Physics Chapter 18 1 CHAPTER 18:Electric current and direct-current circuits (7 Hours) www.kmph.matrik.edu.my Physics Chapter 18 2 At the end of this chapter, students should be able to: Describe microscopic model of current. Define and use electric current formulae,Learning Outcome: 18.1Electrical conduction (1 hour) dtdQI =www.kmph.matrik.edu.my Physics Chapter 18 3 18.1.1 Electric current, I Consider a simple closed circuit consists of wires, a battery and a light bulb as shown in Figure 18.1. 18.1 Electrical conduction Area, A eFEIFigure18.1 Physics Chapter 18 4 From the Figure 18.1, Direction of electric field or electric current :Positive to negative terminal Direction of electron flows :Negative to positive terminal The electron accelerates because of the electric force acted on it. is defined as the total (nett) charge, Q flowing through the area per unit time, t. Mathematically, tQI =dtdQI =OR instantaneous current average current Physics Chapter 18 5 It is a base and scalar quantities. The S.I. unit of the electric current is the ampere (A). 1 ampere of current is defined as one coulomb of charge passing through the surface area in one second. OR 1s C 1second 1coulomb 1ampere 1= =Note: If the charge move around a circuit in the same direction at all times, the current is called direct current (dc), which is produced by the battery. Physics Chapter 18 6 is defined as the current flowing through a conductor per unit cross-sectional area. Mathematically, It is a vector quantity. Its unit is ampere per squared metre (A m2) The direction of current density, J always in the same direction of the current I. e.g. in Figure 18.2. 18.1.2 Current density, J AIJ =where current electric : Iconductor the of area sectional - cross : AImaxJ0 = JArea, A Figure18.2 Physics Chapter 18 7 In metal the charge carrier is free electrons and a lot of free electrons are available in it. They move freely and randomly throughout the crystal lattice structure of the metal but frequently interact with the lattices. When the electric field is applied to the metal, the freely moving electron experience an electric force and tend to drift with constant average velocity (called drift velocity) towards a direction opposite to the direction of the field as shown in Figure 18.3. Then the electric current is flowing in the opposite direction of the electron flows. 18.1.3 Electrical conduction in metal EIdvdvFigure18.3 Note: The magnitude of the drift velocity is much smaller than the random velocities of the free electron. Physics Chapter 18 8 Consider a metal rod of length L and cross-sectional area A, which is applied to the electric field as shown in Figures 18.4. Suppose there are n free electrons (charge carrier) per unit volume in the metal rod, thus the number of free electron, N is given by 18.1.4 Drift velocity of charges, vd EJIdvdvLA Figure18.4 VNn = AL V =and ALNn = nAL N =Physics Chapter 18 9 The total charge Q of the free electrons that pass through the area A along the rod is The time required for the electron moving along the rod is SinceNe Q =( )e nAL Q =tLv =ddvLt =then the drift velocity vd is given by tQI =( )ddnAevvLe nALI =||.|

\|=nAeIv =dJAI=and OR where electron the of charge : eDefinition Density ofthe free electron neJv =delectron free of number: ne unit volum percarrier) (chargePhysics Chapter 18 10 A silver wire carries a current of 3.0 A. Determine a. the number of electrons per second pass through the wire, b. the amount of charge flows through a cross-sectional area of thewire in 55 s. (Given charge of electron, e = 1.60 1019 C) Solution : a. By applying the equation of average current, thus b. Given, thus the amount of charge flows is given by Example 18.1 : A 0 . 3 = ItQI =( )tN1910 60 . 10 . 3=1 19s electrons 10 88 . 1 =tNandNe Q =tNeI =s 55 = tIt Q =( )55 0 . 3 = Q C 165 = QPhysics Chapter 18 11 A copper wire of radius 900 m carries a current of 17 mA. The wire contains 8.49 1028 free electrons per cubic meter. Determine a. the magnitude of the drift velocity in the wire, b. the current density in the wire. (Given charge of electron, e = 1.60 1019 C) Solution : a. By applying the equation of the drift velocity, thus b. The current density is given by Example 18.2 : 3 28 3 6m 10 49 . 8 A; 10 17 m; 10 900 = = = n I rnAeIv =d( )( ) ( )1926 283d10 60 . 1 10 900 10 49 . 810 17 =v1 7ds m 10 92 . 4 = vand 2r A=e r nIv2dt=2rIJ =( )2 3263m A10 68 . 610 90010 17 ==JPhysics Chapter 18 12 A high voltage transmission line with a diameter of 3.00 cm and a length of 100 km carries a steady current of 1500 A. If the conductor is copper wire with a free charge density of 8.49 1028 electrons m-3, calculate the time taken by one electron to travel the full length of the line. (Given charge of electron, e = 1.60 1019 C) Solution : By applying the equation of the drift velocity, thus Therefore the time taken by one electron to travel the line isExample 18.3 : nAeIv =d( )( )( ) ( )1922 28d10 60 . 1 10 00 . 3 10 49 . 81500 4 =v1 4ds m 10 56 . 1 = vand 42dA =e d nIv2d4t=dvLt =s 10 41 . 610 56 . 110 100843 ==tA; 1500 m; 10 100 m; 10 00 . 33 2= = =I L d3 28m 10 49 . 8 = nPhysics Chapter 18 13 Explain how electrical devices can begin operating almost immediately after you switch on, even though the individual electrons in the wire may take hours to reach the device. Solution : Example 18.4 : Each electron in the wire affects its neighbours by exerting a force on them, causing them to move. When electrons begin to move out of a battery or source their motion sets up a propagating influence that moves through the wire at nearly the speed of light, causing electrons everywhere in the wire begin to move. Physics Chapter 18 14 At the end of this chapter, students should be able to: Define and use resistivity, State and use Ohms law. Learning Outcome: 18.2Resistivity and Ohms law ( hour) IR V =lRA =www.kmph.matrik.edu.my Physics Chapter 18 15 18.2.1 Resistance, R is defined as a ratio of the potential difference across an electrical component to the current passing through it. Mathematically, It is a measure of the components opposition to the flow of the electric charge. It is a scalar quantity and its unit is ohm (O ) or V A1 In general, the resistance of a metallic conductor increases with temperature. 18.2 Resistivity and Ohms law IVR =where (voltage) difference potential : Vcurrent : INote: If the temperature of the metallic conductor is constant hence its resistance also constant. (18.1) Physics Chapter 18 16 Resistivity, is defined as the resistance of a unit cross-sectional area per unit length of the material. Mathematically, It is a scalar quantity and its unit is ohm meter (O m) It is a measure of a materials ability to oppose the flow of an electric current. It also known as specific resistance. Resistivity depends on the type of the material and on the temperature. A good electric conductors have a very low resistivities and good insulators have very high resistivities. 18.2.2 Resistivity and conductivity lRA =where material the of length: larea sectional - cross : A(18.2) Physics Chapter 18 17 From the eq. (18.2), the resistance of a conductor depends on the length and cross-sectional area. Table 18.1 shows the resistivity for various materials at 20 C. Conductivity, o is defined as the reciprocal of the resistivity of a material. Mathematically, It is a scalar quantity and its unit is O1 m1. Material Resistivity, ( O m) Silver1.59 108 Copper1.68 108 Aluminum2.82 108 Gold2.44 108 Glass10101014 Table18.1 1= (18.3) Physics Chapter 18 18 Two wires P and Q with circular cross section are made of the same metal and have equal length. If the resistance of wire P is three times greater than that of wire Q, determine the ratio of their diameters. Solution : Given Example 18.5 : l l l = = = =Q P Q P;Q P3R R =and AlR =3PQ=ddQQ QPP P3Al Al =and 42dA =||.|

\|=2Q2P434dldlOR 31QP=ddPhysics Chapter 18 19 Whena potential difference of 240 Visappliedacross a wire that is 200 cmlong andhasa0.50 mmradius, the currentdensity is 7.14 109 A m2. Calculate a. the resistivity of the wire, b. the conductivity of the wire. Solution : a. From the definition of resistance, thus b. The conductivity of the wire is given by Example 18.6 : IVR =m 10 68 . 18O =where AlR =JAVAl= ( )910 14 . 724000 . 2= 1 1 78m 10 95 . 510 68 . 11 O == m; 10 50 . 0 m; 00 . 2 V; 2403 = = = r l V2 9m A10 14 . 7 = Jand JA I =1=Physics Chapter 18 20 States that the potential difference across a metallic conductor is proportional to the current flowing through it if its temperature is constant. Mathematically, Ohms law also can be stated in term of electric field E and current density J. Consider a uniform conductor of length l and cross-sectional area A as shown in Figure 18.5. 18.2.3 Ohms law (18.4) I V where conductor a of resistance : Rwhere constant = TThenIR V =Figure18.5 EI AlPhysics Chapter 18 21 A potential difference V is maintained across the conductor sets up by an electric field E and this field produce a current I that is proportional to the potential difference. If the field is assumed to be uniform, the potential difference V is related to the field through the relationship below : From the Ohms law, Ed V = El V =IR V = JA I = where |.|

\|=AlJA ElAlR =andJ E =1=andOR E J =(18.5) Physics Chapter 18 22 Figures 18.6a, 18.6b, 18.6c and 18.6d show the potential difference V against current I graphs for various materials. VI0Gradient, m = R Figure18.6a : metal VI0Figure18.6b : semiconductor Physics Chapter 18 23 VI0Figure18.6c : carbon VI0Figure18.6d : electrolyte Note: Some conductors have resistances which depend on the currents flowing through them are known as Ohmic conductors and are said to obey Ohms law. Meanwhile, non-ohmic conductors are the conductors where their resistance depend only of the temperature. Physics Chapter 18 24 Acopperwirecarriesacurrentof10.0A.Thecrosssectionofthe wire is a square of side 2.0 mm and its length is 50 m. The density of the free electron in the wire is 8.0 1028 m3. Determine a. the current density, b. the drift velocity of the electrons, c. the electric field intensity between both end of the wire, d. the potential difference across the wire, e. the resistance of the wire. (Given the resistivity of copper is 1.68 108O m and charge of electron, e = 1.60 1019 C) Solution : a. The current density is given by Example 18.7 : ; m 10 0 . 8 m; 10 0 . 2 A; 0 . 103 28 3 = = = n a Im 50 = lAIJ =2a A = and2aIJ =( )2 623m A10 5 . 210 0 . 20 . 10 == JPhysics Chapter 18 25 Solution : d. By using the equation of drift velocity, thus c. The electric field intensity is; m 10 0 . 8 m; 10 0 . 2 A; 0 . 103 28 3 = = = n a Im 50 = lnAeIv =d( )( ) ( )1923 28d10 60 . 1 10 0 . 2 10 0 . 80 . 10 = v1 4ds m 10 95 . 1 = vand 2a A =e naIv2d =J E =( )( )6 810 5 . 2 10 68 . 1 =E1C N 042 . 0= EPhysics Chapter 18 26 Solution : d. By applying the relationship between uniform E and V, hence e. From the ohms law, therefore ; m 10 0 . 8 m; 10 0 . 2 A; 0 . 103 28 3 = = = n a Im 50 = lEl V =( )( ) 50 042 . 0 = VV 1 . 2 = VIR V =R 0 . 10 1 . 2 =O = 21 . 0 RPhysics Chapter 18 27 Exercise 18.1 : 1.A block in the shape of a rectangular solid has a cross-sectional area of 3.50 cm2 across its width, a front to rear length of 15.8 cm and a resistance of 935 O. The material of which the block is made has 5.33 1022 electrons m3. A potential difference of 35.8 V is maintained between its front and rear faces. Calculate a.the current in the block, b.the current density in the block, c.the drift velocity of the electron, d.the magnitude of the electric field in the block. (Fundamentals of Physics,6th edition, Halliday, Resnick & Walker, Q24, p.631) ANS. :3.83 102 A;109 A m2; 1.28 102 m s1; 227 V m1

Physics Chapter 18 28 2. Figure 18.7 shows a rod in is made of two materials. Each conductor has a square cross section and 3.00 mm on a side. Thefirst material has a resistivity of 4.00 103 O m and is 25.0 cm long, while the second material has a resistivity of 6.00 103 O m and is 40.0 cm long. Determine the resistance between the ends of the rod.(Physics for scientists and engineers,6th edition,Serway&Jewett, Q24, p.853) ANS. :378 O Figure18.7 Physics Chapter 18 29 3.A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire. Determine a.the current in the copper and silver wires. b.the magnitude of the electric field in copper and silver wires. c. the potential difference between the ends of the silver section of wire. (Given (silver) is 1.47 108 O m and (copper) is 1.72 108 O m) (University physics,11th edition, Young&Freedman, Q25.56, p.976)ANS. :45 A; 2.76 V m1, 2.33 V m1; 2.79 V Physics Chapter 18 30 At the end of this chapter, students should be able to: Explain the effect of temperature on electrical resistance in metals and superconductors Define and use temperature coefficient of resistivity,o. Apply resistance : Learning Outcome: 18.3Variation of resistance with temperature (1 hour) ( ) | |0 01 T T R R + = owww.kmph.matrik.edu.my T AA=0Physics Chapter 18 31 18.3.1 Effect of temperature on resistance Metal When the temperature increases, the number of free electrons per unit volume in metal remains unchanged. Metal atoms in the crystal lattice vibrate with greater amplitude and cause the number of collisions between the free electrons and metal atoms increase. Hence the resistance in the metal increases. Superconductor Superconductor is a class of metals and compound whose resistance decreases to zero when they are below the critical temperature Tc. 18.3Variation of resistance withtemperature Physics Chapter 18 32 Table 18.2 shows the critical temperature for various superconductors. When the temperature of the metal decreases, its resistance decreases to zero at critical temperature. Superconductor have many technological applications such as magnetic resonance imaging (MRI) magnetic levitation of train faster computer chips powerful electric motors and etc Material Tc( K) Lead7.18 Mercury4.15 Tin3.72 Aluminum1.19 Zinc0.88 Table18.2 Video 18.1 Video 18.2 Physics Chapter 18 33 is defined as a fractional increase in resistivity of a conductor per unit rise in temperature.OR Since A = 0 then The unit of ois C1 OR K 1. From the equation (18.7), the resistivity of a conductors varies approximately linearly with temperature. 18.3.2 Temperature coefficient of resistivity, o T AA=0where yresistivit in the change : A0change uretemperat : T T T = Ayresistivit initial :0y resistivit final : where ( ) T A + = 10(18.6) (18.7) Physics Chapter 18 34 From the definition of resistivity, thus then the equation (18.7) can be expressed as Table 18.3 shows the temperature coefficients of resistivity for various materials. R ( ) T R R A + = 10(18.8) where resistance initial :0Rresistance final : RMaterial o (C1) Silver4.10 103 Mercury0.89 103 Iron6.51 103 Aluminum4.29 103 Copper6.80 103 Table18.3 Physics Chapter 18 35 Figures 18.8a, 18.8b, 18.8c and 18.8d show the resistance R against temperature T graphs for various materials. RT00RcTFigure18.8a : metal Figure18.8b : semiconductor RT0RT0Figure18.8c : superconductor RT0Figure18.8d : carbon Physics Chapter 18 36 A copper wire has a resistance of 25 mO at 20 C. When the wire is carryingacurrent,heatproducedbythecurrentcausesthe temperature of the wire to increase by 27 C. a. Calculate the change in the wires resistance. b. If its original current was 10.0 mA and the potential differenceacross wire remains constant, what is its final current? (Given thetemperaturecoefficientofresistivityforcopperis 6.80 103 C1) Solution : a. By using the equation for temperature variation of resistance, thus Example 18.8 : C 27 C; 20 ; 10 25030 = A = O =T T R( ) T R R A + = 10R R R A = 0andO = A10 59 . 43RT R R R A = 0 0T R R A = A0( )( )( ) 27 10 80 . 6 10 253 3 = ARPhysics Chapter 18 37 Solution : b. GivenBy using the equation for temperature variation of resistance, thus C 27 C; 20 ; 10 25030 = A = O =T T R( ) T R R A + = 100IVR =and whereIVR =A 10 0 . 1030 = I( ) T IVIVA + = 10( )( )( ) | | 27 10 80 . 6 110 0 . 101 133 +=IA 10 45 . 83 = IPhysics Chapter 18 38 At the end of this chapter, students should be able to: Define emf, c Explain the difference between emf of a battery and potential difference across the battery terminals. Apply voltage, Learning Outcome: 18.4Electromotive force (emf), potential difference and internal resistance ( hour) Ir V =www.kmph.matrik.edu.my Physics Chapter 18 39 18.4.1 Emf, cand potential difference, V Consider a circuit consisting of a battery (cell) that is connected by wires to an external resistor R as shown in Figure 18.9. 18.4Electromotive force (emf), potentialdifference and internal resistance I Battery (cell) A B rRIFigure18.9 Physics Chapter 18 40 A current I flows from the terminal A to the terminal B. For the current to flow continuouslyfrom terminal A to B, a source of electromotive force (e.m.f.), c is required such as battery to maintained the potential difference between point A and point B. Electromotive force (emf),c is defined as the energy provided by the source (battery/cell) to each unit charge that flows through the external and internal resistances. Terminal potential difference (voltage), V is defined as the work done in bringing a unit (test) charge from the negative to the positive terminals of the battery through the external resistance only. The unit for both e.m.f. and potential difference are volt (V). When the current I flows naturally from the battery there is an internal drop in potential difference (voltage) equal to Ir. Thus the terminal potential difference (voltage), V is given by Physics Chapter 18 41 then Equation (18.9) is valid if the battery (cell) supplied the current to the circuit where For the battery without internal resistance or if no current flows in the circuit (open circuit), then equation (18.9) can be written as Ir V =(18.9) and IR V =( ) r R I + =(18.10) wheree.m.f. : (voltage) difference potential terminal : VrOR difference potential indrop internal : V Irresistance external total : R(battery) cell a of resistance internal : r V < V =Physics Chapter 18 42 is defined as the resistance of the chemicals inside thebattery (cell) between the poles and is given by The value of internal resistance depends on the type of chemical material in the battery. The symbol of emf and internal resistance in the electrical circuit are shown in Figures 18.10a and 18.10b. 18.4.2 Internal resistance of a battery, r IVrwhen the cell (battery) is used. where resistance internal across difference potential :rVcircuit in the current: IrOR rFigure18.10aFigure18.10b Physics Chapter 18 43 A battery has an emf of 9.0 V and an internal resistance of 6.0 O.Determine a. the potential difference across its terminals when it is supplying acurrent of 0.50 A, b. the maximum current which the battery could supply. Solution :a. GivenBy applying the expression for emf, thus b. The current is maximum when the total external resistance, R =0,therefore Example 18.9 : O = = 0 . 6 V; 0 . 9 r A 50 . 0 = IV 0 . 6 = V( )( ) 0 . 6 50 . 0 0 . 9 + =VIr V + =A 5 . 1max= I( ) 0 . 6 0 0 . 9max+ = I( ) r R I + =Physics Chapter 18 44 Acarbatteryhasan emf of12.0Vandaninternalresistanceof 1.0 O. The external resistor of resistance 5.0 O is connected in series with the battery as shown in Figure 18.11. Determinethereadingoftheammeterandvoltmeterifbothmeters are ideal. Example 18.10 : RV rA Figure18.11 Physics Chapter 18 45 Solution : By applying the equation of e.m.f., the current in the circuit is Therefore the reading of the ammeter is 2.0 A. The voltmeter measures the potential difference across the terminals of the battery equal to the potential difference across the total external resistor, thus its reading is O = O = = 0 . 5 ; 0 . 1 V; 0 . 12 R r IR V =A 0 . 2 = I( ) r R I + =( ) 0 . 1 0 . 5 0 . 12 + = I( )( ) 0 . 5 0 . 2 = VV 10 = VPhysics Chapter 18 46 At the end of this chapter, students should be able to: Apply electrical energy, Learning Outcome: 18.5Electrical energy and power ( hour) VI P =VIt W =and power, www.kmph.matrik.edu.my Physics Chapter 18 47 18.5.1 Electrical energy, E Consider a circuit consisting of a battery that is connected by wires to an electrical device (such as a lamp, motor or battery being charged) as shown in Figure 18.12 where the potential different across that electrical device is V. 18.5 Electrical energy and power Figure18.12 Electrical device AB V IIPhysics Chapter 18 48 A current I flows from the terminal A to the terminal B, if it flows for time t, the charge Q which it carries from B to A is given by Then the work done on this charge Q from B to A (equal to the electrical energy supplied) is If the electrical device is passive resistor (device which convert all the electrical energy supplied into heat), the heat dissipated H is given by QV W =It Q =VIt E W = =(18.11) VIt W H = =OR Rt I H2=(18.12) Physics Chapter 18 49 is defined as the energy liberated per unit time in the electrical device. The electrical power P supplied to the electrical device is given by When the electric current flows through wire or passive resistor, hence the potential difference across it is then the electrical power can be written as It is a scalar quantity and its unit is watts (W). 18.5.2 Power, P tVIttWP = =IV P =(18.13) IR V =R I P2=OR RVP2=(18.14) Physics Chapter 18 50 InFigure18.13,abatteryhasanemfof12Vandaninternal resistance of 1.0 O. Determine a. the rate of energy transferredto electrical energy in the battery, b. the rate of heat dissipated in the battery, c. the amount ofheatlossinthe5.0 O resistor if the current flowsthrough it for 20 minutes. Example 18.11 : Figure18.13 RrPhysics Chapter 18 51 Solution : The current in the circuit is given by a. The rate of energy transferred to electrical energy (power) in thebattery is b. The rate of heat dissipated due to the internal resistance is c. GivenThe amount of heat loss in the resistor isO = O = = 0 . 5 ; 0 . 1 V; 0 . 12 R r I P =A 0 . 2 = I( ) r R I + =( ) 0 . 1 0 . 5 0 . 12 + = I( )( ) 0 . 12 0 . 2 = P W 24 = Pr I P2= ( ) ( ) 0 . 1 0 . 22= P W 0 . 4 = P( ) s 1200 60 20 = = tRt I H2= ( ) ( )1200 0 . 5 0 . 22= HJ 10 4 . 24 = HPhysics Chapter 18 52 Cells in series Consider two cells are connected in series as shown in Figure 18.14. The total emf, c and the total internal resistance, r are given by 18.5.3 Combination of cells 1r2r12Figure18.14 2 1r r r + =2 1 + =and (18.15) (18.16) Note: If one cell, e.m.f. c2 say, is turned round in opposition to the others, then but the total internal resistance remains unaltered. 2 1 =Physics Chapter 18 53 Cells in parallel Consider two equal cells are connected in parallel as shown in Figure 18.15. The total emf, c and the total internal resistance, r are given by 1r1r11Figure18.15 1 11 1 1r r r+ =1 =and (18.17) (18.18) Note: If different cells are connected in parallel, there is no simple formula for the total emf and the total internal resistance where Kirchhoffs laws have to be used. Physics Chapter 18 54 Exercise 18.2 : 1.A wire of unknown composition has a resistance of 35.0 O when immersed in the water at 20.0 C. When the wire is placed in the boiling water, its resistance rises to 47.6 O. Calculate the temperature on a hot day when the wire has a resistance of 37.8 O. (Physics,7th edition, Cutnell & Johnson, Q15, p.639) ANS. :37.8 C 2.a.A battery of emf 6.0 V is connected across a 10 O resistor. If the potential difference across the resistor is 5.0 V, determine i.the current in the circuit, ii. the internal resistance of the battery. b.When a 1.5 V dry cell is short-circuited, a current of 3.0 A flows through the cell. What is the internal resistance of the cell? ANS. :0.50 A, 2.0 O; 0.50 O Physics Chapter 18 55 3.An electric toy of resistance 2.50 O is operated by a dry cell of emf 1.50 V and an internal resistance 0.25 O. a.What is the current does the toy drawn? b.If the cell delivers a steady current for 6.00 hours, calculate the charge pass through the toy. c.Determine the energy was delivered to the toy. ANS. :0.55 A; 1.19 104 C; 16.3 kJ 4.A wire 5.0 m long and 3.0 mm in diameter has a resistance of 100 O. A 15 V of potential difference is applied across the wire. Determine a.the current in the wire, b.the resistivity of the wire, c.the rate at which heat is being produced in the wire. (College Physics,6th edition, Wilson, Buffa & Lou, Q75, p.589) ANS. :0.15 A; 1.40 104 O m; 2.30 W Physics Chapter 18 56 At the end of this chapter, students should be able to: Deduce and calculate effective resistance of resistors in series and parallel. Learning Outcome: 18.6Resistors in series and parallel (1 hour) www.kmph.matrik.edu.my Physics Chapter 18 57 18.6.1 Resistors in series The symbol of resistor in an electrical circuit can be shown in Figure 18.16. Consider three resistors are connected in series to the battery as shown in Figure 18.17. 18.6 Resistors in series and parallel OR RRFigure18.16 1R2R3RV1V2V3VIIFigure18.17 Physics Chapter 18 58 Characteristics of resistors in series The same current I flows through each resistor where Assuming that the connecting wires have no resistance, the total potential difference, V is given by From the definition of resistance, thus Substituting for V1, V2 , V3 and V in the eq. (5.19) gives (18.19) (18.20) 3 2 1I I I I = = =3 2 1V V V V + + =;2 2IR V= ;3 3IR V=;1 1IR V=effIR V =3 2 1 effIR IR IR IR + + =3 2 1 effR R R R + + =where resistance t) (equivalen effective :effRPhysics Chapter 18 59 V1R3R2R Consider three resistors are connected in parallel to the battery as shown in Figures 18.18a and 18.18b. 18.6.2 Resistors in parallel II2I1I3I1V2V3VV1R3R2RII1I3I2IFigure18.18a Figure18.18b 2V3V1VPhysics Chapter 18 60 Characteristics of resistors in parallel There same potential difference, V across each resistor where The charge is conserved, therefore the total current I in the circuit is given by From the definition of resistance, thus Substituting for I1, I2 , I3 and I in the eq. (18.21) gives (18.21) (18.22) 3 2 1V V V V = = =3 2 1I I I I + + =;22RVI = ;33RVI =;11RVI =effRVI =3 2 1 effRVRVRVRV+ + =3 2 1 eff1 1 1 1R R R R+ + =Physics Chapter 18 61 For the circuit in Figure 18.19, calculate a. the effective resistance of the circuit, b. the current passes through the 12 O resistor, c. the potential difference across 4.0 O resistor, d. the power delivered by the battery. The internal resistance of the battery may be ignored. Example 18.12 : Figure18.19 O 0 . 4O 0 . 2V 0 . 8O 12Physics Chapter 18 62 Solution : a. The resistors R1 and R2 are in series, thus R12 is Since R12 and R3 are in parallel, therefore Reff is given byV 0 . 8 ; 0 . 2 ; 12 ; 0 . 43 2 1= O = O = O = V R R R1RV2R3R12RV3RO = 1612R2 1 12R R R + = 12 0 . 412+ = R3 12 eff1 1 1R R R+ =21161 1eff+ =RO = 78 . 1effRPhysics Chapter 18 63 Solution : b. Since R12 and R3 are in parallel, thus Therefore the current passes through R2 is given by c. Since R1 and R2 are in series, thus Hence the potential difference across R1 is d. The power delivered by the battery isV 0 . 8 ; 0 . 2 ; 12 ; 0 . 43 2 1= O = O = O = V R R RA 50 . 02 = IV 0 . 83 12= = = V V V12122RVI =A 50 . 02 1= = I I1 1 1R I V=V 0 . 21 = V160 . 82 = I( ) 0 . 4 50 . 01 = Veff2RVP =( )78 . 10 . 82= P W 0 . 36 = PPhysics Chapter 18 64 For the circuit in Figure 18.20, calculate the effective resistance between the points A and B. Solution : ; 20 ; 10 ; 0 . 5 ; 0 . 54 3 2 1O = O = O = O = R R R RO = 105RExample 18.13 : Figure18.20O 0 . 5O 10O 10ABO 0 . 5O 202R3R5RAB1R4R3R5RAB12R4RPhysics Chapter 18 65 Solution : R1 and R2 are connected in series, thus R12 is 2 1 12R R R + =; 20 ; 10 ; 0 . 5 ; 0 . 54 3 2 1O = O = O = O = R R R RO = 105RO = + = 10 0 . 5 0 . 512R5RAB123R4RSinceR12andR3areconnectedin parallel , thus R123 is given by 3 12 1231 1 1R R R+ =O = 0 . 5123R101101 1123+ =R5RAB1234RR123andR4areconnectedinseries, thus R1234 is given by 4 123 1234R R R + =O = 251234R20 0 . 51234+ = RPhysics Chapter 18 66 Solution : SinceR1234andR5areconnectedinparallel,thereforethe effective resistance Reff is given by ; 20 ; 10 ; 0 . 5 ; 0 . 54 3 2 1O = O = O = O = R R R RO = 105R5 1234 eff1 1 1R R R+ =O = 14 . 7effR101251 1eff+ =RABeffRPhysics Chapter 18 67 Exercise 18.3 : 1.Determine the equivalent resistances of the resistors in Figures 18.21, 18.22 and 18.23. ANS. : 0.80 O; 2.7 O; 8.0 O O 0 . 2O 0 . 2O 0 . 2O 0 . 2Figure18.21Figure18.22 O 0 . 6O 0 1O 0 . 6O 0 . 4O 18O 16O 0 . 8O 0 . 9O 16O 0 . 6O 20Figure18.23 Physics Chapter 18 68 2. The circuit in Figure 18.24 includes a battery with a finite internal resistance, r = 0.50 O. a.Determine the current flowing through the 7.1 O and 3.2 O resistors. b.How much current flows through the battery? c.What is the potential difference between the terminals of the battery? (Physics,3th edition, James S. Walker, Q39, p.728) ANS. : 1.1 A, 0.3 A; 1.4 A; 11.3 V O 0 . 1V 12rO 1 . 7O 8 . 5O 5 . 4O 2 . 3Figure18.24 Physics Chapter 18 69 3. Four identical resistors are connected to a battery as shown in Figure 18.25. When the switch is open, the current through the battery is I0.a.When the switch is closed, will the current through the battery increase, decrease or stay the same? Explain. b.Calculate the current that flows through the battery when the switch is closed, Give your answer in terms of I0. (Physics,3th edition, James S. Walker, Q45, p.728) ANS. : U think Figure18.25 RRRRPhysics Chapter 18 70 At the end of this chapter, students should be able to: State and use Kirchhoffs Laws.Learning Outcome: 18.7Kirchhoffs laws (1 hours) www.kmph.matrik.edu.my Physics Chapter 18 71 18.7.1 Kirchhoffs first law (junction or current law) statesthe sum of the currents entering any junctions in a circuit must equal the sum of the currents leaving that junction.OR For example : 18.7 Kirchhoffs laws =out inI I(18.23) A B2I1I5I4I3I3I3 2 1I I I = +5 4 3I I I + = =out inI IFigure18.26 Physics Chapter 18 72 statesin any loop, the sum of emfs is equal to the sum of the products of current and resistance. OR In any loop, Sign convention For emf, c: 18.7.2 Kirchhoffs second law (loop or voltage law) = c IR(18.24) +direction of loop + - - + direction of loop Physics Chapter 18 73 For product of IR: Choose and labeling the current at each junction in the circuit given. Choose any one junction in the circuit and apply the Kirchhoffs first law. Choose any two closed loops in the circuit and designate a direction (clockwise OR anticlockwise) to travel around the loop in applying the Kirchhoffs second law. Solving the simultaneous equation to determine the unknown currents and unknown variables. IR +direction of loop IRIR IRdirection of loop 18.7.3 Problem solving strategy (Kirchhoffs Laws) Physics Chapter 18 74 For example, Consider a circuit is shown in Figure 18.27a. At junction A or D (applying the Kirchhoffs first law) : 1R3R1EDF2R23CAB1I1I1I1I2I2I3I 3I3I3ILoop 1 Loop 2 Loop 3 Figure18.27a =out inI I3 2 1I I I + =(1) Physics Chapter 18 75 For the closed loop (either clockwise or anticlockwise), apply the Kirchhoffs second law. From Loop 1 Figure18.27b (2) FEDAF 11REDF2R2A1I1I1I1I2I2ILoop 1 1 1 2 2 2 1R I R I + = + = c IRPhysics Chapter 18 76 From Loop 2 Figure18.27c (3) ABCDA 23RD2R3CAB2I2I3I 3I3I3ILoop 2 3 3 2 2 3 2R I R I = = c IRPhysics Chapter 18 77 From Loop 3 By solving equation (1) and any two equations from the closed loop, hence each current in the circuit can be determined. Figure18.27d (4) FECBF 1R3R1E F3CB1I1I1I1I3I 3I3I3ILoop 3 1 1 3 3 3 1R I R I + = +Note: From the calculation, sometimes we get negative value of current. This negative sign indicates that the direction of the actual current is opposite to the direction of the current drawn. Physics Chapter 18 78 For the circuit in Figure 18.28, Determine the current and its direction in the circuit. Example 18.14 : Figure18.28 O 1 . 15O .22 6O 50 . 8O 2 , V 1.5 1O 4 , V 5.0 1Physics Chapter 18 79 Solution : By applying the Kirchhoffs 2nd law, thus = IR A 74 . 0 = II I I I I 4 50 . 8 2 22 . 6 1 . 15 5 . 11 0 . 15 + + + + = +O 1 . 15O .22 6O 50 . 8O 2 , V 1.5 1O 4 , V 5.0 1Loop 1 IIII(anticlockwise) Physics Chapter 18 80 For the circuit in Figure 18.29, determinea. the currents I1, I2 and I, b. the potential difference across the 6.7 O resistor, c. the power dissipated from the 1.2 O resistor. Example 18.15 : Figure18.29 O 8 . 9O 9 . 3V .0 9 V 2 1O 7 . 6O .2 1I1I2IPhysics Chapter 18 81 Solution : a. At junction A, by using the Kirchhoffs 1st law, thus By using the Kirchhoffs 2nd law,From Loop 1: =out inI II I I = +2 1O 8 . 9O 9 . 3V .0 9 V 2 1O 7 . 6O .2 11I2II1I2IIA B Loop 1 Loop 2 (1) = IR 1 18 . 9 2 . 1 9 . 3 12 I I I + + =12 2 . 1 7 . 131= + I I(2) Physics Chapter 18 82 Solution : a. From Loop 2: By solving the simultaneous equations, we get b. The potential difference across the 6.7 O resistor is given by c. The power dissipated from the 1.2 O resistor is = IR I I 2 . 1 7 . 6 0 . 92 + =0 . 9 2 . 1 7 . 62= + I I(3) A75 . 1 A; 03 . 1 A; 72 . 02 1= = = I I IR I V2=( ) 7 . 6 03 . 1 = VV 90 . 6 = VR I P2=( ) ( ) 2 . 1 75 . 12= P W 68 . 3 = PPhysics Chapter 18 83 Exercise 18.4 : 1.For a circuit in Figure 18.30, Given c1= 8V, R2= 2 O, R3= 3 O, R1 = 1 O and I = 3 A. Ignore the internal resistance in each battery. Calculatea. the currents I1 and I2. b. the emf, c2. ANS. : 1.0 A, 4.0 A; 17 V Figure18.30 3R12R21I2II1RPhysics Chapter 18 84 2. Determine the current in each resistor in the circuit shown in Figure 18.31. (College Physics,6th edition, Wilson, Buffa & Lou, Q57, p.619) ANS. : 3.75 A; 1.25 A; 1.25 A Figure18.31 O 0 . 4O 0 . 4V .0 5V .0 5O .0 4V 0 1Physics Chapter 18 85 At the end of this chapter, students should be able to: Explain the principle of a potential divider. Apply equation of potential divider, Learning Outcome: 18.8Potential divider ( hour) VR RRV||.|

\|+=2 111www.kmph.matrik.edu.my Physics Chapter 18 86 A potential divider produces an output voltage that is a fraction of the supply voltage V. This is done by connecting two resistors in series as shown in Figure 18.32. Since the current flowing through each resistor is the same, thus 18.8 Potential divider V1V1RI2V2RI2 1 effR R R + =effRVI =and 2 1R RVI+=Figure18.32 Physics Chapter 18 87 Therefore, the potential difference (voltage) across R1 is given by Similarly, Resistance R1 and R2 can be replaced by a uniform homogeneous wire as shown in Figure 18.33. Figure18.33 1 1IR V=VR RRV||.|

\|+=2 111VR RRV||.|

\|+=2 122(18.25) (18.26) VI2l1lIBAC2V1VPhysics Chapter 18 88 The total resistance, RAB in the wire is Since the current flowing through the wire is the same, thus AlR =CB AC ABR R R + =AlAlR2 1AB+ =and ABRVI =( )2 1l lAVI+=( )2 1 ABl lAR + =Physics Chapter 18 89 Therefore, the potential difference (voltage) across the wire with length l1 is given by Similarly, AC 1IR V=( )|.|

\|((((

+=All lAVV12 11Vl llV||.|

\|+=2 111(18.27) Vl llV||.|

\|+=2 122(18.28) Note: From Ohms law, l V |.|

\|= =AlI IR VPhysics Chapter 18 90 For the circuit in Figure 18.34,a. calculate the output voltage. b. If a voltmeter of resistance 4000 O is connected across the output,determine the reading of the voltmeter. Example 18.16 : Figure18.34 O 000 4V 2 1O 000 8outVPhysics Chapter 18 91 Solution : a. The output voltage is given by b. The connection between the voltmeter and 4000 O resistor isparallel, thus the equivalent resistance is Hence the new output voltage is given by Therefore the reading of the voltmeter is 2.4 V. V 12 ; 4000 ; 80002 1= O = O = V R RVR RRV||.|

\|+=2 12outV 0 . 4out = V4000140001 1eq+ =R124000 80004000out|.|

\|+= VO = 2000eqRV 4 . 2out = V122000 80002000out|.|

\|+= VPhysics Chapter 18 92 At the end of this chapter, students should be able to: Explain principles of potentiometer and Wheatstone Bridge and their applications. Use related equations such as Learning Outcome: 18.9Potentiometer and Wheatstone bridge ( hour) x321RRRR=llRRx x=www.kmph.matrik.edu.my Physics Chapter 18 93 18.9.1 Potentiometer Consider a potentiometer circuit is shown in Figure 18.35. The potentiometer is balanced when the jockey (sliding contact) is at such a position on wire AB that there is no current through the galvanometer. Thus 18.9 Potentiometer and Wheatstone bridge Figure18.35 (Driver cell -accumulator) Jockey VBACxVIG +- II IGalvanometer reading = 0 Physics Chapter 18 94 When the potentiometer in balanced, the unknown voltage (potential difference being measured) is equal to the voltage across AC. Potentiometer can be used to compare the emfs of two cells. measure an unknown emf of a cell. measure the internal resistance of a cell. Compare the emfs of two cells In this case, a potentiometer is set up as illustrated in Figure 18.36, in which AB is a wire of uniform resistance and J is a sliding contact (jockey) onto the wire. An accumulator X maintains a steady current I through the wire AB. AC xV V =Physics Chapter 18 95 Initially, a switch S is connected to the terminal (1) and the jockey moved until the emf c1 exactly balances the potential difference (p.d.) from the accumulator (galvanometer reading is zero) at point C. Hence Figure18.36 XBAIG I(2) (1) 2S I I1CJ D1l2lPhysics Chapter 18 96 After that, the switch S is connected to the terminal (2) and the jockey moved until the emf c2 balances the p.d. from the accumulatorat point D. Hence AC 1V =AC ACIR V =where AlR1AC =and 1 1lAI|.|

\|= (1) then AD 2V =AD ADIR V =where AlR2AD =and 2 2lAI|.|

\|= (2) then Physics Chapter 18 97 By dividing eq. (1) and eq. (2) then Measure an unknown emf of a cell By using the same circuit shown in Figure 18.36, the value of unknown emf can be determined if the cell c1 is replaced with a standard cell. A standard cell is one in which provides a constant and accurately known emf. Thus the emf c2 can be calculated by using the equation (18.29). 2121ll=2121lAIlAI|.|

\||.|

\|=(18.29) Physics Chapter 18 98 Measure the internal resistance of a cell Consider a potentiometer circuit as shown in Figure 18.37. Figure18.37 BAIG I10lCJ SRrIIPhysics Chapter 18 99 An accumulator of emf c maintains a steady current I through the wire AB. Initially, a switch S is opened and the jockey J moved until the emf c1 exactly balances the emf cfrom the accumulator (galvanometer reading is zero) at point C. Hence After the switch S is closed, the current I1 flows through the resistance box R and the jockey J moved until the galvanometer reading is zero (balanced condition) at point D as shown in Figure 18.38. AC 1V =AC ACIR V = where AlR0AC =and 0 1lAI|.|

\|= (1) then Physics Chapter 18 100 Figure18.38 BAIG I1J SRrII1IDl1I1I1I1IPhysics Chapter 18 101 Hence From the equation of emf, ADV V =AD ADIR V =where AlR =ADand lAIV|.|

\|= (2) then r I V 1 1+ =11IV r=RVI =1and RVV r|.|

\|=1 (3) Physics Chapter 18 102 By substituting eqs. (1) and (2) into the eq. (3), we get The value of internal resistance, r is determined by plotting the graph of 1/lagainst 1/R . Rearranging eq. (4) : Rll lr|.|

\|=0Rllr|.|

\| = 10 (4) c x m y + =Then compare with 0 01 1 1l R lrl+||.|

\|=Physics Chapter 18 103 Therefore the graph is straight line as shown in Figure 18.39. 0, Gradientlrm =01lR1l10Figure18.39 Physics Chapter 18 104 CellsAandBandcentre-zerogalvanometerGareconnectedtoa uniform wire OS using jockeys X and Y as shown in 18.40. a. the potential difference across OY when OY = 75.0 cm, b. the potential difference across OY when Y touches S and thegalvanometer is balanced, c. the internal resistance of the cell A, d. the emf of cell A. Example 18.17 : Figure18.40 ASOG BX Y The length of the uniform wire OS is 1.00 m and its resistance is 12 O. When OY is 75.0 cm, the galvanometer does not show any deflection when OX= 50.0 cm. If Y touches the end S of the wire, OX = 62.5 cm when the galvanometer is balanced. The emf of the cell B is 1.0 V. Calculate Physics Chapter 18 105 Solution : a. Given When G = 0 (balance condition), thus

V 0 . 1 ; 12 m; 00 . 1B OS OS= O = = R lm 50 . 0 m; 75 . 0OX1 OY1= = l lASOG BX Y 0 =OY1lOX1lSince wire OS is uniform thus OSOS1 OXOX1RllR||.|

\|=and O =|.|

\|= 0 . 6 1200 . 150 . 0OX1RO =|.|

\|= 0 . 9 1200 . 175 . 0OY1RB OX1 V =OX1 1 OX1R I V =and 1I1I1I1I1IB OX1 1 R I = ( ) 0 . 1 0 . 61= IA 17 . 01 = IPhysics Chapter 18 106 Solution : a. Therefore the potential difference across OY is given by b. Given V 0 . 1 ; 12 m; 00 . 1B OS OS= O = = R lOY1 1 OY1R I V =( ) 0 . 9 17 . 0OY1 = VV 53 . 1OY1 = Vm 625 . 0 m; 00 . 1OX2 OY2= = l lASOG BX Y 0 =OY2lOX2l2I2I2I2I2ISince wire OS is uniform thus OSOS2 OXOX2RllR||.|

\|=and O =|.|

\|= 5 . 7 1200 . 1625 . 0OX2RO =|.|

\|= 12 1200 . 100 . 1OY2RPhysics Chapter 18 107 Solution : b. When G = 0 (balance condition), thus

Therefore the potential difference across OY is given by c. The emf of cell A is given by For case in the question (a) : V 0 . 1 ; 12 m; 00 . 1B OS OS= O = = R lB OX2 V =OX2 2 OX2R I V =and B OX2 2 R I = ( ) 0 . 1 5 . 72= IA 13 . 02 = IOY2 2 OY2R I V =( )12 13 . 0OY2 = VV 56 . 1OY2 = V( ) r R I + =A) (1 OY 1 Ar R I + =( ) r + = 0 . 9 17 . 0A (1) Physics Chapter 18 108 Solution : c. For case in the question (b) : (1) = (2): d. The emf of cell A isV 0 . 1 ; 12 m; 00 . 1B OS OS= O = = R l) (2 OY 2 Ar R I + =( ) r + = 12 13 . 0A (2) ( ) ( ) r r + = + 12 13 . 0 0 . 9 17 . 0O = 65 . 0 r( ) r + = 0 . 9 17 . 0A( ) 65 . 0 0 . 9 17 . 0A+ = V 64 . 1A = Physics Chapter 18 109 It is used to measured the unknown resistance of the resistor. Figure 18.41 shows the Wheatstone bridge circuit consists of a cell of emf c (accumulator), a galvanometer , know resistances (R1, R2 and R3) and unknown resistance Rx. The Wheatstone bridge is said to be balanced when no current flows through the galvanometer. 18.9.2 Wheatstone bridge BA G CD1R2R3RxR0 =II2I1I2I1IFigure18.41 Physics Chapter 18 110 Hence then Therefore Since Dividing gives 1 CB ACI I I = =2 DB ADI I I = =and Potential at C = Potential at D AD ACV V =BD BCV V = and IR V =3 2 1 1R I R I =thus and x 2 2 1R I R I =x 23 22 11 1R IR IR IR I=312xRRRR||.|

\|=(18.30) Physics Chapter 18 111 The application of the Wheatstone bridge is Metre Bridge. Figure 18.42 shows a Metre bridge circuit. The metre bridge is balanced when the jockey J is at such a position on wire AB that there is no current through the galvanometer. Thus the current I1 flows through the resistance Rx and R but current I2 flows in the wire AB. = 0Accumulator Jockey Thick copper strip (Unknown resistance) (resistance box) Wire of uniform resistance xRAG BRJ2l1lFigure18.42 II1I2I1IPhysics Chapter 18 112 LetVx : p.d. across Rx and V : p.d. across R, At balance condition, By applying Ohms law, thus Dividing gives AJ xV V =JBV V =and AJ 2 x 1R I R I =JB 2 1R I R I = and AlR1AJ =JB 2AJ 21x 1R IR IR IR I= whereand AlR2JB =|.|

\||.|

\|=AlAlRR21xRllR||.|

\|=21x(18.31) Physics Chapter 18 113 An unknown length of platinum wire 0.920 mm in diameter is placed as the unknown resistance in a Wheatstone bridge as shown in Figure 18.43. Resistors R1 and R2 have resistance of 38.0 O and 46.0 O respectively. Balance is achieved when the switch closed and R3 is 3.48 O. Calculate thelengthof the platinumwireifitsresistivityis10.6 108 O m.Example 18.18 : Figure18.43 Physics Chapter 18 114 Solution : At balance condition, the ammeter reading is zero thus the resistance of the platinum wire is given by From the definition of resistivity, thus ; 0 . 46 ; 0 . 38 m; 10 920 . 02 13O = O = =R R d; m 10 6 . 10 ; 48 . 383 = O = R123xRRRR=O = 21 . 4xR0 . 380 . 4648 . 3x=RlA Rx=42dAt= and ld R42xt=( ) ( )l 410 920 . 0 21 . 410 6 . 10238 t= m 4 . 26 = lPhysics Chapter 18 115 Exercise 18.5 : 1.In Figure 18.44, PQ is a uniform wire of length 1.0 m and resistance 10.0 O. ANS. : 0.50 V; 7.5 O; 25.0 cm; 25.0 cm 2S1PQG 2T 1R2R1SFigure18.44 c1 is an accumulator of emf 2.0 V and negligible internal resistance. R1 is a 15 O resistor and R2 is a 5.0 O resistor when S1 and S2 open, galvanometer G is balanced when QT is 62.5 cm. When both S1 and S2 are closed, the balance length is 10.0 cm. Calculate a. the emf of cell c2. b. the internal resistance of cell c2. c. the balance length QT when S2

is opened and S1 closed. d. the balance length QT when S1

is opened and S2 closed. Physics Chapter 18 116 R2.The circuit shown in Figure 18.45 is known as a Wheatstone bridge. Determine the value of the resistor R such that the current through the 85.0 O resistor is zero. (Physics,3th edition, James S. Walker, Q93, p.731) ANS. : 7.50 O Figure18.45 Physics Chapter 18 117 3.A potentiometer with slide-wire of length 100 cm and resistance of 5.0 O, is connected to a driver cell of emf 2.0 V and negligible internal resistance. Calculate a.the length of the potentiometer wire needed to balance a potential difference of 1.5 V, b.the resistance which must be connected in series with the slide-wire to give a potential difference of 7.0 mV across the whole wire, c.the emf c of a dry cell which is balanced by 80 cm of the wire, setup as in part (b). ANS. : 75.0 cm; 1424 O; 5.6 mV Physics Chapter 18 118 Next Chapter CHAPTER 19 : Magnetic field www.kmph.matrik.edu.my