Topic_5_Electric_current_student_questio.pdf

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1、1 IB Physics:Question Book Topic 5:Electric current Name:Physics class:2 5.1:Electric potential difference,current and resistance 1.Two isolated spheres X and Y of unknown materials are touching one another as shown below.Sphere Y is negatively charged and sphere X is earthed.The earth connection is

2、 removed from sphere X and then the spheres are separated as shown below.Sphere X is found to be positively charged and sphere Y remains negatively charged.Which of the following describes the nature of the materials from which the spheres are made?Sphere X Sphere Y A.Insulator Insulator B.Insulator

3、 Conductor C.Conductor Insulator D.Conductor Conductor(1)2.Which one of the following is a correct definition of electric potential difference between two points?A.The power to move a small positive charge between the two points.B.The work done to move a small positive charge between the two points.

4、C.The power per unit charge to move a small positive charge between the two points.D.The work done per unit charge to move a small positive charge between the two points.(1)sphere X,earthedsphere Y,negatively chargedsphere X,positively chargedsphere Y,negatively charged3 3.A proton and an alpha part

5、icle are accelerated from rest from the positively charged plate X to the negatively charged plate Y.At the mid-point between the plates,the proton has a kinetic energy EK.At this point,the alpha particle has a kinetic energy A.B.EK.C.2EK.D.4EK.(1)4.The electron volt is defined as A.a unit of energy

6、 exactly equal to 1.6 1019 J.B.a fraction of the ionization energy of atomic hydrogen.C.the energy gained by an electron when it moves through a potential difference of 1.0 V.D.the energy transfer when 1.0 C of charge moves through a potential difference of 1.0 V.(1)5.The definition of the unit of c

7、urrent,the ampere,is based on A.the force per unit length on a conductor in a uniform magnetic field.B.the force per unit length on parallel current-carrying conductors.C.the charge per unit time delivered by a cell of e.m.f.1.0 V.D.the charge passing a point per unit time in an electrical circuit.(

8、1)+protonalpha particlemid-pointYX2KE6.1314 6.In the circuit below,n charge carriers pass the point P in a time t.Each charge carrier has charge q.The current in the circuit is given by the expression A.B.C.D.nqt.(1)7.The graph below shows the current/voltage characteristics of a filament lamp.The r

9、esistance of the filament at 4.0 V is A.250.B.4 000.C.8 000.D.64 000.(1)Ptqtnqnqt2416800246810Voltage/VCurrent/?10 A5 8.The variation with potential difference V of the current I in an electric lamp is shown below.At point P,the current is Ip,the potential difference is Vp and the gradient of the ta

10、ngent to the curve is G.What is the resistance of the lamp at point P?A.B.G C.D.(1)9.The graph below shows the variation with voltage V of the current I in three resistors X,Y and Z.Which of the following corresponds to resistors for which the resistance increases with increasing current?A.X only B.

11、Z only C.X and Z D.Y and Z(1)IIpp00VVPG1ppVIppIVI0V0XYZ6 10.A heater has a resistance R when the potential difference across it is 12 V.In the circuit below,it is connected in series with a 36 V supply and a resistor S.To ensure that the potential difference across the heater is 12 V,the resistance

12、of the resistor S should be A.B.C.D.2R.(1)11.This question is about electrical components.(a)In the space below,draw a circuit diagram that could be used to determine the current-voltage(I-V)characteristics of an electrical component X.(2)36 VS2R32R23Rcomponent X7 The graph below shows the I-V chara

13、cteristics for the component X.The component X is now connected across the terminals of a battery of e.m.f.6.0 V and negligible internal resistance.(b)Use the graph to determine(i)the current in component X.(1)(ii)the resistance of component X.(2)6420?02468V/VI/A8 A resistor R of constant resistance

14、 2.0 is now connected in series with component X as shown below.(c)(i)On the graph above,draw the I-V characteristics for the resistor R.(2)(ii)Determine the total potential difference E that must be applied across component X and across resistor R such that the current through X and R is 3.0 A.(2)(

15、Total 9 marks)12.This question is about a filament lamp.(a)On the axes below,draw a sketch-graph to show the variation with potential difference V of the current I in a typical filament lamp(the IV characteristic).(Note:this is a sketch-graph;you do not need to add any values to the axes).(1)(b)(i)E

16、xplain how the resistance of the filament is determined from the graph.(1)(ii)Explain whether the graph you have sketched indicates ohmic behaviour or non-ohmic behaviour.EXR2.0VI009 .(1)A filament lamp operates at maximum brightness when connected to a 6.0 V supply.At maximum brightness,the current

17、 in the filament is 120 mA.(c)(i)Calculate the resistance of the filament when it is operating at maximum brightness.(1)(ii)You have available a 24 V supply and a collection of resistors of a suitable power rating and with different values of resistance.Calculate the resistance of the resistor that

18、is required to be connected in series with the supply such that the voltage across the filament lamp will be 6.0 V.(2)(Total 6 marks)13.The element of an electric heater has a resistance R when in operation.What is the resistance of a second heater that has a power output three times as large at the

19、 same operating voltage?A.B.C.3R D.9R(1)9R3R10 5.2:Electric current 1.The current in the circuit shown below is constant when the switch is closed.The energy transfer in the internal resistance r of the battery is 15 J when a charge of 40 C passes through it.For the same amount of charge,45 J of ene

20、rgy is transferred in the resistor R.Which of the following gives the e.m.f.of the battery?A.B.C.D.(1)2.A battery is connected in series with a resistor R.The battery transfers 2 000 C of charge completely round the circuit.During this process,2 500 J of energy is dissipated in the resistor R and 1

21、500 J is expended in the battery.The e.m.f.of the battery is A.2.00 V.B.1.25 V.C.0.75 V.D.0.50 V.(1)RrV4015V4030V4045V406011 3.A cell of e.m.f.E and internal resistance r is connected to a variable resistor.A voltmeter is connected so as to measure the potential difference across the terminals of th

22、e cell.Which one of the following is the correct circuit diagram of the arrangement?(1)4.This question is about electrical energy and associated phenomena.A cell of electromotive force(e.m.f.)E and internal resistance r is connected in series with a resistor R,as shown below.The cell supplies 8.1 10

23、3 J of energy when 5.8 103 C of charge moves completely round the circuit.The current in the circuit is constant.A.C.B.D.EEEErrrrVVVVE12 (i)Calculate the e.m.f.E of the cell.(2)(ii)The resistor R has resistance 6.0.The potential difference between its terminals is 1.2 V.Determine the internal resist

24、ance r of the cell.(3)(iii)Calculate the total energy transfer in the resistor R.(2)(iv)Describe,in terms of a simple model of electrical conduction,the mechanism by which the energy transfer in the resistor R takes place.(5)(Total 12 marks)13 5.This question is about e.m.f.and internal resistance.A

25、 dry cell has an e.m.f.E and internal resistance r and is connected to an external circuit.There is a current I in the circuit when the potential difference across the terminals of the cell is V.(a)State expressions,in terms of E,V,r and I where appropriate,for(i)the total power supplied by the cell

26、.(1)(ii)the power dissipated in the cell.(1)(iii)the power dissipated in the external circuit.(1)(b)Use your answers to(a)to derive a relationship between V,E,I and r.(2)VrEI14 The graph below shows the variation of V with I for the dry cell.1.61.51.41.31.21.11.00.900.800.700.600.500.400.300.200.100

27、.0V/V0.00.100.200.300.400.500.600.700.800.901.01.11.21.3I/A15 (c)Complete the diagram below to show the circuit that could be used to obtain the data from which the graph was plotted.(3)(d)Use the graph,explaining your answers,to(i)determine the e.m.f.E of the cell.(2)(ii)determine the current in th

28、e external circuit when the resistance R of the external circuit is very small.(2)(iii)deduce that the internal resistance r of the cell is about 1.2.(3)16 (e)The maximum power dissipated in the external circuit occurs when the resistance of the external circuit has the same value as the internal re

29、sistance of the cell.Calculate the maximum power dissipation in the external circuit.(3)(Total 18 marks)6.This question compares the electrical properties of two 12 V filament lamps.A lamp is designed to operate at normal brightness with a potential difference of 12 V across its filament.The current

30、 in the filament is 0.50 A.(a)For the lamp at normal brightness,calculate(i)the power dissipated in the filament.(1)(ii)the resistance of the filament.(1)In order to measure the voltage-current(V-I)characteristics of a lamp,a student sets up the following electrical circuit.12 V battery17 (b)On the

31、circuit above,add circuit symbols showing the correct positions of an ideal ammeter and an ideal voltmeter that would allow the V-I characteristics of this lamp to be measured.(2)The voltmeter and the ammeter are connected correctly in the circuit opposite.(c)Explain why the potential difference acr

32、oss the lamp(i)cannot be increased to 12 V.(2)(ii)cannot be reduced to zero.(2)An alternative circuit for measuring the V-I characteristic uses a potential divider.(d)(i)Draw a circuit that uses a potential divider to enable the V-I characteristics of the filament to be found.(3)18 (ii)Explain why t

33、his circuit enables the potential difference across the lamp to be reduced to zero volts.(2)The graph below shows the V-I characteristic for two 12 V filament lamps A and B.(e)(i)Explain why these lamps do not obey Ohms law.(2)(ii)State and explain which lamp has the greater power dissipation for a

34、potential difference of 12 V.(3)12000.51.0current/APotentialdifference/Vlamp Alamp B19 The two lamps are now connected in series with a 12 V battery as shown below.(f)(i)State how the current in lamp A compares with that in lamp B.(1)(ii)Use the V-I characteristics of the lamps to deduce the total c

35、urrent from the battery.(4)(iii)Compare the power dissipated by the two lamps.(2)(Total 25 marks)lamp Alamp B12 V battery20 7.The diagrams below show combinations X,Y and Z of three resistors,each resistor having the same resistance.Which one of the following shows the resistances of the combination

36、s in increasing order of magnitude?lowest highest A.Y X Z B.Z X Y C.X Y Z D.Z Y X(1)8.In the circuits below,the cells each have the same e.m.f.and zero internal resistance.All the resistors have the same resistance.Which of the following gives the current through the cells in order of increasing mag

37、nitude?Lowest current Highest current A.X Y Z B.Z X Y C.Y Z X D.Y X Z(1)combination Xcombination Zcombination YCircuit XCircuit YCircuit Z21 9.In the circuit below,which meter is not correctly connected?A.1 B.2 C.3 D.4(1)10.Which one of the following shows a correct circuit,using ideal voltmeters an

38、d ammeters,for measuring the I-V characteristic of a filament lamp?(1)AVAV1234AAAAA.C.B.D.VVVV22 11.A battery of e.m.f.E and negligible internal resistance is connected to three resistors,each of resistance R,a voltmeter and a switch,as shown below.The voltmeter has infinite resistance.What are the

39、readings on the voltmeter when the switch is open and when it is closed?Switch open Switch closed A.0 less than E B.0 E C.E less than E D.E E(1)ERRRV23 12.In the circuit shown,the voltmeter has a resistance of 20 k and the battery has an e.m.f.of 6.0 V and negligible internal resistance.The reading

40、on the voltmeter is A.2.0 V.B.3.0 V.C.4.0 V.D.6.0 V.(1)13.This question is about electric circuits.Susan sets up the circuit below in order to measure the current-voltage(I-V)characteristic of a small filament lamp.The supply is a battery that has an e.m.f.of 3.0 V and the ammeter and voltmeter are

41、considered to be ideal.The lamp is labelled by the manufacturer as“3 Volts,0.6 Watts”.V6.0 V10 k20 k20 kAV3.0 VS24 (a)(i)Explain what information this labelling provides about the normal operation of the lamp.(2)(ii)Calculate the current in the filament of the lamp when it is operating at normal bri

42、ghtness.(2)Susan sets the variable resistor to its maximum value of resistance.She then closes the switch S and records the following readings.Ammeter reading=0.18 A Voltmeter reading=0.60 V She then sets the variable resistor to its zero value of resistance and records the following readings.Ammete

43、r reading=0.20 A Voltmeter reading=2.6 V(b)(i)Explain why,by changing the value of the resistance of the variable resistance,the potential difference across the lamp cannot be reduced to zero or be increased to 3.0 V.(2)(ii)Determine the internal resistance of the battery.(3)25 (c)Calculate the resi

44、stance of the filament when the reading on the voltmeter is(i)0.60 V.(1)(ii)2.6 V.(1)(d)Explain why there is a difference between your answers to(c)(i)and(c)(ii).(2)(e)Using the axes below,draw a sketch-graph of the I-V characteristic of the filament of the lamp.(Note:this is a sketch-graph;you do n

45、ot need to add any values to the axis.)(1)IV0026 The diagram below shows an alternative circuit for varying the potential difference across the lamp.The potential divider XZ has a potential of 3.0 V across it.When the contact is at the position Y,the resistance of XY equals the resistance of YZ whic

46、h equals 12.The resistance of the lamp is 4.(f)Calculate the potential difference across the lamp.(4)(Total 18 marks)3.0 VXYZ27 Answers 5.1:Electric potential energy,current and resistance 1.C 1 2.D 1 3.C 1 4.C 1 5.B 1 6.B 1 7.A 1 8.D 1 9.A 1 10.D 1 11.(a)component X,battery,ammeter all in series an

47、d including means of varying current;with voltmeter in parallel across component X;2(b)(i)4.0 A;1(ii)use of R=,and not gradient of graph;resistance=1.5 ;2(c)(i)straight-line through origin,quadrants 1 or 3 or both;correct gradient,i.e.passes through V=4.0 V,I=2.0 A;2(ii)p.d.s across X and across R w

48、ill be 3.7 V(0.1V)and 6.0 V;Award 0 if only one p.d.is correct.total p.d.=9.7 V;2 9 12.(a)IV28 1 Any reasonable curve in the right direction.(b)(i)from the value of V/I at any point on the curve;1 Do not accept just“from V/I”.(ii)non-ohmic because the resistance(V/I at each point)is not constant/OWT

49、TE;1(c)(i)50;1(ii)recognize that the voltage must divide in the ratio 3:1;to give R=150;2#Or answer could be solved via the current.6 1.B 1 5.2:Electric current 1.D 1 2.A 1 3.B 1 4.(i)use of e.m.f.=energy/charge;=1.4 V;2 Award 0 for formula E=seen or implied even if answer is numerically correct.VI0

50、0)108.5()101.8(33QF29 (ii)p.d.across internal resistance=0.2 V;OR current=0.2 A;resistance r=6.0;total resistance=7.0;=1.0;internal resistance=7 6=1.0;3 Accept any other valid route.(iii)idea of use of ratio of resistances;energy transfer=6/7 8.1 103 =6.9(4)103 J;2 Accept any other valid route.(iv)c