (e) How long would it take a current of 6 A to transfer a charge of 4000 C? 2. A certain quantity of charge does 3000 J of work when passing through a resistor which has a potential difference of 10 V across its ends. It takes 2 minutes for the charge to be transferred from one point to the other. Calculate the quantity of charge, the current and the resistance of the resistor. 3. The rating on a torch bulb is 0.3 A, 4 V. How many electrons are removed from the negative terminal of a 4 V battery per minute when the bulb is connected across it? © Dublin School of Grinds 8. A current of 4 A flows through the filament of a car headlight bulb when it is connected to a 12 V battery. Find the resistance of the filament. 9. Find the potential difference across a 6 Ω resistor when it is carrying a current of 5 A. 10. At a certain temperature, the current through a conductor is 3 A when the p.d. across it is 24 V. Find the resistance of the conductor. When the temperature of the conductor is raised, the same p.d. causes a current of 2 A to flow through it. Find the increase in its resistance. Resistivity 11. Calculate the resistance of a circular wire made of Silver of length 25 m and crosssectional area of 3.14 × 10−6 m2. 12. Calculate the length of a platinum wire of cross-sectional area of 6.28 × 10−6 m2 whose total resistance is 5 Ω. 13. Find the average cross-sectional area of a conductor whose resistance is 10 Ω, whose length is 5 m and is made of a material whose resistivity is 2 × 10−8 Ω m. Page 50 Kieran Mills & Tony Kelly 14. A circular wire has a diameter of 1 mm. If a 100 m length of it has a resistance of 10 Ω calculate the resistivity of the material in the wire. 15. What is the resistance of a Copper wire of length 50 cm and mean diameter 0.1 mm? Resistors in Series and Parallel 23.Calculate (a) the current, I (b) the potential difference between A and B, VAB, and the potential difference between B and C, VBC. 12 V 16. Calculate the diameter of a Copper wire 2.5 m long if its resistance is 0.1 Ω. 17. What is the length of the Copper wire in a galvanometer coil if its diameter is 0.1 mm and the resistance of the coil is 100 Ω? 18. Calculate the cross-sectional area of a 10 m length of Copper wire which carries a current of 4.8 A when a potential difference of 12 V is maintained across its ends. If the wire has a circular cross-section, calculate its diameter. 19. A wire of circular cross-sectional area has a diameter of 3 mm. If the resistance of 5 m of it is 0.05 Ω calculate its resistivity. A 22. The resistance of a wire is 0.8 Ω. Its length is 21 cm and diameter 4 mm. Calculate the resistivity of the material. 12 C 24. If a current of 0.5 A flows what is the value of R? What is VAB? What value of R would cause a current of 0.25 A to flow? 12 V A 7 R B 12 25. Calculate (a) the current, I and (b) the current flowing in each resistor. 9V 20. A wire 100 m long and 2 mm in diameter has a resistivity of 4.8 × 10−8 Ω m. What is the resistance of the wire? 21. An Aluminium bar 2.5 m long has a rectangular cross-section 1 cm by 5 cm. (a) What is its resistance? (b) What would be the length of an Iron wire 15 mm in diameter having the same resistance? B 4 I I 6 10 26.Calculate: (a) the total current I, (b) the potential difference between A and B, (c) the current flowing in the 4 Ω resistor. I 9V I 4 3 6 9 © Dublin School of Grinds Page 51 B A 8 Kieran Mills & Tony Kelly 36. Calculate the maximum safe current for (a) a 2 MΩ resistor of power 5 W and (b) a 1 kΩ resistor of power 2 W. 27. Calculate the current I leaving the battery, the potential difference between A and B, VAB, and the current passing through the 6 Ω resistor. 37. A 10 Ω resistor dissipates 22.5 W as heat. Calculate (a) the current, and (b) the potential difference across the resistor. 9V 6 A 38. What is the maximum current you could safely pass through a 500 Ω, 5 W resistor? 4 B 2 3 1 Wheatstone/Metre Bridge 28. In a balanced Wheatstone bridge R1 = 4 Ω, R2 = 6 Ω and R3 = 15 Ω. Find the value of R4. 30. In a balanced Wheatstone bridge R1 = 10R2 and R3 = 4 Ω. Find the value of R4. 42. A cell of EMF 5 V passes a steady current of 0.1 A for 40 minutes. What is the energy stored in the cell? 31. A metre bridge is balanced when l = 42 cm and l = 58 cm. If R = 50 Ω calculate the value of R . 2 40. Calculate the resistance of a coil which produces 1500 J of energy when a current of 2 A flows through it for 90 s. 41. Two lamps, one taking 40 W and the other taking 100 W when connected separately to a 200 V supply, are now connected in series to the 200 V supply. Find the resistance of each lamp and the total power in Watts taken by the two lamps. 29. In a balanced Wheatstone bridge R1 = 10 Ω, R2 = 20 Ω and R4 = 60 Ω. Find the value of R3. 1 39. A 50 Ω resistor can safely pass a current of 0.1 A. Calculate the power rating of the resistor. 1 2 43. Show that the power dissipated when a potential difference of V volts is 3V 2 connected across AB is P = . 2R Power/Domestic electricity 32. Calculate the current flowing through a 1 kW heater when it is connected to the mains supply (220 V). Find also the resistance of the element of the heater. A 33. A current of 0.26 A flows through a bulb when connected to a 220 V supply. What is the wattage and resistance of the bulb? 34. What is the maximum current you could safely pass through a 50 Ω, 2 W resistor? 35. A small heater with a resistance of 100 Ω is connected to a 220 V supply. Calculate the current and the heat energy it gives off per minute. © Dublin School of Grinds R R R R R R B 44. Each of the 2 Ω resistors can dissipate a maximum of 18 W without becoming excessively heated. What is the maximum power the circuit can dissipate? Page 52 2 2 2 Kieran Mills & Tony Kelly 45. Two lamps, marked ‘60 W, 120 V’ and ‘40 W, 120 V’ are connected in series across a 120 V line. What power is consumed in each lamp? 46. Two resistors are rated 20 W, 6 V and 30 W, 6 V. Calculate the power consumed by each resistor when they are connected in series to a 6 V source. 54. In a model of a power line, a 12 V A.C. supply of negligible resistance is connected by wires of total resistance 4 Ω to a lamp of resistance 6 Ω. Calculate: (a) the current flowing in the wires, (b) the power loss in the wires, (c) the voltage drop along the wires, (d) the power converted in the lamp. 47. A current of 3.6 A flows through a resistor R for 30 s. when a potential difference of 4.5 V is maintained across it. Calculate the quantity of charge passed in this time, the number of electrons that this charge corresponds to and the work done by these electrons. Also calculate the resistance of R and the power dissipated in it. (Charge on electron e = 1.6×10−19 C). 48. A lamp has a power rating of 100 W. It operates on 230 V mains. What current does it draw? What size fuse should be put in its plug, a 3 A or a 13 A? 49. An electric cooker has four 500 W plates, a 2 kW grill and a 3 kW oven. It operates on 230 V mains. Is a 40 A fuse suitable for this cooker? 50. What fuse should be used in a plug connected to a 2.5 kW kettle connected to the mains (220 V), a 3 A or a 13 A? 51. A 2000 W appliance operates for 3 hours. How many kilowatt-hours of energy does it use? 52. A 75 W lamp is operating for 40 minutes. How many kilowatt-hours of energy does it use? 53. Calculate the cost of using a 4.2 kW immersion heater for four and a half hours, a 60 W bulb for 6 hours, a 1250 W toaster for 20 minutes and a 2.8 kW kettle for 10 minutes if one unit costs 8 cent. © Dublin School of Grinds Page 53 Kieran Mills & Tony Kelly Answers -5 6 1. (a) 9 × 10 J; (b) 6000 J; (c) 1.6 × 10 V; (d) 1080 C; (e) 666.7 s 30. 0.4 Ω 31. 69.05 Ω 2. 300 C, 2.5 A, 4 Ω 20 3. 1.125 × 10 32. 4.55 A, 48.4 Ω 33. 57.2 W, 846.2 Ω 4. 240 C, 1.5 × 10 5. 5 Ω 34. 0.2 A 35. 2.2 A, 29 040 J 6. 60 V 7. 2.3 A 36. (a) 1.58 mA, (b) 45 mA 37. (a) 1.5 A, (b) 15 V 8. 3 Ω 38. 0.1 A 39. 0.5 W 21 9. 30 V 10. 8 Ω, 4 Ω 11.0.127 Ω 40. 4.17 Ω 12.285.5 m 42. 1200 J 13.10 m2 -8 14.7.85 × 10 Ω m 44. 27 W 45. 9.6 W, 14.4 W 15.1.08 Ω 16.0.74 mm 46. 7.2 W, 4.8 W 47. 108 C, 6.75×1020, 486 J, 1.25 Ω, 16.2 W 17. 46 m −8 18.6.8 × 10 m2, 0.294 mm 48. 0.43 A, 3 A 49. 30.4 A, Yes −8 19.7.07 × 10 Ω m 20.1.53 Ω 50. 13 A fuse ( I = 11.36 A ). 41. 1000 Ω, 400 Ω, 28.57 W -8 21. (a) 1.4 × 10 Ω, (b) 17.7 cm −5 22.4.8 × 10 Ω m −4 23. (a) 0.75 A (b) 3 V, 9 V 24.5 Ω, 2.5 V, 29 Ω 51. 6 kW-hr 52. 0.05 kW-hr 53. 1.61 euro 54. (a) 1.2 A, (b) 5.76 W, (c) 4.8 V, (d) 8.64 W 25. (a) 2.4 A, (b) 1.5 A, 0.9 A 26. (a) 2.1 A, (b) 5.6 V, (c) 1.4 A 27.1.7 A, 2.55 V, 0.425 A 28. 22.5 Ω 29. 30 Ω © Dublin School of Grinds Page 54 Kieran Mills & Tony Kelly Electrical Circuits Revision This set of questions is excellent if you feel you need a deeper understanding of electrical circuits. Question 6: Find the potential difference (voltage) between A and B. Question 1: Find the current I. V = 10 V V=6V 2 A B Question 2: Find the current I. 4V V = 100 V Question 7: Find the potential difference (voltage) between A and B. V = 10 V 4 Question 3: Find the current I. A B V=3V 4V Question 8: Find the potential difference (voltage) between A and B. 3 V=6V Question 4: Find the current I. V=2V V=2V A B 10 Question 9: Find the potential difference (voltage) between A and B. Question 5: Find the current I. V = 10 V 2 V = 10 V 3 A B 6V © Dublin School of Grinds Page 55 Kieran Mills & Tony Kelly Question 10: Find the potential difference (voltage) between A and B. V = 12 V A B Question 14: Find (a) the current I, (b) the potential difference between A and B, (c) the potential difference between B and C, (d) the potential at A, (e) the potential at B, (f) the potential at C. V = 12 V 2V 4V Question 11: Find the potential difference (voltage) between A and B. V = 10 V A B 2 C 4 Question 15: If the current os 0.25 A, find (a) the value of R, (b) the potential difference between A and B. V=6V B A 4V Question 12: Find the potential difference (voltage) between A and B. V = 10 V 6 A R B 4 Question 16: Find (a) the current I, (b) the current in the 6 W resistor, (b) the current in the 12 W resistor. V=6V B A I I 2V 6 2V Question 13: Find the potential difference (voltage) between A and B. 12 V = 10 V A B 3V © Dublin School of Grinds 4V Page 56 Kieran Mills & Tony Kelly Question 17: Find (a) the current in the 8 W resistor, (b) the current in the 12 W resistor, (c) the current in R, (d) the potential difference across R, (e) the value of R. Question 19: Find (a) the current in the 8 W resistor, (b) the current in the 4 W resistor, (c) the current in the 2 W resistor, (d) the current in R, (e) the voltage across the 2 W resistor, (f) the voltage across the R resistor, (g) the value of R. V=6V V = 12 V 4V 8 8 R 4 12 2 2V Question 18: Find (a) the current in the 3 W resistor, (b) the voltage across the 6 W resistor, (c) the current in the 6 W resistor, (d) the current in the 4 W resistor, (e) the current in R, (f) the voltage across the 4 W resistor, (g) the voltage across the R resistor, (h) the value of R. Answers 1. 5 A 2. 25 A 3. 1 A 4. 0·4 A 5. 2 A 6. 2 V 7. 6 V 8. 6 V 9. 4 V 10. 6 V 11. 6 V 12. 6 V 13. 3 V 14. (a) 2 A, (b) 4 V, (c) 8 V, (d) 12 V, (e) 8 V, (f) 0 V 15. (a) 14 W, (b) 3·5 V 16. (a) 1·5 A, (b) 1 A, (c) 0·5 A 17. (a) 14 A, (b) 16 A, (c) 125 A, (d) 4 V, (e) 9·6 W 18. (a) 1 A, (b) 3 V, (c) 0·5 A, (d) 0·5 A, (e) 0·5 A, (f) 2 V, (g) 1 V 19. (a) 0·5 A, (b) 2 A, (c) 1·5 A, V=6V 6 3 4 R 3V © Dublin School of Grinds R Page 57 (d) 1·5 A, (e) 3 V, (f) 1 V, (g) 23 W Kieran Mills & Tony Kelly Chapter 5 • Electromagnetism 5.1 Magnetism Magnetism is produced by the motion of electric charge. This statement can be illustrated as follows: When S is closed, the compass needle deflects showing the presence of a nearby magnetic field. When S is opened, the compass returns to its normal orientation of pointing to magnetic north. The nearby magnetic field has disappeared. Compass S Explanation: An electric current is simply a flow of electric charges. Once the charges are in motion a magnetic field exists. The above effect is called the Oersted effect. Horse-shoe magnet The magnetic fields near the ends of a bar magnet and horse-shoe magnet are strong. These are the poles of the magnet. N S Bar Magnet N S Magnets exhibit both attractive and repulsive forces. A magnet is said to have a North and a South pole. If the N-pole of one magnet is brought near the N-pole of another magnet, repulsion occurs. The same is true of two S-poles. If opposite poles are brought close together, attraction occurs. S Like poles repel; unlike poles attract. N N S Repulsion S N S N Attraction Electromagnets and their uses Used in scrap-yards to lift metal scrap, electric motors and electromagnetic relays. Magnetic Fields A piece of iron close to the magnet will experience a force due to its magnetic field. The field is described by a number of lines called magnetic field lines or flux lines. Definition: The space around a magnet is a store of magnetic energy that is called the magnetic field. Definition: The direction of a magnetic field at any point is taken to be the direction of the force on a N-pole placed at that point. © Dublin School of Grinds Page 58 Kieran Mills & Tony Kelly A simple experiment which describes such a line is as follows: A magnetised needle is pushed through a cork and floats on water with its N-pole uppermost. When the needle is held near the N-pole of the magnet and then released, it is repelled and travels towards the S-pole along a curved path which represents the direction of the magnetic field. A magnetic field line moves from the North pole to the South pole. Demonstration: Magnet Field due to a Bar Magnet Fields can be examined using iron filings or a plotting compass. The magnetic field lines form closed loops leaving the N-pole and entering the S-pole where they are directed through the magnet and out from the N-pole again. This is the field pattern between two bar magnets of opposite poles. Demonstration: Magnet Field due to a Current 1. Field due to a current in a straight wire A straight wire carrying current is passed through a piece of cardboard and the field is examined using ironfilings or a compass. The pattern of the filings on the card shows that the field lines are circles centred about the wire. 2. Field due to a short circular coil or loop The field is similar to that produced by a short bar magnet and the coil acts as if it has a N-pole on one face and a S-pole on the other. It behaves as a magnetic dipole. 3. Field due to a solenoid A solenoid is a long coil made up of a number of turns of wire. The field is similar to a long bar magnet. The magnetic field can be made much stronger by inserting a piece of iron in the centre of the solenoid. This is the basis of the electromagnet. © Dublin School of Grinds Page 59 Kieran Mills & Tony Kelly Earth’s magnetic field The earth is surrounded by a giant magnetic field with a magnetic north and south. The true magnetic north, about which the earth spins, is not quite in the same place as the magnetic north. The angle between them is called the angle of declination. Using a magnetic compass and having knowledge of the magnetic declination allow mariners to navigate around the world. 5.2 Current in a Magnetic Field A current flowing in a conductor sets up a magnetic field around it (Oersted Effect). If a current carrying conductor (c.c.c.) is placed inside the permanent magnetic field of a horse-shoe magnet, there will be a force between the two magnetic fields causing the c.c.c. to experience a force. Demonstration: Force acting on a c.c.c. in a magnetic field A wire carrying current (c.c.c.) is passed at right angles to the magnetic field of a horse shoe magnet. The c.c.c. experiences a force pulling it down or pushing it up depending on the direction of the current. If the current is increased using the rheostat, the force on the wire increases. N S S Fleming’s left-hand rule works out the direction of the thrust or force on the c.c.c. Fleming’s left-hand rule: Place the forefinger, centre finger and the thumb of the left hand mutually at right angles. If the forefinger then points in the direction of the field (North to South) and the centre finger in the direction of the current (+ to −), the thumb will point in the direction of the thrust. Thumb (Thrust) Forefinger (Field) Centre finger (Current) By experiment it is found that the force produced depends on the following factors: 1. The size of the current I : F ∝ I. 2. The length of the conductor in the field: F ∝ l. 3. The strength of the magnetic field of the magnet. The magnetic field strength is also called the magnetic flux density and it is denoted by B. Combining these results, F = BIl is obtained. © Dublin School of Grinds Page 60 Kieran Mills & Tony Kelly F = BIl F: Force on the wire (N) B: Magnetic field intensity of the magnet (Tesla, T) I: Current (A) l: Length of wire in the magnetic field (m) Formulae and Tables Book: Page 62 Force on a current-carrying conductor (Electricity) Note: The above equation is true for a c.c.c. perpendicular to the field. The force is less for other angles and is zero when the c.c.c. is parallel to the field. Example 1: A straight wire of length 2 m carrying a current of 4 A experiences a force of 10 N when placed perpendicular to a uniform magnetic field. What is the magnetic flux density at the wire? Solution Example 2: A rectangular loop of wire is free to rotate about an axis parallel to its longer sides. The plane of the loop is parallel to a uniform magnetic field of magnetic flux density B. The dimensions of the loop are 20 cm by 15 cm and the direction of B is perpendicular to the longer sides. The magnitude of B is 0.44 T and a current of 2.6 A flows around the loop. Calculate: (a) the magnitude of the force acting on one of the longer sides of the loop, (b) the moment of force about the axis. Solution © Dublin School of Grinds Page 61 Axis 20 cm B 15 cm Kieran Mills & Tony Kelly Force on a moving charge in a magnetic field There is already a formula for the force on a c.c.c. in a magnetic field. However, a conductor carrying a current is simply a number of charged particles moving in a wire. Proof: Force on a Moving Charge in a Magnetic Field Consider a length l of conductor containing n charged particles each of charge q and average drift velocity v. Each particle will take the same time t to travel the distance l. l v Total Charge Q = nq Q l F = BIl = B l = Bnq = nBqv t t For a single charge, n = 1 ⇒ F = Bqv Hence, the force on a single charge is: Formulae and Tables Book: Page 62 F = Bqv F: Force on the charge (N) B: Magnetic field intensity Force on a charged particle (Electricity) of the magnet (T) q: Quantity of the charge (C) v: Velocity of the charge (m s−1) Example 3: An electron of charge 1.6 × 10−19 C enters a uniform magnetic field of flux density 3 T and moves at right angles to the field. If the force on the electron is 1.5 × 10−18 N calculate the speed of the electron. Solution © Dublin School of Grinds Page 62 Kieran Mills & Tony Kelly Example 4: A charge of 2 × 10−19 C travelling at 4.2 × 105 m s-1 enters a uniform magnetic field of flux density 2.5 T and moves at right angles to the field. Calculate the force acting on the charge and the length of radius of the path followed by the charge, given that the mass of the charged particle is 3·5 × 10-27 kg. Solution B v q F r r Defining the Ampere When currents flow through two conductors placed near each other they either attract or repel. This is because each current carrying conductor is surrounded by a magnetic field. If the currents flow in opposite directions, the wires repel each other. If the currents flow in the same direction, they attract each other. Such a phenomenon is used to define the unit of current, the Ampere (or Amp, for short). Demonstration: Defining the Ampere Set up the apparatus shown with a battery connected to two parallel strips of aluminium. A rheostat controls the current through the circuit. A wooden block keeps each half of the alumimum foil separated. When the circuit is turned on, the aluminium foils repel each other. This is the principle on which the definition of the Ampere is based. © Dublin School of Grinds Page 63 Rheostat Aluminium strip I F I F Wooden block Kieran Mills & Tony Kelly Definition: The Ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 m apart in a vacuum, causes each to exert a force of 2 × 10−7 N per metre length on the other. Electromagnetism A current carrying conductor in a magnetic field experiences a force. This is the principle behind the operation of the motor, electromagnetic relay and the loudspeaker. 5.3 Electromagnetic Induction An electric current flowing in a conductor produces a magnetic field. An electric current is a flow of charge (a changing electric field). When the charge ceases to flow the magnetic field disappears. A changing electric field produces a magnetic field. What about the opposite effect? Could a changing magnetic field produce an electric field? The following experiment showed this to be true: A bar magnet is thrust inside a solenoid connected to a galvanometer. The galvanometer needle deflects showing that a current flowed in it. The current only flows when the magnet moves or when there is a changing magnetic field. To cause a current to flow in the solenoid an EMF or voltage must have been generated inside the solenoid. v G The conclusion from this experiment must be that: A changing magnetic field produces an electric field. Therefore, an EMF is generated and a current flows. This is electromagnetic induction or the generation of electrical energy. Mechanical energy in the movement of the magnet has been converted into electrical energy. Electromagnetic Induction Electromagnetic Induction is the principle on which the electrical generator is based. Generators are used in power stations, alternators in cars and dynamos on bicycles. © Dublin School of Grinds Page 64 Kieran Mills & Tony Kelly Magnetic Flux Surrounding a magnet there is a magnetic field. This is the space around a magnet where magnetic forces can be experienced. The field can be represented by a number of lines called flux lines. These lines are directed from the North to the South pole. The number of flux lines that are drawn is determined by the strength of the magnet. Therefore, a strong magnet would have many flux lines close together (a high flux density). The number of flux lines per unit cross-sectional area is equal to the strength B of the magnetic field. B is called the magnetic flux density. The total magnetic flux is denoted by F (phi). ∴B = F = BA Φ A F: Magnetic Flux (Weber, Wb) B: Magnetic Field Strength or Magnetic Flux Density (T) A: Area (m2) Formulae and Tables Book: Page 62 Magnetic flux (Electricity) Example 5: A magnetic field of strength 1.5 T is directed through a cross-sectional area of 0.8 m2. Find the magnetic flux through the area when the field lines are perpendicular to the area. Solution Definition: The magnetic flux F is the number of magnetic field lines passing through a certain area A whose surface is perpendicular to the field lines. Definition: The magnetic field strength B is the number of flux lines per unit cross-sectional area. © Dublin School of Grinds Page 65 Kieran Mills & Tony Kelly Cause and Effect Faraday carried out an experiment to find out the factors on which the size of the EMF depended. He moved the magnet into the solenoid. The effect is that the galvanometer needle deflects. What is changing in the circuit to cause this effect? The number of flux lines linking the circuit is changing with time as the magnet approaches. v G Effect: An EMF e is produced in the circuit. Cause: There is a change in flux linking the circuit. Definition: Electromagnetic induction is the production of an EMF in a circuit when the magnetic flux linking the circuit changes. Faraday found that the faster the magnet moves, the greater the deflection on the galvanometer. In other words, the size of the EMF generated depends on the rate at which the flux lines cut or link the circuit. dΦ He showed experimentally that: ε ∝ dt He also found that the size of EMF depended on the number of turns of wire N in the solenoid the more turns, the greater the EMF. Faraday’s formula now becomes: ε =N dΦ dt e: N: dF: dt: EMF or Voltage induced in the coil (V) Number of turn in the coil Change in the magnetic flux linking the circuit (Wb) Time over which this change takes place (s) Definition: Faraday’s Law of electromagnetic induction states that the size of the EMF, |e|, induced in a circuit is directly proportional to the rate of change of magnetic flux linking the circuit. Lenz’s Law When a current flows in the coil it sets up its own magnetic field. This magnetic field opposes the magnetic field of the incoming magnet. It is the energy expended by the moving magnet in overcoming this repulsive force that is converted to electrical energy. When the magnet is withdrawn current flows in the opposite direction. This sets up a magnetic field in the opposite direction which attracts the moving magnet. N S N S N S S N Definition: Lenz’s Law states that the EMF induced in a circuit acts in such a direction as to oppose the change producing it. Faraday’s Law and Lenz’s Law are the Laws of Electromagnetic Induction. © Dublin School of Grinds Page 66 Kieran Mills & Tony Kelly Demonstration: Faraday’s Law Set up the circuit shown with a solenoid attached to a galvanometer. The galvanometer measures the induced current and hence, the EMF in the circuit. A magnet is moved into the solenoid. v G The faster the magnet is moved into the solenoid the larger the deflection on the galvanometer showing that the induced EMF is directly proportional to the rate of change of magnetic flux linking the circuit. If the number of turns of wire are increased, the galvanometer deflection increases for a magnet moving at the same speed. Demonstration: Lenz’s Law Set up the circuit shown with a solenoid attached to a galvanometer. As the magnet is moved into the solenoid there is a force repelling its movement. As the magnet is withdrawn there is a force attracting the magnet. S N An EMF is induced in the circuit which acts in such a direction as to oppose the change producing it. N N S N S S In many problems on electromagnetic induction, it is convenient to combine two formulae: F = BA ε =N dΦ dt ε= BAN t e: EMF or Voltage induced (V) B: Magnetic Flux Density (T) A: Area (m2) N: Number of turns t: Time (s) Example 6: A planar loop of wire consists of a single turn and has a cross-sectional area of 50 cm2. The plane of the loop is perpendicular to a magnetic field that increases uniformly in magnitude from 1.5 T to 3.5 T in 2.5 s. What is the resulting induced current if the loop has a total resistance of 10 Ω. Solution © Dublin School of Grinds Page 67 Kieran Mills & Tony Kelly 5.4 Alternating Current (AC) Direct current (D.C) is the current produced by batteries. The current is constant flowing in the same direction. Alternating current (A.C.) or alternating EMF is one which varies periodically with time in direction and magnitude. It results from the way electricity is produced by generators. The time for a complete cycle is the periodic time T. This is one complete revolution of the coil. The number of complete cycles per second is the frequency f. The relationship between f and T is given by: 1 f = T f: Frequency (Hz) T: Periodic time (s) Direct current (DC) Current (A) I 0 Time (s) Alternating current (AC) Current (A) Time for 1 cycle, T Io 0 Io Time (s) The value of A.C. varies from one instant to another. An average current called the root mean square value (RMS) of the current is taken. This works out to be: I RMS I = 0 2 Io: Peak value of the current (A) IRMS: Root mean square value of the current (A) Formulae and Tables Book: Page 62 Alternating current (Electricity) Definition: The root mean square (RMS) current is the value of steady direct current which would dissipate the same heat at the same rate as alternating current in a given resistance. Example 7: The AC flowing in a wire of resistance 10 Ω produces heat at the rate of 80 W. Find: (a) the RMS value of the current, (b) the RMS voltage, (c) the peak value of the voltage across the wire. Solution National Grid and Alternating Current (AC) AC is produced by generators and power stations. Using transformers, it is fed into the national grid at high voltages (EHT) to reduce energy losses. © Dublin School of Grinds Page 68 Kieran Mills & Tony Kelly Demonstration: Using an Oscilloscope to show AC A signal generator produces AC. If it is connected to a Cathode Ray Oscilloscope (CRO), a picture of the AC wave is seen. By adjusting the frequency control, more waves per second will be produced. By adjusting the amplitude control, the height of the waves changes. CRO Signal Generator Amplitude Control Frequency Control 5.5 Mutual and Self Induction Iron core Mutual Induction An alternating current is fed into coil A (primary coil). The changing current produces a changing magnetic field which passes through an iron core. This changing magnetic field induces a current in coil B (secondary coil). The iron core has the effect of localising the flux lines so that most of them cut through coil B. It prevents flux leakage. Coil A Coil B AC G Primary Coil Secondary Coil Definition: Any arrangement of two coils such that a current is induced in one when the current is changed in the other is said to possess mutual inductance. Demonstration: Mutual Induction Set up the circuit. As the switch is opened and closed in the primary circuit, the galvanometer needle deflects back and forth about the zero position. Coil A Coil B G S Primary Coil © Dublin School of Grinds Page 69 Secondary Coil Kieran Mills & Tony Kelly A.C. with Inductors Consider an A.C. source connected to a coil of wire called an inductor. The current increases first of all in one direction and then in the opposite direction. As a result the magnetic flux around the inductor is constantly changing. This changing magnetic flux, according to Faraday’s Law, induces an EMF in the coil itself called a back EMF. This back EMF opposes the EMF from the source. Therefore, the current never gets very high and the bulb is dim. AC S Inductor Iron core If an iron core is inserted inside the coil, this has the effect of concentrating the magnetic field around the coil. As a result a larger back EMF is induced in the coil and the bulb gets dimmer. Definition: Self induction is the production of a back EMF in a coil whenever the current in that coil changes. Demonstration: Self Induction The high EMF of self-induction can be demonstrated by connecting a Neon discharge lamp across the terminals of an electromagnet. A Neon lamp requires at least 180 V to cause it to glow, yet it flashes every time the current in the coil is switched off. S Example 8: A coil consists of 200 turns of wire of total resistance 400 Ω and is connected to an AC supply. Over a certain time period of 1 ms, the flux threading each turn of the coil increased by 4 × 10-4 Wb. Calculate (a) the average induced EMF over the 1 m s period, (b) the average current in the coil if the average applied voltage over the 1 m s period is 100 V. Solution © Dublin School of Grinds Page 70 Kieran Mills & Tony Kelly Uses of inductors • Dimmer switches in stage lighting • Tuning circuits in radios • Smoothing slight variations in DC from power units The Transformer A transformer changes (transforms) an alternating potential difference (voltage) from one value to another of greater or smaller value using the mutual induction principle. Construction Two coils, called the primary and secondary coils, are wound on a complete soft iron core. The purpose of the iron core is to ensure that the magnetic flux produced by current flowing in the primary completely links the windings of the secondary (there is no flux leakage). Principle of Operation: The transformer is based on the mutual induction principle where a change of current in one coil induced a change of current in another coil. Primary Coil Secondary Coil VP VS Iron Core Transformer Formulae VP: Primary voltage (V) VS: Secondary voltage (V) NP: No. of turns in the primary NS: No. of turns in the secondary VP N P = VS N S Formulae and Tables Book: Page 62 Transformer (Electricity) In a transformer that is 100% efficient: Power into primary = Power out of secondary. (Voltage × Current) = (Voltage × Current) P VP × I P = VS × I S S VP: Primary voltage (V) VS: Secondary voltage (V) IP: Primary current (A) IS: Secondary current (A) Step-up transformer: If the secondary coil has more turns than the primary, the alternating voltage produced across the secondary coil will be greater than across the primary. Step-down transformer: If the secondary coil has fewer turns than the primary, the alternating voltage produced across the secondary coil will be less than across the primary. © Dublin School of Grinds Page 71 Step-up NP < NS Step-down NP > NS Kieran Mills & Tony Kelly Uses of Transformers • Generating stations generate electricity between 20 kV and 30 kV. This is stepped up to 220 kV or 400 kV for distribution on the national grid. It is then stepped down to 230 V at substations for home comsumption. All of this is achieved by transformers. • Computers, TVs and radios all contain transformers for supplying the correct voltages to their electronic parts. Example 9: The primary/secondary turns ratio in a transformer is 8:1. The power input to the primary is 80 W. If a current of 2.5 A flows in the secondary coil, calculate (a) the potential difference across the secondary coil, (b) the potential difference across the primary coil and (c) the current flowing in the primary coil. Assume 100% efficiency. Solution Answers to Examples Example 1: 1·25 T Example 2: (a) 0·2288 N, (b) 1·72 × 10-2 N m Example 3: 3·125 m s-1 Example 4: 2·1 × 10-13 N, 2·94 × 10-3 m Example 5: 1·2 Wb Example 6: 0·4 mA Example 7: (a) 2·83 A, (b) 28·3 V, (c) 40 V Example 8: (a) 80 V, (b) 0·05 A Example 9: (a) 32 V, (b) 256 V, (c) 0·3125 A © Dublin School of Grinds Page 72 Kieran Mills & Tony Kelly Numerical Problems 1. A straight horizontal wire of length 50 cm and carrying a current of 5 A is placed at right angles to a horizontal magnetic field of flux density 0.8 T. Calculate the force on the wire. 2. A straight piece of wire of length 3 m carrying a current of 2 A experiences a force of 12 N when placed perpendicular to the magnetic field. Calculate the magnetic flux density. 3. A wire 50 cm long and carrying a current of 25 A is at right angles to a magnetic field of flux density 5 × 10−4 T. What is the magnitude of the force on the wire? 4. Calculate the force acting on a conductor of length 40 cm which is carrying a current of 3 A and which is placed perpendicular to a uniform magnetic field of flux density 5.2 T. 5. A straight piece of wire of length 0.5 m and carrying a current of 4 A experiences a force of 5 N when placed perpendicular to a uniform magnetic field of magnetic flux density B. Find the value of B. 6. A straight wire of length 1 m carrying a current of 3 A experiences a force of 9 N when placed perpendicular to a uniform magnetic field. What is the magnetic flux density at the wire? 7. Calculate the force on a charge of 1.6 × 10−19 C travelling at 2 × 107 m s−1 at right angles to a magnetic field of uniform flux density 2.5 T. What path will the charge now follow? 8. A particle of charge 2 × 10−3 C moving at 100 m s−1 moves at right angles to a uniform magnetic field of flux density 3 T. What is the force on the charge? 9. An electron of charge 1.6 × 10−19 C enters a uniform magnetic field of flux density © Dublin School of Grinds 2 T and moves at right angles to the field. If the force on the electron is 2 × 10−18 N calculate the speed of the electron. 10. A proton moves in a plane perpendicular to a magnetic field B. If the proton has a speed that is 0.1% of the speed of light, what value of B will result in a magnetic force that is 106 times the weight of the proton? (Charge of proton = 1.6 × 10−19 C; Speed of light c = 3 × 108 m s−1; Mass of proton = 1.67 × 10−27 kg; Acceleration due to gravity g = 9.8 m s−2 ) 11. A proton in an accelerator enters a magnetic field of flux density 0.5 T with a velocity of 4.6 × 107 m s−1 at right angles to the field. Calculate the radius of the path which the proton follows in the field. (Charge of proton = 1.6 × 10−19 C; Mass of proton = 1.67 × 10−27 kg ) 12. A Helium ion enters a magnetic field of flux density 0.22 T with a velocity of 1.6 × 107 ms−1 at right angles to the field. As it leaves the field it’s velocity makes an angle of 45o with the velocity it had on entering the field. Calculate the distance travelled by the ion in the field. (Charge on Helium ion = 3.2 × 10−19 C; Mass of Helium ion = 6.7 × 10−27 kg ) 13. A magnetic field of strength 0.4 T is directed through a cross-sectional area of 0.4 m2. Find the magnetic flux through the area when the field lines are perpendicular to the area. 14. The flux through a single turn loop is changing at the rate of 2 Wb per second. Find the induced EMF in the loop. 15. The magnetic flux through a coil of 100 turns of wire changes from 0.4 Wb to zero in 0.1 s. Calculate the average EMF induced in the coil. Page 73 Kieran Mills & Tony Kelly 16. A coil of 500 turns of wire is wound around an electromagnet. The magnetic flux of the electromagnet falls from 0.02 Wb to zero in 1/50 second, when the current to the electromagnet is switched off. Calculate the EMF induced in the coil. 22. Domestic electricity is supplied at a RMS voltage of 230 V. Find the maximum value of the voltage in any one cycle. 17. At one instant the magnetic flux in the core of an electromagnet is 1.5 × 10−3 Wb. One thousandth of a second later, the magnetic flux is 1.35 × 10−3 Wb. Calculate the average EMF induced in a coil of 5,000 turns wound tightly around the electromagnet. 24. The AC flowing in a wire of resistance 10 Ω produces heat at the rate of 60 W. Find: (a) the RMS value of the current, (b) the RMS voltage, (c) the peak value of the voltage across the wire. 18. A planar loop of wire consists of a single turn and has a cross-sectional area of 100 cm2. The plane of the loop is perpendicular to a magnetic field that increases uniformly in magnitude from 0.5 T to 2.5 T in 1.5 s. What is the resulting induced current if the loop has a total resistance of 4 Ω? 25. The primary/secondary turns ratio in a transformer is 10:1. The power input to the primary is 100 W. If a current of 2 A flows in the secondary coil, calculate (a) the potential difference across the secondary coil, (b) the potential difference across the primary coil and (c) the current flowing in the primary coil. Assume 100% efficiency. 19. A loop of wire is in a magnetic field of flux 0.1 Wb. What EMF is produced when the knot is drawn up tight in 0.2 s? 26. An input of 200 kW to a transformer produces an output of 50 A at 3 kV. Calculate the efficiency of the transformer. 20. The plane of a rectangular coil of dimensions 10 cm by 8 cm is perpendicular to the direction of a magnetic field. If the coil has 50 turns and a total resistance of 12 Ω, at what rate must the magnitude of the field change in order to induce a current of 5 mA in the windings of the coil? 21. A flexible loop has a radius A of 12 cm and is in a magnetic field of strength 0.15 T. The loop is grasped at points A and B and B stretched until it closes. If it takes 0.2 s to close the loop, find the average induced EMF in it during this time. © Dublin School of Grinds 23. The peak value of an AC current is 10 A. Find its RMS value. 27. A step-up transformer operates from a source of 110 V and delivers 2 A at 770 V in the secondary. If there are 50 turns in the primary how many turns are in the secondary? 28. The input of a transformer is 120 V at 3.2 A and the output is 15 V at 24 A. What is the efficiency of the transformer? 29. The primary of a transformer is connected to a 120 V supply. The turns ratio of secondary/primary is 4.5:1. A heater is connected in the secondary and uses 1.5 kW of power. What current flows in the secondary? What current flows in the primary assuming 98% efficiency? Page 74 Kieran Mills & Tony Kelly Answers 1. 2 N 2. 2 T 3. 6.25 × 10−3 N 4. 6.24 N 5. 2.5 T 6. 3 T 7. 8 × 10−12 N 8. 0.6 N 9. 6.25 m s−1 10. 3.4 × 10−7 T 11. 96 cm 12. 1.2 m 13. 0.16 Wb 14. 2 V 15. 400V 16. 500 V 17. 750V 18. 3·3 × 10-3 A 19. 0.5 V 20. 0.15 T s−1 21. 0.034 V 22. 325.3 V 23. 7.07 A 24. (a) 2.45 A, (b) 24.5 V, (c) 34.6 V 25. (a) 50 V, (b) 500 V, (c) 0.2 A 26. 75% 27. 350 28. 93.75% 29. 2.78 A, 12.75 A © Dublin School of Grinds Page 75 Kieran Mills & Tony Kelly Leaving Cert Questions Chapter 2: Electric Fields Theoretical Section B Coulomb’s law, electric fields [2005 Question 10]........................................................78 Electric fields, Van de Graff generator [2007 Question 8].............................................80 Coulomb’s law, gold-leaf electroscope [2011 Question 9]............................................82 Van de Graff generator, GLE, electric fields [2015 Question 8]....................................84 STS: (Lightning) [2002 Question 11]............................................................................86 Coulomb’s law, electric fields [2003 Question 12(c)]...................................................88 Electric fields, point discharge [2010 Question 12(d)]..................................................89 Coulomb’s law, electric fields [2013 Question 12(c)]...................................................90 Chapter 3: Capacitance Theoretical Section B Parallel plate capacitor [2006 Question 12(b)]..............................................................91 Parallel plate capacitor, gold leaf electroscope [2008 Question 12(d)].........................92 Touchscreens (capacitors, electric fields) [2014 Question 9]........................................93 Chapter 4: Electric Current Experimental Section A Exp. E1: Joule’s law [2003 Question 4].........................................................................95 Exp. E1: Joule’s law [2006 Question 4].........................................................................97 Exp. E1: Joule’s law [2014 Question 4].........................................................................99 Exp. E2: Resistivity of a metallic conductor [2004 Question 4]...................................101 Exp. E2: Resistivity of a metallic conductor [2009 Question 4]...................................102 Exp. E3: Resistance vs. temperature for a metallic conductor [2008 Question 4]........103 Exp. E3: Resistance vs. temperature for a metallic conductor [2015 Question 4]........105 Exp. E4: Resistance vs. temperature for a thermistor [2010 Question 4]......................107 Exp. E5[B]: Voltage vs. current for a filament bulb [2005 Question 4]........................109 Exp. E5[B]: Voltage vs. current for a filament bulb/wire [2013 Question 4]................111 Exp. E5[C]: Voltage vs. current for an ionic solution [2002 Question 4]......................113 © Dublin School of Grinds Page 76 Kieran Mills & Tony Kelly Exp. E5[C]: Voltage vs. current for an ionic solution [2011 Question 4]......................115 Exp. E5[D]: Voltage vs. current for a semiconductor diode [2007 Question 4]............117 Exp. E5[D]: Voltage vs. current for a semiconductor diode [2012 Question 4]............119 Theoretical Section B Power and resistivity [2002 Question 8]........................................................................121 Ampere, conduction in materials [2003 Question 8].....................................................123 Electrical circuit with a capacitor [2004 Question 8]....................................................125 STS: Domestic Electricity [2004 Question 11].............................................................127 Electrical circuit with a thermistor [2005 Question 9]...................................................129 Metre bridge, resistance and resistivity [2007 Question 9]...........................................131 Resistivity, power rating [2008 Question 7]..................................................................133 Electrical circuit with a capacitor [2009 Question 9]....................................................135 Electrical circuit with a hairdryer [2010 Question 8]....................................................137 Electrical circuit, Wheatstone bridge [2012 Question 9]...............................................139 Wheatstone bridge, resistivity (incl. Doppler Effect) [2014 Question 10]....................141 Semiconductor diodes [2004 Question 12 (d)]..............................................................143 Semiconductor diodes [2009 Question 12 (b)]..............................................................144 Resistivity, power rating [2011 Question 12 (c)]...........................................................145 Chapter 5: Electromagnetism Theoretical Section B Ampere, AC/DC, power rating [2006 Question 9]........................................................146 Electromagnetic induction [2008 Question 8]...............................................................148 Inductors, diodes and capacitors, resistance and resistivity [2013 Question 8].............150 Electromagnetic induction, inductors [2002 Question 12 (c)].......................................152 Electromagnetic induction, Lenz’s law [2003 Question 12 (d)]....................................153 Electromagnetic induction, Lenz’s law [2004 Question 12 (c)]....................................154 Electromagnetic induction [2005 Question 12 (b)].......................................................155 Electromagnetic induction [2007 Question 12 (c)]........................................................156 Electromagnetic induction, Lenz’s law [2014 Question 12 (d)]....................................157 © Dublin School of Grinds Page 77 Kieran Mills & Tony Kelly LC 2005: Coulomb's law, electric fields 10. Define electric field strength. State Coulomb’s law of force between electric charges. (12) Why is Coulomb’s law an example of an inverse square law? (6) Give two differences between the gravitational force and the electrostatic force between two electrons. (6) Describe an experiment to show an electric field pattern. © Dublin School of Grinds Page 78 (12) Kieran Mills & Tony Kelly B 10 mm electron Calculate the electric field strength at the point B, which is 10 mm from an electron. What is the direction of the electric field strength at B? A charge of 5 µC is placed at B. Calculate the electrostatic force exerted on this charge. (permittivity of free space = 8.9 × 10–12 F m–1; charge on the electron = 1.6 × 10–19 C) © Dublin School of Grinds Page 79 Kieran Mills & Tony Kelly (20) LC 2007: Electric fields, Van de Graff generator 8. Define electric field strength and give its unit of measurement. (9) Describe how an electric field pattern may be demonstrated in the laboratory. (12) The dome of a Van de Graff generator is charged. The dome has a diameter of 30 cm and its charge is 4 C. A 5 μC point charge is placed 7 cm from the surface of the dome. Calculate: (i) the electric field strength at a point 7 cm from the dome (ii) the electrostatic force exerted on the 5 μC point charge. © Dublin School of Grinds Page 80 (15) Kieran Mills & Tony Kelly All the charge resides on the surface of a Van de Graff generator’s dome. Explain why. Describe an experiment to demonstrate that total charge resides on the outside of a conductor. Give an application of this effect. (20) (permittivity of free space = 8.9 × 10–12 F m–1) © Dublin School of Grinds Page 81 Kieran Mills & Tony Kelly LC 2011: Coulomb's law, gold-leaf electroscope 9. (a) State Coulomb’s law. (6) Two identical spherical conductors on insulated stands are placed a certain distance apart. One conductor is given a charge Q while the other conductor is given a charge 3Q and they experience a force of repulsion F. The two conductors are then touched off each other and returned to their original positions. What is the new force, in terms of F, between the spherical conductors? (18) (b) Q 3Q Draw a labelled diagram of an electroscope. Why should the frame of an electroscope be earthed? Describe how to charge an electroscope by induction. © Dublin School of Grinds Page 82 (15) Kieran Mills & Tony Kelly (c) How does a full-body metal-foil suit protect an operator when working on high voltage power lines? Describe an experiment to investigate the principle by which the operator is protected. © Dublin School of Grinds Page 83 (17) Kieran Mills & Tony Kelly LC 2015: Van de Graff generator, GLE, electric fields 8. Define electric field strength. (6) Both Van de Graaff generators and gold leaf electroscopes are used to investigate static electricity in the laboratory. Draw a labelled diagram of a gold leaf electroscope. Describe how it can be given a negative charge by induction. © Dublin School of Grinds Page 84 (20) Kieran Mills & Tony Kelly A Van de Graaff generator can be used to demonstrate point discharge. Explain, with the aid of a labelled diagram, how point discharge occurs. Describe an experiment to demonstrate point discharge. (18) The polished spherical dome of a Van de Graaff generator has a diameter of 40 cm and a charge of +3.8 µC. What is the electric field strength at a point 4 cm from the surface of the dome? © Dublin School of Grinds Page 85 (12) Kieran Mills & Tony Kelly LC 2002: STS: Lightning Question 11. Read the following passage and answer the accompanying questions. Benjamin Franklin designed the lightning conductor. This is a thick copper strip running up the outside of a tall building. The upper end of the strip terminates in one or more sharp spikes above the highest point of the building. The lower end is connected to a metal plate buried in moist earth. The lightning conductor protects a building from being damaged by lightning in a number of ways. During a thunderstorm, the value of the electric field strength in the air can be very high near a pointed lightning conductor. If the value is high enough, ions, which are drawn towards the conductor, will receive such large accelerations that, by collision with air molecules, they will produce vast additional numbers of ions. Therefore the air is made much more conducting and this facilitates a flow of current between the air and the ground. Thus, charged clouds become neutralised and lightning strikes are prevented. Alternatively, in the event of the cloud suddenly discharging, the lightning strike will be conducted through the copper strip, thus protecting the building from possible catastrophic consequences. Raised umbrellas and golf clubs are not to be recommended during thunderstorms for obvious reasons. On high voltage electrical equipment, pointed or roughly-cut surfaces should be avoided. (Adapted from “Physics – a teacher’s handbook”, Dept. of Education and Science.) (a) Why is a lightning conductor made of copper? (7) (b) What is meant by electric field strength? (7) (c) Why do the ions near the lightning conductor accelerate? (7) (d) How does the presence of ions in the air cause the air to be more conducting? (7) (e) How do the charged clouds become neutralised? (7) © Dublin School of Grinds Page 86 Kieran Mills & Tony Kelly (h) What are the two ways in which a lightning conductor prevents a building from being damaged by lightning? (7) (g) Why are raised umbrellas and golf clubs not recommended during thunderstorms? (7) (h) Explain why pointed surfaces should be avoided when using high voltage electrical equipment. (7) © Dublin School of Grinds Page 87 Kieran Mills & Tony Kelly LC 2003: Coulomb's law, electric fields 12. (6) (c) State Coulomb’s law of force between electric charges. Define electric field strength and give its unit. (9) How would you demonstrate an electric field pattern? (9) The diagram shows a negative charge – Q at a point X. Copy the diagram and show on it the direction of the electric field strength at Y. (4) X Y Q © Dublin School of Grinds Page 88 Kieran Mills & Tony Kelly LC 2010: Electric fields, point discharge 12. (d) Define electric field strength and give its unit of measurement. (9) The diagram shows a negative charge of 2 μC, positioned 25 cm away from a positive charge of 5 μC. 10 cm –2 μC 15 cm 5 μC P Copy the diagram into your answerbook and show on it the direction of the electric field at point P. Calculate the electric field strength at P. (15) Under what circumstances will point discharge occur? (4) (permittivity of free space = 8.9 × 10–12 F m–1) © Dublin School of Grinds Page 89 Kieran Mills & Tony Kelly LC 2013: Coulomb's law, electric fields 12. (c) Define the unit of charge, the coulomb. State Coulomb’s law. (9) Calculate the force of repulsion between two small spheres when they are held 8 cm apart in a vacuum. Each sphere has a positive charge of +3 µC. (9) +3 µC 8 cm +3 µC Copy the diagram above and show on it the electric field generated by the charges. Mark on your diagram a place where the electric field strength is zero. © Dublin School of Grinds Page 90 Kieran Mills & Tony Kelly (10) LC 2006: Parallel plate capacitor 12. (b) List the factors that affect the capacitance of a parallel plate capacitor. (6) The plates of an air filled parallel plate capacitor have a common area of 40 cm2 and are 1 cm apart. The capacitor is connected to a 12 V d.c. supply. Calculate (i) the capacitance of the capacitor; (ii) the magnitude of the charge on each plate. (15) What is the net charge on the capacitor? Give a use for a capacitor. (7) (permittivity of free space = 8.85 × 10–12 F m–1) © Dublin School of Grinds Page 91 Kieran Mills & Tony Kelly LC 2008: Parallel plate capacitor, gold-leaf electroscope 12. (d) Define capacitance. (6) Describe how an electroscope can be charged by induction. (10) How would you demonstrate that the capacitance of a parallel plate capacitor depends on the distance between its plates? (12) © Dublin School of Grinds Page 92 Kieran Mills & Tony Kelly LC 2014 Touchscreens (capacitors, electric fields) 9. Most modern electronic devices contain a touchscreen. One type of touchscreen is a capacitive touchscreen, in which the user’s finger acts as a plate of a capacitor. Placing your finger on the screen will alter the capacitance and the electric field at that point. Explain the underlined terms. (12) Describe an experiment to demonstrate an electric field pattern. (12) Two parallel metal plates are placed a distance d apart in air. The plates form a parallel plate capacitor with a capacitance of 12 µF. A 6 V battery is connected across the plates. Calculate (i) the charge on each plate and (ii) the energy stored in the capacitor. © Dublin School of Grinds Page 93 Kieran Mills & Tony Kelly (12) While the battery is connected the distance d is increased by a factor of three. Calculate the new capacitance. (4) A capacitor and a battery are both sources of electrical energy. State two differences between a capacitor and a battery. (6) Touchscreens also contain two polarising filters. What is meant by polarisation of light? (6) Give one application of capacitors, other than in touchscreens. (4) © Dublin School of Grinds Page 94 Kieran Mills & Tony Kelly LC 2003: Experiment E1 (Joule's law) 4. In an experiment to verify Joule’s law, a heating coil was placed in a fixed mass of water. The temperature rise ∆θ produced for different values of the current I passed through the coil was recorded. In each case the current was allowed to flow for a fixed length of time. The table shows the recorded data. I/A 1.5 2.0 2.5 3.0 3.5 4.0 4.5 ∆θ / °C 3.5 7.0 10.8 15.0 21.2 27.5 33.0 Describe, with the aid of a labelled diagram, how the apparatus was arranged in this experiment. (12) Using the given data, draw a suitable graph on graph paper and explain how your graph verifies Joule’s law. (18) Explain why the current was allowed to flow for a fixed length of time in each case. (5) Apart from using insulation, give one other way of reducing heat losses in the experiment. (5) © Dublin School of Grinds Page 95 Kieran Mills & Tony Kelly Graph I/A 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Dq / oC 3.5 7.0 10.8 15.0 21.2 27.5 33.0 I 2 /A2 © Dublin School of Grinds Page 96 Kieran Mills & Tony Kelly LC 2006: Experiment E1 (Joule's law) 4. In an experiment to verify Joule’s law a student passed a current through a heating coil in a calorimeter containing a fixed mass of water and measured the rise in temperature Δθ for a series of different values of the current I. The student allowed the current to flow for three minutes in each case. Δθ K 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 I2 A2 Describe, with the aid of a labelled diagram, how the student arranged the apparatus. (12) Why was a fixed mass of water used throughout the experiment? (6) The student drew a graph, as shown. Explain how this graph verifies Joule’s law. (7) © Dublin School of Grinds Page 97 Kieran Mills & Tony Kelly Given that the mass of water in the calorimeter was 90 g in each case, and assuming that all of the electrical energy supplied was absorbed by the water, use the graph to determine the resistance of the heating coil. The specific heat capacity of water is 4200 J kg–1 K–1. (15) © Dublin School of Grinds Page 98 Kieran Mills & Tony Kelly LC 2014: Experiment E1 (Joule's law) 4. In an experiment to verify Joule’s law, a fixed mass of water was heated in an insulated cup. , the highest temperature reached, was recorded for different values of current, I. In each case the current flowed for 4 minutes and the initial temperature of the water was 20.0 °C. The recorded data is shown in the table. I (A) 1.0 1.5 2.0 2.5 3.0 3.5 q (oC) 22.0 24.5 28.5 34.0 38.5 45.5 (12) Draw a labelled diagram of the apparatus used in the experiment. Draw a suitable graph to verify Joule’s law. Explain how the graph verifies Joule’s law. Use your graph to estimate the highest temperature of the water when a current of 1.6 A flows through the coil for 4 minutes. Explain why a fixed mass of water was used. © Dublin School of Grinds Page 99 (18) (6) (4) Kieran Mills & Tony Kelly Graph I (A) 1.0 1.5 2.0 2.5 3.0 3.5 q (oC) 22.0 24.5 28.5 34.0 38.5 45.5 I 2 (A2) © Dublin School of Grinds Page 100 Kieran Mills & Tony Kelly LC 2004: Experiment E2 (Resistivity of a metallic conductor) 4. The following is part of a student’s report of an experiment to measure the resistivity of nichrome wire. “The resistance and length of the nichrome wire were found. The diameter of the wire was then measured at several points along its length.” The following data was recorded. resistance of wire length of wire diameter of wire = 32.1 Ω = 90.1 cm = 0.19 mm, 0.21 mm, 0.20 mm, 0.21 mm, 0.20 mm Name an instrument to measure the diameter of the wire and describe how it is used. (12) Why was the diameter of the wire measured at several points along its length? (6) Using the data, calculate a value for the resistivity of nichrome. (15) Give two precautions that should be taken when measuring the length of the wire. (7) © Dublin School of Grinds Page 101 Kieran Mills & Tony Kelly LC 2009: Experiment E2 (Resistivity of a metallic conductor) 4. In an experiment to measure the resistivity of nichrome, the resistance, the diameter and appropriate length of a sample of nichrome wire were measured. The following data were recorded: resistance of wire length of wire average diameter of wire = 7.9 = 54.6 cm = 0.31 mm Describe the procedure used in measuring the length of the sample of wire. (6) Describe the steps involved in finding the average diameter of the wire. (15) Use the data to calculate the resistivity of nichrome. (15) The experiment was repeated on a warmer day. What effect did this have on the measurements? © Dublin School of Grinds Page 102 (4) Kieran Mills & Tony Kelly LC 2008: Experiment E3 (Resistance vs. temperature for a metallic conductor) 4. A student investigated the variation of the resistance R of a metallic conductor with its temperature θ. The student recorded the following data. q / oC 20 30 40 50 60 70 80 R/W 4.6 4.9 5.1 5.4 5.6 5.9 6.1 Describe, with the aid of a labelled diagram, how the data was obtained. Draw a suitable graph to show the relationship between the resistance of the metal conductor and its temperature. (9) (12) Use your graph to: (i) estimate the resistance of the metal conductor at a temperature of –20 oC; (ii) estimate the change in resistance for a temperature increase of 80 oC; (iii) explain why the relationship between the resistance of a metallic conductor and its temperature is not linear. © Dublin School of Grinds Page 103 (19) Kieran Mills & Tony Kelly Graph q / oC 20 30 40 50 60 70 80 R/W 4.6 4.9 5.1 5.4 5.6 5.9 6.1 © Dublin School of Grinds Page 104 Kieran Mills & Tony Kelly LC 2015: Experiment E3/E5[B] (Resistance vs. temperature for a metallic conductor) In an experiment to measure the variation of the resistance R of a metallic conductor with its temperature , a student recorded the following data. q (oC) 15 20 30 40 50 60 80 100 R (W) 6∙0 6∙2 6∙5 6∙8 7∙2 7∙5 8∙2 8∙8 Using the recorded data, plot a graph to show the variation of the resistance of the metallic conductor with its temperature. Use your graph to estimate (i) the rate of change of resistance with respect to temperature for the metallic conductor (ii) the resistance of the metallic conductor when it is immersed in melting ice. (20) Answer this question on the next page. The student then completed an experiment to establish the relationship between current and voltage for the thin metallic filament of a bulb. Data were recorded and the following graph plotted. 80 Current (mA) 4. 60 40 20 0 0 1 2 3 4 5 Voltage (V) Describe, with the aid of a circuit diagram, how the student carried out this second experiment. Use the findings of the first experiment to explain the shape of the graph in the second experiment. © Dublin School of Grinds Page 105 Kieran Mills & Tony Kelly (20) Graph q (oC) 15 20 30 40 50 60 80 100 R (W) 6∙0 6∙2 6∙5 6∙8 7∙2 7∙5 8∙2 8∙8 © Dublin School of Grinds Page 106 Kieran Mills & Tony Kelly LC 2010: Experiment E4 (Resistance vs. temperature for a thermistor) 4. In an experiment to investigate the variation of the resistance R of a thermistor with its temperature θ, a student measured its resistance at different temperatures. The table shows the measurements recorded. θ/°C 20 30 40 50 60 70 80 R/Ω 2000 1300 800 400 200 90 40 Draw a labelled diagram of the apparatus used. (9) How was the resistance measured? (6) Describe how the temperature was varied. (6) Using the recorded data, plot a graph to show the variation of the resistance of a thermistor with its temperature. In this investigation, why is the thermistor usually immersed in oil rather than in water? © Dublin School of Grinds Page 107 (4) Kieran Mills & Tony Kelly Graph θ/°C 20 30 40 50 60 70 80 R/Ω 2000 1300 800 400 200 90 40 Use your graph to estimate the average variation of resistance per kelvin in the range 45 °C – 55 °C. © Dublin School of Grinds Page 108 (15) Kieran Mills & Tony Kelly LC 2005: Experiment E5[B] (Voltage vs. current for a filament bulb) 4. A student investigated the variation of the current I flowing through a filament bulb for a range of different values of potential difference V. Draw a suitable circuit diagram used by the student. Describe how the student varied the potential difference. © Dublin School of Grinds Page 109 (16) Kieran Mills & Tony Kelly The student drew a graph, as shown, using data recorded in the experiment. I mA 100 90 80 70 60 50 40 30 20 10 0 1 2 3 4 5 With reference to the graph, 6 V V (i) explain why the current is not proportional to the potential difference; (ii) calculate the change in resistance of the filament bulb as the potential difference increases from 1 V to 5 V. Give a reason why the resistance of the filament bulb changes. © Dublin School of Grinds Page 110 (18) (6) Kieran Mills & Tony Kelly LC 2013: Experiment E5[B] (Voltage vs. current for a filament bulb/metallic conductor) 4. A student was asked to investigate the variation of current with potential difference for a thin metallic conductor. The student set up a circuit using appropriate equipment. The student recorded the values of the current I passing through the conductor for the corresponding values of potential difference V. The recorded data are shown in the table. V/V 1.0 2.0 3.0 4.0 5.0 6.0 I/A 0.17 0.34 0.50 0.64 0.77 0.88 Draw and label the circuit diagram used by the student. Name the device in the circuit that is used to vary the potential difference across the conductor. Explain how the student used this device to vary the potential difference. (18) Use the data in the table to draw a graph on graph paper to show the variation of current with potential difference. Use your graph to find the value of the resistance of the conductor when the current is 0.7 A. (15) © Dublin School of Grinds Page 111 Kieran Mills & Tony Kelly Graph V/V 1.0 2.0 3.0 4.0 5.0 6.0 I/A 0.17 0.34 0.50 0.64 0.77 0.88 Explain the shape of your graph. © Dublin School of Grinds (7) Page 112 Kieran Mills & Tony Kelly LC 2002: Experiment E5[C] (Voltage vs. current for an ionic solution) 4. In an experiment to investigate the variation of current I with potential difference V for a copper sulfate solution, the following results were obtained. V /V 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 I /mA 24 48 79 102 120 143 185 195 215 263 Draw a diagram of the apparatus used in this experiment, identifying the anode and the cathode. (12) Draw a suitable graph on graph paper to show how the current varies with the potential difference. (12) Draw a sketch of the graph that would be obtained if inactive electrodes were used in this experiment. © Dublin School of Grinds Page 113 Kieran Mills & Tony Kelly (7) Graph V /V 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 I /mA 24 48 79 102 120 143 185 195 215 263 Using your graph, calculate the resistance of the copper sulfate solution. (Assume the resistance of the electrodes is negligible.) © Dublin School of Grinds (9) Page 114 Kieran Mills & Tony Kelly LC 2011: Experiment E5[C] (Voltage vs. current for an ionic solution) 4. A student investigated the variation of the current I through an electrolyte as the potential difference V across the electrolyte was changed. The electrolyte used was a solution of copper sulfate. The electrodes used were made of copper. The student recorded the following data: V/V 0 1 2 3 4 5 6 I / mA 0 30 64 93 122 160 195 Draw a suitable circuit diagram for this investigation and label the components. How was the potential difference changed during the investigation? (12) Draw a suitable graph to show the relationship between the current and the potential difference in this investigation. Use your graph to calculate the resistance of the electrolyte. (18) What was observed at the electrodes as current flowed through the electrolyte? (10) © Dublin School of Grinds Page 115 Kieran Mills & Tony Kelly Graph V/V 0 1 2 3 4 5 6 I / mA 0 30 64 93 122 160 195 © Dublin School of Grinds Page 116 Kieran Mills & Tony Kelly LC 2007: Experiment E5[D] (Voltage vs. current for a semiconductor diode) 4. The following is part of a student’s report of an experiment to investigate of the variation of current I with potential difference V for a semiconductor diode. I put the diode in forward bias as shown in the circuit diagram. I increased the potential difference across the diode until a current flowed. I measured the current flowing for different values of the potential difference. I recorded the following data. V/V 0.60 0.64 0.68 0.72 0.76 0.80 I / mA 2 4 10 18 35 120 Draw a circuit diagram used by the student. How did the student vary and measure the potential difference? (15) Draw a graph to show how the current varies with the potential difference. Estimate from your graph the junction voltage of the diode. (12) The student then put the diode in reverse bias and repeated the experiment. What changes did the student make to the initial circuit? Draw a sketch of the graph obtained for the diode in reverse bias. (13) © Dublin School of Grinds Page 117 Kieran Mills & Tony Kelly Graph V/V 0.60 0.64 0.68 0.72 0.76 0.80 I / mA 2 4 10 18 35 120 © Dublin School of Grinds Page 118 Kieran Mills & Tony Kelly LC 2012: Experiment E5[D] (Voltage vs. current for a semiconductor diode) 4. The following is part of a student’s report on an experiment to investigate the variation of the current I with potential difference V for a semiconductor diode. “I set up the apparatus as shown in the circuit diagram. I measured the current flowing through the diode for different values of the potential difference. I recorded the following data.” V/V 0 0.50 0.59 0.65 0.68 0.70 0.72 I / mA 0 3.0 5.4 11.7 17.4 27.3 36.5 Draw a circuit diagram used by the student. How did the student vary and measure the potential difference? (15) Using the data, draw a graph to show how the current varies with the potential difference for the semiconductor diode. Does the resistance of the diode remain constant during the investigation? Justify your answer. (18) The student continued the experiment with the connections to the semiconductor diode reversed. What adjustments should be made to the circuit to obtain valid readings? (7) © Dublin School of Grinds Page 119 Kieran Mills & Tony Kelly Graph V/V 0 0.50 0.59 0.65 0.68 0.70 0.72 I / mA 0 3.0 5.4 11.7 17.4 27.3 36.5 © Dublin School of Grinds Page 120 Kieran Mills & Tony Kelly LC 2002: Power and resistivity 8. Define (i) power, (ii) resistivity. (12) Describe an experiment that demonstrates the heating effect of an electric current. (12) The ESB supplies electrical energy at a rate of 2 MW to an industrial park from a local power station, whose output voltage is 10 kV. The total length of the cables connecting the industrial park to the power station is 15 km. The cables have a diameter of 10 mm and are made from a material of resistivity 5.0 × 10–8 Ω m. Calculate (i) the total resistance of the cables; © Dublin School of Grinds (15) Page 121 Kieran Mills & Tony Kelly (ii) the current flowing in the cables; (6) (iii) the rate at which energy is “lost” in the cables. (6) (5) Suggest a method of reducing the energy “lost” in the cables. © Dublin School of Grinds Page 122 Kieran Mills & Tony Kelly LC 2003: Ampere, conduction in materials 8. Define the unit of current, i.e. the ampere. (9) Describe an experiment to demonstrate the principle on which the definition of the ampere is based. (15) Various materials conduct electricity. Draw a graph to show the relationship between current and voltage for each of the following conductors: (i) a metal at constant temperature (ii) an ionic solution with inactive electrodes (iii) a gas. © Dublin School of Grinds (18) Page 123 Kieran Mills & Tony Kelly How would the graph for the metal differ if its temperature were increasing? (7) How would the graph for the ionic solution differ if its concentration were reduced? (7) © Dublin School of Grinds Page 124 Kieran Mills & Tony Kelly LC 2004: Electrical circuit with a capacitor 8. Define (i) potential difference, (ii) capacitance. (12) Describe an experiment to demonstrate that a capacitor can store energy. (12) © Dublin School of Grinds Page 125 Kieran Mills & Tony Kelly The circuit diagram shows a 50 µF capacitor connected in series with a 47 kΩ resistor, a 6 V battery and a switch. When the switch is closed the capacitor starts to charge and the current flowing at a particular instant in the circuit is 80 µA. 6V 47 kΩ 50 µF Calculate (i) the potential difference across the resistor and hence the potential difference across the capacitor when the current is 80 µA; (ii) the charge on the capacitor at this instant; (iii) the energy stored in the capacitor when it is fully charged. Describe what happens in the circuit when the 6 V d.c. supply is replaced with a 6 V a.c. supply. © Dublin School of Grinds Page 126 Kieran Mills & Tony Kelly (27) (5) LC 2004: STS: Domestic electricity 11. Read the following passage and answer the following questions. Your home is supplied with electricity at 230 volts, 50 Hertz. At the electrical supply intake position is your main consumer unit or fuse board. At that position you will find your main switch. Your sockets, immersion group and bathroom heater (or shower) are protected by Residual Current Devices (RCD) installed in your fuse board. These provide a high degree of safety on these circuits and it is important that they are tested at least every 3 months. The final circuits are protected by Miniature Circuit Breakers (MCB). It is advisable to contact your local ESB about cheaper night tariffs, these could make significant savings to your electricity bill. Storage heaters may be used to avail of these cheaper rates. Each plug top contains a small cartridge fuse. Cartridge fuses are supplied with a rating of 1A, 2A, 3A, 5A and 13A. A fuse should never be replaced by anything other than a suitable fuse. (Adapted from “Home Safety”, Register of Electrical Contractors of Ireland. RECI) (a) Name and give the colour of the wire that should be connected to the fuse in a standard three-pin plug. (7) (b) Explain why replacing a fuse with a piece of aluminium foil is dangerous. (7) (c) A table lamp has a power rating of 100 W. What is the most suitable fuse for the lamp? (7) © Dublin School of Grinds Page 127 Kieran Mills & Tony Kelly (d) Some electrical appliances are supplied with two-pin plugs. Why is an earth wire not required in these devices? (7) (e) Sketch a voltage-time graph of the 230 V supply. (7) (f) Explain how a Residual Current Device (RCD) operates. (7) (g) Give one advantage of a Residual Current Device (RCD) over a Miniature Circuit Breaker (MCB). (7) (h) Storage heaters have a large heat capacity. Explain why. © Dublin School of Grinds Page 128 (7) Kieran Mills & Tony Kelly LC 2005: Electrical circuit with a thermistor 9. (12) Define (i) potential difference, (ii) resistance. Two resistors, of resistance R1 and R2 respectively, are connected in parallel. Derive an expression for the effective resistance of the two resistors in terms of R1 and R2 . © Dublin School of Grinds Page 129 Kieran Mills & Tony Kelly (12) 750 Ω Α 300 Ω In the circuit diagram, the resistance of the thermistor at room temperature is 500 Ω. At room temperature, calculate (i) the total resistance of the circuit; (ii) the current flowing through the 750 Ω resistor. (18) As the temperature of the room increases, explain why (iii) the resistance of the thermistor decreases; (iv) the potential at A increases. (14) © Dublin School of Grinds Page 130 Kieran Mills & Tony Kelly LC 2007: Metre bridge, resistance and resistivity 9. Define (i) resistance, (ii) resistivity. (12) A metre bridge was used to measure the resistance of a sample of nichrome wire. The diagram indicates the readings taken when the metre bridge was balanced. The nichrome wire has a length of 220 mm and a radius of 0.11 mm. nichrome wire 20 Ω 718 mm 282 mm Calculate: (i) the resistance of the nichrome wire (ii) the resistivity of nichrome. © Dublin School of Grinds Page 131 (18) Kieran Mills & Tony Kelly Sketch a graph to show the relationship between the temperature and the resistance of the nichrome wire as its temperature is increased. (6) What happens to the resistance of the wire: (i) as its temperature falls below 0oC? (ii) as its length is increased? (iii) if its diameter is increased? (11) Name another device, apart from a metre bridge, that can be used to measure resistance. Give one advantage and one disadvantage of using this device instead of a metre bridge. © Dublin School of Grinds Page 132 (9) Kieran Mills & Tony Kelly LC 2008: Resistivity, power rating 7. Define resistivity and give its unit of measurement. (9) An electric toaster heats bread by convection and radiation. What is the difference between convection and radiation as a means of heat transfer? (8) A toaster has a power rating of 1050 W when it is connected to the mains supply. Its heating coil is made of nichrome and it has a resistance of 12 Ω . The coil is 40 m long and it has a circular cross-section of diameter 2.2 mm. Calculate: (i) the resistivity of nichrome; (ii) the heat generated by the toaster in 2 minutes if it has an efficiency of 96%. © Dublin School of Grinds Page 133 (18) Kieran Mills & Tony Kelly The toaster has exposed metal parts. How is the risk of electrocution minimised? (9) When the toaster is on, the coil emits red light. Explain, in terms of movement of electrons, why light is emitted when a metal is heated. (12) © Dublin School of Grinds Page 134 Kieran Mills & Tony Kelly LC 2009: Electrical circuit with a capacitor 9. Define (i) potential difference, (ii) capacitance. (12) A capacitor stores energy. Describe an experiment to demonstrate that a capacitor stores energy. (14) © Dublin School of Grinds Page 135 Kieran Mills & Tony Kelly The ability of a capacitor to store energy is the basis of a defibrillator. During a heart attack the chambers of the heart fail to pump blood because their muscle fibres contract and relax randomly. To save the victim, the heart muscle must be shocked to reestablish its normal rhythm. A defibrillator is used to shock the heart muscle. A 64 F capacitor in a defibrillator is charged to a potential difference of 2500 V. The capacitor is discharged through electrodes attached to the chest of a heart attack victim. Calculate (i) the charge stored on each plate of the capacitor; (ii) the energy stored in the capacitor; (iii) the average current that flows through the victim when the capacitor discharges in a time of 10 ms; (iv) the average power generated as the capacitor discharges. (30) © Dublin School of Grinds Page 136 Kieran Mills & Tony Kelly LC 2010: Electrical circuit with a hairdryer 8. A hair dryer with a plastic casing uses a coiled wire as a heat source. When an electric current flows through the coiled wire, the air around it heats up and a motorised fan blows the hot air out. What is an electric current? Heating is one effect of an electric current. Give two other effects of an electric current. (12) The diagram shows a basic electrical circuit for a hair dryer. (i) Describe what happens: (a) when switch A is closed and the rheostat is adjusted (b) when switch A and switch B are closed. (9) A B fuse M 230 V fan (ii) The maximum power generated in the heating coil is 2 kW. (a) What is the initial resistance of the coil? (b) Calculate the current that flows through the coil when the dryer is turned on. (9) © Dublin School of Grinds Page 137 Kieran Mills & Tony Kelly coil (iii) A length of nichrome wire of diameter 0.17 mm is used for the coil. Calculate the length of the coil of wire. (18) (iv) Explain why the current through the coil would decrease if the fan developed a fault and stopped working. (8) (resistivity of nichrome = 1.1 × 10– 6 Ω m ) © Dublin School of Grinds Page 138 Kieran Mills & Tony Kelly LC 2012: Electrical circuit, Wheatstone bridge 9. Define resistance. (i) Two resistors of resistance R1 and R2 are connected in series. Derive an expression for the effective resistance of the two resistors in terms of R1 and R2. (18) (ii) Two 4 Ω resistors are connected in parallel. Draw a circuit diagram to show how another 4 Ω resistor could be arranged with these two resistors to give an effective resistance of 6 Ω. (iii) A fuse is a resistor used as a safety device in a circuit. How does a fuse operate? © Dublin School of Grinds Page 139 Kieran Mills & Tony Kelly (9) (11) A Wheatstone bridge circuit is used to measure the resistance of an unknown resistor R. The bridge ABCD is balanced when X = 2.2 kΩ, Y = 1.0 kΩ and Z = 440 Ω. B +V R A X C Y Z 0V D (iv) What test would you use to determine that the bridge is balanced? (v) What is the resistance of the unknown resistor R? (vi) When the unknown resistor R is covered by a piece of black paper, the bridge goes out of balance. What type of resistor is it? Give a use for this type of resistor. © Dublin School of Grinds Page 140 Kieran Mills & Tony Kelly (6) (6) (6) LC 2014: Wheatstone bridge, resistivity (incl. Doppler Effect) 10. Blood pressure can be measured in many ways. One technique uses the Doppler effect; another uses strain gauges contained in Wheatstone bridges. What is the Doppler effect? Explain, with the aid of labelled diagrams, how the Doppler effect occurs. (18) An ambulance siren emits a sound of frequency 750 Hz. When the ambulance is travelling towards an observer, the frequency detected by the observer is 820 Hz. What is the speed of the ambulance? (12) (4) State two other practical applications of the Doppler effect. © Dublin School of Grinds Page 141 Kieran Mills & Tony Kelly The resistance of the conductor in a strain gauge increases when a force is applied to it. Strain gauges can act as the resistors in a Wheatstone bridge, and any change in their resistance can then be detected. How would an observer know that a Wheatstone bridge is balanced? 5.1 Ω ?Ω 11.9 Ω 40.5 Ω (4) (6) What is the resistance of the unknown resistor? Write an expression for the resistance of a wire in terms of its resistivity, length and diameter. The radius of a wire is doubled. What is the effect of this on the resistance of the wire? (speed of sound in air = 340 m s–1) © Dublin School of Grinds (12) Page 142 Kieran Mills & Tony Kelly LC 2004: Semiconductor diodes 12. (d) A p-n junction is formed by taking a single crystal of silicon and doping separate but adjacent layers of it. A depletion layer is formed at the junction. (i) (ii) What is doping? Explain how a depletion layer is formed at the junction. (15) The graph shows the variation of current I with potential difference V for a p-n junction in forward bias. Explain, using the graph, how the current varies with the potential difference. Why does the p-n junction become a good conductor as the potential difference exceeds 0.6 V? (13) I 0.6 V © Dublin School of Grinds Page 143 V Kieran Mills & Tony Kelly LC 2009: Semiconductor diodes 12. (b) A semiconductor diode is formed when small quantities of phosphorus and boron are added to adjacent layers of a crystal of silicon to increase its conduction. Explain how the presence of phosphorus and boron makes the silicon a better conductor. (6) What happens at the boundary of the two adjacent layers? (9) Describe what happens at the boundary when the semiconductor diode is (i) forward biased, (ii) reverse biased. (9) Give a use of a semiconductor diode. (4) © Dublin School of Grinds Page 144 Kieran Mills & Tony Kelly LC 2011: Resistivity, power rating 12. (c) List the factors that affect the heat produced in a current-carrying conductor. (7) An electric cable consists of a single strand of insulated copper wire. The wire is of uniform cross-sectional area and is designed to carry a current of 20 A. To preserve the insulation, the maximum rate at which heat may be produced in the wire is 2.7 W per metre length. Calculate (i) the maximum resistance per metre of the wire (ii) the minimum diameter of the wire. (21) (resistivity of copper = 1.7 × 10–8 m) © Dublin School of Grinds Page 145 Kieran Mills & Tony Kelly LC 2006: Ampere, AC/DC, power rating 9. What is an electric current? Define the ampere, the SI unit of current. (12) Describe an experiment to demonstrate the principle on which the definition of the ampere is based. (15) Sketch a graph to show the relationship between current and time for (i) alternating current; (ii) direct current. © Dublin School of Grinds Page 146 Kieran Mills & Tony Kelly (9) The peak voltage of the mains electricity is 325 V. Calculate the rms voltage of the mains? (6) What is the resistance of the filament of a light bulb, rated 40 W, when it is connected to the mains? (9) Explain why the resistance of the bulb is different when it is not connected to the mains. © Dublin School of Grinds Page 147 (5) Kieran Mills & Tony Kelly LC 2008: Electromagnetic induction 8. What is electromagnetic induction? State the laws of electromagnetic induction. (18) A bar magnet is attached to a string and allowed to swing as shown in the diagram. A copper sheet is then placed underneath the magnet. Explain why the amplitude of the swings decreases rapidly. (12) N S What is the main energy conversion that takes place as the magnet slows down? © Dublin School of Grinds Page 148 (6) Kieran Mills & Tony Kelly A metal loop of wire in the shape of a square of side 5 cm enters a magnetic field of flux density 8 T. The loop is perpendicular to the field and is travelling at a speed of 5 m s–1. (i) How long does it take the loop to completely enter the field? (ii) What is the magnetic flux cutting the loop when it is completely in the magnetic field? (iii) What is the average emf induced in the loop as it enters the magnetic field? © Dublin School of Grinds Page 149 (20) Kieran Mills & Tony Kelly LC 2013: Inductors, diodes and capacitors, resistance and resistivity 8. (a) The diagram shows a circuit used in a charger for a mobile phone. Name the parts labelled F, G and H. F (9) G X H AC input Y Describe the function of G in this circuit. (6) Sketch graphs to show how voltage varies with time for (i) the input voltage (ii) the output voltage, VXY. (12) The photograph shows the device H used in the circuit. Use the data printed on the device to calculate the maximum energy that it can store. © Dublin School of Grinds Page 150 Kieran Mills & Tony Kelly (9) (b) Electricity generating companies transmit electricity over large distances at high voltage. Explain why high voltage is used. (6) A 3 km length of aluminium wire is used to carry a current of 250 A. The wire has a circular cross-section of diameter 18 mm. (i) Calculate the resistance of the aluminium wire. (ii) Calculate how much electrical energy is converted to heat energy in the wire in ten minutes. (resistivity of aluminium = 2.8 × 10 −8 Ω m) © Dublin School of Grinds Page 151 Kieran Mills & Tony Kelly (14) LC 2002: Electromagnetic induction, inductors 12. (c) What is meant by electromagnetic induction? (6) State Lenz’s law of electromagnetic induction. (6) In an experiment, a coil was connected in series with an ammeter and an a.c. power supply as shown in the diagram. Explain why the current was reduced when an iron core was inserted in the coil. (12) A Give an application of the principle shown by this experiment. © Dublin School of Grinds Page 152 (4) Kieran Mills & Tony Kelly LC 2003: Electromagnetic induction, Lenz's law 12. (d) State the laws of electromagnetic induction. (12) A small magnet is attached to a spring as shown in the diagram. The magnet is set oscillating up and down. Describe the current flowing in the circuit. (6) spring S N A coil If the switch at A is open, the magnet will take longer to come to rest. Explain why. © Dublin School of Grinds Page 153 Kieran Mills & Tony Kelly (10) LC 2004: Electromagnetic induction, Lenz's law 12. (c) What is electromagnetic induction? Describe an experiment to demonstrate electromagnetic induction. (15) A light aluminium ring is suspended from a long thread as shown in the diagram. When a strong magnet is moved away from it, the ring follows the magnet. Explain why. What would happen if the magnet were moved towards the ring? N © Dublin School of Grinds (13) S Page 154 Kieran Mills & Tony Kelly LC 2005: Electromagnetic induction 12. (b) Define magnetic flux. (6) State Faraday’s law of electromagnetic induction. (6) A square coil of side 5 cm lies perpendicular to a magnetic field of flux density 4.0 T. The coil consists of 200 turns of wire. (i) What is the magnetic flux cutting the coil? (9) (ii) The coil is rotated through an angle of 90o in 0.2 seconds. Calculate the magnitude of the average e.m.f. induced in the coil while it is being rotated. (7) © Dublin School of Grinds Page 155 Kieran Mills & Tony Kelly LC 2007: Electromagnetic induction 12. (c) State Faraday’s law of electromagnetic induction. Describe an experiment to demonstrate Faraday’s law. (6) (12) A resistor is connected in series with an ammeter and an ac power supply. A current flows in the circuit. The resistor is then replaced with a coil. The resistance of the circuit does not change. What is the effect on the current flowing in the circuit? Justify your answer. (10) © Dublin School of Grinds Page 156 Kieran Mills & Tony Kelly LC 2014: Electromagnetic induction, Lenz's law 12. (d) State Faraday’s law of electromagnetic induction. (6) Describe an experiment to demonstrate Faraday’s law. (9) A hollow copper pipe and a hollow glass pipe, with identical dimensions, were arranged as shown in the diagram. A student measured the time it took a strong magnet to fall through each cylinder. It took much longer for the magnet to fall through the copper pipe. Explain why. (13) © Dublin School of Grinds Page 157 Kieran Mills & Tony Kelly Kieran Mills 6th Year Physics Higher Level Kieran Mills has 21 years’ experience teaching Physics at the very highest level. He has co-written several books including Leaving Certificate Sample Papers. He is also the co-author of the Physics Experiment Book, the bible of the examinable experiment element worth 30% of the exam, which is also given free of charge to all Physics students at The Dublin School of Grinds. Kieran has also appeared on RTE Radio as an expert contributor on Physics. OUR EXPERT TEACHERS