PHYSICS CHAPTER 51 CHAPTER 5: Electric current and direct-current circuits (7 Hours) PHYSICS CHAPTER 5 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: 5.1 Electrical conduction (1 hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s dt dQ I = PHYSICS CHAPTER 5 3 5.1.1 Electric current, I Consider a simple closed circuit consists of wires, a battery and a light bulb as shown in Figure 5.1. 5.1 Electrical conduction Area, A e F E I Figure 5.1 PHYSICS CHAPTER 5 4 From the Figure 5.1, 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, t Q I = dt dQ I = OR instantaneous current average current PHYSICS CHAPTER 5 5 It is a base and scalar quantities. The S.I. unit of the electric current is the ampere (A). Its dimension is given by 1 ampere of current is defined as one coulomb of charge passing through the surface area in one second. OR | | A = I 1 s C 1 second 1 coulomb 1 ampere 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 5 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 m ÷2 ) The direction of current density, J always in the same direction of the current I. e.g. in Figure 5.2. 5.1.2 Current density, J A I J = where current electric : I conductor the of area sectional - cross : A I max J 0 = J Area, A Figure 5.2 PHYSICS CHAPTER 5 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 5.3. Then the electric current is flowing in the opposite direction of the electron flows. 5.1.3 Electrical conduction in metal E I d v d v Figure 5.3 Note: The magnitude of the drift velocity is much smaller than the random velocities of the free electron. PHYSICS CHAPTER 5 8 Consider a metal rod of length L and cross-sectional area A, which is applied to the electric field as shown in Figures 5.4. 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 5.1.4 Drift velocity of charges, v d E J I d v d v L A Figure 5.4 V N n = AL V = and AL N n = nAL N = PHYSICS CHAPTER 5 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 Since Ne Q = ( )e nAL Q = t L v = d d v L t = then the drift velocity v d is given by t Q I = ( ) d d nAev v L e nAL I = | | . | \ | = nAe I v = d J A I = and OR where electron the of charge : e Definition Density of the free electron ne J v = d electron free of number : n e unit volum per carrier) (charge PHYSICS CHAPTER 5 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 the wire in 55 s. (Given charge of electron, e = 1.60 × 10 ÷19 C) Solution : a. By applying the equation of average current, thus b. Given , thus the amount of charge flows is given by Example 1 : A 0 . 3 = I t Q I = ( ) t N 19 10 60 . 1 0 . 3 ÷ × = 1 19 s electrons 10 88 . 1 ÷ × = t N and Ne Q = t Ne I = s 55 = t It Q = ( )55 0 . 3 = Q C 165 = Q PHYSICS CHAPTER 5 11 A copper wire of radius 900 µm carries a current of 17 mA. The wire contains 8.49 × 10 28 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 × 10 ÷19 C) Solution : a. By applying the equation of the drift velocity, thus b. The current density is given by Example 2 : 3 28 3 6 m 10 49 . 8 A; 10 17 m; 10 900 ÷ ÷ ÷ × = × = × = n I r nAe I v = d ( ) ( ) ( ) 19 2 6 28 3 d 10 60 . 1 10 900 10 49 . 8 10 17 ÷ ÷ ÷ × × × × = π v 1 7 d s m 10 92 . 4 ÷ ÷ × = v and 2 πr A= e r n I v 2 d t = 2 πr I J = ( ) 2 3 2 6 3 m A 10 68 . 6 10 900 10 17 ÷ ÷ ÷ × = × × = π J PHYSICS CHAPTER 5 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 × 10 28 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 × 10 ÷19 C) Solution : By applying the equation of the drift velocity, thus Therefore the time taken by one electron to travel the line is Example 3 : nAe I v = d ( ) ( ) ( ) ( ) 19 2 2 28 d 10 60 . 1 10 00 . 3 10 49 . 8 1500 4 ÷ ÷ × × × = π v 1 4 d s m 10 56 . 1 ÷ ÷ × = v and 4 2 πd A = e d n I v 2 d 4 t = d v L t = s 10 41 . 6 10 56 . 1 10 100 8 4 3 × = × × = ÷ t A; 1500 m; 10 100 m; 10 00 . 3 3 2 = × = × = ÷ I L d 3 28 m 10 49 . 8 ÷ × = n PHYSICS CHAPTER 5 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 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 5 14 At the end of this chapter, students should be able to: Define and use resistivity formulae, State Ohm’s law. Apply formulae, Learning Outcome: 5.2 Resistivity and Ohm’s law (1 hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s IR V = l RA ρ = PHYSICS CHAPTER 5 15 5.2.1 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 component’s opposition to the flow of the electric charge. It is a scalar quantity and its unit is ohm (O ) or V A ÷1 In general, the resistance of a metallic conductor increases with temperature. 5.2 Resistivity and Ohm’s law I V R = where (voltage) difference potential : V current : I Note: If the temperature of the metallic conductor is constant hence its resistance also constant. (5.1) PHYSICS CHAPTER 5 16 Resistivity, µ 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 material’s 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. 5.2.2 Resistivity and conductivity l RA ρ = where material the of length : l area sectional - cross : A (5.2) PHYSICS CHAPTER 5 17 From the eq. (5.2), the resistance of a conductor depends on the length and cross-sectional area. Table 5.1 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 O ÷1 m ÷1 . Material Resistivity, µ ( O m) Silver 1.59 × 10 ÷8 Copper 1.68 × 10 ÷8 Aluminum 2.82 × 10 ÷8 Gold 2.44 × 10 ÷8 Glass 10 10 ÷10 14 Table 5.1 ρ σ 1 = (5.3) PHYSICS CHAPTER 5 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 5 : l l l ρ ρ ρ = = = = Q P Q P ; Q P 3R R = and A ρl R = 3 P Q = d d Q Q Q P P P 3 A l ρ A l ρ = and 4 2 πd A = | | . | \ | = 2 Q 2 P 4 3 4 πd ρl πd ρl OR 3 1 Q P = d d PHYSICS CHAPTER 5 19 When a potential difference of 240 V is applied across a wire that is 200 cm long and has a 0.50 mm radius, the current density is 7.14 × 10 9 A m ÷2 . 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 6 : I V R = m 10 68 . 1 8 O × = ÷ ρ where A ρl R = JA V A ρl = ( ) 9 10 14 . 7 240 00 . 2 × = ρ 1 1 7 8 m 10 95 . 5 10 68 . 1 1 ÷ ÷ ÷ O × = × = σ m; 10 50 . 0 m; 00 . 2 V; 240 3 ÷ × = = = r l V 2 9 m A 10 14 . 7 ÷ × = J and JA I = ρ σ 1 = PHYSICS CHAPTER 5 20 States that the potential difference across a metallic conductor is proportional to the current flowing through it if its temperature is constant. Mathematically, Ohm’s 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 5.5. 5.2.3 Ohm’s law (5.4) I V · where conductor a of resistance : R where constant = T Then IR V = Figure 5.5 E I A l PHYSICS CHAPTER 5 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 Ohm’s law, Ed V = El V = IR V = JA I = where | . | \ | = A ρl JA El A ρl R = and ρJ E = σ ρ 1 = and OR σE J = (5.5) PHYSICS CHAPTER 5 22 Figures 5.6a, 5.6b, 5.6c and 5.6d show the potential difference V against current I graphs for various materials. V I 0 Gradient, m = R Figure 5.6a : metal V I 0 Figure 5.6b : semiconductor PHYSICS CHAPTER 5 23 V I 0 Figure 5.6c : carbon V I 0 Figure 5.6d : electrolyte Note: Some conductors have resistances which depend on the currents flowing through them are known as Ohmic conductors and are said to obey Ohm’s law. Meanwhile, non-ohmic conductors are the conductors where their resistance depend only of the temperature. PHYSICS CHAPTER 5 24 A copper wire carries a current of 10.0 A. The cross section of the 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 × 10 28 m ÷3 . 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 × 10 ÷8 O m and charge of electron, e = 1.60 × 10 ÷19 C) Solution : a. The current density is given by Example 7 : ; m 10 0 . 8 m; 10 0 . 2 A; 0 . 10 3 28 3 ÷ ÷ × = × = = n a I m 50 = l A I J = 2 a A = and 2 a I J = ( ) 2 6 2 3 m A 10 5 . 2 10 0 . 2 0 . 10 ÷ ÷ × = × = J PHYSICS CHAPTER 5 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 . 10 3 28 3 ÷ ÷ × = × = = n a I m 50 = l nAe I v = d ( )( ) ( ) 19 2 3 28 d 10 60 . 1 10 0 . 2 10 0 . 8 0 . 10 ÷ ÷ × × × = v 1 4 d s m 10 95 . 1 ÷ ÷ × = v and 2 a A = e na I v 2 d = J E µ = ( )( ) 6 8 10 5 . 2 10 68 . 1 × × = ÷ E 1 C N 042 . 0 ÷ = E PHYSICS CHAPTER 5 26 Solution : d. By applying the relationship between uniform E and V, hence e. From the ohm’s law, therefore ; m 10 0 . 8 m; 10 0 . 2 A; 0 . 10 3 28 3 ÷ ÷ × = × = = n a I m 50 = l El V = ( )( ) 50 042 . 0 = V V 1 . 2 = V IR V = R 0 . 10 1 . 2 = O = 21 . 0 R PHYSICS CHAPTER 5 27 Exercise 5.1 : 1. A block in the shape of a rectangular solid has a cross- sectional area of 3.50 cm 2 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 × 10 22 electrons m ÷3 . 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,6 th edition, Halliday, Resnick & Walker, Q24, p.631) ANS. : 3.83 × 10 ÷2 A; 109 A m ÷2 ; 1.28 × 10 ÷2 m s ÷1 ; 227 V m ÷1 PHYSICS CHAPTER 5 28 Exercise 5.1 : 2. Figure 5.7 shows a rod in is made of two materials. Each conductor has a square cross section and 3.00 mm on a side. The first material has a resistivity of 4.00 × 10 –3 O m and is 25.0 cm long, while the second material has a resistivity of 6.00 × 10 –3 O m and is 40.0 cm long. Determine the resistance between the ends of the rod. (Physics for scientists and engineers,6 th edition,Serway&Jewett, Q24, p.853) ANS. : 378 O Figure 5.7 PHYSICS CHAPTER 5 29 Exercise 5.1 : 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 µ (silver) is 1.47 × 10 ÷8 O m and µ (copper) is 1.72 × 10 ÷8 O m) (University physics,11 th edition, Young&Freedman, Q25.56, p.976) ANS. : 45 A; 2.76 V m ÷1 , 2.33 V m ÷1 ; 2.79 V PHYSICS CHAPTER 5 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 explain temperature coefficient of resistivity,o. Apply formulae : Learning Outcome: 5.3 Variation of resistance with temperature (1 hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s ( ) | | 0 0 1 T T R R ÷ + = o PHYSICS CHAPTER 5 31 5.3.1 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 T c . 5.3 Variation of resistance with temperature PHYSICS CHAPTER 5 32 Table 5.2 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 T c ( K) Lead 7.18 Mercury 4.15 Tin 3.72 Aluminum 1.19 Zinc 0.88 Table 5.2 PHYSICS CHAPTER 5 33 is defined as a fractional increase in resistivity of a conductor per unit rise in temperature. OR Since Aµ = µ÷µ 0 then The unit of o is °C ÷1 OR K ÷ 1 . From the equation (5.7), the resistivity of a conductors varies approximately linearly with temperature. 5.3.2 Temperature coefficient of resistivity, o T ρ ρ α A A = 0 where y resistivit in the change : ρ A 0 change ure temperat : T T T ÷ = A y resistivit initial : 0 ρ y resistivit final : ρ where ( ) T α ρ ρ A + = 1 0 (5.6) (5.7) PHYSICS CHAPTER 5 34 From the definition of resistivity, thus then the equation (5.7) can be expressed as Table 5.3 shows the temperature coefficients of resistivity for various materials. R ρ · ( ) T α R R A + = 1 0 (5.8) where resistance initial : 0 R resistance final : R Material o (°C ÷1 ) Silver 4.10 × 10 ÷3 Mercury 0.89 × 10 ÷3 Iron 6.51 × 10 ÷3 Aluminum 4.29 × 10 ÷3 Copper 6.80 × 10 ÷3 Table 5.3 PHYSICS CHAPTER 5 35 Figures 5.8a, 5.8b, 5.8c and 5.8d show the resistance R against temperature T graphs for various materials. R T 0 0 R c T Figure 5.8a : metal Figure 5.8b : semiconductor R T 0 R T 0 Figure 5.8c : superconductor R T 0 Figure 5.8d : carbon PHYSICS CHAPTER 5 36 A copper wire has a resistance of 25 mO at 20 °C. When the wire is carrying a current, heat produced by the current causes the temperature of the wire to increase by 27 °C. a. Calculate the change in the wire’s resistance. b. If its original current was 10.0 mA and the potential difference across wire remains constant, what is its final current? (Given the temperature coefficient of resistivity for copper is 6.80 × 10 ÷3 °C ÷1 ) Solution : a. By using the equation for temperature variation of resistance, thus Example 8 : C 27 C; 20 ; 10 25 0 3 0 = A = O × = ÷ T T R ( ) T α R R A + = 1 0 R R R A = ÷ 0 and O × = A ÷ 10 59 . 4 3 R T α R R R A = ÷ 0 0 T α R R A = A 0 ( )( )( ) 27 10 80 . 6 10 25 3 3 ÷ ÷ × × = AR PHYSICS CHAPTER 5 37 Solution : b. Given By using the equation for temperature variation of resistance, thus C 27 C; 20 ; 10 25 0 3 0 = A = O × = ÷ T T R ( ) T α R R A + = 1 0 0 I V R = and where I V R = A 10 0 . 10 3 0 ÷ × = I ( ) T α I V I V A + = 1 0 ( ) ( )( ) | | 27 10 80 . 6 1 10 0 . 10 1 1 3 3 ÷ ÷ × + × = I A 10 45 . 8 3 ÷ × = I PHYSICS CHAPTER 5 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 formulae, Learning Outcome: 5.4 Electromotive force (emf), potential difference and internal resistance (½ hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s Ir ε V ÷ = PHYSICS CHAPTER 5 39 5.4.1 Emf, c and 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 5.9. 5.4 Electromotive force (emf), potential difference and internal resistance I Battery (cell) A B r ε R I Figure 5.9 PHYSICS CHAPTER 5 40 A current I flows from the terminal A to the terminal B. For the current to flow continuously from 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 5 41 then Equation (5.9) 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 (5.9) can be written as Ir ε V ÷ = (5.9) and IR V = ( ) r R I ε + = (5.10) where e.m.f. : ε (voltage) difference potential terminal : V r OR difference potential in drop internal : V Ir resistance external total : R (battery) cell a of resistance internal : r ε V < ε V = PHYSICS CHAPTER 5 42 is defined as the resistance of the chemicals inside the battery (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 5.10a and 5.10b. 5.4.2 Internal resistance of a battery, r I V r when the cell (battery) is used. where resistance internal across difference potential : r V circuit in the current : I r ε OR r ε Figure 5.10a Figure 5.10b PHYSICS CHAPTER 5 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 a current of 0.50 A, b. the maximum current which the battery could supply. Solution : a. Given By applying the expression for emf, thus b. The current is maximum when the total external resistance, R =0, therefore Example 9 : O = = 0 . 6 V; 0 . 9 r ε A 50 . 0 = I V 0 . 6 = V ( )( ) 0 . 6 50 . 0 0 . 9 + =V Ir V ε + = A 5 . 1 max = I ( ) 0 . 6 0 0 . 9 max + = I ( ) r R I ε + = PHYSICS CHAPTER 5 44 A car battery has an emf of 12.0 V and an internal resistance of 1.0 O. The external resistor of resistance 5.0 O is connected in series with the battery as shown in Figure 5.11. Determine the reading of the ammeter and voltmeter if both meters are ideal. Example 10 : R V ε r A Figure 5.11 PHYSICS CHAPTER 5 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 ε IR V = A 0 . 2 = I ( ) r R I ε + = ( ) 0 . 1 0 . 5 0 . 12 + = I ( )( ) 0 . 5 0 . 2 = V V 10 = V PHYSICS CHAPTER 5 46 At the end of this chapter, students should be able to: Apply formula, Learning Outcome: 5.5 Electrical energy and power (½ hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s VI P = VIt W = and PHYSICS CHAPTER 5 47 5.5.1 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 5.12 where the potential different across that electrical device is V. 5.5 Electrical energy and power Figure 5.12 Electrical device A B V I I PHYSICS CHAPTER 5 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 = = (5.11) VIt W H = = OR Rt I H 2 = (5.12) PHYSICS CHAPTER 5 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). 5.5.2 Power, P t VIt t W P = = IV P = (5.13) IR V = R I P 2 = OR R V P 2 = (5.14) PHYSICS CHAPTER 5 50 In Figure 5.13, a battery has an emf of 12 V and an internal resistance of 1.0 O. Determine a. the rate of energy transferred to electrical energy in the battery, b. the rate of heat dissipated in the battery, c. the amount of heat loss in the 5.0 O resistor if the current flows through it for 20 minutes. Example 11 : Figure 5.13 R ε r PHYSICS CHAPTER 5 51 Solution : The current in the circuit is given by a. The rate of energy transferred to electrical energy (power) in the battery is b. The rate of heat dissipated due to the internal resistance is c. Given The amount of heat loss in the resistor is O = O = = 0 . 5 ; 0 . 1 V; 0 . 12 R r ε Iε P = A 0 . 2 = I ( ) r R I ε + = ( ) 0 . 1 0 . 5 0 . 12 + = I ( )( ) 0 . 12 0 . 2 = P W 24 = P r I P 2 = ( ) ( ) 0 . 1 0 . 2 2 = P W 0 . 4 = P ( ) s 1200 60 20 = = t Rt I H 2 = ( ) ( )1200 0 . 5 0 . 2 2 = H J 10 4 . 2 4 × = H PHYSICS CHAPTER 5 52 Cells in series Consider two cells are connected in series as shown in Figure 5.14. The total emf, c and the total internal resistance, r are given by 5.5.3 Combination of cells 1 r 2 r 1 ε 2 ε Figure 5.14 2 1 r r r + = 2 1 ε ε ε + = and (5.15) (5.16) Note: If one cell, e.m.f. c 2 say, is turned round ‘in opposition’ to the others, then but the total internal resistance remains unaltered. 2 1 ε ε ε ÷ = PHYSICS CHAPTER 5 53 Cells in parallel Consider two equal cells are connected in parallel as shown in Figure 5.15. The total emf, c and the total internal resistance, r are given by 1 r 1 r 1 ε 1 ε Figure 5.15 1 1 1 1 1 r r r + = 1 ε ε = and (5.17) (5.18) Note: If different cells are connected in parallel, there is no simple formula for the total emf and the total internal resistance where Kirchhoff’s laws have to be used. PHYSICS CHAPTER 5 54 Exercise 5.2 : 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,7 th 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 5 55 Exercise 5.2 : 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 × 10 4 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,6 th edition, Wilson, Buffa & Lou, Q75, p.589) ANS. : 0.15 A; 1.40 × 10 ÷4 O m; 2.30 W PHYSICS CHAPTER 5 56 At the end of this chapter, students should be able to: Deduce effective resistance of resistors in series and parallel. Calculate effective resistance of resistors in series and parallel. Learning Outcome: 5.6 Resistors in series and parallel (1 hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s PHYSICS CHAPTER 5 57 5.6.1 Resistors in series The symbol of resistor in an electrical circuit can be shown in Figure 5.16. Consider three resistors are connected in series to the battery as shown in Figure 5.17. 5.6 Resistors in series and parallel OR R R Figure 5.16 1 R 2 R 3 R V 1 V 2 V 3 V I I Figure 5.17 PHYSICS CHAPTER 5 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 V 1 , V 2 , V 3 and V in the eq. (5.19) gives (5.19) (5.20) 3 2 1 I I I I = = = 3 2 1 V V V V + + = ; 2 2 IR V = ; 3 3 IR V = ; 1 1 IR V = eff IR V = 3 2 1 eff IR IR IR IR + + = 3 2 1 eff R R R R + + = where resistance t) (equivalen effective : eff R PHYSICS CHAPTER 5 59 V 1 R 3 R 2 R Consider three resistors are connected in parallel to the battery as shown in Figures 5.18a and 5.18b. 5.6.2 Resistors in parallel I I 2 I 1 I 3 I 1 V 2 V 3 V V 1 R 3 R 2 R I I 1 I 3 I 2 I Figure 5.18a Figure 5.18b 2 V 3 V 1 V PHYSICS CHAPTER 5 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 I 1 , I 2 , I 3 and I in the eq. (5.21) gives (5.21) (5.22) 3 2 1 V V V V = = = 3 2 1 I I I I + + = ; 2 2 R V I = ; 3 3 R V I = ; 1 1 R V I = eff R V I = 3 2 1 eff R V R V R V R V + + = 3 2 1 eff 1 1 1 1 R R R R + + = PHYSICS CHAPTER 5 61 For the circuit in Figure 5.19, 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 12 : Figure 5.19 O 0 . 4 O 0 . 2 V 0 . 8 O 12 PHYSICS CHAPTER 5 62 Solution : a. The resistors R 1 and R 2 are in series, thus R 12 is Since R 12 and R 3 are in parallel, therefore R eff is given by V 0 . 8 ; 0 . 2 ; 12 ; 0 . 4 3 2 1 = O = O = O = V R R R 1 R V 2 R 3 R 12 R V 3 R O = 16 12 R 2 1 12 R R R + = 12 0 . 4 12 + = R 3 12 eff 1 1 1 R R R + = 2 1 16 1 1 eff + = R O = 78 . 1 eff R PHYSICS CHAPTER 5 63 Solution : b. Since R 12 and R 3 are in parallel, thus Therefore the current passes through R 2 is given by c. Since R 1 and R 2 are in series, thus Hence the potential difference across R 1 is d. The power delivered by the battery is V 0 . 8 ; 0 . 2 ; 12 ; 0 . 4 3 2 1 = O = O = O = V R R R A 50 . 0 2 = I V 0 . 8 3 12 = = = V V V 12 12 2 R V I = A 50 . 0 2 1 = = I I 1 1 1 R I V = V 0 . 2 1 = V 16 0 . 8 2 = I ( ) 0 . 4 50 . 0 1 = V eff 2 R V P = ( ) 78 . 1 0 . 8 2 = P W 0 . 36 = P PHYSICS CHAPTER 5 64 For the circuit in Figure 5.20, calculate the effective resistance between the points A and B. Solution : ; 20 ; 10 ; 0 . 5 ; 0 . 5 4 3 2 1 O = O = O = O = R R R R O = 10 5 R Example 13 : Figure 5.20 O 0 . 5 O 10 O 10 A B O 0 . 5 O 20 2 R 3 R 5 R A B 1 R 4 R 3 R 5 R A B 12 R 4 R PHYSICS CHAPTER 5 65 Solution : R 1 and R 2 are connected in series, thus R 12 is 2 1 12 R R R + = ; 20 ; 10 ; 0 . 5 ; 0 . 5 4 3 2 1 O = O = O = O = R R R R O = 10 5 R O = + = 10 0 . 5 0 . 5 12 R 5 R A B 123 R 4 R Since R 12 and R 3 are connected in parallel , thus R 123 is given by 3 12 123 1 1 1 R R R + = O = 0 . 5 123 R 10 1 10 1 1 123 + = R 5 R A B 1234 R R 123 and R 4 are connected in series , thus R 1234 is given by 4 123 1234 R R R + = O = 25 1234 R 20 0 . 5 1234 + = R PHYSICS CHAPTER 5 66 Solution : Since R 1234 and R 5 are connected in parallel , therefore the effective resistance R eff is given by ; 20 ; 10 ; 0 . 5 ; 0 . 5 4 3 2 1 O = O = O = O = R R R R O = 10 5 R 5 1234 eff 1 1 1 R R R + = O = 14 . 7 eff R 10 1 25 1 1 eff + = R A B eff R PHYSICS CHAPTER 5 67 Exercise 5.3 : 1. Determine the equivalent resistances of the resistors in Figures 5.21, 5.22 and 5.23. ANS. : 0.80 O; 2.7 O; 8.0 O O 0 . 2 O 0 . 2 O 0 . 2 O 0 . 2 Figure 5.21 Figure 5.22 O 0 . 6 O 0 1 O 0 . 6 O 0 . 4 O 18 O 16 O 0 . 8 O 0 . 9 O 16 O 0 . 6 O 20 Figure 5.23 PHYSICS CHAPTER 5 68 2. The circuit in Figure 5.24 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,3 th edition, James S. Walker, Q39, p.728) ANS. : 1.1 A, 0.3 A; 1.4 A; 11.3 V O 0 . 1 V 12 r O 1 . 7 O 8 . 5 O 5 . 4 O 2 . 3 Figure 5.24 PHYSICS CHAPTER 5 69 3. Four identical resistors are connected to a battery as shown in Figure 5.25. When the switch is open, the current through the battery is I 0 . 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 I 0 . (Physics,3 th edition, James S. Walker, Q45, p.728) ANS. : U think Figure 5.25 ε R R R R PHYSICS CHAPTER 5 70 At the end of this chapter, students should be able to: State and use Kirchhoff’s Laws. Learning Outcome: 5.7 Kirchhoff’s laws (1 hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s PHYSICS CHAPTER 5 71 5.7.1 Kirchhoff’s first law (junction or current law) states the algebraic sum of the currents entering any junctions in a circuit must equal the algebraic sum of the currents leaving that junction. OR For example : 5.7 Kirchhoff’s laws ¿ ¿ = out in I I (5.23) A B 2 I 1 I 5 I 4 I 3 I 3 I 3 2 1 I I I = + 5 4 3 I I I + = ¿ ¿ = out in I I Figure 5.26 PHYSICS CHAPTER 5 72 states in any closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of current and resistance. OR In any closed loop, Sign convention For emf, c: 5.7.2 Kirchhoff’s second law (loop or voltage law) ¿ ¿ = c IR (5.24) ε + ε direction of loop + - ε ÷ - ε + direction of loop PHYSICS CHAPTER 5 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 Kirchhoff’s 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 Kirchhoff’s second law. Solving the simultaneous equation to determine the unknown currents and unknown variables. IR + direction of loop I R IR ÷ I R direction of loop 5.7.3 Problem solving strategy (Kirchhoff’s Laws) PHYSICS CHAPTER 5 74 For example, Consider a circuit is shown in Figure 5.27a. At junction A or D (applying the Kirchhoff’s first law) : 1 R 3 R 1 ε E D F 2 R 2 ε 3 ε C A B 1 I 1 I 1 I 1 I 2 I 2 I 3 I 3 I 3 I 3 I Loop 1 Loop 2 Loop 3 Figure 5.27a ¿ ¿ = out in I I 3 2 1 I I I + = (1) PHYSICS CHAPTER 5 75 For the closed loop (either clockwise or anticlockwise), apply the Kirchhoff’s second law. From Loop 1 Figure 5.27b (2) FEDAF 1 ε 1 R E D F 2 R 2 ε A 1 I 1 I 1 I 1 I 2 I 2 I Loop 1 1 1 2 2 2 1 R I R I ε ε + = + ¿ ¿ = c IR PHYSICS CHAPTER 5 76 From Loop 2 Figure 5.27c (3) ABCDA 2 ε 3 R D 2 R 3 ε C A B 2 I 2 I 3 I 3 I 3 I 3 I Loop 2 3 3 2 2 3 2 R I R I ε ε ÷ = ÷ ¿ ¿ = c IR PHYSICS CHAPTER 5 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. Figure 5.27d (4) FECBF 1 R 3 R 1 ε E F 3 ε C B 1 I 1 I 1 I 1 I 3 I 3 I 3 I 3 I Loop 3 1 1 3 3 3 1 R I R I ε ε + = + 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 5 78 For the circuit in Figure 5.28, Determine the current and its direction in the circuit. Example 14 : Figure 5.28 O 1 . 15 O .22 6 O 50 . 8 O 2 , V 1.5 1 O 4 , V 5.0 1 PHYSICS CHAPTER 5 79 Solution : By applying the Kirchhoff’s 2 nd law, thus ¿ ¿ = IR ε A 74 . 0 = I I I I I I 4 50 . 8 2 22 . 6 1 . 15 5 . 11 0 . 15 + + + + = + O 1 . 15 O .22 6 O 50 . 8 O 2 , V 1.5 1 O 4 , V 5.0 1 Loop 1 I I I I (anticlockwise) PHYSICS CHAPTER 5 80 For the circuit in Figure 5.29, determine a. the currents I 1 , I 2 and I, b. the potential difference across the 6.7 O resistor, c. the power dissipated from the 1.2 O resistor. Example 15 : Figure 5.29 O 8 . 9 O 9 . 3 V .0 9 V 2 1 O 7 . 6 O .2 1 I 1 I 2 I PHYSICS CHAPTER 5 81 Solution : a. At junction A, by using the Kirchhoff’s 1 st law, thus By using the Kirchhoff’s 2 nd law, From Loop 1: ¿ ¿ = out in I I I I I = + 2 1 O 8 . 9 O 9 . 3 V .0 9 V 2 1 O 7 . 6 O .2 1 1 I 2 I I 1 I 2 I I A B Loop 1 Loop 2 (1) ¿ ¿ = IR ε 1 1 8 . 9 2 . 1 9 . 3 12 I I I + + = 12 2 . 1 7 . 13 1 = + I I (2) PHYSICS CHAPTER 5 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 ε I I 2 . 1 7 . 6 0 . 9 2 + = 0 . 9 2 . 1 7 . 6 2 = + I I (3) A 75 . 1 A; 03 . 1 A; 72 . 0 2 1 = = = I I I R I V 2 = ( ) 7 . 6 03 . 1 = V V 90 . 6 = V R I P 2 = ( ) ( ) 2 . 1 75 . 1 2 = P W 68 . 3 = P PHYSICS CHAPTER 5 83 Exercise 5.4 : 1. For a circuit in Figure 5.30, Given c 1 = 8V, R 2 = 2 O, R 3 = 3 O, R 1 = 1 O and I = 3 A. Ignore the internal resistance in each battery. Calculate a. the currents I 1 and I 2 . b. the emf, c 2 . ANS. : 1.0 A, 4.0 A; 17 V Figure 5.30 3 R 1 ε 2 R 2 ε 1 I 2 I I 1 R PHYSICS CHAPTER 5 84 Exercise 5.4 : 2. Determine the current in each resistor in the circuit shown in Figure 5.31. (College Physics,6 th edition, Wilson, Buffa & Lou, Q57, p.619) ANS. : 3.75 A; 1.25 A; 1.25 A Figure 5.31 O 0 . 4 O 0 . 4 V .0 5 V .0 5 O .0 4 V 0 1 PHYSICS CHAPTER 5 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: 5.8 Potential divider (½ hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s V R R R V | | . | \ | + = 2 1 1 1 PHYSICS CHAPTER 5 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 5.32. Since the current flowing through each resistor is the same, thus 5.8 Potential divider V 1 V 1 R I 2 V 2 R I 2 1 eff R R R + = eff R V I = and 2 1 R R V I + = Figure 5.32 PHYSICS CHAPTER 5 87 Therefore, the potential difference (voltage) across R 1 is given by Similarly, Resistance R 1 and R 2 can be replaced by a uniform homogeneous wire as shown in Figure 5.33. Figure 5.33 1 1 IR V = V R R R V | | . | \ | + = 2 1 1 1 V R R R V | | . | \ | + = 2 1 2 2 (5.25) (5.26) V I 2 l 1 l I B A C 2 V 1 V PHYSICS CHAPTER 5 88 The total resistance, R AB in the wire is Since the current flowing through the wire is the same, thus A ρl R = CB AC AB R R R + = A ρl A ρl R 2 1 AB + = and AB R V I = ( ) 2 1 l l A ρ V I + = ( ) 2 1 AB l l A ρ R + = PHYSICS CHAPTER 5 89 Therefore, the potential difference (voltage) across the wire with length l 1 is given by Similarly, AC 1 IR V = ( ) | . | \ | ( ( ( ¸ ( ¸ + = A ρl l l A ρ V V 1 2 1 1 V l l l V | | . | \ | + = 2 1 1 1 (5.27) V l l l V | | . | \ | + = 2 1 2 2 (5.28) Note: From Ohm’s law, l V · | . | \ | = = A ρl I IR V PHYSICS CHAPTER 5 90 For the circuit in Figure 5.34, 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 16 : Figure 5.34 O 000 4 V 2 1 O 000 8 out V PHYSICS CHAPTER 5 91 Solution : a. The output voltage is given by b. The connection between the voltmeter and 4000 O resistor is parallel, 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 ; 8000 2 1 = O = O = V R R V R R R V | | . | \ | + = 2 1 2 out V 0 . 4 out = V 4000 1 4000 1 1 eq + = R 12 4000 8000 4000 out | . | \ | + = V O = 2000 eq R V 4 . 2 out = V 12 2000 8000 2000 out | . | \ | + = V PHYSICS CHAPTER 5 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: 5.9 Potentiometer and Wheatstone bridge (½ hour) w w w . k m p h . m a t r i k . e d u . m y / p h y s i c s x 3 2 1 R R R R = l l R R x x = PHYSICS CHAPTER 5 93 5.9.1 Potentiometer Consider a potentiometer circuit is shown in Figure 5.35. 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 5.9 Potentiometer and Wheatstone bridge Figure 5.35 (Driver cell -accumulator) Jockey V B A C x V I G + - I I I Galvanometer reading = 0 PHYSICS CHAPTER 5 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 5.36, 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 x V V = PHYSICS CHAPTER 5 95 Initially, a switch S is connected to the terminal (1) and the jockey moved until the emf c 1 exactly balances the potential difference (p.d.) from the accumulator (galvanometer reading is zero) at point C. Hence Figure 5.36 X B A I G I (2) (1) 2 ε S I I 1 ε C J D 1 l 2 l PHYSICS CHAPTER 5 96 After that, the switch S is connected to the terminal (2) and the jockey moved until the emf c 2 balances the p.d. from the accumulator at point D. Hence AC 1 V ε = AC AC IR V = where A ρl R 1 AC = and 1 1 l A ρI ε | . | \ | = (1) then AD 2 V ε = AD AD IR V = where A ρl R 2 AD = and 2 2 l A ρI ε | . | \ | = (2) then PHYSICS CHAPTER 5 97 By dividing eq. (1) and eq. (2) then Measure an unknown emf of a cell By using the same circuit shown in Figure 5.36, the value of unknown emf can be determined if the cell c 1 is replaced with a standard cell. A standard cell is one in which provides a constant and accurately known emf. Thus the emf c 2 can be calculated by using the equation (5.29). 2 1 2 1 l l ε ε = 2 1 2 1 l A ρI l A ρI ε ε | . | \ | | . | \ | = (5.29) PHYSICS CHAPTER 5 98 Measure the internal resistance of a cell Consider a potentiometer circuit as shown in Figure 5.37. Figure 5.37 ε B A I G I 1 ε 0 l C J S R r I I PHYSICS CHAPTER 5 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 c 1 exactly balances the emf c from the accumulator (galvanometer reading is zero) at point C. Hence After the switch S is closed, the current I 1 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 5.38. AC 1 V ε = AC AC IR V = where A ρl R 0 AC = and 0 1 l A ρI ε | . | \ | = (1) then PHYSICS CHAPTER 5 100 Figure 5.38 ε B A I G I 1 ε J S R r I I 1 I D l 1 I 1 I 1 I 1 I PHYSICS CHAPTER 5 101 Hence From the equation of emf, AD V V = AD AD IR V = where A ρl R = AD and l A ρI V | . | \ | = (2) then r I V ε 1 1 + = 1 1 I V ε r ÷ = R V I = 1 and R V V ε r | . | \ | ÷ = 1 (3) PHYSICS CHAPTER 5 102 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/l against 1/R . Rearranging eq. (4) : R l l l r | . | \ | ÷ = 0 R l l r | . | \ | ÷ = 1 0 (4) c x m y + = Then compare with 0 0 1 1 1 l R l r l + | | . | \ | = PHYSICS CHAPTER 5 103 Therefore the graph is straight line as shown in Figure 5.39. 0 , Gradient l r m = 0 1 l R 1 l 1 0 Figure 5.39 PHYSICS CHAPTER 5 104 Cells A and B and centre-zero galvanometer G are connected to a uniform wire OS using jockeys X and Y as shown in 5.40. a. the potential difference across OY when OY = 75.0 cm, b. the potential difference across OY when Y touches S and the galvanometer is balanced, c. the internal resistance of the cell A, d. the emf of cell A. Example 17 : Figure 5.40 A S O G B X 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 5 105 Solution : a. Given When G = 0 (balance condition), thus V 0 . 1 ; 12 m; 00 . 1 B OS OS = O = = ε R l m 50 . 0 m; 75 . 0 OX1 OY1 = = l l A ε S O G B ε X Y 0 = OY1 l OX1 l Since wire OS is uniform thus OS OS 1 OX OX1 R l l R | | . | \ | = and O = | . | \ | = 0 . 6 12 00 . 1 50 . 0 OX1 R O = | . | \ | = 0 . 9 12 00 . 1 75 . 0 OY1 R B OX1 ε V = OX1 1 OX1 R I V = and 1 I 1 I 1 I 1 I 1 I B OX1 1 ε R I = ( ) 0 . 1 0 . 6 1 = I A 17 . 0 1 = I PHYSICS CHAPTER 5 106 Solution : a. Therefore the potential difference across OY is given by b. Given V 0 . 1 ; 12 m; 00 . 1 B OS OS = O = = ε R l OY1 1 OY1 R I V = ( ) 0 . 9 17 . 0 OY1 = V V 53 . 1 OY1 = V m 625 . 0 m; 00 . 1 OX2 OY2 = = l l A ε S O G B ε X Y 0 = OY2 l OX2 l 2 I 2 I 2 I 2 I 2 I Since wire OS is uniform thus OS OS 2 OX OX2 R l l R | | . | \ | = and O = | . | \ | = 5 . 7 12 00 . 1 625 . 0 OX2 R O = | . | \ | = 12 12 00 . 1 00 . 1 OY2 R PHYSICS CHAPTER 5 107 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 . 1 B OS OS = O = = ε R l B OX2 ε V = OX2 2 OX2 R I V = and B OX2 2 ε R I = ( ) 0 . 1 5 . 7 2 = I A 13 . 0 2 = I OY2 2 OY2 R I V = ( )12 13 . 0 OY2 = V V 56 . 1 OY2 = V ( ) r R I ε + = A ) ( 1 OY 1 A r R I ε + = ( ) r ε + = 0 . 9 17 . 0 A (1) PHYSICS CHAPTER 5 108 Solution : c. For case in the question (b) : (1) = (2): d. The emf of cell A is V 0 . 1 ; 12 m; 00 . 1 B OS OS = O = = ε R l ) ( 2 OY 2 A r R I ε + = ( ) r ε + = 12 13 . 0 A (2) ( ) ( ) r r + = + 12 13 . 0 0 . 9 17 . 0 O = 65 . 0 r ( ) r ε + = 0 . 9 17 . 0 A ( ) 65 . 0 0 . 9 17 . 0 A + = ε V 64 . 1 A = ε PHYSICS CHAPTER 5 109 It is used to measured the unknown resistance of the resistor. Figure 5.41 shows the Wheatstone bridge circuit consists of a cell of emf c (accumulator), a galvanometer , know resistances (R 1 , R 2 and R 3 ) and unknown resistance R x . The Wheatstone bridge is said to be balanced when no current flows through the galvanometer. 5.9.2 Wheatstone bridge ε B A G C D 1 R 2 R 3 R x R 0 = I I 2 I 1 I 2 I 1 I Figure 5.41 PHYSICS CHAPTER 5 110 Hence then Therefore Since Dividing gives 1 CB AC I I I = = 2 DB AD I I I = = and Potential at C = Potential at D AD AC V V = BD BC V V = and IR V = 3 2 1 1 R I R I = thus and x 2 2 1 R I R I = x 2 3 2 2 1 1 1 R I R I R I R I = 3 1 2 x R R R R | | . | \ | = (5.30) PHYSICS CHAPTER 5 111 The application of the Wheatstone bridge is Metre Bridge. Figure 5.42 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 I 1 flows through the resistance R x and R but current I 2 flows in the wire AB. = 0 Accumulator Jockey Thick copper strip (Unknown resistance) (resistance box) Wire of uniform resistance x R A ε G B R J 2 l 1 l Figure 5.42 I I 1 I 2 I 1 I PHYSICS CHAPTER 5 112 Let V x : p.d. across R x and V : p.d. across R, At balance condition, By applying Ohm’s law, thus Dividing gives AJ x V V = JB V V = and AJ 2 x 1 R I R I = JB 2 1 R I R I = and A ρl R 1 AJ = JB 2 AJ 2 1 x 1 R I R I R I R I = where and A ρl R 2 JB = | . | \ | | . | \ | = A ρl A ρl R R 2 1 x R l l R | | . | \ | = 2 1 x (5.31) PHYSICS CHAPTER 5 113 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 5.43. Resistors R 1 and R 2 have resistance of 38.0 O and 46.0 O respectively. Balance is achieved when the switch closed and R 3 is 3.48 O. Calculate the length of the platinum wire if its resistivity is 10.6 × 10 ÷8 O m. Example 18 : Figure 5.43 PHYSICS CHAPTER 5 114 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 . 0 2 1 3 O = O = × = ÷ R R d ; m Ω 10 6 . 10 ; 48 . 3 8 3 ÷ × = O = ρ R 1 2 3 x R R R R = O = 21 . 4 x R 0 . 38 0 . 46 48 . 3 x = R l A R ρ x = 4 2 d A t = and l d R ρ 4 2 x t = ( ) ( ) l 4 10 920 . 0 21 . 4 10 6 . 10 2 3 8 ÷ ÷ × t = × m 4 . 26 = l PHYSICS CHAPTER 5 115 Exercise 5.5 : 1. In Figure 5.44, 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 2 S 1 ε P Q G 2 ε T 1 R 2 R 1 S Figure 5.44 c 1 is an accumulator of emf 2.0 V and negligible internal resistance. R 1 is a 15 O resistor and R 2 is a 5.0 O resistor when S 1 and S 2 open, galvanometer G is balanced when QT is 62.5 cm. When both S 1 and S 2 are closed, the balance length is 10.0 cm. Calculate a. the emf of cell c 2 . b. the internal resistance of cell c 2 . c. the balance length QT when S 2 is opened and S 1 closed. d. the balance length QT when S 1 is opened and S 2 closed. PHYSICS CHAPTER 5 116 R 2. The circuit shown in Figure 5.45 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,3 th edition, James S. Walker, Q93, p.731) ANS. : 7.50 O Exercise 5.5 : Figure 5.45 PHYSICS CHAPTER 5 117 Exercise 5.5 : 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 5 118 Next Chapter… CHAPTER 6 : Magnetic field
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