Pra-U STPM Physics Penggal 2 2019 CB039249b

Physics Term 2 STPM Chapter 14 Electric Current 14 95 3. The resistance of a piece of pure silicon falls rapidly as the temperature rises because A the ratio of positive to negative charge carriers increases. B random motions of the charge carriers are reduced. C the total number of charge carriers increases with temperature. D the speed of the charge carriers increases. 4. The diagram below shows two copper wires X and Y with different thickness connected in series with a battery. The current in the wires is large enough to cause Y to be hotter than X. X Y Which of the following is correct? Density of Mean time interval conduction between collision electron of free electrons A Same in X and Y Smaller in X than Y B Same in X and Y Smaller in Y than X C Greater in X than Y Smaller in Y than X D Greater in Y than X Same in X and Y 5. When a metal rod is heated, the electrical conductivity of the metal A decreases because the mean time between collisions of conduction electron increases. B increases because the mean time between collisions of conduction electron increases. C decreases because the mean time between collisions of conduction electron decreases. D increases because the mean time between collisions of conduction electrons decreases. 6. The resistance of a piece of copper wire increases when its temperature increases because A the number of free electrons per unit volume increases. B the drift velocity of the free electrons decreases. C the mean velocity of the free electrons increases. D the acceleration of the free electron increases. 7. When the temperature of an intrinsic semiconductor increases, its resistivity decreases because A the density of charge carriers increases. B the amplitude of oscillations of semiconductor atoms increases. C the thermal energy of the semiconductor atoms increases. D the rate of collision between the charge carriers and semiconductors atoms increases. 8. Certain materials, such as mercury, become superconductor at low temperatures. Which of the following graphs shows the correct variation of resistance R with temperature T in kelvin for a volume of mercury? A R T / K 0 C R T / K 0 B R T / K 0 D R T / K 0 9. A wire of cross sectional area, A carries a current, I. The conductivity of the material of the wire is σ 7HICH EXPRESSION IS CORRECT for the electric field strength E in the wire? A E = I –– A B E = I –– Aσ C E = Iσ –– A D E = σ –– IA

Physics Term 2 STPM Chapter 14 Electric Current 14 96 Important Formulae 1. Current, I = nAve, current density, J = I –– A = nve 5. The copper coil of an electric motor carries a current of 50.0 A under normal working conditions. The coil will be damaged if the RATE OF HEAT GENERATED EXCEEDS  7 M–1. What should be the minimum diameter of the copper wire to avoid overheating? (Resistivity of copper = 2.0 u 10–8 :m). A 1.41 mm C 3.99 mm B 2.82 mm D 5.64 mm 6. A piece of semiconductor has a cross-sectional area A. There are n1 free electrons per unit volume and n2 holes per unit volume. The drift velocities of the free electrons and holes are v1 and v2 respectively when the current is I. 7HICH EXPRESSION FOR I is correct? A I = n1 Av1 e B I = n2 Av2 e C I = n1 Av1 e+ n2 Av2 e D I = n1 Av1 e – n2 Av2 e 7. The resistance of a piece of nichrome wire is R. What is the resistance of another piece of nichrome wire of twice the length but half the diameter? A 0.5R C 4R B 2R D 8R 8. A conductor of length l and uniform crosssectional area A is made of material of conductivity σ. What is the resistance of the conductor? A —1 σ B —– A σl C —– σA l D —–l σA 9. A nichrome wire of diameter 0.58 mm carries a current of 5.0 A. What is the current density in the wire? A 8.6 u 103 A m–2 C 4.7 u 107 A m–2 B 1.7 u 106 A m–2 D 1.9 u 107 A m–2 1. The resistivity of a metal increases when the temperature increase is due to the increase in A the number of free electrons B the amplitude of vibration of metal ions C the drift velocity of free electrons D the mean free path of free electrons 2. The resistance of a piece of wire of radius r is R. 4HE OUTER SURFACE OF THE WIRE IS OXIDISED 4HE THICKNESS OF THE OXIDISED SURFACE IS ONE TENTH THE RADIUS AND ITS RESISTIVITY IS SIX TIMES THE resistivity of the metal. What is the resistance of this piece of wire after its outer surface is OXIDISEDfi A 32.8R C 31.6R B 32.8R D 1.18R 3. A lamp lights up immediately when it is switched on. This is because A the drift velocity of the free electrons is large. B the random velocity of the free electrons is large. C all the free electrons drift simultaneously D the free electrons travel freely. 4. The following are statements about electrical conduction in metal. Which of the following statement is not correct? A The instantaneous speeds of all the free electrons are equal. B The random speed of the free electrons increases with increase in temperature. C The mean free path of the free electrons decreases with increase in temperature. D The increase in kinetic energy of the free electrons in the electric fi eld is lost due to collisions with the lattice atoms. 2. Electrical conductivity, σ = —J E = ne2 ——τ m 3. Resistivity, ρ = —– RA l STPM PRACTICE 14

Physics Term 2 STPM Chapter 14 Electric Current 14 97 10. A piece of nichrome wire of rectangular cross sectional area 1.00 mm u0.0500 mm and length 2.50 m is used as the heating element of a kettle rated 2.00 A, 300 W. What is the resistivity of the nichrome? A 2.40 u 10-7: m B 1.50 u 10-6:m C 4.20 u 10-6: m D 8.50 u 10-5: m 11. A battery is connected across the points P and Q of a uniform wire which is earthed at Q as shown below. – + P Q Which graph shows correctly the variation of current density J along PQ? A C 0 J P Q 0 J P Q B D 0 J P Q 0 J P Q 12. A potential difference V is applied across a length l of a wire made of material of conductivity σ. Which of the following gives the current density J? A —σl V C —– σV l B —– Vl σ D —– V σl 13. Two wires of the same material X and Y are of the same length but the diameter of X is half that of Y. The wires are connected in parallel to a battery. What is the fraction of the total current passes through X? A —1 5 C —3 4 B — 1 4 D — 4 5 14. The diagram shows a conducting material in the form of a rectangular block with dimensions x, 2x and 3x. 3x 2x 2x 3x x x If the electrical resistance between the two opposite shaded faces is R, what is the resistance between the top and bottom faces of the block? A —R 9 C —2 3 R B —– 2R 9 D —3 2 R 15. A wire 2.00 m long, of uniform cross-sectional area 3.00 mm2 , has conductance of 0.75 S. What is the resistivity of the material of the wire? (1 S = 1 : –1) A 1.1 u 10–6 : m C 4.5 u 10–6 : m B 2.0 u 10–6 : m D 8.0 u 10–6 : m 16. 7HICH OF THE FOLLOWING CORRECTLY EXPLAINS WHY the current in a metallic conductor increases when the potential difference across the conductor is increased? A The mean time between collisions for free electrons decreases. B The acceleration of the free electrons increases. C The rise in temperature increases the speed of the free electrons. D There is a smaller chance of an electron colliding with the metal ions. 17. A /N THE SAME AXES SKETCH GRAPHS TO SHOW THE VARIATION WITH TEMPERATURE IN + FOR the resistivity of (i) copper, (ii) semiconductor, (iii) superconductor.

Physics Term 2 STPM Chapter 14 Electric Current 14 98 (b) A copper wire with diameter of 3.0 mm carries a current of 10.0 A. The free electron density in copper is 8.5 u 1028 m–2. Calculate the drift velocity of free electrons in copper. 18. (a) (i) A lamp is marked 240 V, 60 W. %XPLAIN THE MARKINGS (ii) What is meant by drift velocity of free electrons in a metal? (b) In copper, one atom contributes one free electron. The density of copper is 8900 kg m–3, and its mass of one mole copper is 0.0635 kg. A copper wire of diameter 2.6 mm and length 50.0 cm carries a current of 5.8 A. (Resistivity of copper = 1.69 u 10–8 : m). Calculate (i) the density number of free electrons, (ii) the current density, (iii) the drift velocity of free electrons in the copper wire, (iv) the resistance of the wire. 19. The speed of electrons in an electron-beam of a cathode-ray tube is 6.0 u 107 m s–1. If the current in the electron-beam is 2.0 mA, find the number of electrons in 0.20 m length of the beam. 20. The number of free electrons in copper is 1.0 u 1029 m–3. The resistivity of copper is 1.72 u 10–8 : m. Calculate the drift velocity of free electrons in copper when a potential difference of 10 mV is applied across a copper wire of length 10.0 cm. 21. When a certain potential difference is applied across a uniform wire of length l and radius r, the current in the wire is I. If the same potential difference is applied across another wire of the same material but of length 2l and radius 1 –– 2 r, what is the current in the wire? 22. The diagram below shows a hollow copper cylinder formed by winding bare copper wire ON A 06# PIPE OF EXTERNAL RADIUS  CM 4HE coils are wound close together so that there is perfect electrical contact between successive turns. 1.5 mm 6.0 cm The bare copper wire has a square cross section of sides 1.5 mm. If there are 300 turns, calculate the resistance between opposite ends of the copper cylinder. (Resistivity of copper = 1.69 u 10–8 : m) 23. A uniform wire of length 25.0 cm has a resistance of 75 :. What is length of the wire after it is stretched so that its resistance becomes 147 :? Assume that the volume of the wire and the resistivity of its material remain constant. 24. Calculate the length of wire of cross-sectional 0.50 mm2 and resistivity 2.5 u 10–7 : m which is required to make a heater rated at 72 W, 12 V. If the resistance of the wire increases with increase in temperature, how does the power change with time when the heater is USED TO MELT A MIXTURE OF ICE AND WATER AND bring it to boiling point? 25. (a) Define resistivity of a material. (b) Sketch graphs to show the dependence of resistivity U on the temperature θ # (i) for a metal, (ii) for a semiconductor. (c) (i) A metal of resistivity U is used to make an electric cable of cross-sectional area A $ERIVE AN EXPRESSION FOR THE resistance r per unit length of the cable in terms of U and A. (ii) Some electrical appliances are connected to the electrical supply by long cables. Give two reasons why the connecting cable should have a low value of r, the resistance per unit length. 26. (a) The current density J, electric field strength E and conductivity V of the material of a wire is given by J = VE. I 3HOW THAT THE EXPRESSION IS consistent with Ohm’s law.

Physics Term 2 STPM Chapter 14 Electric Current 14 99 II 7RITE ANOTHER EXPRESSION FOR J in terms of the drift velocity v of free electrons. Hence derive the EXPRESSION V = ne2 ——–τ m where n is the number density of free electrons, τ mean time between collisions and m is the mass of electron. III 5SE THE EXPRESSION TO EXPLAIN THE effect of temperature increase on the conductivity of semiconductors. (b) At room temperature, the resistivity of silicon is 2.1 u 10–3 : m, and number density of charge carriers is 3.0 u 1022 m–3. Calculate the mean free time between collisions of the charge carriers. 27. (a) Electric conduction in a metal can be EXPLAINED IN TERMS OF MOTION OF FREE electrons. Give an estimated value at room temperature for (i) the mean speed of free electrons, (ii) the drift velocity of free electrons in a typical metal. B 3HOW THAT THE EXPRESSION FOR CURRENT density J = σE where σ is electrical conductivity, and E the electric fi eld, is equivalent to Ohm’s law. Hence, deduce the relation between the resistance R of a wire with its crosssectional area A. (c) A wire of length 1.50 m and crosssectional area 1.2 mm2 carries a current of 5.0 A when a potential difference of 0.24 V is applied across the wire. (i) Calculate the drift velocity of the free electrons, if the number of free electrons is 1.5 u 1029 m–3. (ii) What is the force on each electron? (iii) Calculate the resistivity of the material of the wire. 1 1. (a) Q = It = (0.20)(5 u 60) = 60 C (b) n(1.60 u 10–19) = 0.20 A n = 1.25 u 1018 s–1 2 1. A 2. C: I = nAve, n is large. 3. B 4. C 3 1. (a) Mass = Vρ M Number of free e- = number of Cu atoms = (Vρ M )(6.02 u1023) = 8.48 u (b) (i) Drift velocity is caused by the electric force on the free electrons due to the potential GLͿHUHQFHDSSOLHGDFURVVWKHFRSSHUFXEH (ii) I = nAve v = 50 _ 8.48 u (0.010)3 +(0.010)2 (1.60 u 10–19) m s-1 = 3.69 u m s-1 2. (a) Number of free e- that passes a cross section of wire s-1 = n(Av) Current, I = (nAv)e 1 s = nAve (b) (i) Current density = I A = 1.41 u A m-2 (ii) v = 1 nAe = 1.03 u m s-1 4 1. C 2. 2. B: R = ρl A , l = 100 m 3. C 4. D 5. D flۙ  (b) 1.18 cm 6. 2 u 109 ۙP 8. 4.00 u 10–3 ۙ 7. 2.0 A 9. 1.50 u 10–2ۙP 5 1. D 2. D 3. C 4. B 5. C 6. B 7. A 8. C 9. B: J = I A = σE E = I Aσ ANSWERS

Physics Term 2 STPM Chapter 14 Electric Current 14 100 STPM PRACTICE 14 1. B 2. D: R = ρl —– Sr2 Oxidised wire, Rcore = ρl —–—— S(0.9r) 2 = 1.23R Rsurface = (6ρ)l —–———— S[r2 – (0.9r)2 ] = 31.6R —–1 R = ——–– 1 1.23R + ——–– 1 31.6R R = 1.18R 3. C 4. A 5. D: Power = I 2 R = I2 1 ρl A 2, For 1 m, A = πd2 4 = ρI 2 2.0 = (2.0 u 10–8)(50)2 2.0 , d = 5.64 mm 6. C: Total charge transferred s–1 = sum of charge s-1 transferred due to free electrons and holes, 7. D: R = ρl A , R1 = ρ(2l) (A/4) = 8R 8. D 9. D: Current density, J = —I A = —————— 5.0 S(0.29 u 10–3)2 A m–2 = 1.9 u 107 A m–2 10. B: P = I 2 R and R = ρl A Hence, ρ = PA I 2 l = 1.50 u 10-6ۙP 11. A 12. C 13. A 14. A 15. B 16. B 17. (a) Resistivity (ii) Semiconductor (iii) Superconductor (i) Copper 0 T(K) (b) I = nAve v = 10 (8.5 u 1028)S(1.5 u 10–3)2 (1.60 u 10–19) = 1.04 u 10–4 m s–1 18. (a) (i) Power dissipated from the lamp is 60 W when the lamp is connected to a potential    GLͿHUHQFHRI9 (ii) (refer to text) (b) (i) density number of free electrons, n = = Number of atoms in 1 m3 (8900 kg) of Cu = 1——— 8900 0.0635 2(6.02 u 1023) = 8.44 u 1028 (ii) Current density, J = —I A = ————––— 5.8 S(1.3 u 10–3)2 A m–2 = 1.1 u 107 A m–2 (iii) I = nAve v = J —ne = 1.1 u 107 ————––—————— (8.44 u 1028)(1.60 u 10–19) m s–1 = 8.1 u 10–4 m s–1 (iv) R = ρl —– A = (1.69 u 10–8 ————––——— )(0.50) S(1.3 u 10–3) 2 ۙ = 1.6 u 10–3ۙ 19. 4.17 u 107 I = nAve Number of free electrons = n(Al) 20. 3.63 u 10–4 m s–1 J = σE, I = nAve 21. —1 8 I I = —V R , R = ρl —–R 22. 1.34 u 10–5ۙ R = ρl —–A 23. FP 9ROXPH  A1 l1 = A2 l2 R = ρl —–A 24. 4.0 m P = V 2 —–R V FRQVWDQW9 P decreases R increases 25. (a) Refer page 87 (b) (i) U 0 T /°C (ii) U 0 T /°C (c) (i) R = ρ —l A Resistance per unit length r = —R l = ρ —A   LL ‡ 9ROWDJHGURSDFURVVFDEOHORZHU    ‡ +HDWGLVVLSDWHGIURPFDEOHUHGXFHG – cable does not get heated up 26. (a) (iii) When temperature increases, ֆ increases. Reasons: n increases (b) ֆ = —1 ρ = ne2 —––τ m τ = —––– m nρe2 = (9.11 u 10–31) ——————————————— (3.0 u 1022)(2.1 u 10–3)(1.60 u 10–19) 2 = 5.6 u 10–13 s 27. (a) (i) 105 to 106 m s–1 (ii) 10–5 to 10–4 m s–1 (b) (i) Substitute J = —I A , E = —V l to obtain —V I (ii) J = σE —I A = σ(—V l ), Hence R = —V I = —l ֆ$ (c) (i) 1.74 u 10–4 m s–1 use I = nAve (ii) 2.56 u 10–20 N F = eE, E = —V l (iii) 3.84 u 10–8ۙP ρ= —– RA l , R = —V I

15 101 CHAPTER DIRECT CURRENT CIRCUITS 15 101 Concept Map 'LUHFW&XUUHFW&LUFXLWV 3RWHQWLRPHWHUDQG :KHDWVWRQH%ULGJH ,QWHUQDO5HVLVWDQFH .LUFKKRII¶V/DZV 3RWHQWLDO'LYLGHU Use of shunts and multipliers Bilingual Keywords ,SLJ[YVTV[P]LMVYJLLTM!+H`HNLYHRLSLR[YPRKNL 0U[LYUHSYLZPZ[HUJL!9PU[HUNHUKHSHT 2PYJOOVMM»ZSH^!/\R\T2PYJOOVMM 4\S[PWSPLY!7LUKHYHI 7V[LU[PHSKP]PKLY!7LTIHOHNPRL\WH`HHU 7V[LU[PVTL[LY!4L[LYRL\WH`HHU :O\U[!7LTPYH\ ;OLYTVJV\WSL!;LYTVNHUKPUNHU >OLH[Z[VUL)YPKNL!1HTIH[HU>OLH[Z[VUL INTRODUCTION 1. Electrostatics is about stationary charges. 2. What is the relationship between electric charge and electricity? 3. A Van de Graaff generator can be used to collect electric charges. When the charged dome of a Van de Graaff generator is connected by a wire to Earth via a microammeter, a current is detected. 4. This shows that current is a fl ow of charged particles. 5. In metals these charged particles are free electrons.

15 102 15.1 Internal Resistance Students should be able to: ‡ H[SODLQWKHHIIHFWVRILQWHUQDOUHVLVWDQFHRQWKHWHUPLQDOSRWHQWLDOGLIIHUHQFHRIDEDWWHU\LQDFLUFXLW Learning Outcome Electromotive Force (e.m.f.) 1. Current is the rate of fl ow of charge. Current, I = dQ dt 2. For current to fl ow in a circuit, there must be a source which provides the charge, and also the energy to drive the charge round the circuit. 3. The source of electrical energy is known as the source of electromotive force (e.m.f.). Examples of sources of e.m.f. are batteries, generators and solar cells. 4. The electromotive force (e.m.f.) E of a source is the energy supplied by the source to each coulomb of charge that fl ows from the source. E.m.f. is measured in volt (V). 5. The e.m.f. of a dry cell is 1.5 V means that each coulomb of charge from the dry cell is supplied with 1.5 J of energy. This energy is dissipated when the coulomb of charge fl ows round the complete circuit. 6. Alternatively, the e.m.f. of a sources equal to the total energy dissipated in a closed circuit when one coulomb of charge fl ows. Info Physics 7KHEDWWHU\XVHGLQDFDULVWKHOHDGDFLGDFFXPXODWRU,WFRQVLVWVRIVL[FHOOVHDFKRIHPI97KHOHDGDFLGDFFXPXODWRULV XVHGEHFDXVHLWKDVYHU\ORZLQWHUQDOUHVLVWDQFH+HQFHLWLVDEOHWRSURYLGHWKHODUJHFXUUHQWUHTXLUHGWRVWDUWWKHHQJLQH7KH EDWWHU\LVFRQWLQXRXVO\FKDUJHGE\FXUUHQWSURGXFHGLQWKHDOWHUQDWRUZKHQWKHHQJLQHLVUXQQLQJ Electrical Resistance, R 2007/P1/Q30, 2011/P2/Q6 1. Energy is dissipated when charge fl ows in an electrical conductor. 2. An electrical conductor has resistance. For current to fl ow through an electrical conductor, there must be a potential difference across the conductor. 3. The potential difference across a conductor is defi ned as the energy required to drive a charge of one coulomb through the conductor. Potential difference is measured in volt (V). 4. Alternatively, the potential difference across a conductor is the energy dissipated from the conductor when one coulomb of charge fl ows through the conductor. 2014/P2/Q7, 2015/P2/Q18(b), 2016/P2/Q7 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 103 5. The resistance R of a conductor is the ratio of the potential difference V across the conductor to the current I in it. Resistance, R = V I 6. Resistance is measured in ohms (:). The resistance of a conductor is one ohm (1 :) if the current in the conductor is 1 A when a potential difference of 1 V is applied across the conductor. Internal Resistance 1. The e.m.f. of a cell can be measured approximately by connecting a high resistance voltmeter across the cell as shown in Figure 15.1, with the switch S open. The voltmeter reading is approximately the e.m.f. E of the cell. 2. When the switch S in the circuit shown is closed so that the cell is connected to a resistor of resistance R, the voltmeter shows a lower reading, V. 3. The loss in potential difference = (E – V). This loss of potential difference represents the energy required to drive one coulomb of charge through the materials of the cell. 4. If I is the current in the circuit, then ——– E – V I = r, the internal resistance of the cell. 5. The internal resistance r of a cell is the resistance due to the chemical materials in the cell. 6. From ——– E – V I = r E = V + Ir Also, V = IR .............................. { Hence, E = I(R + r) .............................. | —{ |: —E V = ——– R + r R —E V = 1 + —r R 7. An experimental method to determine the internal resistance of a cell is to use the circuit shown in Figure 15.1. With the switch S closed, the voltmeter reading V is noted for difference values of the resistance R. A graph of —1 V against —1 R is plotted as shown in Figure 15.2. 1 V 0 1 r – – 1 R – – Figure 15.2 Figure 15.1 V Cell S R Info Physics 7KH PRVW FRPPRQ W\SH RI UHVLVWRU XVHG LQ HOHFWURQLFFLUFXLWVLVWKHFDUERQUHVLVWRU VIDEO *VUK\J[VYZ HUK 0UZ\SH[VYZ Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 104 From the equation —E V = 1 + r R , when —1 V = 0, 1 + —r R = 0 —1 R = – —1 r The intercept on the —1 R -axis when —1 V = 0 gives – — 1 r . Hence, the internal resistance r can be calculated. 8. From the equation E = V + Ir, V = E when I = 0 Hence the e.m.f. of a cell is the potential difference across the terminals of the cell in open-circuit, I = 0. 9. From E = V + Ir V = E – Ir When we measure the e.m.f. of a cell connected by a high resistance voltmeter across the terminals of the cell, the voltmeter reading, V = E – Ir < E, the e.m.f. For a voltmeter of high resistance, the current I in the voltmeter would be small, hence V < E, e.m.f. of cell. Arrangement of Cells Cells in Series 1. Cells are connected in series when a potential difference greater than the e.m.f. of a cell is required. 2. When n identical cells each of e.m.f. E and internal resistance r are connected in series (Figure 15.3) to form a battery, (a) the effective e.m.f. = nE (b) the effective internal resistance = nr (c) the maximum current = —– nE nr = —E r is obtained when the terminals of the battery is short-circuited. Cells in Parallel 1. When n identical cells each of e.m.f. E and internal resistance r are connected in parallel (Figure 15.4) to form a battery, (a) the effective e.m.f. = E (b) the effective internal resistance = —r n (c) the maximum current obtained when the terminals of the battery are short-circuited = —– E —r n = —– nE r Hence a large maximum current is obtained compared to the cells connected in series. Figure 15.3 n cells Figure 15.4 n cells Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 105 Maximum Power 1. The power dissipated from a resistor connected to a cell depends on the value of the resistance R of the resistor. 2. When a resistor of resistance R is connected to a cell of e.m.f. E and internal resistance r, the current I = ——– E R + r . 3. The power dissipated as heat from the resistor is P = IV (V = IR) = I2 R (I = ——– E R + r ) = (——– E R + r ) 2 R .............................. { 4. To determine the value of R when the power P is maximum, we differentiate equation { with respect to R. —– dP dR = E2 (R + r)2 – 2E2 R(R + r) ——————————–– (R + r)4 The power P is maximum when —– dP dR = 0. Hence, E2 (R + r)2 – 2E2 R(R + r) = 0 (R + r) – 2R = 0 R = r The power dissipated from a resistor connected to a cell is maximum when the resistance of the resistor equals the internal resistance of the cell. Example 1 A generator supplies a fi xed amount of power to an electrical installation by power cables of fi nite resistance. Show that the power lost as heat in the cables is proportional to —– 1 V 2 , where V is the potential difference at the generator output. This result suggests that power losses can be reduced indefi nitely by a suitable choice of generator potential difference. Discuss briefl y whether this is practicable. Solution: Let P = power supplied by generator Using P = IV, I = —P V If R = total resistance of the cables. Rate of heat loss from the cables = I2 R = (—P V ) 2 R = P2 ——R V 2  v —– 1 V 2 since P and R are constants To reduce power loss indefi nitely, or to zero would mean increasing the potential difference V to an infi nite value. The very strong electric fi eld between the cables would result in the insulation of the air breaking down. Figure 15.5 R V I E,r Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 106 Example 2 Two almost identical lead-acid accumulator batteries P and Q are connected with opposing polarities in a circuit which includes an ammeter, a millivoltmeter and a resistor of resistance R as shown in the diagram. (a) Initially the switch S1 is closed and S2 left open. With switch S1 remaining closed, S2 is then closed. Describe and explain how the meters readings change when both S1 and S2 are closed. (b) When both the switches S1 and S2 are closed, the millivoltmeter reading changes by 20 mV, and the ammeter reads 5.0 A. (i) Calculate the internal resistance of the battery Q. You may assume that the resistance of the millivoltmeter to be much larger than R. (ii) Why is this assumption reasonable and necessary? (iii) Why is the two-battery method capable of giving greater accuracy for the internal resistance of the accumulator compared to of technique of the one-battery, ammeter, resistor and voltmeter? Solution: (a) With S1 closed and S2 open, ammeter reading = 0, because there is no current in ammeter Millivoltmeter reading = E1 – E2 where E1 = e.m.f. of battery P E2 = e.m.f. of battery Q With both S1 and S2 closed, ammeter reading changes from zero to a defi nite value because a current fl ows in the ammeter. millivoltmeter reading = E1 – V2 where V2 = potential difference across Q, V2 < E2 . Hence, the millivoltmeter reading (E1 – V2 ) increases. (b) (i) Change in millivoltmeter reading = 20 mV (E1 – V2 ) – (E1 – E2 ) = 20 mV E2 – V2 = 20 mV Using E – V = Ir, Internal resistance of Q, r = E2 – V ———–2 I = 20 u 10–3 ———–– 5.0 = 4.0 × 10–3 Ω (ii) The assumption that the resistance of the millivoltmeter is much larger than R is reasonable because the millivoltmeter reading is equal to the difference in e.m.f. (E1 – E2 ) when the millivoltmeter resistance is large. (iii) The two-battery technique is more accurate than the one-battery technique as shown in the fi gure because the internal resistance of the lead-acid accumulator is very small, 4.0 × 10–3 Ω. If the one-battery, ammeter, resistor and voltmeter technique (fi gure) is used, there would be no difference between the voltmeter readings when the switch S is open and when S is closed. R S2 P Q S1 A + – – + mV R Q S V A Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 107 Quick Check 1 1. A battery is connected in series with a 2 Ω resistor and a switch as shown below. A voltmeter connected across the battery reads 12 V when the switch is open and 8 V when the switch is closed. V 2 Ω What is the internal resistance of the battery? A —2 3 Ω C —4 3 Ω B 1 Ω D 4 Ω 2. A cell of e.m.f. 1.5 V and internal resistance 1.0 Ω maintains a constant current of 0.5 A in an external circuit. Which of the following statements about the energy in the circuit when an electric charge of 2.0 C fl ows in the external circuit is correct? A The thermal energy dissipated in the cell is 0.5 J. B The total energy dissipated in the external circuit is 2.0 J. C The energy provided by the cell is 1.5 J. D The thermal energy generated in the external circuit is 1.0 J. 3. A cell of e.m.f. 12 V and internal resistance 3 Ω is connected to two 6 Ω resistors X and Y as shown below. X Y What is the current in the resistor Y? A 0.5 A C 2.0 A B 1.0 A D 4.0 A 4. A cell which has an e.m.f. ε and an internal resistance r is connected to a resistor of resistance R. What is the potential difference between the terminals of the cell? A ε C ——– r R + r ε B —r R ε D ——– R R + r ε 5. A 18 Ω resistor and three identical cells each of e.m.f. 1.5 V and internal resistance 6 Ω are connected as shown in the diagram. I 18 8 The current in the resistor is A 0.075 A B 0.083 A C 0.24 A D 0.36 A 6. Differentiate between the e.m.f. of a cell, and the potential difference across the terminals of a cell. Give one situation when the potential difference across the terminals of a cell (a) is equal to the e.m.f. of the cell, (b) is greater than the e.m.f. of the cell. 7. A cell is of e.m.f. 1.5 V and internal resistance 0.5 Ω. (a) Calculate the power supplied to a 2.5 Ω resistor which is connected to the cell. (b) What is the value of the resistance of the external resistor when the power supplied is maximum? 8. A source of e.m.f. E with an internal resistance r is connected to a load of resistance R. (a) What is the current in the load? (b) Find the power P dissipated from the load. (c) What is the value of R, when the power P is maximum? Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 108 9. Three identical cells each of e.m.f. 1.5 V and constant internal resistance 2.0 Ω are connected in series with a resistor R = 4.0 Ω as shown in circuit (a) and then as in circuit (b). Circuit (a) Circuit (b) R R What is the value of the ratio power dissipated from R in circuit (a) power dissipated from R in circuit (b) ? 10. (a) What is meant by the internal resistance of a cell? (b) In the circuit shown, the voltmeter may be assumed to have infi nite resistance, but the resistance of the ammeter is not negligible. When the resistance R of the variable resistor is varied, the corresponding voltmeter reading V and ammeter reading are as shown in the table. Battery V R A 20.0 8 R/: V/V I/A 1.00 2.86 1.40 2.00 3.59 1.23 3.00 4.12 1.11 4.00 4.54 1.03 5.00 4.80 0.95 7.00 5.29 0.84 9.00 5.62 0.77 (i) Explain why the voltmeter reading decreases as the current increases. (ii) Plot a graph of V against I and use it to determine a value for the internal resistance of the battery. (iii) For what value of R in the table above is the power output from the battery greatest? (iv) Use your answer in (iii) to estimate the resistance of the ammeter. 15.2 Kirchhoff’s Laws Students should be able to: ‡ VWDWHDQGDSSO\.LUFKKRIIҋVODZV Learning Outcome 1. Kirchhoff’s laws are used to solve problems involving complex circuits. 2. Kirchhoff’s fi rst law state that: When steady currents fl ow in a network of conductors, the algebraic sum of the currents entering any point in the network is zero. ¦I = 0 3. Kirchhoff’s second law state that: In any closed path in a network, the algebraic sum of the products of current and resistance equals the algebraic sum of e.m.f. in the loop. ¦(IR) = ¦E 4. The fi rst law is based on the principle of conservation of charge since the rate of charge entering a point in a circuit must be equal to the rate of charge leaving the point. If currents entering a point are considered positive, then currents leaving the point are considered negative. 2010/P2/Q12, 2013/P2/Q27, 2014/P2/Q8, 2016/P2/Q9, 2017/P2/Q7, 2018/P2/Q19 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 109 5. The second law is based on the principle of conservation of energy. ¦(IR) represents the algebraic sum of potential differences which is equal to the total energy dissipated when a charge of one coulomb fl ows round a closed loop. This energy must be equal to the total energy provided by the sources of e.m.f. in the loop to drive the charge of one coulomb round the loop, i.e. ¦E. Example 3 Two cells A and B of e.m.f. 6.0 V and 4.0 V, and internal resistance 3.0 Ω and 2.0 Ω respectively are connected to a 8.0 Ω resistor as shown in the circuit. Calculate the potential difference across the 8 Ω resistor. E Exam Tips ÷ 6KRZWKHGLUHFWLRQRIWKHORRS ÷ 'L UHF WLRQ R I DQ HP I , SRLQWVDZD\IURPWKHSRVLWLYH WHUPLQDO ÷ &XUUHQW0DQGWKHHPI,DUHFRQVLGHUHGSRVLWLYHLIWKH\DUHLQWKHVDPH GLUHFWLRQDVWKDWRIWKHORRSWDNHQ Solution: Using Kirchhoff’s fi rst law for the point F, I1 + I2 – I = 0 I = I1 + I2 .......................... { Using Kirchhoff’s second law for the loop CFGDC ∑(IR) = ∑E 3I1 – 2I2 = 6 – 4 3I1 – 2I2 = 2 .......................... | For the loop FHJGF, 2I2 + 8(I) = 4 2I2 + 8(I1 + I2 ) = 4 8I1 + 10I2 = 4 .......................... } | × 5 : 15I1 – 10I2 = 10 23I1 = 14 I1 = —– 14 23 = 0.609A From |, 2I2 = 3I1 – 2 = 3(0.609) – 2 I2 = –0.0865 A Hence, I = I1 + I2 = 0.609 – 0.0865 = 0.5225 A Potential difference across 8 Ω resistor, V = IR = 0.5225 × 8 = 4.18 V A D G H J F C I2 I1 B I E1 = 6 V, 3 Ω E2 = 4 V, 2 Ω 8 Ω A D G H J F C I2 I1 B I E1 = 6 V, 3 Ω E2 = 4 V, 2 Ω 8 Ω Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 110 Example 4 In the circuit shown in the diagram, fi nd the electric potential at point P. Sketch a graph to show the variation of electric potential around the circuit starting from the point E. P E I I 4 Ω 5 Ω E1 = 3 V r1 = 1 Ω E2 = 2 V r2 = 2 Ω 6 Ω 2 Ω Solution: Using Kirchhoff’s second law, I(6 + 1 + 4 + 5 + 2 + 2) = 3 + 2 20I = 5 I = 0.25 A The electric potential at E = 0 since the point E is connected to the earth. The current in the circuit is in the clockwise direction. From E to P, in the direction of the current, the cell E1 raises the electric potential by E1 = 3.0 V, but the drop in potential difference = I(6 + 1 + 4) = 0.25 × 11 = 2.75 V Hence the electric potential at P = 3.0 – 2.75 = 0.25 V The variation of the electric potential V round the circuit from the point E is as shown below. E P E I V 6 Ω 4 Ω 5 Ω 2 Ω 0.5 V 1.0 V –1.0 V 0.25 V 1.5 V 1.25 V –1.5 V I = 0.25 A 2 1 0 –1 –2 E1 = 3 V E2 = 2 V r1 = 1 Ω r2 = 2 Ω Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 111 Example 5 A uniform piece of wire is bent to form a cubic frame as shown. If the resistance of each side of the cube is R, deduce the effective resistance between the corner A and H of the cube in terms of R. Solution: 6 – B I I 6 – I 3 – I 3 – I 3 – I 6 – I 6 – 6 I – I 6 – I 3 – I 3 – I I 3 – I A D J F H G C I I R1 VAH A H In the fi gure shown above, if the current entering the point A = I, then currents in AB, AD, and AF = —1 3 I (using Kirchhoff’s fi rst law) Similarly, current in the other sides of the cube are as shown in the fi gure. If R1 = effective resistance between A and H. Then, VAH = IR1 Also, VAH = sum of potential differences between AD, DC, and CH. = VAD + VDC + VCH Hence, IR1 = —1 3 IR + —1 6 IR + —1 3 IR = — 5 6 IR Effective resistance, R1 = — 5 6 R B G C F H D J A Quick Check 2 1. Two cells of e.m.f. 1.50 V and 6.0 V are connected to resistors of resistance 2.0 : and 4.0 : as shown in the circuit. ȍ ȍ 1.5 V 6.0 V X Y What is the potential difference across XY? A 1.5 V C 3.0 V B 2.0 V D 4.5 V 2. The diagrams below show four different ways in which currents I1 , I 2 , I 3 and I 4 are combined at a junction. For which of these junctions is I1 + I2 = I3 + I4 ? A I1 I2 I3 I 4 C I1 I2 I3 I 4 B I1 I2 I3 I 4 D I1 I2 I3 I 4 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 112 3. The circuit shows three resistors of resistance 2 Ω, 1 Ω and 3 Ω connected in series to a 6 V battery. The point Q is earthed. P Q 2 Ω 1 Ω 3 Ω 6 V R S Which graph represents the change in electric potential along the line PQRS? A C B D 0 –2 –4 –6 6 4 2 P V /V QRS 0 –2 –4 –6 6 4 2 P V /V Q S R 0 –2 –4 –6 6 4 2 P V /V QRS 0 –2 –4 –6 6 4 2 P V /V QRS 4. A 6.0 V battery is connected to four resistors as shown in the diagram. 6.0 V 3 1 9 1 9 1 9 1 Y X The point Y is earthed. What is the potential at the point X? A - 2.0 V B - 0.40 V C + 1.5 V D + 4.5 V 5. (a) State Kirchhoff’s laws. Explain how each law is based on a fundamental physics principle. 3 8 3 8 1 V + 4 8 2 V Ia Ib Ic Id 2 V 2 V 3 8 2 8 2 8 + + Figure (a) Figure (b) (b) Use the laws to deduce (i) values of Ia and Ib in Figure (a), (ii) values of Ic and Id in Figure (b). 6. The circuit shows a battery A of e.m.f. 7.2 V, internal resistance 1.5 Ω and a battery B of e.m.f. E and internal resistance 2.0 Ω connected to three resistors of resistance 5.0 Ω, 10.0 Ω and R. The currents in the circuit are as shown. 0.3 A 7.2 V 5.0 8 10.0 8 0.40 A 1.5 8 E 2.0 8 A B R I Calculate (a) the resistance R, (b) the current I, (c) the e.m.f. E. Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 113 Resistors in Series 1. Kirchhoff’s laws can be used to derive expressions for the effective resistance of resistors connected in series and in parallel. I II I V R1 R2 R3   I I V R   Ÿ (a) (b) Figure 15.6 2. Figure 15.6(a) shows three resistors of resistances R1 , R2 and R3 connected in series with a voltage V. Figure 15.6(b) shows the equivalent resistor of resistance R which is the effective resistance of the three resistors in series. 3. Applying Kirchhoff’s first law, the currents in all the resistors are the same, I. 4. Applying Kirchhoff’s second law to the circuit, E = ∑(IR) V = IR1 + IR2 + IR3 = I(R1 + R2 + R3 ) .......................... { 5. For the equivalent resistor in Figure 15.6(b), V = IR From equation {: V = I(R1 + R2 + R3 ) Hence, IR = I(R1 + R2 + R3 ) Effective resistance, R = R1 + R2 + R3 Resistors in Parallel 2009/P1/Q29, 2009/P2/Q6, 2012/P1/Q28, 2015/P2/Q17, 2018/P2/Q8 1. Figure 15.7(a) shows the three resistors connected in parallel to a voltage V. Figure 15.7(b) shows the equivalent circuit with R, the effective resistance of the three resistors in Figure 15.7(a). (a) (b) I A B X Y I V R2 R1 R3 I 3 I 3 I 1 I 2 I 1   I A B I V R   Figure 15.7 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 114 2. Applying Kirchhoff’s first law at the point A, I = I1 + I2 + I3 3. Applying Kirchhoff’s second law, E = ∑(IR) s TO THE LOOP 8!R1 BYX: V = I1 R1 which gives I1 = V R1 s TO THE LOOP 8!R2 BYX: V = I2 R2 which gives I2 = V R2 s TO THE LOOP 8!R3 BYX: V = I3 R3 which gives I3 = V R3 4. From I = I1 + I2 + I3 = V R1 + V R2 + V R3 I V = 1 R1 + 1 R2 + 1 R3 .......................... { 5. For the equivalent circuit in Figure 15.7(b), V = IR gives I V = 1 R .......................... | Equating { and |,: I V = 1 R = 1 R1 + 1 R2 + 1 R3 Hence, 1 R = 1 R1 + 1 R2 + 1 R3 6. If n resistors each of resistance R are connected in parallel, the effective resistance is — R n . 7. Two resistors of resistance R1 and R2 when connected in parallel act as a current divider (Figure 15.8). The effective resistance R is given by —1 R = —1 R1 + —1 R2 = R2 + R ——–—1 R1 R2 R = R1 R ——–—2 R1 + R2 8. The potential difference across R1 = potential difference across R2 = potential difference across R I1 R1 = I2 R2 = I ( R1 R ——–—2 R1 + R2 ) Hence, I1 = ( ——–— R2 R1 + R2 ) I and I2 = ( ——–— R1 R1 + R2 ) I Figure 15.8 R1 R2 V I 1 I I I 2 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 115 Example 6 The diagram shows part of a circuit. Find the current I1 and I2 . Solution: Using I1 = ( R ——–—2 R1 + R2 ) I Similarly, I2 = ( ——–– 5 5 + 10 )3 = ( ——–– 10 5 + 10 )3 = 1 A = 2 A Example 7 The circuit shows a 5 : and 10 : resistor connected in series to a 12 battery of negligible internal resistance. (a) What is the potential difference across the 10 : resistor? (b) A voltmeter of resistance 500 : is connected across the 10 : resistor. (i) What is the voltmeter reading? (ii) Find the percentage error in the reading. (c) Give one advantage and one disadvantage of using a voltmeter of resistance 1 000 : to measure the potential difference across the 10 : resistor. Solution: (a) The potential difference across (5 + 10) : is 12 V. Hence, the potential difference across 10 :is 10 —– 15 u 12 V V = 8.0 V (b) (i) When the voltmeter of resistance 500 : is connected in parallel to the 10 : resistor, the effective resistance R is given by — 1 R = —– 1 10 + —–– 1 500 R = 500 ————u10 500 + 10 = 9.804 : The voltmeter reading, V c = ————– 9.804 5 + 9.804 u 12 = 7.947 V (ii) Error in voltmeter reading = (8.000 – 7.947) V = 0.053 V Percentage error in voltmeter reading = ——–– 0.053 8.000 u 100% = 0.663% I 1 I = 3 A I2 5 Ω 10 Ω 12 V 5 Ω 10 Ω Exam Tips &DOFXODWH WKH YDOXH RI9 DQG= ’ WR  VLJQLðFDQW ðJXUHLQRUGHUWRREWDLQPRUHDFFXUDWHYDOXHRIWKH HUURULQYROWPHWHUUHDGLQJ 2018/P2/Q8 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 116 (c) If a voltmeter of 1 000 : resistance is used, —1 R = —– 1 10 + —–—1 – 1 000 R = 1 000 ————– u10 1 010 = 9.900 : Voltmeter reading, Vc = ————–– 9.900 (5 + 9.900) u 12 = 7.973 V Percentage error of voltmeter reading = (8.000 – 7.973) ——————–– 8.000 u 100% = 0.338% Hence, the reading is more accurate. The disadvantage is that the voltmeter is less sensitive since it has a large voltage value for the full-scale defl ection. Example 8 The circuit shows two lamps L1 and L2 rated 3 V, 0.25 A and 9 V, 0.05 A respectively light up with normal brightness. (a) Calculate the values of the resistance R1 and R2 . (b) What happens to the lamp L1 , if lamp L2 fuses? (c) What happens to the lamp L2 , if lamp L1 fuses? Solution: (a) Since both the lamps L1 and L2 light up with normal brightness, the potential difference across L1 is 3.0 V, and L2 is 6.0 V. Potential difference across R1 = 12.0 – 6.0 = 6.0 V Current in R1 = 0.25 + 0.05 = 0.30 A Hence, R1 = ——6.0 0.30 = 20 : Total potential difference across L1 and R2 = 6.0 V 3.0 + V2 = 6.0 V Potential difference across R2 , V2 = (6.0 – 3.0) V = 3.0 V Current in R2 = current in L1 = 0.25 A Using V = IR, R2 = ——3.0 0.25 = 12 : R1 R2 12 V d.c. L2 6 V, 0.05 A + – L1 3 V, 0.25 A Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 117 (b) If lamp L2 fuses, total resistance in the circuit = R1 + R2 + resistance of L1 = 20 + 12 + ——3 0.25 = 44 : Current in circuit = current in lamp L1 = ——12.0 44 = 0.27 A > 0.25 A which is slightly more than the working current of lamp L1 , hence the lamp L1 becomes brighter. (c) If lamp L1 fuses, total resistance in the circuit = R1 + resistance of L2 = 20 + ——6 0.05 = 140 : Current in circuit = current in lamp L2 = —— 12.0 140 = 0.086 A > 0.05 A The current in the lamp L2 increases by 72%. The lamp L2 lights up very bright and then fuses. Quick Check 3 1. Five resistors and a battery of e.m.f. 12.0 V are connected in the circuit shown below. The internal resistance of the battery is negligible. ȍ ȍ ȍ 12.0 V ȍ I ȍ What is the value of the current I from the battery? A 0.67 A C 1.5 A B 1.2 A D 2.4 A 2. A 6.0 V battery which has an internal resistance of 0.20 : is connected to a 4.0 : resistor. What is the ratio of the power dissipated in the external resistor to the power supplied by the battery? A 0.67 C 0.80 B 0.75 D 0.95 3. A uniform piece of wire of resistance 12 : is bent into the shape of an equilateral triangle. What is the effective resistance between any two corners of the triangle? A 2.0 : C 4.0 : B 2.7 : D 6.0 : 4. The diagram shows a network of three resistors. Two of the resistors have the same resistance R and the other has a resistance of 5.0 :. 5.0 8 Y R X R Z The effective resistance between Y and Z is 2.5 :. What is the effective resistance between X and Y? A 0.21 : C 1.9 : B 0.53 : D 2.5 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 118 5. Four resistors are connected as shown. 1 8 3 8 2 8 4 8 Q R P S Between which two points is the effective resistance (a) maximum (b) minimum? (a) maximum (b) minimum A PQ SP B QR PQ C RS QR D SP RS 6. Five identical resistors each of resistance R are connected as shown. R R R R R X Y What is the effective resistance between X and Y? A —2 7 R C —2 3 R B —5 8 R D —3 4 R 7. Three resistors are connected as shown and the points X and Y are connected to a battery. R1 R2 R3 X Y I 1 I 2 I 3 The ratio I —1 I2 is A R —–2 R1 C R2 + R ———3 R2 B R —–3 R2 D R2 + R ———3 R3 8. Two lamps P and Q rated 12 V, 24 W and 12 V, 36 W respectively are connected in series with a 24 V battery. Which of the following statements is true? A Resistance of P is less than that of Q. B The potential difference across P is bigger than that across Q. C The power dissipated from P is less than that from Q. D The same current is in both P and Q, and the potential differences across P and Q are the same. 9. Three resistors X, Y, Z of the same resistance are connected to a battery of negligible internal resistance as shown in the circuit. X Y Z If the battery supplies a total power of 12 W, what is the power dissipated as heat from the resistor Z? A 2 W C 6 W B 4 W D 8 W 10. In the circuit shown below, the power dissipated as heat from the 4.0 : resistor is 4 W. 6 V 4 8 R 6 8 What is the value of the resistance R? A 2 : C 6 : B 3 : D 8 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 119 11. In the circuit shown below, a potential difference of 3.0 V is applied across XY. X Y What is the value of the current in the 5 : resistor? A —8 3 A C —3 5 A B —3 4 A D —3 8 A 12. The circuit shown is used to determine the resistance of a voltmeter. R is a standard 100 : resistor. The milliammeter reading is 30 mA and the voltmeter reading is 2.7 V. R V mA What is the resistance of the voltmeter? A 900 : C 2 700 : B 1 000 : D 3 000 : 15.3 Potential Divider Students should be able to: ‡ H[SODLQDSRWHQWLDOGLYLGHUDVDVRXUFHRIYDULDEOHYROWDJH ‡ H[SODLQWKHXVHVRIVKXQWVDQGPXOWLSOLHUV Learning Outcomes 1. A potential divider is used to obtain a fraction of a voltage supply. 2. Basically a potential divider consists of two resistors in series. A rheostat connected as shown in Figure 15.9 may be used as a potential divider. Figure 15.9 R2 I I V ’ R1 Rheostat V R1 R2 I I V V ’ 3. The current, I = —–—V –– R1 + R2 . Hence the potential difference across R2 , V c = IR2 = ( R2 –—–— R1 + R2 ) V = ( Resistance R ———————2 Total resistance ) u (Voltage supply) 2010/P1/Q29, 2011/P1/Q29, 2013/P2/Q9, 2017/P2/Q8 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 120 4. By moving the sliding contact of the rheostat, the voltage V’ can be varied. Hence the potential divider becomes a source of variable voltage. Example 9 In the circuit shown, the e.m.f. of the battery is 6.0 V and its internal resistance is negligible. 100 Ω 6 V 50 Ω 100 Ω 50 Ω (a) What is the effective resistance in the circuit? (b) Calculate the current in each resistor. Solution: The circuit given can be redrawn as shown. 6 V R1 = 100 Ω R2 = 50 Ω R3 = 100 Ω R4 = 50 Ω I1 I2 I3 I4 (a) The effective resistance R of the 3 resistors (50 :, 100 :, and 50 :) in parallel is given by — 1 R = —– 1 50 + —–– 1 100 + —– 1 50 = —–– 5 100 R = 20 : Hence, effective resistance in the circuit = (100 + 20) : = 120 : (b) The current from the battery = current in R1 I1 = —–– 6 120 = 0.050 A The potential difference across the 3 resistors in parallel = potential difference across the effective resistor R = I1 R = 0.050 u 20 = 1.0 V Hence, current I2 = —–– 1.0 50 = 0.020 A current I3 = —–– 1.0 100 = 0.010 A current I4 = —–– 1.0 50 = 0.020 A Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 121 Example 10 Five 10 : resistors are connected as shown in the diagram. Calculate the effective resistance between the points A and B. Solution: When the points A and B are connected to a voltage supply, the points C and D are each at the midpoint between two equal resistors of 10 : each. Hence, the potential at points C and D are equal, and there is no current in the resistor between C and D. Hence, the resistor does not contribute any effective resistance in the circuit. The effective resistance is due to two (10 + 10) : resistance in parallel. —1 R = —– 1 20 + —– 1 20 = —– 1 10 R = 10 : Example 11 All the resistors shown in the circuit are identical. Calculate (a) the potential difference between C and F, VCF, (b) the potential difference between D and F, VDF. Solution: The circuit given can be redrawn as shown. (a) The potential difference VAB = VXY = 100 V Using the principle of the potential divider, VFB = ( R –—–– R + R)100 = 50 V VCY = ( R + R –—––—— R + R + R )100 = 66.7 V Hence, VCF = VCY – VFB = 66.7 – 50 = 16.7 V (b) VDY = ( R –—––—— R + R + R )100 = 33.3 V VDF = VDY – VFB = 33.3 – 50 = –16.7 V The negative sign shows that the potential at D is lower than that at F. C D A B 10 Ω 10 Ω 10 Ω 10 Ω 10 Ω A R R F X R R R Y C D B 100 V 100 V A R C R D R B R R F Exam Tips 5HGUDZ WKH FLUFXLW WR VKRZ WKDW UHVLVWRUVDUHLQSDUDOOHODQGLQVHULHV Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 122 Example 12 Four resistors of resistance R1 , R2 , R3 and R4 are connected to a battery of e.m.f. V and negligible resistance as shown in the circuit. Deduce an expression for the potential difference between the points B and D. Solution: Using the principle of the potential divider, VCB = ( R1 –—–— R1 + R2 )V VCD = ( R3 –—–— R3 + R4 )V VBD = VCB – VCD = ( R2 –—–— R1 + R2 )V – ( R3 –—–— R3 + R4 )V = ( R2 –—–— R1 + R2 – R3 –—–— R3 + R4 )V Alternative solution VBD = VBA – VDA = ( R1 –—–— R1 + R2 – R4 –—–— R3 + R4 )V R2 R3 R R4 1 C A B V D R2 R3 R4 R1 C A B V D Quick Check 4 1. A rheostat is used as a potential divider. In which circuit is the rheostat correctly connected? A External circuit Rheostat C External circuit Rheostat B External circuit Rheostat D External circuit Rheostat 2. A 2.0 V battery of negligible resistance is connected to the terminal X and Z of each of the arrangements of resistors shown below. A high resistance voltmeter is connected across the terminals Y and Z. What is the reading of the voltmeter? 20 8 10 8 X Y Z 2.0 V V 20 8 10 8 X Y Z 2.0 V V (a) (b) Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 123 20 8 10 8 X Y Z 2.0 V V 20 8 10 8 X Y Z 2.0 V V (c) (d) 20 8 6 8 X Y Z 2.0 V 10 8 V (e) 3. The circuit below shows a battery of e.m.f. V and negligible internal resistance connected to a resistor of resistance R. R X V Y Two voltmeters of resistance R and 10R respectively are connected one at a time across the terminals X and Y. What are the voltmeters readings in terms of V ? Uses of Shunt and Multiplier 2015/P2/Q7, 2017/P2/Q8 Conversion of Milliammeter to Ammeter 1. Moving-coil ammeters and voltmeters used in school laboratories are basically milliammeters which are converted into voltmeters or ammeters. 2. The coil of the moving-coil milliammeter are made of fi ne wire and only small currents can fl ow in the coil. 3. To convert a milliammeter into an ammeter to measure larger currents, a suitable resistance known as a shunt is connected in parallel with the milliammeter. 4. A shunt is a conductor with low resistance connected in parallel with the miliammeter to divert a large fraction of the current. Example 13 A milliammeter has a full-scale defl ection of 5 mA and a resistance of 20 :. Calculate the resistance of the shunt required to convert the milliammeter into an ammeter with full-scale defl ection of 5.0 A. Solution: The maximum safe current in the milliammeter is 5 mA = 0.005 A. Hence, the current deviated to the shunt of resistance R = 5.0 – 0.005 = 4.995 A Since the shunt and milliammeter are in parallel, potential difference = potential difference across shunt across milliammeter 4.995 u R = 0.005 u 20 R = 0.02002 : R Milliammeter 20 Ω 5.0 A 5 mA 4.995 A Shunt Exam Tips 7KHUHVLVWDQFHRIWKHVKXQWLVYHU\VPDOO FRPSDUHGWRUHVLVWDQFHRIPLOOLDPPHWHU Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 124 Conversion of Milliammeter to Voltmeter 1. Since the maximum safe current in milliammeter is of the order of magnitude of a few milliampere, the maximum safe potential difference across it is small. 2. To convert a milliammeter to measure voltages up to a few volts, a high resistance resistor, known as a multiplier is connected in series with the milliammeter. 3. A multiplier is a resistor of high resistance connected in series with the milliammeter so that the maximum potential difference across the milliammeter plus multiplier is increased. Example 14 A milliammeter of full-scale defl ection 5 mA and resistance 20 : is to be converted into a voltmeter for measuring voltage up to 10 V. Calculate the resistance of the multiplier. Solution: The multiplier of resistance R is connected in series with the milliammeter. The same current fl ows in the milliammeter and multiplier. Total potential difference across milliammeter and R V = 0.0050(20 + R) 10 = 0.10 + 0.0050R R = ——– 9.90 0.05 = 1 980 : Example 15 A dual-range d.c. voltmeter has a common negative terminal and two positive terminals marked ‘3 V’ and ’30 V’. It consists of a moving coil meter of resistance 50 :. The maximum defl ection on the scale is produced by a current of 1.0 mA. Draw a diagram to show how the meter is connected to three terminals, and calculate the values of the additional components required. Solution: R1 R + 3 V 2 1 mA – 3 V 30 V + 30 V 1 mA Meter 1.0 mA 50 Ω The resistors of resistance R1 and R2 are connected in series with the meter as shown above. R V1 V2 20 Ω Milliammeter 10 V Multiplier I = 5 mA I Exam Tips 5HVLVWDQFH RI PXOWLSOLHU LV YHU\ KLJK FRPSDUHGWRUHVLVWDQFHRIPLOOLDPPHWHU Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 125 When a current of 1.0 mA fl ows, potential difference between +3 V and negative terminal = (0.001)(R1 + 50) 3 = 0.001R1 + 0.050 R1 = ——– 2.95 0.001 = 2 950 : Potential difference between +30 V and negative terminal = (0.001)(R2 + R1 + 50) 30 = (0.001)(R2 + 2 950 + 50) = 0.001R2 + 3.0 R2 = ——– 27.0 0.001 = 2.7 u 104 : Example 16 Figure (a) shows the circuit of a simple ohmmeter. A milliammeter of resistance r, which has a full-scale defl ection of 5 mA is connected to a 1.5 V cell of negligible internal resistance and a 297 : resistor. mA 0 5 1 4 2 3 Figure (a) Figure (b) r A B mA ffiȍ 1.5 V The resistor to be measured is connected between the terminals A and B. When A and B are short-circuited, the meter gives full-scale defl ection. (a) Find the resistance r of the meter. (b) Find the resistance when connected between A and B gives half-full defl ection. (c) Figure (b) shows the scale of the meter in mA. Copy the fi gure and mark on the fi gure the scale 0 :, 300 :, 900 : and 1 200 :. Solution: (a) When AB is short-circuited, resistance between A and B is zero. Using V = IR, 1.5 = (5 u 10–3)(297 + r) r + 297 = 300 r = 3 : (b) Let R = resistance between A and B. When the defl ection is half-full, I = 1 ––2 u 5 mA = 2.5 mA Using V = IR, 1.5 = (2.5 u 10–3)(3 + 297 + R) (300 + R) = 600 : R = 300 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 126 (c) When R = 900 :, current, I = —————–— 1.5 900 + 3 + 297 = 1.25 mA When R = 1 200 :, I = —————–—– 1.5 1 200 + 3 + 297 = 1.00 mA 0 1 1 200 Ω 900 Ω 2 300 Ω mA 3 4 5 0 Ω Quick Check 5 1. The full-scale defl ection of an ammeter of negligible resistance is 1 mA. The meter is to be used as a voltmeter with a full-scale defl ection of 10 V. The correct circuit for the conversion is to connect A a resistance of 10 : in series with the meter. B a resistance of 104 : in series with the meter. C a resistance of 100 : in parallel with the meter. D a resistance of 10 : in parallel with the meter. 2. A galvanometer of resistance R and fullscale defl ection 3.0 mA is to be converted to measure current up to 300 mA. The resistance required is A —– R 99 connected in parallel with the galvanometer. B 99R connected in series with the galvanometer. C —– 9R 100 connected in parallel with the galvanometer. D —– R 104 connected in parallel with the galvanometer. 3. A moving-coil meter of resistance 100 : gives a maximum defl ection when the current in it is 100 μA. If the meter is to have a fullscale defl ection of 1 mA, the resistance of the shunt is A 9 : B 10 : C 100 —– 9 : D 900 : 4. A voltmeter is connected across a resistor in a circuit. Which statement is correct? A The current in the voltmeter is very small. B The same current fl ows in the resistor and voltmeter. C The high resistance of the voltmeter causes the current in the circuit to decrease. D The low resistance of the voltmeter causes the current in the circuit to increase. 5. A dual-range d.c. voltmeter has a common negative terminal and two positive terminals of +10 V and +3 V. The resistance of the voltmeter between the negative terminal and +3 V terminal is 1 000 :. What is the resistance between the negative terminal and +10 V terminal? A 333 : C 3 000 : B 1 000 : D 3 333 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 127 6. The diagram shows a dual-range d.c. milliammeter which has a common negative terminal A and positive terminals B and C. 50 Ω 20 mA 2 mA A C B R2 R1 The indicator is a moving-coil meter of resistance 50 : which gives a full-scale defl ection with a current of 1.0 mA. (a) When the instrument is connected into a circuit using terminals A and B, full-scale defl ection is obtained with a current of 2.0 mA. Find the total shunt resistance R1 + R2 . (b) When connection is made with the terminals A and C, a current of 20 mA gives full-scale deflection. Find the resistance R2 . 15.4 Potentiometer and Wheatstone Bridge Students should be able to: ‡ H[SODLQWKHZRUNLQJSULQFLSOHVRIDSRWHQWLRPHWHUDQGLWVXVHV ‡ H[SODLQWKHZRUNLQJSULQFLSOHVRID:KHDWVWRQHEULGJHDQGLWVXVHV ‡ VROYHSUREOHPVLQYROYLQJSRWHQWLRPHWHUDQG:KHDWVWRQHEULGJH Learning Outcomes Potentiometer 1. A potentiometer is used to measure potential difference accurately. 2. Figure 15.10 shows a potentiometer which consists of a uniform slide wire AB usually of length 1.0 m long. A driver cell D, usually a lead-acid accumulator is connected across the wire AB. 3. The potential difference V to be measured is connected to terminals X and Y. The jockey J is then moved along the wire AB until the centre-zero galvanometer is balanced. If l = AJ, the balanced length, potential difference, V = potential difference across AJ = I(lr) V vl where I = current in slide wire r = resistance per unit length of wire. Figure 15.10 l + – Switch Driver cell Slide wire Centre-zero galvanometer A X V Y J I B D G 2014/P2/Q17, 2015/P2/Q8, 2018/P2/Q9 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 128 Comparing e.m.f.s 1. The circuit shown in Figure 15.11 is used to compare the e.m.f.s of two cells. 2. The two-way switch S is fi rst connected to P and the jockey J moved along the slide wire AB until the galvanometer is balanced. Then e.m.f. E1 = potential differences across AX E1 v l 1 , the balanced length. 3. The switch S is then closed at Q, and the balanced length l 2 obtained. E.m.f., E2 v l 2 —– E1 E2 = l —1 l 2 4. The experiment is then repeated by changing the resistance of the rheostat. 5. If E1 is a standard cell of known e.m.f., then the e.m.f. E2 = E1( l —2 l 1 ) Example 17 G + + Rheostat X J B + D l1 A E1 E2 Q P S ( ) Figure 15.11 Cells A and B and a centre-zero galvanometer G are connected to a uniform wire OS using sliders X and Y as shown in the diagram. The length of the uniform wire OS is 1.00 m and its resistance is 12 :. When OY is 75.0 cm, the galvanometer does not show any defl ection when OX = 50.0 cm. If Y touches the end S of the wire, OX = 62.5 cm when the galvanometer is balanced. The e.m.f. of cell B is 1.0 V. Calculate (a) the potential difference across OY when OY = 75.0 cm, (b) the potential difference across OS when Y touches S and the galvanometer is balanced, (c) the internal resistance of cell A, (d) the e.m.f. of cell A. Solution: (a) When OX = 50.0 cm, the galvanometer is balanced. Hence e.m.f. of cell B v 50.0 cm 1.0 V v 50.0 cm Potential difference across OY, V1 v 75.0 cm V —–1 1.0 = ——– 75.0 50.0 V1 = 1.50 V (b) When OX = 62.5 cm, e.m.f. of cell B, 1.0 V v 62.5 cm Potential difference across OS, V2 v 100.0 cm V —–2 1.0 = 100.0 ——– 62.5 V2 = 1.60 V A X B O Y S G Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 129 (c) When OY = 75.0 cm, V1 = 1.50 Resistance of 75.0 cm of wire = ——– 75.0 100.0 u 12 = 9.0 : Using —E V = ——– R + r R for cell A, —– E 1.5 = ——–– 9.0 + r 9.0 ........................ {(r = internal resistance of A) When the slider is at S, OS = 100.0 cm R = 12 : and V = 1.60 V then —–– E 1.60 = ——–– 12 + r 12 ........................ | —{ | : —–– 1.60 1.50 = (——–– 9.0 + r 12 + r )—– 12 9.0 4.80(12 + r) = 6.00(9.0 + r) r = —–———– 57.6 – 54.0 1.2 = 3.0 : (d) From —– E 1.5 = ——–– 9.0 + r 9.0 , E = ( ———–– 9.0 + 3.0 9.0 )1.5 = 2.0 V Determination of Internal Resistance 1. The Figure 15.12 shows a cell of e.m.f. E and internal resistance r and a resistance box connected to the potentiometer. 2. With the switch S open, the balanced length l 0 for the e.m.f. of the cell is fi rst obtained. Then e.m.f. E  v l 0 , E = kl0 k = constant 3. With a suitable value of R in the resistance box, the switch S is closed, and the new balanced length l is obtained, The potential difference across R, V = kl  ?— E V = kl —–0 kl = l —0 l 4. The experiment is then repeated for different values of R, and the corresponding balanced length l is obtained. From the equation —E V = ——– R + r R , l —0 l = 1 + — r R A graph of —1 l against — 1 R is then plotted. (Figure 15.13) From the graph, when — 1 l = 0, — 1 R = – — 1 r Hence, the internal resistance r of the cell can be calculated from the intercept on the —1 R – axis. Figure 15.12 G + Resistance box X J B + A E, R r S l0 Figure 15.13 x x x x x x x 1 l 1 0 r – – 1 R – – Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 130 Example 18 In the potentiometer circuit shown, the cell E has a constant e.m.f. of 2.0 V and negligible internal resistance. The uniform wire AB is 100 cm long and has a resistance of 5.0 :. The e.m.f. of cell X is 1.5 V and its internal resistance is 0.8 :. Calculate (a) the length AP when the defl ection of the galvanometer G is zero, (b) the length AP when the galvanometer is balanced when a 1.0 : resistor is connected in series with cell E, (c) the balanced length when the 1.0 : resistor is shifted from E and connected in series with cell X, (d) the balanced length when the 1.0 : resistor is then connected in parallel with cell X. Solution: (a) When the galvanometer is balanced, potential difference across AP = e.m.f. of cell X. ——AP 100 u 2.0 = 1.5 AP = 75.0 cm (b) When the 1.0 : resistor is connected in series with E, the current in the slide wire AB I = ————– 2.0 (5.0 + 1.0)A = —1 3 A When the galvanometer is balanced, potential difference across AP = e.m.f. of cell X I u (——AP 100 u 5.0) = 1.5 — 1 3 u (——AP 100 u 5.0) = 1.5 AP = 90.0 cm (c) When the 1.0 : resistor is in series with the cell X, the potentiometer measures the e.m.f. of cell X as in (a) above. When the galvanometer is balanced, there is no current in the galvanometer and 1.0 : resistor. Potential difference across AP = e.m.f. of cell X —— AP 100 u 2.0 = 1.5 AP = 75.0 cm (d) When the 1.0 : resistor is in parallel with the cell X, current fl ows from cell X and the 1.0 : resistor. The potentiometer now measures the potential difference V across the 1.0 : resistor. From the equation —V E = ——– R R + r , V = (———— 1.0 1.0 + 0.8)1.5 = 0.833 V Potential difference across AP = Potential difference V across 1.0 : resistor —— AP 100 u 2.0 = 0.833 AP = 41.7 cm B P X A E G Exam Tips 'HFLGHIRUHDFKFDVHZKHWKHU WKHSRWHQWLRPHWHULVPHDVXULQJ ÷ HPI,RU ÷ SRWHQWLDOGLIIHUHQFH= Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 131 Quick Check 6 1. A potentiometer has a slide wire XY of length l and resistance R. The driver cell is of e.m.f. E and internal resistance r, and is in series with a resistor of resistance r. With a cell P in the circuit, the null point is found to be —1 3 from the end X as shown in the diagram below. r E, r P X Y 3 l l – G The e.m.f. of cell P is A ——–— ER 3(R + r) B ——–—– Er 3(R + 2r) C ——–—– ER 3(R + 2r) D E ——–—– (R + 2r) 3R 2. In the potentiometer circuit shown, the galvanometer pointer defl ects in the same direction when the jockey touches the ends P or Q of the slide wire. R1 R2 E1 E2 P Q G The most likely reason is A the resistance R1 is too low. B the resistance R2 is too high. C the e.m.f. E1 is much smaller than the e.m.f. E2 . D the internal resistance of the cell of e.m.f. E2 is higher. 3. Two cells of e.m.f.s E1 and E2 and negligible internal resistances are connected with two variable resistors as shown in the diagram below. E1 E2 P Q G Galvanometer When the galvanometer shows no defl ection, the values of the resistances are P and Q. What is the value of the ratio E —2 E1 ? A —P Q C ——– Q P + Q B ——– P P + Q D ——– P + Q P 4. A standard cell of e.m.f. 1.50 V is used to fi nd the potential difference across a uniform wire XY as shown in the fi gure. When the sliding contact is at S, the balanced length is l1 from X and l 2 and Y. There is no current in the galvanometer. X Y l1 l2 S G 1.50 V What is the potential difference across XY? A 1.50( l —1 – l 2 )V C 1.50( l 1 + l —–—2 – l 1 )V B 1.50( l —2 – l 1 )V D 1.50( l 1 + l —–—2 – l 2 )V Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 132 5. In the circuit shown below, the e.m.f.s of the cells E1 and E2 are 12 V and 4 V respectively, and the internal resistance of cell E is 4 :. E1 E2 8 8 10 8 G If the reading of the galvanometer G is zero, what is the internal resistance of cell E2 ? A 4 : C 8 : B 6 : D 12 : 6. In the potentiometer circuit shown, PQ is a uniform wire of length 1.0 m and resistance 10.0 :. E is an accumulator of e.m.f. 2.0 V and negligible internal resistance. R is a 15 : resistor and r is a 5.0 : resistor. S1 S2 E2 R T P E r Q G When S1 and S2 open, galvanometer G is balanced when QT is 62.5 cm. When both S1 and S2 are closed, the balanced length is 10.0 cm. Calculate (a) the e.m.f. of cell E2 , (b) the internal resistance of cell E2 , (c) the balanced length QT when S2 is opened and S1 closed, (d) the balanced length QT when S1 is opened and S2 closed. To Measure the e.m.f. of a Thermocouple 1. A thermocouple consists of two wires of different materials such as copper and iron joined at two junctions. 2. When there is a temperature difference between the two junctions, an e.m.f. is produced (Figure 15.14). The magnitude of the e.m.f. increases as the temperature increases until a maximum value. The maximum value of the e.m.f. depends on the types of metal of the two wires. 3. For a copper-iron thermocouple with a temperature difference of 100°C, the e.m.f. of the thermocouple is about 4 mV. 4. If a potentiometer with a uniform slide wire AB of length 100 cm connected to a driver cell of e.m.f. 2.0 V and negligible internal resistance is used, then potential difference across 100 cm of wire is 2.0 V. The balanced length for an e.m.f. of 4 mV would be given by l = —— 100 2.0 u (4 u 10–3) = 0.2 cm which is rather small and the percentage error in l would be rather large. Figure 15.14 Cold junction Cu Fe Cu 0°C 100°C Hot junction Figure 15.15 A B 2 V 100 cm Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 133 5. To increase the accuracy in the measurement of the balanced length, the balanced length must be increased. This is done by connecting a suitable resistance R in series with the slide wire AB as shown in Figure 15.16. 6. Suppose that the maximum e.m.f. of a thermocouple is 4 mV. Resistance of slide wire AB of length 1 m, r = 5.0 :, e.m.f. of driver cell = 2.0 V. If R = resistance required in series with slide wire, then 2.0 = I(R + 5.0) .............................. { If the balanced length for the maximum e.m.f. of 4 mV is 100 cm, then 4 u 10–3 = I u (5.0) .............................. | { | : ——–— 2.0 4 u 10–3 = I(R + 5.0) ——–—— I u 5.0 R = 2 495 : If l = balanced length when the e.m.f. of the thermocouple is e, then e = I(lr) r = resistance per unit length of slide wire ——e 2.0 = I(lr) ——–— I(R + r) = 5.0 : m–1 e = ——–– 2.0 lr (R + r) For example, if r = 5 : m–1, and l = 36.0 cm = 0.360 m R = 2 495 : then e.m.f. of thermocouple, e = 2.0 u (0.360)(5) ——–————— 2 495 + 5 = 1.44 mV Example 19 The potentiometer circuit shown is used to measure the e.m.f. of a thermocouple. The slide wire AB is 1.00 m long and has a resistance of 5.5 :. The resistance R is 2 000 :. The e.m.f. of the driver cell D is 2.0 V. If the balanced length AP is 65.2 cm, what is the e.m.f. of the thermocouple? Solution: If I = current in R and slide wire AB, then 2.0 = I(R + r) r = resistance of AB = I(2 000 + 5.5) .............................. { If e = e.m.f of thermocouple, then e = Ilr = I(0.652)(5.5) .............…................. | —| { : —–e 2.0 = ————— 0.652 u 5.5 2 005.5 e = 3.6 u 10–3 V Figure 15.16 D I A l I B 2.0 V R e G Thermocouple Thermocouple D A B P R G Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 134 Example 20 The circuit shown is used to measure the e.m.f. of a thermocouple. The length of the slide wire AB is 1.00 m and its resistance is 5.00 :. The resistance R2 is 1 000 : and the value of the resistance R1 is adjusted so that a balance point P can be obtained with the two-way switch S at Y. When R1 = 1 040 :, and the switch S is at Y, the balanced length AP = 21.1 cm. When S is at X, the balanced length AP = 68.5 cm. If the e.m.f. of the cell E1 is 1.018 V, calculate the e.m.f. of the thermocouple. Solution: Let I = current in R2 , R1 and the slide wire AB. When the switch S is at Y, e.m.f. of E1 = potential difference across R1 and AP 1.018 = I (R1 + l 1 r) = I (1 040 + 0.211 u 5.00) .............................. { When the switch S is at X, e.m.f. of thermocouple, e = potential difference across AP e = I (l 2 r) = I (0.685 u 5.00) .............................. | | { : ——–—e 1.018 = —————————— 0.685 u 5.00 (1 040 + 0.211 u 5.00) e = 3.35 u 10–3 V Quick Check 7 R2 R1 E1 = 1.018 V A e Y X S P B G 1. A potentiometer is to be calibrated with a standard cell using the circuit shown below. R L M J P Galvanometer Standard cell The balance point is found to be near L. To improve accuracy, the balance point should be nearer to M. This may be achieved by A replacing the galvanometer with one of lower resistance B replacing the potentiometer wire with one of higher resistance C putting a shunt resistance in parallel with the galvanometer D increasing the resistance R 2. The diagram below shows a potentiometer circuit used to measure the e.m.f. of a thermocouple. A P B R G 6.00 mV Thermocouple 0.600 m 2.00 V The metre wire AB has a resistance of 5.0 : and the driver cell has an e.m.f. of 2.00 V. If the balanced length AP = 0.600 m when measuring an e.m.f. of 6.00 mV, what is the value of the resistance R? A 195 : C 995 : B 495 : D 1 995 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 135 3. The diagram shows a potentiometer circuit with l as the balanced length. P R l Q G Which of the following steps would give a balanced length bigger than l ? A Connect a standard resistor in parallel with the cell Q. B Substitute cell Q with one of smaller e.m.f.. C Substitute the resistor R with one of higher resistance. D Connect a standard resistor in series with cell Q. 4. The slide wire of a potentiometer has a resistance of 5.0 :. A cell of e.m.f. 2.0 V and negligible internal resistance is used as a driver cell. Draw a labelled circuit diagram to show how the potentiometer may be modified to measure the e.m.f. of up to 2.5 mV from a copper-constantant thermocouple. Calculate a suitable value for the resistor which should be placed in series with the slide wire. 5. In the potentiometer circuit shown below, YZ is a slide wire of length 100 cm and resistance 5.00 :. The standard cell P has an e.m.f. of 1.018 V. When switch T is connected to N, the balanced length is 72.6 cm. When T is connected to M, the balanced length becomes 67.4 cm. Y Q M T P N X G 10.0 Ω Explain why the currents in the 10.0 : resistor and the slide wire YZ are the same when the galvanometer is balanced. Calculate the e.m.f. of cell Q. Comparing Resistances 2007/P1/Q32 1. A potentiometer can be used to compare the resistances of two resistors by measuring the potential differences across the resistors when the same current is in both of them. 2. This method is most suitable for comparing low resistances, because when the potentiometer is balanced no current flows in the wires used to connect the resistor to the potentiometer. Hence the resistance of the connecting wires has no effect on the measured resistance. 3. The circuit is set up as shown in Figure 15.17. First the potential difference V1 across the resistor of resistance R1 is connected to the potentiometer. If l 1 is the balanced length, then V1 = potential difference across balanced length l 1 I1 R1 = kl1 (k = constant) 4. The connecting wires from R1 to the potentiometer are disconnected, and then connected across the resistor of resistance R2 . If l 2 = balanced length, then V2 = potential difference across balanced length l 2 I1 R2 = kl2 V —1 – V2 = I1 R — 1 –—I1 R2 = kl —1 –– kl2 R —1 – R2 = l —1 – l 2 A B G G + + l1 l2 I1 I1 R1 R2 V1 V2 Figure 15.17 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 136 Example 21 The circuit shown is used to compare the resistance R of an unknown resistor with a 100 : resistor. The balanced lengths l are 400 mm and 588 mm when X is connected to Y, and to Z respectively. The length of the slide wire is 1.00 m. What is the value of R? l X Z 100 Ω Y R Solution: If I = current in both the 100 : resistor and R, when X is connected to Y, potential difference across 100 : resistor = potential difference across 400 mm I(100) = k u 400 (k = constant) When X is connected to Z, potential difference across (100 + R) : = potential difference across 588 mm I (100 + R) = 588k I(100 + R) —–———— I(100) = —588 –—400 R = 47 : To Measure Current 1. The current in a standard resistor can be measured by using a potentiometer to measure the potential difference across the standard resistor. 2. Figure 15.18 shows how a potentiometer is used to calibrate an ammeter. 3. The two-way switch is connected to a standard cell of known e.m.f. E0 . If l 0 is the balanced length, then E0 = kl0 (k = constant) 4. The two-way switch is then connected to Y, the potentiometer now measures the potential difference V across the standard resistor of resistance R. If l 1 is the balanced length, then V = I1 R = kl1 I1 = current in R I1 R—–– E0 = l —1 – l 0 I1 = ( l —1 – l 0 ) E —0 – R 5. By changing the resistance of the rheostat, different values of the current I 1 are obtained and the ammeter calibrated. l0 I1 I1 E0 A B X Y R + Rheostat + + A G Figure 15.18 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 137 Quick Check 8 1. In the potentiometer circuit shown, the balance point on the slide wire can be obtained when X is connected to Y. When X is connected to Z, no balance point is obtainable on the slide wire. Z P Q X R Y This is most likely due to A the value of the resistance R is too small. B the resistance of the slide wire is too high. C the current in P and Q is not steady. D the value of the resistance of Q is much bigger than that of P. 2. In the potentiometer circuit shown, the length of the slide wire AB is 1.00 m. The e.m.f. of cell E1 is 2.0 V and its internal resistance is negligible. E2 is a cell of e.m.f. 1.1 V and its internal resistance is 1.0 :. R1 = 1.0 : and R2 = 2.0 :. The two-way switch S1 can be connected to X or Y. E1 S1 S2 R2 R1 E2 B X Y A G Calculate the balanced length for each of the positions of switch S1 , when (a) S2 is open, (b) S2 is closed. Wheatstone Bridge 1. Charles Wheatstone invented the Wheatstone Bridge in 1843. It is used to measure resistance. 2. Figure 15.19 is the circuit of a Wheatstone bridge. P and Q are two standard resistors with known resistance so that the value of the ratio —P Q can be fi xed, say —P Q = 1. 3. R is the resistor whose resistance is to be measured, and S is a standard variable resistor of known resistance. 4. The switch K is closed and the value of S is adjusted until the centrezero galvanometer is balanced. 5. When the galvanometer is balanced, there is no current between the points B and D. Hence, the potentials at B and D are equal. 6. If I1 = current in P, then current in Q is also I1 . Similarly the current in R = current in S = I2 . 7. For the resistors P and R, A is a common terminal and the points B and D have the same potential. Hence, potential difference across P = potential difference across R Figure 15.19 A R K S C Q B P D G I1 I2 I2 I1 2017/P2/Q9 Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 138 I1 P = I2 R .............................. { Similarly, potential difference across Q = potential difference across S I1 Q = I2 S .............................. | —{ | : I1 P —– I1 Q = I2 R—– I2 S — P Q = — R S R = P (—Q)S 8. To measure small resistance, the ratio of —P Q is fixed at say —1 10 or —– 1 100, hence R = —1 10 S or R = —– 1 100 S respectively. 9. To measure large resistance, the ratio of —P Q is fixed at say 10, so R = 10S. Metre Bridge 1. A simplified version of the Wheatstone bridge is the metre bridge (Figure 15.20(a)). Corresponding points of the metre bridge and Wheatstone bridge in Figure 15.20 are marked by the same alphabets. Figure 15.20 (a) Metre bridge (b) Wheatstone bridge G (100 – l ) K R S D A l C B P Q J ( ) A C R S D P Q B G 2. The unknown resistor R is first connected to the left-hand gap, and the standard resistor S is in the right-hand gap of the metre bridge. The key K is closed and use the jockey J to touch points along the slide wire AC until the centre-zero galvanometer is balanced. The balanced length l cm is noted. Then the resistance P = lr and Q = (100 – l)r where r = resistance per cm length of the slide wire AC. — R S = — P Q = ———— lr (100 – l)r —R S = ———– l (100 – l) R = (——–– l 100 – l )S Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 139 3. A number of precautions have to be taken when using the metre bridge. (a) The value of the standard resistor S must be chosen so that the balance point is in the middle third of the slide wire (from 30 cm mark to 70 cm mark). Hence the percentage error in the measurement of l and (100 – l ) would be of the same order. (b) The jockey J should not be dragged along the slide wire, otherwise the slide wire would become non-uniform. (c) The experiment is repeated with the unknown resistor in the right gap of the metre bridge. This is to eliminate end errors. The resistance at the two ends of the slide wire differs due to different amounts of solder used. Example 22 In the metre bridge circuit shown, the resistances of X and Y are 5.0 : and 3.0 : respectively. When a shunt is connected across X, the balanced length is 0.527 m from A. Calculate the resistance of the shunt. Solution: If R = effective resistance of 5.0 : resistor and shunt in parallel Using —R S = ————–— l cm (100 – l ) cm, —– R 3.0 = —–———– 0.527 (1 – 0.527) R = ——— 0.527 0.473 u 3.0 = 3.342 : Quick Check 9 X Y G A B If R1 = resistance of shunt, ——– 1 3.342 = —– 1 5.0 + —1 R1 —1 R1 = ——– 1 3.342 – —– 1 5.0 R1 = 10.08 : 1. The diagram shows a Wheatstone bridge circuit. ȍ G X ȍ ȍ The galvanometer G is balanced. What is the resistance of the resistor X? A 2 : C 10 : B 5 : D 15 : 2. The centre-zero galvanometer, G of the Wheatstone bridge shown in the diagram below is balanced. 15 1 60 1 10 1 20 1 G R What is the value of the resistance R? A 20 : C 40 : B 30 : D 60 : 3. In the circuit shown, what is the ammeter reading? Explain your answer. 10 1 50 1 50 1 10 1 P Q A Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 140 Important Formulae 1. Current, I = —dQ–– dt 2. Resistance, R = —V – I , V = IR 3. In series: R = R1 + R2 + R3 In parallel: —1 – R = —1 – R1 + —1 – R2 + —1 – R3 4. Potential divider, V1 = ( R —–—1 –– R1 + R2 )V 5. E.m.f. E and internal resistance, r E = V + IR = I(R + r) —E – V = —–— R + r – R = 1 + —r – R STPM PRACTICE 15 1. A bulb which is rated 60 W, 240 V is connected to a 110 V supply. What is the power dissipated? A 12.6 W C 27.5 W B 15.0 W D 60.0 W 2. The variation of potential difference V across a conductor with current I is as shown in the graph below. 0 0.2 1.0 2.0 3.0 4.0 5.0 0.4 0.6 0.8 1.0 V / V I / A What is the power dissipated from the conductor when the current is 0.5 A? A 0.156 W B 0.312 W C 0.625 W D 1.25 W 3. Three resistors of resistance 2.0 :, 3.0 : and 5.0 :are connected to two cells of negligible internal resistance. The current from the cells are 3.2 A and 1.4 A as shown in the diagram. 2.0 Ÿ 1.4 A 5.0 Ÿ 3.0 Ÿ 3.2 A What is the potential difference across the 3.0 :resistor? A 5.4 V C 7.0 V B 6.4 V D 13.8 V 4. A potentiometer is used to determine the resistance R of a resistor as shown in the circuit. The slide wire XY is 100.0 cm long. The centre-zero galvanometer G is connected to the point P. It is balanced when the slider is at the 35.0 cm mark. When connected to the point Q, the galvanometer G is balanced when the slider is at the 63.0 cm mark. 20 Ÿ 63.0 cm X Y Q P R 0 35.0 cm G G What is the value of the resistance R? A 16 : C 37 : B 36 : D 65 : Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 141 5. In the circuit shown, a cell of e.m.f. E1 and internal resistance r1 and another cell of e.m.f. E2 and internal resistance r2 are connected to resistors of resistances R1 and R2 . E1 , r 1 E2 , r 2 R2 R1 I 1 I 2 S R P Q Applying Kirchhoff’s law to the loop PQRS, which equation is correct? A E1 – E2 = I1 (r1 + R1 ) + I2 (r2 + R2 ) B E1 – E2 = I1 (r1 + R1 ) – I2 (r2 + R2 ) C E1 + E2 = I1 (r1 + R1 ) + I2 (r2 + R2 ) D E1 + E2 = I1 (r1 + R1 ) – I2 (r2 + R2 ) 6. Which of the following is correct about the resistance of a voltmeter, and the resistance of an ammeter? Resistance Resistance of voltmeter of ammeter A Low Low B Low High C High High D High Low 7. Two fixed resistors and a rheostat are connected in a circuit as shown. A1 A2 A3 How do the readings of the ammeters vary when the resistance of the rheostat is increased? A1 A2 A3 A Decreases Decreases Decreases B Increases Decreases Decreases C Decreases Increases Decreases D Decreases Increases Increases 8. Two resistors of resistances 1.0 : and 2.0 : are connected to a 3.0 V cell with internal resistance 2.0 :. 2.0 Ÿ 1.0 Ÿ 3.0 V, 2.0 Ÿ What is the current in the 1.0 :resistor? A 0.375 A C 0.750 A B 0.600 A D 1.00 A 9. Three resistors are connected to a 12.0 V battery as shown in the diagram. 6.0 Ÿ 8.0 Ÿ 12.0 V 4.0 V R X What is the potential at the terminal X, and the value of the resistance R? Potential at X R A 3.0 V 8.0 : B 6.0 V 10.0 : C 7.0 V 2.0 : D 9.0 V 10.0 : 10. Part of a circuit is shown in the diagram. 2.0 Ÿ I 0.5 A 2.8 A 1.5 A 0.8 A What is the value of the current I? A 1.0 A C 3.0 A B 2.6 A D 4.1 A 11. A battery of e.m.f. 6.0 V and internal resistance 2.0 : is connected to a resistor of resistance 10.0 :. What is the potential difference across the terminals of the battery, and the power lost due to the internal resistance? Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 142 Terminal potential Power lost difference A 6.0 V 2.50 W B 5.0 V 0.50 W C 4.0 V 2.50 W D 1.0 V 0.50 W 12. A galvanometer with a resistance of 500 : and full-scale deflection of 20 mA is converted into a dual-scale ammeter as shown in the diagram.  15 A 5 A R1 R2 G What are the values of the resistance of R1 and R2 ? R1 R2 A 0.500 : 1.50 : B 0.669 : 1.34 : C 1.00 : 1.50 : D 1.50 : 0.50 : 13. Four resistors are connected as shown in the diagram below. X Y 1.0 Ω 2.0 Ω 2.0 Ω 2.0 Ω What is the effective resistance between the points X and Y? A 1.0 : C 2.8 : B 2.0 : D 3.4 : 14. Two cells of negligible internal resistance are connected to three resistors as shown in the diagram below. P 9.0 Ω 3.0 Ω 3.0 Ω 1.5 V 6.0 V Q What is the potential difference between the points P and Q? A 0.9 V C 3.3 V B 2.7 V D 3.6 V 15. A voltmeter has a full-scale deflection of 3.00 V and a resistance of 1200 :. To convert the voltmeter to have a full-scale deflection of 12.0 V, a resistor of A 3600 : is connected in series with the voltmeter. B 3600 : is connected in parallel with the voltmeter. C 4800 : is connected in series with the voltmeter. D 4800 : is connected in parallel with the voltmeter. 16. The circuit shown below is used to determine the e.m.f. of a cell E. The sliding contact of the variable resistor is moved until the pointer of the centre-zero galvanometer G is not deflected. X Y E 8.0 Ω 2.0 V G When the centre-zero galvanometer is balanced, the resistance of the variable resistor between the points X and Y is 4.0 :. What is the e.m.f. E? A 0.67 V C 1.20 V B 0.80 V D 1.50 V Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 143 17. The meter in the circuit shown below has an uncalibrated linear scale. With the circuit as shown, the scale reading is 20. Meter 5 8 2 995 8 2 000 8 1.5 V X Y When another 2 000 : resistor is connected across XY, the scale reading is A 10 C 28 B 25 D 40 18. In the diagram shown, the variable resistor R can be adjusted from zero to 107 :. 10 Ω 104 Ω 0 A107 Ω P R Q What are the approximate limits for the resistance between P and Q? A 10 : to 104 : B 10 : to 107 : C 10 : to 1011 : D 104 : to 107 : 19. A cell of e.m.f. 2.0 V and negligible internal resistance is connected to four resistors as shown in the diagram. 2.0 V 15 8 5 8 5 8 5 8 P Q V What is the reading of the high resistance voltmeter connected between the points P and Q? A 0 C 0.75 V B 0.50 V D 1.00 V 20. In which of the following arrangements of resistors does the meter X, which has a resistance of 2 :, give the largest reading when the same potential difference is applied between points P and Q? A C P Q 2 Ω 1 Ω X P Q 1 Ω 2 Ω X B D P Q 1 Ω 2 Ω X P Q 2 Ω 1 Ω X 21. The diagram shows a potential divider circuit. By adjusting the position of the contact X, the potential difference obtainable between the terminals P and Q can be varied. 12 8 12 V 36 8 P Q X What are the limits of this potential difference? A 0 and 9 V B 3 and 12 V C 0 and 12 V D 9 and 12 V 22. The diagram below shows part of a circuit. The potential difference across the 2.0 : resistor is 6.0 V. The cell of e.m.f. E = 1.5 V has an internal resistance r = 0.5 :. The current I flows in the direction shown. A B 2.0 Ω I 1.0 Ω E = 1.5 V r ȍ If the potential at A is 9.0 V, what is the potential at the point B? A – 3.0 V B 0 V C 18 V D 21 V Physics Term 2 STPM Chapter 15 Direct Current Circuits

15 144 23. A moving-coil galvanometer has resistance of 40 : and gives full-scale deflection when a potential difference of 0.20 V is applied across its terminals. A shunt of resistance 0.025 : is connected in parallel to the galvanometer to change it into an ammeter. Which of the following is the deflection of the ammeter when there is a current of 2.0 A in the ammeter? A 1 ––8 full-scale C 1 ––2 full-scale B 1 ––4 full-scale D 3 ––4 full-scale 24. A cell E is connected to a high resistance voltmeter, a 10 : resistor and a switch S as shown in Figure (a). The graph in Figure (b) shows the variation of the voltmeter reading V with time t. S E 10 Ω V t V / V S closed 0 2.4 2.0 Figure (a) Figure (b) What is the internal resistance of cell E? A 1.6 : C 4.0 : B 2.0 : D 7.2 : 25. In the circuit shown, the current in the 2 : resistor is 3 A. What are the values of the current I from the power supply, and the voltage V? 3 A 2 8 6 8 V + – I Supply 1.5 8 A B C D I/A 3 4 4 12 V/V 10.5 9 12 18 26. Four resistors and a capacitor are connected in a circuit as shown in the diagram below. 12.0 F 6.0 V I 5.0 Ÿ 3.5 Ÿ 2.5 Ÿ 4.0 Ÿ Calculate the current I. 27. A power station generates power at a voltage of 2.0 kV and current 120 A. The voltage is stepped up to 360 kV and transmitted along cables of resistance 5.0 : km–1. (a) What is the current in the transmission cables? (b) What is the power lost per kilometre of cable? 28. (a) (i) State Kirchhoff’s laws for an electric circuit. (ii) Name the fundamental principles which form the basis of Kirchhoff’s laws. (b) In the circuit shown below, the cell with e.m.f. 4.0 V has an internal resistance of 2.0 :, while the other cells have no internal resistance. 3.0 V 4.0 V, 2.0 Ÿ 6.0 V ABC FED I 2 I 3 I 1 2.0 Ÿ 6.0 Ÿ 3.0 Ÿ 4.0 Ÿ Calculate (i) the currents I1 , I2 and I3 , (ii) the potential difference across AB, (iii) the power dissipated from the 6.0 : resistor. Physics Term 2 STPM Chapter 15 Direct Current Circuits