Topic 5 Electricity 31.1.2018 [140 marks] · Topic 5 Electricity 31.1.2018 [140 marks] 1. The graph...

Topic 5 Electricity 31.1.2018 [140 marks]

1. [1 mark]The graph shows the variation of current with potential difference for a filament lamp.

What is the resistance of the filament when the potential difference across it is 6.0 V?A. 0.5 mΩB. 1.5 mΩC. 670 ΩD. 2000 Ω

MarkschemeC

Examiners report[N/A]

2. [1 mark]An electron is accelerated through a potential difference of 2.5 MV. What is the change in kinetic energy of the electron?

A. 0.4µJ

B. 0.4 nJ

C. 0.4 pJ

D. 0.4 fJ

MarkschemeC


3. [1 mark]A cell is connected in series with a resistor and supplies a current of 4.0 A for a time of 500 s. During this time, 1.5 kJ of energy isdissipated in the cell and 2.5 kJ of energy is dissipated in the resistor.

What is the emf of the cell?

A. 0.50 V

B. 0.75 V

C. 1.5 V

D. 2.0 V

MarkschemeD


4. [1 mark]An electron travelling at speed v perpendicular to a magnetic field of strength B experiences a force F.

What is the force acting on an alpha particle travelling at 2v parallel to a magnetic field of strength 2 B?

A. 0

B. 2F

C. 4F

D. 8F

MarkschemeA


5. [1 mark]The diagram shows two equal and opposite charges that are fixed in place.

At which points is the net electric field directed to the right?

A. X and Y only

B. Z and Y only

C. X and Z only

D. X, Y and Z

MarkschemeC


6. [1 mark]A wire has variable cross-sectional area. The cross-sectional area at Y is double that at X.

At X, the current in the wire is I and the electron drift speed is v. What is the current and the electron drift speed at Y?

MarkschemeB


7. [1 mark]A circuit contains a cell of electromotive force (emf) 9.0 V and internal resistance 1.0 Ω together with a resistor of resistance 4.0 Ωas shown. The ammeter is ideal. XY is a connecting wire.

What is the reading of the ammeter?

A. 0 A

B. 1.8 A

C. 9.0 A

D. 11 A

MarkschemeC


8. [1 mark]A positively-charged particle moves parallel to a wire that carries a current upwards.

What is the direction of the magnetic force on the particle?

A. To the left

B. To the right

C. Into the page

D. Out of the page

MarkschemeA


9. [1 mark]Electrons, each with a charge e, move with speed v along a metal wire. The electric current in the wire is I.

Plane P is perpendicular to the wire. How many electrons pass through plane P in each second?

A.

B.

C.

D.

MarkschemeD


e

I

ve

I

I

ve

I

e

10. [1 mark]Positive charge is uniformly distributed on a semi-circular plastic rod. What is the direction of the electric field strength at point S?

MarkschemeB


11. [1 mark]The diagram shows the path of a particle in a region of uniform magnetic field. The field is directed into the plane of the page.

This particle could be

A. an alpha particle.

B. a beta particle.

C. a photon.

D. a neutron.

MarkschemeA


12. [1 mark]A –5µC charge and a +10µC charge are a fixed distance apart.

Where can the electric field be zero?

A. position I only B. position II only C. position III only D. positions I, II and III

MarkschemeC


13. [1 mark]An electrical circuit is shown with loop X and junction Y.

What is the correct expression of Kirchhoff’s circuit laws for loop X and junction Y?

MarkschemeA


14. [1 mark]A cell of emf 4V and negligible internal resistance is connected to three resistors as shown. Two resistors of resistance 2Ω areconnected in parallel and are in series with a resistor of resistance 1Ω.

What power is dissipated in one of the 2Ω resistors and in the whole circuit?

MarkschemeD


15. [1 mark]A wire carrying a current is at right angles to a uniform magnetic field of strength B. A magnetic force F is exerted on the wire.Which force acts when the same wire is placed at right angles to a uniform magnetic field of strength 2B when the current is ?

A.

B.

C. F

D. 2F

MarkschemeB


II

4

F

4

F

2

16. [1 mark]A 12V battery has an internal resistance of 2.0Ω. A load of variable resistance is connected across the battery and adjusted to haveresistance equal to that of the internal resistance of the battery. Which statement is correct for this circuit?

A. The current in the battery is 6A. B. The potential difference across the load is 12V. C. The power dissipated in the battery is 18W. D. The resistance in the circuit is 1.0Ω.

MarkschemeC


17. [1 mark]Three fixed charges, +Q, –Q and –2Q, are at the vertices of an equilateral triangle. What is the resultant force on an electron at thecentre of the triangle?

MarkschemeB


18. [1 mark]The graph shows the variation of current I in a device with potential difference V across it.

What is the resistance of the device at P?

A. zero

B. 0.1Ω

C. 10Ω

D. infinite

MarkschemeC


19. [1 mark]A circuit consists of a cell of electromotive force (emf) 6.0V and negligible internal resistance connected to two resistors of 4.0Ω.

The resistance of the ammeter is 1.0 Ω. What is the reading of the ammeter?

A. 2.0A

B. 3.0A

C. 4.5A

D. 6.0A

MarkschemeA


20. [1 mark]A wire carrying a current I is placed in a region of uniform magnetic field B, as shown in the diagram.

The direction of the field B is out of the page and the length of the wire is L. What is correct about the direction and magnitude of the forceacting on the wire?

MarkschemeA


21. [1 mark]A circuit consists of a cell of electromotive force (emf) 6.0V and negligible internal resistance connected to two resistors of 4.0Ω.

The ammeter has resistance equal to 1.0Ω and the voltmeter is ideal. What are the readings of the ammeter and the voltmeter?

MarkschemeC


22. [1 mark]A cylindrical resistor of length is made from a metal of mass . It has a resistance .

Two resistors, each of length and mass , are then created from the same volume of the metal.

What is the resistance of the two resistors when connected in parallel?

A.

B.

C.

D.

MarkschemeC

l m R

2l m

2

R

2R

4R

8R

Examiners reportThink proportionality.

There were a few G2 comments suggesting that this question was too complex and took too much time, but this is only the case ifcandidates reach for equations before considering proportionality.

A simple sketch will show that if the new resistors are placed side by side (ie in parallel) then the new length is twice the previous length(leading to a doubling of the resistance) and half its cross-sectional area (leading to a further doubling of the resistance). So the correctresponse is C.

23. [1 mark]Three resistors of resistance are connected in parallel across a cell of electromotive force (emf) that has a negligible internalresistance. What is the rate at which the cell supplies energy?

A.

B.

C.

D.

MarkschemeD


R V

V 2

3R

V 2

9R

9V 2

R

3V 2

R

24. [1 mark]A charge and a charge are a distance apart. P is a distance from the charge on the straight line joining the charges.

What is the magnitude of the electric field strength at P?

A.

B.

C.

D.

MarkschemeB

Examiners reportAlthough it is convenient to use k in the equation for coulombic attraction, candidates need to be able to unpack this as .

+3 C−4 C x x

+3 C

1πε0x2

12πε0x2

14πε0x2

17πε0x2

14πε0

25. [1 mark]An electron is moving parallel to a straight current-carrying wire. The direction of conventional current in the wire and the direction ofmotion of the electron are the same. In which direction is the magnetic force on the electron?

MarkschemeD

Examiners reportThe conventional current is opposite in direction to the electron flow. So here we have essentially anti-parallel currents and thecandidates should know that such currents keep well away from each other, ie they repel, leaving D as the correct response.

26. [1 mark]A filament lamp and a semiconducting diode have the voltage–current (–) characteristics shown and are connected in parallel.

What is the resistance of the lamp and the resistance of the diode when the current in each device is 2.0 A?

MarkschemeD


V

I

27. [1 mark]Four resistors are connected as shown.

What is the total resistance between X and Y?

A. 3 Ω

B. 4 Ω

C. 6 Ω

D. 24 Ω

MarkschemeA


28. [1 mark]What is the definition of electric current?

A. The ratio of potential difference across a component to the resistance of the component

B. The power delivered by a battery per unit potential difference

C. The rate of flow of electric charge

D. The energy per unit charge dissipated in a power supply

MarkschemeC


29. [1 mark]The diagram shows a circuit used to investigate internal resistance of a cell.

The variable resistor R is adjusted and the values of potential difference V across the cell and current I are recorded. Which graph shows thevariation of V with I?

MarkschemeB


30a. [1 mark]

A heater in an electric shower has a power of 8.5 kW when connected to a 240 V electrical supply. It is connected to the electricalsupply by a copper cable.

The following data are available:

Length of cable = 10 mCross-sectional area of cable = 6.0 mmResistivity of copper = 1.7 × 10 Ω m

Calculate the current in the copper cable.

MarkschemeI «= » =35«A»


2

–8

8.5×103

240

30b. [2 marks]Calculate the resistance of the cable.

MarkschemeR =

= 0.028 «Ω»

Allow missed powers of 10 for MP1.


1.7×10−8×106.0×10−6

30c. [3 marks]Explain, in terms of electrons, what happens to the resistance of the cable as the temperature of the cable increases.

Markscheme«as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions

OR

increased collisions of electrons

drift velocity decreases «so current decreases»

«as V constant so» R increases

Award [0] for suggestions that the speed of electrons increases so resistance decreases.


30d. [4 marks]The heater changes the temperature of the water by 35 K. The specific heat capacity of water is 4200 J kg K .

Determine the rate at which water flows through the shower. State an appropriate unit for your answer.

–1 –1

Markschemerecognition that power = flow rate × cΔT

flow rate « »

= 0.058 «kg s »

kg s / g s / l s / ml s / m s

Allow MP4 if a bald flow rate unit is stated. Do not allow imperial units.


= powercΔT

= 8.5×103

4200×35

–1

−1 −1 −1 −1 3 −1

31a. [2 marks]State the quark structures of a meson and a baryon.

MarkschemeMeson: quark-antiquark pairBaryon: 3 quarks


31b. [2 marks]

A possible decay of a lambda particle ( ) is shown by the Feynman diagram.

Explain which interaction is responsible for this decay.

MarkschemeAlternative 1

strange quark changes «flavour» to an up quark

changes in quarks/strangeness happen only by the weak interaction

Alternative 2

Strangeness is not conserved in this decay «because the strange quark changes to an up quark»

Strangeness is not conserved during the weak interaction

Do not allow a bald answer of weak interaction.


Λ0

[1 mark]31c. Draw arrow heads on the lines representing and d in the .u π−

Markschemearrows drawn in the direction shown

Both needed for [1] mark.


31d. [1 mark]Identify the exchange particle in this decay.

MarkschemeW

Do not allow W or W .


−

+

31e. [1 mark]Outline one benefit of international cooperation in the construction or use of high-energy particle accelerators.

Markschemeit lowers the cost to individual nations, as the costs are shared

international co-operation leads to international understanding OR historical example of co-operation OR co-operation always allowsscience to proceed

large quantities of data are produced that are more than one institution/research group can handle co-operation allows effective analysis

Any one.


32a. [3 marks]

A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires areprotected by an insulator.

The copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is asignificant electric current only in the copper wires.

Markschemewhen an electric field is applied to any material «using a cell etc» it acts to accelerate any free electrons

electrons are the charge carriers «in copper»

Accept “free/valence/delocalised electrons”.

metals/copper have many free electrons whereas insulators have few/no free electrons/charge carriers


–8

32b. [2 marks]

The cable consists of 32 copper wires each of length 35 km. Each wire has a resistance of 64 Ω. The resistivity of copper is 1.7 x 10 Ωm.

Calculate the radius of each wire.

Markschemearea = «= 9.3 x 10 m »


–8

1.7×10−3×35×103

64–6 2

32c. [2 marks]There is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W.

Markscheme«resistance of cable = 2Ω»

power dissipated in cable = 730 x 2 «= 1.07 MW»

power loss per meter or 30.6 «W m »

Allow [2] for a solution where the resistance per unit metre is calculated using resistivity and answer to (b)(i) (resistance per unit lengthof cable =5.7 x 10 m)


2

= 1.07×10−6

35×103–1

–5

33. [2 marks]Calculate the power dissipated in the cable.

Markschemepower = «35 × 0.028» = 34 «W»

Allow 35 – 36 W if unrounded figures for R or I are used.Allow ECF from (a)(i) and (a)(ii).


2

34a. [2 marks]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) generator to an electricalload. The copper wires are protected by an insulator.

The cable consists of 32 copper wires each of length 35 km. Each wire has a resistance of 64 Ω. The cable is connected to the acgenerator which has an output power of 110 MW when the peak potential difference is 150 kV. The resistivity of copper is 1.7 x 10 Ωm.

output power = 110 MW

Calculate the radius of each wire.

Markschemearea = «= 9.3 x 10 m »

radius = « » 0.00172 m


–8

1.7×10−3×35×103

64–6 2

√ =9.3×10−6

π

34b. [1 mark]Calculate the peak current in the cable.

MarkschemeI « » = 730 « A »


peak =Ppeak

Vpeak

34c. [3 marks]Determine the power dissipated in the cable per unit length.

Markschemeresistance of cable identified as « » 2 Ω

seen in solution

plausible answer calculated using «plausible if in range 10 W m to 150 W m when quoted answers in (b)(ii) used» 31 «W m »

Allow [3] for a solution where the resistance per unit metre is calculated using resistivity and answer to (a) (resistance per unit length ofcable = 5.7 x 10 m )

Award [2 max] if 64 Ω used for resistance (answer x32).

An approach from

or VI using 150 kV is incorrect (award [0]), however allow this approach if the pd across the cable has been calculated (pd dropped

across cable is 1.47 kV).


=6432

a power35000

2I2

35000–1 –1 –1

–5

V 2

R

34d. [1 mark]

To ensure that the power supply cannot be interrupted, two identical cables are connected in parallel.

Calculate the root mean square (rms) current in each cable.

Markscheme« » = 260 «A»


response to (b)(ii)

2√2

34e. [2 marks]The two cables in part (c) are suspended a constant distance apart. Explain how the magnetic forces acting between the cablesvary during the course of one cycle of the alternating current (ac).

Markschemewires/cable attract whenever current is in same direction

charge flow/current direction in both wires is always same «but reverses every half cycle»

force varies from 0 to maximum

force is a maximum twice in each cycle

Award [1 max] if response suggests that there is repulsion between cables at any stage in cycle.


34f. [2 marks]

The energy output of the ac generator is at a much lower voltage than the 150 kV used for transmission. A step-up transformer is usedbetween the generator and the cables.

Suggest the advantage of using a step-up transformer in this way.

Markschemehigher voltage gives lower current

«energy losses depend on current» hence thermal/heating/power losses reduced


34g. [1 mark]The use of alternating current (ac) in a transformer gives rise to energy losses. State how eddy current loss is minimized in thetransformer.

Markschemelaminated core

Do not allow “wires are laminated”.


35a. [3 marks]

The graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T.

(i) State how the resistance of T varies with the current going through T.

(ii) Deduce, without a numerical calculation, whether R or T has the greater resistance at I=0.40 A.

Markschemei

R decreases with increasing I

OR

R and I are negatively correlated

Must see reference to direction of change of current in first alternative.Do not allow “inverse proportionality”. May be worth noting any marks on graph relating to 7bii

ii

at 0.4 A: V > V or V = 5.6 V and V = 5.3 V

Award [0] for a bald correct answer without deduction or with incorrect reasoning.

Ignore any references to graph gradients.

so R >R because V = IR / V∝ R «and I same for both»

Both elements must be present for MP2 to be awarded.


T

T

R T R T

R T

35b. [3 marks]Components R and T are placed in a circuit. Both meters are ideal.

Slider Z of the potentiometer is moved from Y to X.

(i) State what happens to the magnitude of the current in the ammeter.

(ii) Estimate, with an explanation, the voltmeter reading when the ammeter reads 0.20 A.

Markschemei

decreasesORbecomes zero at X

ii

realization that V is the same for R and TORidentifies that currents are 0.14 A and 0.06 A

Award [0] if pds 2.8 V and 3.7 V or 1.4 V and 2.6V are used in any way. Otherwise award [1 max] for a bald correct answer. Explanationexpected.

2 V = 2 V OR 2.0 V


–

[1 mark]36a.

A beam of electrons e enters a uniform electric field between parallel conducting plates RS. RS are connected to a direct current (dc)power supply. A uniform magnetic field B is directed into the plane of the page and is perpendicular to the direction of motion of theelectrons.

The magnetic field is adjusted until the electron beam is undeflected as shown.

Identify, on the diagram, the direction of the electric field between the plates.

Markschemedirection indicated downwards, perpendicular to plates

Arrows must be between plates but allow edge effects if shown. Only one arrow is required.


–

36b. [2 marks]The following data are available.

Separation of the plates RS = 4.0 cm

Potential difference between the plates = 2.2 kV

Velocity of the electrons = 5.0×10 m s

Determine the strength of the magnetic field B.

5 –1

Markscheme «Vm »

B = « » 0.11 «T»

ECF applies from MP1 to MP2 due to math error.

Award [2] for a bald correct answer.


E = = 55 000V

d–1

=55 0005×105

36c. [2 marks]The velocity of the electrons is now increased. Explain the effect that this will have on the path of the electron beam.

MarkschemeALTERNATIVE 1

magnetic force increasesORmagnetic force becomes greater than electric force

electron beam deflects “downwards” / towards SOR path of beam is downwards

ALTERNATIVE 2

when v increases, the B required to maintain horizontal path decreases«but B is constant» so path of beam is downwards

Do not apply an ecf from (a).

Award [1 max] if answer states that magnetic force decreases and therefore path is upwards.

Ignore any statement about shape of path

Do not allow “path deviates in direction of magnetic force” without qualification.


37a. [4 marks]

A company designs a spring system for loading ice blocks onto a truck. The ice block is placed in a holder H in front of the spring andan electric motor compresses the spring by pushing H to the left. When the spring is released the ice block is accelerated towards aramp ABC. When the spring is fully decompressed, the ice block loses contact with the spring at A. The mass of the ice block is 55 kg.

Assume that the surface of the ramp is frictionless and that the masses of the spring and the holder are negligible compared to the massof the ice block.

(i) The block arrives at C with a speed of 0.90ms . Show that the elastic energy stored in the spring is 670J.

(ii) Calculate the speed of the block at A.

Markscheme(i)

OR«E =»E +E

OR669 J «E = 669 ≈ 670J»

Award [1 max] for use of g=10Nkg , gives 682 J.

(ii)

If 682J used, answer is 5.0ms .


−1

≪ Eel= ≫ mv2 + mgh12

el P K

≪ Eel= ≫ × 55 × 0.902+55 × 9.8 × 1.212

el

–1

× 55 × v2 = 670J12

v =≪ √ = ≫ 4.9ms−12×67055

–1

37b. [3 marks]Describe the motion of the block

(i) from A to B with reference to Newton's first law.

(ii) from B to C with reference to Newton's second law.

Markscheme(i)no force/friction on the block, hence constant motion/velocity/speed

(ii)force acts on block OR gravity/component of weight pulls down slope

velocity/speed decreases OR it is slowing down OR it decelerates

Do not allow a bald statement of “N2” or “F = ma” for MP1.Treat references to energy as neutral.


37c. [2 marks]On the axes, sketch a graph to show how the displacement of the block varies with time from A to C. (You do not have to putnumbers on the axes.)

Markschemestraight line through origin for at least one-third of the total length of time axis covered by candidate line

followed by curve with decreasing positive gradient

Ignore any attempt to include motion before A.

Gradient of curve must always be less than that of straight line.


37d. [2 marks]The spring decompression takes 0.42s. Determine the average force that the spring exerts on the block.

Markscheme

F=642≈640N

Allow ECF from (a)(ii).


F ≪= ≫=Δp

Δt

55×4.90.42

37e. [2 marks]The electric motor is connected to a source of potential difference 120V and draws a current of 6.8A. The motor takes 1.5s tocompress the spring.

Estimate the efficiency of the motor.

Markscheme«energy supplied by motor =» 120 × 6.8 × 1.5 or 1224 J OR«power supplied by motor =» 120 × 6.8 or 816 We = 0.55 or 0.547 or 55% or 54.7%

Allow ECF from earlier results.


38a. [1 mark]

In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.

State what is meant by an ideal voltmeter.

Markschemeinfinite resistance OR draws no current from circuit/component OR has no effect on the circuit

Do not allow “very high resistance”.


38b. [3 marks]The student adjusts the variable resistor and takes readings from the ammeter and voltmeter. The graph shows the variation of thevoltmeter reading V with the ammeter reading I.

Use the graph to determine

(i) the electromotive force (emf) of the cell.

(ii) the internal resistance of the cell.

Markscheme(i)«vertical intercept = emf» = 8.8 − 9.2 V

(ii)attempt to evaluate gradient of graph=0.80Ω

Accept other methods leading to correct answer, eg using individual data points from graph.

Allow a range of 0.78 – 0.82 Ω.

If ε = I(R + r) is used then the origin of the value for R must be clear.


38c. [1 mark]A connecting wire in the circuit has a radius of 1.2mm and the current in it is 3.5A. The number of electrons per unit volume of thewire is 2.4×10 m . Show that the drift speed of the electrons in the wire is 2.0×10 ms .

Markscheme3.5=2.4×10 ×π(1.2×10 ) ×1.6×10 ×v«⇒v=2.0×10 ms »


28 −3 −4 −1

28 −3 2 −19 −4 −1

38d. [2 marks]The diagram shows a cross-sectional view of the connecting wire in (c).

The wire which carries a current of 3.5A into the page, is placed in a region of uniform magnetic field of flux density 0.25T. The field isdirected at right angles to the wire.

Determine the magnitude and direction of the magnetic force on one of the charge carriers in the wire.

MarkschemeF = «qvB =1.6×10 ×2.0×10 ×0.25 =»8.1×10 N

directed down OR south


–19 –4 –24

39a. [2 marks]Two cells of negligible internal resistance are connected in a circuit.

The top cell has electromotive force (emf) 12V. The emf of the lower cell is unknown. The ideal ammeter reads zero current.

Calculate the emf E of the lower cell.

MarkschemeALTERNATIVE 1correct application of Kirchhoff to at least one loop E=«4.0×2.0=»8.0V

FOR EXAMPLE12 = 2.0I + 4.0I for top loop with loop anticlockwise «but I = I as I = 0»«E =» 8.0 V

ALTERNATIVE 2«recognition that situation is simple potential divider arrangement»pd across 4Ω resistor =

=8V

Award [0] for any answer that begins with the treatment as parallel resistors.


1 2

2 1 3

12×4(2+4)

39b. [3 marks]The diagram shows charge carriers moving with speed v in a metallic conductor of width L. The conductor is exposed to a uniformmagnetic field B that is directed into the page.

(i) Show that the potential difference V that is established across the conductor is given by V=vBL.

(ii) On the diagram, label the part of the conductor where negative charge accumulates.

Markscheme(i)ALTERNATIVE 1equating electric to magnetic force qE=qvB

substituting

«to get given result»

ALTERNATIVE 2

AND work done = force × distance work done = qv=Bqv×L «to get given result»

(ii)some mark indicating lower surface of conductorORindication that positive charge accumulates at top of conductor

Do not allow negative or positive at top and bottom.


E = V

L

V = workdoneQ

40a. [4 marks]

This question is in two parts. Part 1 is about kinematics and Newton’s laws of motion.

Part 2 is about electrical circuits.

Part 1 Kinematics and Newton’s laws of motion

Cars I and B are on a straight race track. I is moving at a constant speed of and B is initially at rest. As I passes B, B starts tomove with an acceleration of .

At a later time B passes I. You may assume that both cars are point particles.

Show that the time taken for B to pass I is approximately 28 s.

Markschemedistances itemized; (it must be clear through use of

or distance I etc)

distances equated;

/ cancel and re-arrange;

substitution shown / 28.1(s) seen;

or

clear written statement that the average speed of B must be the same as constant speed of I;

as B starts from rest the final speed must be ;

so ;

28.1 (s) seen; (for this alternative the method must be clearly described)

or

attempts to compare distance travelled by I and B for 28 s;

I distance ;

B distance ;

deduces that overtake must occur about later;

45 ms−1

3.2 ms−2

sI

t = 2v

a

( )2×453.2

2 × 45

t = =Δv

a903.2

= (45 × 28 =) 1260 (m)

= ( × 3.2 × 282 =) 1255 (m)12

( =) 0.1 s545


40b. [2 marks]Calculate the distance travelled by B in this time.

Markschemeuse of appropriate equation of motion;

1.3 (km);



(1.26 ≈)

40c. [3 marks]B slows down while I remains at a constant speed. The driver in each car wears a seat belt. Using Newton’s laws of motion,explain the difference in the tension in the seat belts of the two cars.

Markschemedriver I moves at constant speed so no net (extra) force according to Newton 1;

driver B decelerating so (extra) force (to rear of car) (according to Newton 1) / momentum/inertia change so (extra) force must bepresent;

(hence) greater tension in belt B than belt I;

Award [0] for stating that tension is less in the decelerating car (B).


40d. [2 marks]

A third car O with mass 930 kg joins the race. O collides with I from behind, moving along the same straight line as I. Before the collisionthe speed of I is and its mass is 850 kg. After the collision, I and O stick together and move in a straight line with an initialcombined speed of .

Calculate the speed of O immediately before the collision.

Markscheme or statement that momentum is conserved;

;

Allow [2] for a bald correct answer.


45 ms−1

52 ms−1

930 × v + 850 × 45 = 1780 × 52

v = 58 (ms−1)

40e. [2 marks]The duration of the collision is 0.45 s. Determine the average force acting on O.

Markschemeuse of force (or any variant, eg: );

; } (must see matched units and value ie: 13 200 without unit gains MP2, 13.2 does not)


Allow use of 58 m s from (c)(i) to give 12 400 (N).


change of momentumtime

930×6.40.45

13.2 × 103 (N)

–1

40f. [2 marks]

This question is in two parts. Part 1 is about kinematics and Newton’s laws of motion.

Part 2 Electrical circuits

The circuit shown is used to investigate how the power developed by a cell varies when the load resistance changes.

The variable resistor is adjusted and a series of current and voltage readings are taken. The graph shows the variation with of thepower dissipated in the cell and the power dissipated in the variable resistor.

An ammeter and a voltmeter are used to investigate the characteristics of a variable resistor of resistance . State how theresistance of the ammeter and of the voltmeter compare to so that the readings of the instruments are reliable.

Markschemeammeter must have very low resistance/much smaller than ;

voltmeter must have very large resistance/much larger than ;

Allow [1 max] for zero and infinite resistance for ammeter and voltmeter respectively.

Allow [1 max] if superlative (eg: very/much/OWTTE) is missing.

R

R

R

R

R

R


40g. [3 marks]Show that the current in the circuit is approximately 0.70 A when .

Markschemepower (loss in resistor)

; } (accept answers in the range of 0.35 to 0.37 (W) – treat value outside this range as ECF (could still lead to 0.7))

;

or ; (allow answers in the range of 0.66 to 0.68 (A).


R = 0.80 Ω

= 0.36 (W)

I 2 × 0.80 = 0.36

I = 0.67 (A) √( )0.360.8

40h. [2 marks]

The cell has an internal resistance.

Outline what is meant by the internal resistance of a cell.

Markschemeresistance of the components/chemicals/materials within the cell itself; } (not “resistance of cell”)

leading to energy/power loss in the cell;


40i. [3 marks]Determine the internal resistance of the cell.

Markschemepower (in cell with 0.7 A) ; } (allow answers in the range of 0.57 W to 0.62 W)

;

; (allow answers in the range of 1.18 to 1.27 ( ))

or

when powers are equal;

;

so which occurs at 1.2(5) ( );

Award [1 max] for bald 1.2(5) ( ).


= 0.58 W

0.72 × r = 0.58

r = 1.2 (Ω) Ω

I 2R = I 2r

r = R Ω

Ω

40j. [2 marks]Calculate the electromotive force (emf) of the cell.

Markscheme;

1.4 (V);

Allow ECF from (e) or (f)(ii).

or

when , power loss ;

;


(E = I(R + r)) = 0.7(0.8 + 1.2)

R = 0 = 1.55

E = (√1.55 × 1.2 =) 1.4 (V)

41a. [3 marks]

Part 2 Power transmissions

The diagram shows the main features of an ideal transformer whose primary coil is connected to a source of alternating current (ac)voltage.

Outline, with reference to electromagnetic induction, how a voltage is induced across the secondary coil.

Markscheme(alternating) pd/voltage across primary coil leads to (alternating) current (in primary coil);

hence there is a changing/alternating magnetic field in primary;

leading to a changing magnetic flux linked to/appearing in secondary;

according to Faraday’s law, an alternating emf is induced in the secondary coil;

Examiners reportIn the second part of this question, candidates showed themselves to be much less confident with ideas of electromagnetic induction.Discussion of induction of the secondary emf in the transformer ought to be well rehearsed and confident from a candidate at this level.Instead, examiners were treated to incomplete and often non-physical descriptions. Standard and expected terminology was rare.Terms such as magnetic flux, linking, emf and induction were either omitted or misused. This is clearly an area where there is muchmisunderstanding.

41b. [2 marks]The primary coil has 25 turns and is connected to an alternating supply with an input voltage of root mean squared (rms) value 12V. The secondary coil has 80 turns and is not connected to an external circuit. Determine the peak voltage induced across the secondary

coil.

Markschemerms secondary voltage ;

peak voltage ; (allow ECF from MP1)


Examiners reportOn the other hand, candidates were well able to cope with the (straightforward) calculation of the induced emf across the secondarycoil. However, a frequent omission was the final conversion to a peak value.

= 38.4 (V)

= (38√2 =) 54 (V)

41c. [1 mark]

A different transformer is used to transmit power to a small town.

The transmission cables from the power station to the transformer have a total resistance of 4.0 Ω. The transformer is 90% efficient andsteps down the voltage to 120 V. At the time of maximum power demand the effective resistance of the town and of the cables from thetransformer to the town is 60 mΩ.

Calculate the current in the cables connected to the town

Markscheme; (30 A is a common and incorrect answer)

Examiners reportThis whole sequence of calculations was poorly done. Candidates appear never to have considered the problem of energy loss in thecabling between energy generator and consumer. This ought to be straightforward work for the candidates, but it proved not to be.Solutions as a whole were confused with little clear explanation of what was going on. Candidates were evaluating the wrong quantitieswithout realising it and then misapplying them later in the question. As an example, simply labelling a quantity as “Power =…” isunhelpful. What power is being referred to by the candidate? The argument must be clear in order that an examiner can award errorcarried forward.

(Is = ) = 2.0 k(A)12060×10−3

41d. [2 marks]Calculate the power supplied to the transformer.

Markschemepower (supplied to town) or ; (allow ECF from (f)(i))

power (supplied to transformer) (30 A in (f)(i) leads to 4 kW)


= 2.0 × 103 × 120 2.4 × 105

= ( =) 2.67 × 105 (W);}2.4×105

0.9


41e. [2 marks]Determine the input voltage to the transformer if the power loss in the cables from the power station is 2.0 kW.

Markscheme;

;

Allow ECF from (f)(i) and (f)(ii).

30 A and 4 kW earlier leads to 179 V.



Ip = √ = 22.4 (A)2×103

4.0

V = = = 12 k(V)P

I

2.67×105

22.4

Printed for Jyvaskylan Lyseon lukio

© International Baccalaureate Organization 2018

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

41f. [2 marks]Outline why laminating the core improves the efficiency of a transformer.

Markschemelaminations increase resistance / reduce current in core material/metal / reduce eddy currents;

thus reducing /power/(thermal) energy/heat losses in the core;

Examiners reportMany had rote learnt the first point about laminations in a transformer core: that the eddy currents are reduced. However, a goodcandidate will realise that the nature of the core must mean that they are never entirely eliminated – complete elimination of the currentswas a common and incorrect answer.

I 2R

Topic 5 Electricity 31.1.2018 [140 marks] · Topic 5 Electricity 31.1.2018 [140 marks] 1. The graph...

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