Electricity & Thermal Physics

1.

a copper wire is 2.0 m long and has a cross-sectional area of 1.00 mm2. it has a p.d. of 0.12 v across it when the current in it is 3.5 a. draw a circuit diagram to show how you would check these voltage and current values.

(3)

calculate the rate at which the power supply does work on the wire. .............................................................................................................................................. .............................................................................................................................................. rate = ......................................................... (2)

copper has about 1.7 × 1029 electrons per metre cubed. calculate the drift speed of the charge carriers in the wire. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. drift speed = ......................................................... (3)

the power from the supply connected to the wire is equal to the total force ft on the electrons multiplied by the drift speed at which the electrons travel. calculate ft. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. ft = ......................................................... (3) (total 11 marks)

2.

a copper wire is 2.0 m long and has a cross-sectional area of 1.00 mm2. it has a p.d. of 0.12 v across it when the current in it is 3.5a. draw a circuit diagram to show how you would check these voltage and current values. circuit showing variable power supply (or fixed but with variable resistor) ammeter in series with labelled wire voltmeter in parallel with wire

(1)

(1)

(1) (3 marks)

ealing, hammersmith and west london college

1

calculate the rate at which the power supply does work on the wire. rate of working or iv = (0.12 v) × (3.5 a) rate = 0.42 w

(1)

(1) (2 marks)

copper has about 1.7 × 1029 electrons per metre cubed. calculate the drift speed of the charge carriers in the wire. using equation i = naqv (1) substitution in v = i/naq

(1)

v = (3.5 a) / (1.7 × 1029 m-3) (1 × 10-6 m2) (1.6 × 10-19c) drift speed = 1.29 × 10-4 m s-1

(1) (3 marks)

the power from the supply connected to the wire is equal to the total force ft on the electrons multiplied by the drift speed at which the electrons travel. calculate ft power = ft × v substitution in ft = power/ v = (0.42 w) / (1.29 × 10-4 m s-1) ft = 3.3 kn (3 marks) [total 11 marks]

3.

a 24 w filament lamp has been switched on for some time. in this situation the first law of thermodynamics, represented by the equation δu = δq + δw, may be applied to the lamp. state and explain the value of each of the terms in the equation during a period of two seconds of the lamp’s operation. δu ....................................................................................................................................... .............................................................................................................................................. .............................................................................................................................................. (2)

δw ...................................................................................................................................... .............................................................................................................................................. .............................................................................................................................................. (2)

δq ....................................................................................................................................... .............................................................................................................................................. .............................................................................................................................................. (2)

ealing, hammersmith and west london college

2

typically, filament lamps have an efficiency of only a few percent. explain what this means and how it is consistent with the law of conservation of energy. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2) (total 8 marks)

4.

a 24 w filament lamp has been switched on for some time. in this situation the first law of thermodynamics, represented by the equation δu = δq + δw, may be applied to the lamp. state and explain the value of each of the terms in the equation during a period of two seconds of the lamp’s operation. δu = 0 (1) because filament temperature is constant

δw

(1) (2 marks)

= 48 j

(1)

work done on the filament by power supply

(1) (2 marks)

δq – = 48 j (1) energy given to (allow ‘lost from’) filament by heating

(1) (2 marks)

typically, filament lamps have an efficiency of only a few percent. explain what this means and how it is consistent with the law of conservation of energy. small proportion out as light

(1)

the rest of the energy heats the surroundings

(1) (2 marks) [total 8 marks]

5.

define the term resistivity. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2)

the resistivity of copper is 1.7 × 10-8 ωm. a copper wire is 0.6 m long and has a cross-sectional area of 1mm2. calculate its resistance. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. resistance = ......................................................... (3) ealing, hammersmith and west london college

3

two such wires as used to connect a lamp to a power supply of negligible internal resistance. the potential difference across the lamp is 12 v and its power is 36 w. calculate the potential difference across each wire. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. potential difference = ......................................................... (3)

draw a circuit diagram of the above arrangement. label the potential differences across the wires, lamp and power supply.

(3) (total 11 marks)

6.

define the term resistivity.

RA ρl orR = A either ρ = l with symbols defined

(1) (1) (2 marks)

the resistivity of copper is 1.7 × 10-8 ω m. a copper wire is 0.6 m long and has a cross-sectional area of 1mm2. calculate its resistance. resistance = (1.7 × 10-8 ωm) resistance =10.2 mω

(1)

(0.6 m) 1 × 10 −6 m 2

(1)

(1) (3 marks)

two such wires as used to connect a lamp to a power supply of negligible internal resistance. the potential difference across the lamp is 12 v and its power is 36 w. calculate the potential difference across each wire. current = 36 w/12 v = 3 (a)

(1)

p.d = (3 a) × (10.2 × 10-3 ω)

(1)

potential difference = 30.6 mv

(1) (3 marks)

ealing, hammersmith and west london college

4

draw a circuit diagram of the above arrangement. label the potential differences across the wires, lamp and power supply. 0 .0 3 V

1 2 .0 6 V

12 V 0 .0 3 V

lamp

(1)

both wires cell

(1)

(1)

(allow 12 v cell and 11.94 v across lamp) (3 marks) [total 11 marks]

7.

a student pours 500 g of water into an aluminium saucepan of mass 1.20 kg, heats it over a steady flame and records the temperature as it heats up. the temperatures are plotted as shown below. 55 T e m p e ra tu re /ºC 50 45 40 35 30 25 20

0

5

10

15

20 25 T im e /m in u te s

calculate the total heat capacity of the saucepan and water. specific heat capacity of water specific heat capacity of aluminium

= =

4200 j kg-1 k-1 900 j kg-1 k-1

.............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. heat capacity = ......................................................... (3) ealing, hammersmith and west london college

5

find the rate of rise of water temperature at the beginning of the heating process. .............................................................................................................................................. .............................................................................................................................................. rate of rise of temperature = ......................................................... (2)

hence find the rate at which energy is supplied to the saucepan and water. .............................................................................................................................................. .............................................................................................................................................. rate of energy supply = ......................................................... (2)

explain why the rate at which the temperature rises slows down progressively as the heating process continues. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2) (total 9 marks)

8.

a student pours 500 g of water into an aluminium saucepan of mass 1.20 kg, heats it over a steady flame and records the temperature as it heats up. the temperatures are plotted as shown below. 55 T e m p e ra tu re /ºC 50 45 40 35 30 25 20

0

5

ealing, hammersmith and west london college

10

15

20 25 T im e /m in u te s

6

calculate the total heat capacity of the saucepan and water. specific heat capacity of water specific heat capacity of aluminium

= =

4200 j kg-1 k-1 900 j kg-1 k-1

heat capacity = (0.500 kg × 4200 j kg-1 k-1) + (1.20 kg × 900 j kg–1 k–1) (1) + (1) heat capacity = 3180 j k–1

(1) (3 marks)

find the rate of rise of water temperature at the beginning of the heating process. (11 K) rate of rise of water temperature = (150 s) (1) rate of rise of temperature = 0.073 k s-1

(1) (2 marks)

hence find the rate at which energy is supplied to the saucepan and water. rate of energy supply = (3180 j k-1) × (0.073 k s-1) rate of energy supply = 0.23 kw

(1)

(1) (2 marks)

explain why the rate at which the temperature rise slows down progressively as the heating process continues. as the temperature of the saucepan increases (1) an increasing fraction of the heat supplied per second goes to the surroundings (1) (2 marks) [total 9 marks]

9.

the circuit shown is used to produce a current-voltage graph for a 12 v, 24 w lamp. R + 20 V –

24 Ω 1 2 V 2 4 W la m p

show on the diagram the correct position for a voltmeter and an ammeter. (2)

calculate the resistance of the lamp in normal operation. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. resistance = ......................................................... (3)

ealing, hammersmith and west london college

7

calculate the value for r which would enable the voltage across the lamp to be varied between 0 v and 12v. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. r = ......................................................... (4) (total 9 marks)

10.

the circuit shown is used to produce a current voltage graph for a 12 v, 24 w lamp. R + 20 V –

24 Ω 1 2 V 2 4 W la m p

V A

show on the diagram the correct position for a voltmeter and an ammeter. (2 marks)

calculate the resistance of the lamp in normal operation. 24 W current i = power / voltage = 12 V = 2a voltage 12 V = 2A resistance = current

resistance = 6 ω

(1)

(1)

(1) (3 marks)

calculate the value for r which would enable the voltage across the lamp to be varied between 0 v and 12v. 6 ω and 24 ω form a parallel pair of resistance 4.8 ω

(1)

(1)

drawing current of 2.5 a

(1)

r = 8 v/2.5 a r = 3.2 ω (4 marks) [total 9 marks]

ealing, hammersmith and west london college

8

11.

you are asked to measure the specific heat capacity of aluminium using a cylindrical block of aluminium which has been drilled out to accept an electrical heater. draw a complete diagram of the apparatus you would use.

(3)

describe how you would carry out the experiment and list the measurements you would take. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (5)

explain how you would calculate the specific heat capacity of aluminium from your measurements. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (3) (total 11 marks)

12.

you are asked to measure the specific heat capacity of aluminium using a cylindrical block of aluminium which has been drilled out to accept an electrical heater. draw a complete diagram of the apparatus you would use. diagram showing heater in aluminium block with suitably-placed thermometer, (1) lagging round the surface of the block and (1) a circuit-diagram with correctly-placed voltmeter, ammeter and power supply (1) (3 marks)

ealing, hammersmith and west london college

9

describe how you would carry out the experiment and list the measurements you would take. measure and record the mass of the block (m) and the initial temperature (θ 1) (1)

(1)

switch on the current and start the clock at the same time. record voltmeter and ammeter readings (i and v).

(1)

stop clock after an appreciable rise in temperature. note time (t). note final temperature of block (θ 2)

(1)

(1) (5 marks)

explain how you would calculate the specific heat capacity of aluminium from your measurements. energy transferred to block = ivt

(1)

increase in internal energy of block = m c (θ 2 – θ 1)

c= specific heat capacity of aluminium

IVt m (θ 2 − θ1 )

(1)

(1) (3 marks) [total 11 marks]

13.

a wire 6.00 m long has a resistivity of 1.72 ? 10–8 ? m and a cross-sectional area of 0.25 mm2 calculate the resistance of the wire. ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… resistance = .............................................… (3)

the wire is made from copper. copper has 1.10 ? 1029 free electrons per metre cubed. calculate the current through the wire when the drift speed of the electrons is 0.093–1 mm s–1. ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… current = ..............................................… (3)

ealing, hammersmith and west london college

10

the wire is cut in two and used to connect a lamp to a power supply. it takes 9 hours for an electron to travel from the power supply to the lamp. explain why the lamp comes on almost as soon as the power supply is connected. ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… (3) (total 9 marks)

14.

a wire 6.00 m long has a resistivity of 1.72 × 10–8 ω m and a cross-sectional area of 0.25 mm2 calculate the resistance of the wire. r

= pl/a (1) (1.72 × 10-8 Ω m) (6.00 m)

= resistance = 0.41a (1)

(0.25 × 10 − 6 m2 )

(1)

the wire is made from copper. copper has 1.10 × 1029 free electrons per metre cubed. calculate the current through the wire when the drift speed of the electrons is 0.093–1 mm s–1. i = m aqv (1) = (1.10 × 1029 m–3) (0.25 × 10–6 m2) × (1.60?10–19 c) (0.093 × 10–3 ms–1) (1) current = 0.41 a (1)

the wire is cut in two and used to connect a lamp to a power supply. it takes 9 hours for an electron to travel from the power supply to the lamp. explain why the lamp comes on almost as soon as the power supply is connected. electrons behave like an incompressible fluid (2) current flow is immediate throughout circuit (1) (allow equivalent explanations) [total 9 marks]

15.

the kinetic theory of gases is based on a number of assumptions. one assumption is that the average distance between the molecules is much larger than the molecular diameter. a second assumption is that the molecules are in continuous random motion. state and explain one observation in support of each assumption. first assumption ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................…

ealing, hammersmith and west london college

11

second assumption ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… ...............................…........................................................................................................… (total 4 marks)

16.

the kinetic theory of gases is based on a number of assumptions. one assumption is that the average distance between the molecules is much larger than the molecular diameter. a second assumption is that the molecules are in continuous random motion. state and explain one observation in support of each assumption. first assumption either large volume change on change of state (1) implies large spacing of gas molecules (1). [or faster diffusion through gases (1) implies more spacing for molecular motion (1)]

(or equivalent)-

second assumption brownian motion in gases (1) smoke particles subject to random knock about from air molecules (1) (or equivalent) [total 4 marks]

17.

define the term resistivity. ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ (2)

a student is asked to measure the resistivity of the alloy nichrome given a nichrome wire known to have a resistance of about two or three ohms. the wire is mounted between two copper clamps, x and y, near the ends of the wire. the power supply is a variable power supply of output 0–5 v. the series resistor is 80 ω.

ealing, hammersmith and west london college

12

complete the following circuit diagram. 80 Ω

+ –

0–5 V

N ic h ro m e w ire ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ (2)

the 80 ω series resistor ensures that the current is kept small. explain why this is important. ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ (2)

a number of measurements were made of the voltage across the wire for different values of the current flowing in it. the following graph was drawn. 0 .1 2 0 .1 0 p .d .a c r o s s w ir e /V 0 .0 8 0 .0 6 0 .0 4 0 .0 2 0

0

ealing, hammersmith and west london college

10

20 30 C u rr e n t/m A

40

50

13

calculate the resistance of the nichrome wire. ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ resistance = (3)

the length of wire between the clamps is 51 cm. the diameter of the nichrome wire is 0.59mm.calculate the resistivity of the nichrome. ...............................…............................................................................................................ ...............................…............................................................................................................ ...............................…............................................................................................................ resistance = (3) (total 12 marks)

18.

definition of resistivity:

Re sis tan ce × area length resistivity =

(2)

2

[give full credit for a correct statement about a 1-meter cube] 80 Ω A

+ –

0–5 V V N ic h ro m e w ir e X

Y

ammeter in correct position (1) voltmeter in correct position (1)

2

explanation: a high current heats the wire (1) and changes the resistivity (1) [allow (1) “to protect ammeter” or similar] resistance of nichrome wire: resistance = v/i (or gradient) = (0.12 v) / (0.050 a) (1) = 2.4 ω (1) 3

ealing, hammersmith and west london college

2

(1)

14

resistivity of nichrome: area = πr2 = 2.73 × 10–7 m2

(1)

(2.4 Ω) (2.73 ×10 -7 m 2 ) (0.51 m) resistivity = 1.29 ×

10–6

ωm

(1)

(1)

3

[no unit penalty] [total 12 marks

19.

a torch has three identical cells, each of e.m.f. 1.5 v, and a lamp which is labelled 3.5 v, 0.3 a. draw a circuit diagram for the torch.

(2)

assume that the lamp is lit to normal brightness and that the connections have negligible resistance. mark on your diagram the voltage across each circuit component and the current flowing in the lamp. (3)

calculate the internal resistance of one of these cells. resistance = .............................………. (3) (total 8 marks)

ealing, hammersmith and west london college

15

20.

diagram of torch circuit: the lamp will light correct circuit

2

[circuit showing one cell only is allowed one mark only unless the cell is labelled 4.5 v. if a resistor is included, allow first mark only unless it is clearly labelled in some way as an internal resistance.]

0 .3 A

3 .5 V

3 .5 V /3 voltage across each circuit component and current in lamp: either 3.5 v/3 shown across the terminals of one cell or 3.5 v across all three cells 3.5 v shown to be across the lamp 0.3 a flowing in the lamp [i.e. an isolated 0.3 a near the lamp does not score] 3 calculation of internal resistance of one of the cells:

3. 5V lost volts = 4.5 v - 3.5 v or 1.5 v – 3 or total resistance = (4.5 v)/0.3 a) = 15 kω internal resistance of one cell = [(1.0 v)/(0.3 a)] ÷ 3 or [(0.33 v) (0.3 a)] or lamp resistance = (3.5 v) / (0.3 a)11.7 ω = 1.1 ω or = (3.3ω)/3 = 1.1 ω 3 [some of these latter marks can be read from the diagram if it is so labelled] [8]

21.

the current i through a metal wire of cross-sectional area a is given by the formula i = naυe

ealing, hammersmith and west london college

16

where e is the electronic charge on the electron. define the symbols n and υ. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (2)

two pieces of copper wire, x and y, are joined end-to-end and connected to a battery by wires which are shown as dotted lines in the diagram. the cross-sectional area of x is double that of y. Y X

in the table below, nx and ny denote the values of n in x and y, and similarly for the other quantities. write in the table the value of each ratio, and alongside it explain your answer. ratio

value

explanation

nY nX IY IX

υY υX (6) (total 8 marks)

ealing, hammersmith and west london college

17

22.

definition of symbols: 3

n

=

number of electrons/carriers per unit volume (per m ) or electron (or carrier) density (1)

υ

=

average (or drift) velocity (or speed) (1)

ratio ny

value

2

explanation

1

same material (1) (1)

1

connected in series/kirchoff’s 1 law/conservation of charge/current is the same (1) (1)

2

a is halved so ν double [accept qualitative, e.g. a ↓ so v ↑, or good analogy] (1) (1)

nx ly lx vy vx

st

6 [accept e.g. ny = nx.....] [no e.c.f ] [nb mark value first, without looking at explanation. if value correct, mark explanation. if value wrong, don’t mark explanation except: if υy/υx = ½ or 1:2, see if explanation is correct physics, and if so give (1). no e.c.f.] [8]

23.

the diagram shows the circuit of a fluorescent light fitting. it consists of a tube, a starter and a ballast resistance of 300 ω.

ealing, hammersmith and west london college

18

the fluorescent tube is filled with gas. it contains two filaments at a and b of resistance 50 ω that heat the gas. I B a lla s t 300 Ω

50 Ω A S ta rte r

230 V

V

s ta r te r

Tube

B 50 Ω

when the light is first turned on, the tube does not conduct but the starter does, drawing a current of 0.50 a from the 230 v supply. calculate the voltages across the ballast resistor and each filament when this current flows. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... voltage across ballast =..................................................... voltage across each filament =......................................... (3)

ealing, hammersmith and west london college

19

mark these voltages on the diagram, and hence calculate the voltage across the starter when the starting current is flowing. mark your answer on the diagram. (2)

the starting current heats the filaments and the gas in the tube but the voltage across the tube is not large enough to make it conduct. however, after a few seconds the starter stops conducting. the voltage across the tube rises and the gas conducts. a current now flows from a to b and the tube lights up. what fundamental change is necessary for a gas, which was an insulator, to be able to conduct? .................................................................................................................................... .................................................................................................................................... (1)

now that the tube is conducting, the voltage across ab is 110 v. calculate the power dissipated in the whole circuit. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... power dissipated =................................. (3)

in a faulty fluorescent lamp the filaments at both ends of the tube glow steadily but the tube does not light up. identify, with a reason, the faulty component. .................................................................................................................................... .................................................................................................................................... (1) (total 10 marks)

24.

calculation of voltages: any use of voltage

=

current x component resistance (1)

ballast

=

150 v (1)

filament

=

25 v (1)

3

voltages on diagram: 3 voltages (150,25,25) marked on diagram near component; ignore units (1) [minimum 150 ÷(1 × 25)] vstarter = 30 v (marked on diagram) (1) fundamental change necessary: (free) charge carriers or free electrons, ionised, particles need to be charged (1) (1) [not t ↑ ]

3

ealing, hammersmith and west london college

20

calculation of power dissipated: vballast

=

230v – 110 v (1)

i

=

120v/300 ω

=

0.40 a (1)

=

230 v × 0.40 a [e.c.f for current]

=

92 w (1)

power

3

faulty component: starter is not breaking the circuit/starter still conducting (1)

1 [10]

25.

the resistors r1 and r2 in circuit (i) are equivalent to a single resistor r in circuit (ii). P

Q R

R

1

2

(i)

R (ii)

prove that r = r1 + r2 .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (3)

in a real circuit it is usually assumed that there is no potential difference between two points, such as p and q in diagram (i), which are on the same connecting lead. explain why this is usually a good approximation. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2)

ealing, hammersmith and west london college

21

in what circumstances might the approximation break down? .............................................................................................................................................. .............................................................................................................................................. (1)

a laboratory lead consists of 16 strands of fine copper wire twisted together. each strand is 30 cm long with a diameter of 0.15 mm. calculate the potential difference across the lead when it is carrying a current of 2.0 a. –8

(the resistivity of copper = 1.7 × 10 ωm) .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. potential difference =................................... (4) (total 10 marks)

26.

proof: v = v1 + v2 v = ir

v1 = ir1

v2 = ir2

substitute and cancel i

v = v1 + v2

(1)

÷i

(1)

sub using r =

(1) 3

explanation of why it is a good approximation: resistance of connecting lead is (very) small

(1)

–1

so i × r(very) small = (very) small p.d./e s do little work so p.d. small/r small

(1)

compared with rest of the circuit so p.d. small 2 circumstances where approximation might break down: if current is large or resistance of rest of circuit is small

(1)

[not high voltage/long lead/thin lead/high resistivity lead/hot lead] 1

ealing, hammersmith and west london college

22

calculation:

ρl use of r = A with a attempted × sectional area correct use of 16

(1)

use of v = ir

(1)

0.036 v

(1)

(1)

4 [10]

27.

the current i flowing through a conductor of cross-sectional area a is given by the formula i = naqν where q is the charge on a charge carrier. give the meanings of n and ν. n ........................................................................................................................................... ν ........................................................................................................................................... (2)

show that the equation is homogeneous with respect to units. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (3)

with reference to the equation, explain the difference between a metal conductor and a plastic insulator. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2) (total 7 marks)

ealing, hammersmith and west london college

23

28.

3

number of carriers or electrons per unit volume / per m /carrier density/electron density (1) [not charge density / concentration] drift velocity or drift speed or average/mean/net/overall velocity (1)

2

[not just velocity; not speed unless drift] –3

m (1) 2

–1

m as m s (1) multiply and reduce to a (1)

3

[base units not needed] [mixed units and symbols could get the third mark] –1 [ma = m loses 1 mark] metal:

M : n la rg e s o th e re is a c u r re n t

n : n in m e ta l m u c h la rg e r (1 )

In s u la to r I : n z e r o ( n e g lig ib le ) /v e r y s m a ll s o le s s c u rre n t (o r z e ro c u rre n t)

C u r r e n t in m e ta l is la rg e r ( 1 ) 2

[ignore anything about v. allow e.g. electron density for n] [7]

ealing, hammersmith and west london college

24

29.

the graph shows the current-voltage characteristic of a semiconductor diode.

100 I/m A 80

60

40

20

0

0

0 .2

0 .4

0 .6

0 .8 V /V

state, with a reason, whether the diode obeys ohm’s law. .............................................................................................................................................. .............................................................................................................................................. (1)

show that when the voltage across the diode is 0.74 v its resistance is about 9 ω. .............................................................................................................................................. .............................................................................................................................................. (2)

when the diode is connected in the following circuit, the voltage across it is 0.74 v.

+ 9 .0 V R

0 V

ealing, hammersmith and west london college

25

calculate the value of the resistance r. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. r = ............................................. (3)

electronic circuit designers often use a simple model of this type of diode. this “model diode” has the following properties: (i)

for any voltage below +0.7 v it does not conduct at all.

(ii)

once the voltage reaches +0.7 v the diode can pass any size of current with no further increase in voltage.

add a second graph to the grid above to show the current-voltage characteristic of this model diode. (2) (total 8 marks)

30.

no, because v is not proportional to i or not straight line through origin / (1) only conducts above 0.5 v / resistance changes

1

use of r = 0.74 / current from graph (1) = 9.25 ω [9.0 – 9.5 ω] [minimum 2 significant figures] (1)

2

calculation of p.d. across r [8.26]

calculation of total resistance[109 – 115]

ratio r: ratio v

e = σir (1)

÷i

– diode resistance [9]

correct substitutions

correct substitutions (1)

103 ω [100 – 106] (1) 3 [if not vertical line, 0/2] (1 ) (1 )

(1 ) (0 )

(1) (0) A n y th in g (g a p , c u rv e , b e lo w a x is )

0 .7

≠ 0 .7

0 .7

(1)(1)

2

[otherwise 0 0 ] [8]

ealing, hammersmith and west london college

26

31.

according to kinetic theory, the pressure p of an ideal gas is given by the equation p=

1 3

ρ〈c2〉

2

where ρ is the gas density and 〈c 〉 is the mean squared speed of the molecules. express ρ in terms of the number of molecules n, each of mass m, in a volume v. .............................................................................................................................................. .............................................................................................................................................. (1)

it is assumed in kinetic theory that the mean kinetic energy of a molecule is proportional to kelvin temperature t. use this assumption, and the equation above, to show that under certain conditions p is proportional to t. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2)

state the conditions under which p is proportional to t. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. (2)

a bottle of gas has a pressure of 303 kpa above atmospheric pressure at a temperature of 0°c. the bottle is left outside on a very sunny day and the temperature rises to 35°c. given that atmospheric pressure is 101 kpa, calculate the new pressure of the gas inside the bottle. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. .............................................................................................................................................. pressure = .................................................. (3) (total 8 marks)

ealing, hammersmith and west london college

27

32.

ρ = nm/v (1)

1 2

2

2

p = 2/3 (n/v) ½ m / ∝ k.e. / m ∝ k.e. (1) [full backwards argument can get 1 / 2; full qualitative argument scores 1 / 2] 2

2

k.e. (or ½ m or ) ∝ t, ∴p ∝ t (1)

2

v constant (1) n constant / for fixed mass / fixed number of moles [not fixed amount] (1) [near ideal conditions, specified, can replace one of the above] [fixed density, 1 mark]

2

see (273, 308) or 404 (1) use p1/t1 = p2/t2 (kelvin temp) (1) = 456 kpa (1) [355 kpa gets 3 marks] [303 kpa → 342 kpa, 101 kpa → 114 kpa gets 2 marks]

3 [8]

33.

the diagram shows two methods of connecting eight heating elements which make up a car rear window heater. the heater is connected to a 12 v car battery. each element used in circuit p has a resistance of 24 ω;each used in circuit s has a resistance of 0.50 ω.

C ir c u it P

+

(a)

8 7 6 5 4 3 2 1

C irc u it S –

8 7 6 5 4 3 2 1

–

+

calculate the current drawn from the battery for each circuit. show your working. circuit p ..................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... circuit s ..................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (5)

ealing, hammersmith and west london college

28

(b)

elements 3 and 4 burn out in each circuit and no longer conduct electricity. what are the new values of the currents in each circuit? circuit p ..................................................................................................................... ..................................................................................................................................... circuit s ..................................................................................................................... ..................................................................................................................................... (2)

(c)

what effect would halving the battery voltage have on the power transfer in circuit p? explain your answer. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) (total 9 marks)

34.

(a)

circuit p:

either

1 1 =Σ Rn use of R 24Ω R= = 3.0Ω 8 12V I= 3Ω = 4.0 a ∴ or

12V = 0.5A i in one element = 24Ω ∴total i = 8 times this = 8 × 0.5 a = 4.0 a

3

circuit s: r = σrn = 8 × 0.5ω 12V = 3. 0 A ∴i = 4Ω (b)

2

circuit p: 24Ω new r = 6 = 4.0ω → new i = 3.0 a circuit s: new current is zero

2

ealing, hammersmith and west london college

29

(c)

12 v down to 6 v → halving i or i = 2 a or new p = (½ v)²/r so new p = ½v × ½ i = ¼ p original

2 [9]

35.

a simple heat engine which drives an electric motor is shown below. the wires connecting the motor to the hot source and the cold sink are made of the same material. the wire linking the source and the sink is made of a different material. M o to r M

Ic e a n d w a te r

H o t w a te r C o ld s in k

H o t so u rc e

state where the energy comes from and to where it goes. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2)

explain which energy transfers are heating and which are working. you may be awarded a mark for the clarity of your answer. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (5)

ealing, hammersmith and west london college

30

explain one way of increasing the efficiency of this heat engine. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2) (total 9 marks)

36.

energy received at/from hot junction/water/source energy goes to cold junction/water/sink and motor

2

transfers at hot/cold junctions - heating because (they are driven by) temperature difference transfer at motor - working as forces are moving charges/there is no temperature difference quality of written communication

5

increase temperature difference reference to correct expression for efficiency

2 [9]

37.

–6

2

29

a copper wire has a cross-sectional area of 0.20 ×10 m . copper has 1.0 × 10 free electrons per cubic metre. –1

calculate the current through the wire when the drift speed of the electrons is 0.94 mm s . ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... current = ................................................. (3) –8

the wire is 4.0 m long. copper has a resistivity of 1.7 × 10 ω m. calculate the resistance of the wire. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... resistance = ............................................ (3)

ealing, hammersmith and west london college

31

calculate the potential difference across the wire. ............................................................................................................................................... potential difference = .............................. (1)

a second wire with the same dimensions is made from a material that has a greater resistivity than copper. explain how, if at all, the current will differ from that in the copper wire when the same p.d. is applied across it. ............................................................................................................................................... ............................................................................................................................................... (2)

the number of free electrons per cubic metre in this wire is the same as that in the copper wire. compare the drift velocities of the free electrons in the two wires. ............................................................................................................................................... ............................................................................................................................................... (1) (total 10 marks)

38.

current: –3 –1 conversion, i.e. 0.94 × 10 m s (1) –19 use of 1.6 × 10 c (1) answer 3.0 a 29 –3 –6 2 –19 –3 –1 1.0 × 10 m × 0.20 × 10 m × 1.6 × 10 c × 0.94 × 10 mm s (1) –3 current = 3.0 a [accept 2.8 a if 0.9 × 10 used.] resistance: ρl recall r = A

3

(1)

substitution: 1.7 × 10 −8 Ω m × 4.0 m 0.20 × 10 -6 m 2 r=

(1)

resistance = 0.34 ω (1)

3

potential difference: potential difference = 3.0 a × 0.34 ω (1) = 1.0 v (1.02 v) [mark for correct substitution of their values or for the answer of 1.0 v]

1

ealing, hammersmith and west london college

32

explanation: (increasing resistivity) increases resistance leads to a smaller current (1)

(1) 2

comparison: drift velocity decreases (in second wire) (1) [allow v1/v2 = i1/i2] [allow e.c.f. answer consistent with their current answer] [resistivity up, current down nd ρ up, i down / 2 (2 mark)]

1

[10]

39.

the contents of a domestic refrigerator are at a constant temperature of 5°c, and the outside surface of the refrigerator is at a constant temperature of 20°c. 5 °C

20 °C

explain how it is possible for the contents of the refrigerator to be at a constant temperature even though energy is continuously flowing in from outside. you may be awarded a mark for the clarity of your answer. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (3)

the equation δu = δq + δw can be applied to the contents of the refrigerator. what is the value of δu? explain your answer. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2)

ealing, hammersmith and west london college

33

what is meant by δq in this equation? ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2)

explain why δw is zero. ............................................................................................................................................... ............................................................................................................................................... (1) (total 8 marks)

40.

explanation: quality of written communication

(1)

1

explanation, any two from: energy [not heat] flows out (1) at the same rate (1) due to motor/heat pump (1)

max 2

δu: δu: δu = 0 (1) temperature (of contents) constant/temperature kept at 5 °c (1)

2

δq: δq: δq is the net energy flowing (1) because of the temperature differences (1) [do not accept “energy due to heating”]

2

δw: δw: no work is done on (or by) the contents or δw must be zero since δu and δq are zero or no mechanical or electrical work done on contents (1)

1 [8]

41.

three resistors r1, r2 and r3 are connected in parallel with each other. they could be replaced by a single resistor of resistance r.

+V

+V I1

R

1

I2

R

2

0V

ealing, hammersmith and west london college

I3

R

I

R

3

0V

34

show that the resistance, r, of the equivalent resistor can be calculated from

1 R1

1 R =

1

1 R2 +

R3 +

............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (3)

ealing, hammersmith and west london college

35

a student has four identical resistors each of resistance 10 ω. she connects them to form the different networks shown below. calculate the equivalent total resistance of each network. ................................................................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... F ir s t n e tw o r k

T o ta l r e s is t a n c e = ...................................................Ω ................................................................................... ................................................................................... ................................................................................... ................................................................................... ...................................................................................

S e c o n d n e tw o rk

T o ta l r e s is t a n c e = ...................................................Ω ................................................................................... ................................................................................... ................................................................................... ................................................................................... ...................................................................................

T h ird n e tw o r k

ealing, hammersmith and west london college

T o ta l r e s is t a n c e = ...................................................Ω

(3)

36

she then connects a battery across the second network and adds meters to make the circuit shown below. a current of 50 ma is drawn from the battery. V2

50 m A A

10 Ω

10 Ω

10 Ω 10 Ω

V1 determine the reading on each of the three meters. reading on ammeter a: ............................................................................................................................................... ammeter reading = .......................................................... ma reading on voltmeter v1: ............................................................................................................................................... voltmeter reading = ........................................................... v reading on voltmeter v2: ............................................................................................................................................... ............................................................................................................................................... voltmeter reading = ........................................................... v (5) (total 11 marks)

42.

show that [in diagram or text} • states p.d. same across each resistor

(1)

• use of i = i1 + i2 + i3 [symbols or words] •

V = V + V + V R R1 R2 R3

(1)

ealing, hammersmith and west london college

(1)

3 37

[i = v / r stated somewhere gains one mark] networks first network: 2.5(ω) (1) second network: 25 (ω) (1) third network: 10 (ω) (1) meter readings ammeter: voltmeter v1: = 0.25 v (1) voltmeter v2: = 1.25 v (1)

3

25 (ma) (1) 25 × 10 or 50 × 5 [ignore powers of 10] (1) 50 × 25 [ignore powers of 10]

(1) 5

nd

[allow full e.c.f. for their resistance for 2 network or their v1 answer] [11]

43.

an electric room heater consists of three heating elements connected in parallel across a power supply. Pow er s u p p ly X

Y

–5

each element is made from a metal wire of resistivity 5.5 × 10 ω m at room temperature. the –7 2 wire has a cross-sectional area 8.0 × 10 m and length 0.65 m. the heater is controlled by two switches, x and y. show that the resistance of one heating element at room temperature is approximately 45 ω. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (3)

ealing, hammersmith and west london college

38

calculate the total resistance of the heater for the following combinations of switches at the moment the switches are closed. switch x

switch y

open

closed

closed

open

closed

closed

resistance of heater / ω

(3)

calculate the maximum power output from the heater immediately it is connected to a 230 v supply. ............................................................................................................................................... ............................................................................................................................................... maximum power = ................................................. (2)

after being connected to the supply for a few minutes the power output falls to a lower steady value. explain why this happens. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2) (total 10 marks)

44.

show that resistance is approximately 45 ω

R= R=

ρl A 5.5 × 10 −5 Ω m × 0.65 m 8.0 ×10 −7

= 44.7 ω [no u.e.] (1) [must see this value and not 45]

ealing, hammersmith and west london college

39

table switch x open closed closed

switch y closed open closed

resistance of heater/ω 22.5/22.35 45/44.7 15/14.9

(1) (1) (1)

[no u.e.]

3

calculation of maximum power P=

V2 R use of equation with 15 ω or their minimum value (1)

= 3526 w,3500 w [full ecf] (1)

2

explanation of power output fall As the temperature of the heater increases resis tan ce (of metals) increases OR as it gets hotter / hot  V2 since v is constant p = R or p = vi and v = ir

[then p ↓ as r ↑] (1)

2

1 or p ∝ R [so p↓ as r↑] [10]

45.

define the term specific latent heat of fusion. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (3)

the graph shows how the temperature of a heated metal sample varies with time. T e m p e r a tu r e /ºC

Room te m p e ra tu re A

ealing, hammersmith and west london college

B

T im e /s

40

during the time interval ab, the metal changes from a solid to a liquid whilst still being heated. explain, in molecular terms, what is happening to the energy being supplied during this time. ............................................................................................................................................... ............................................................................................................................................... (1)

describe, in molecular terms, the main differences between the solid and liquid states. you may illustrate your answer with simple diagrams.

............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... (2) (total 6 marks)

46.

definition of specific latent heat of fusion

energy l = change in mass /energy to change 1 kg (1) during a change of state/solid to liquid (1) at constant temperature/at melting point (1)

3

explanation of time interval ab energy used to break bonds/pull molecules apart/overcome forces of attraction (1)

ealing, hammersmith and west london college

1

41

differences between solid and liquid states marks can be scored for diagrams and/or words any two rows: difference arrangement motion

solid regular array or lattice vibrational motion

spacing

close packed

liquid no regular pattern

(1)

random/brownian free to move around (each other) slightly further apart than in a solid

(1) (1) max 2 [6]

47.

complete the table by selecting a typical value for each physical quantity from the list below. each value may be used once, more than once or not at all. 0.05 ω 6.0 ω 2 kω 10 mω

–1

0.3 mm s –1 30 cm s

–8

2.0 × 10 ωm 15 2.0 × 10 ωm

physical quantity

300 k 3000 k

typical value

resistance of a voltmeter

internal resistance of a car battery internal resistance of an e.h.t. supply

resistivity of an insulator

drift velocity of electrons in a metallic conductor temperature of a working filament bulb (total 6 marks)

48.

table

ealing, hammersmith and west london college

42

physical quantity

typical value

resistance of a voltmeter

10 mω

1

internal resistance of a car battery

0.05ω

1

internal resistance of an eht supply

10 ωm

1

15

resistivity of an insulator

1

2.0 × 10 ωm

drift velocity of electrons in a metallic conductor

0.3 mm s

temperature of a working filament bulb

3000 k

–1

1 1

[mark is lost if 2 or more values are put into one box] [6]

49.

write down the ideal gas equation. .......................................................................................................................................... .......................................................................................................................................... (1)

use your equation to give the unit for the molar gas constant in si base units. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... (2) 2

the pressure p of a gas is related to its density p and the mean square molecular speed 〈c 〉 by the formula

1 ρ c2 p= 3

ealing, hammersmith and west london college

43

use this relationship and the ideal gas equation to show that the average kinetic energy of a molecule is proportional to the kelvin temperature of the gas. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... (3)

sketch a graph to show how the product pv (pressure × volume) varies with temperature in °c for one mole of ideal gas.

comment on any significant features of your graph. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... (4) (total 10 marks)

50.

gas equation pv = nrt [accept symbols or words]

1

molar gas constant unit

PV r = nT –1 –2

p – kg m s

1

3

v–m t–k ) n – mol

ealing, hammersmith and west london college

) 1 )

44

kinetic energy of molecule nm = m

Nm 2 c ρ = 3V 2

nm〈c 〉 = 3nrt

correctly combined the 2 equations

1

density = any mass ÷ volume

1

show that

m c2 kinetic energy =

2

=

3 RT n 2 N

1

sketch graph pv on y axis ) temperature/°c on x axis ) [accept axes reversed and correct graph] straight line graph with negative intercept gradient r

1 1

1

intercept at -273 °c1 [all these marks can be scored on graph] [10]

51.

a cell of negligible internal resistance is connected in series with a microammeter of negligible resistance and two resistors of value 15 kΩ and 25 kΩ. the current is 150 μa. draw a circuit diagram of the arrangement.

(1)

show that the e.m.f. of the cell is 6.0 v. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (2)

ealing, hammersmith and west london college

45

a voltmeter is now connected in parallel with the 25 kΩ resistor. draw a diagram of the new circuit.

(1)

when the voltmeter is connected the reading on the microammeter increases to 170 μa. calculate the resistance of the voltmeter. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ resistance = ......................................... (3) (total 7 marks)

52.

circuit ammeters and two resistors in series (1) [1 mark circuit penalty for line through cell or resistor] cell e.m.f –6 3 e= 150 x 10 (a) x 40 x 10 (ω) total r (1) powers of 10 (1) e = 6.0 (v)

ealing, hammersmith and west london college

1

2

46

new circuit voltmeter in parallel with 25 (kω) resistor (1) resistance of voltmeter 6( V) –6 (total resistance) = 170 × 10 ( A) = (35.3 kω)

(resistance of ll combination) = 35 – 15 kω = (20 ω) [e.c.f. their total resistance] 1 1 1 = + 20 25 RV 1 5–4 = RV 100 rv = 100 kω [108 kω if rt calculated correctly]

1

(1) (1)

(1)

            

alternative route 1: p.d. across 15 kω = 2.55 v (∴p.d. across ll combination = 3.45 v) resistance combination = 20 kω → rv = 100 kω

3 (1) (1) (1)

       [7]

alternative route 2: p.d. across parallel combination = 3.45 v i through 25 kω = 138 µa → rv = 100 kω

ealing, hammersmith and west london college

3 (1) (1) (1)

    

47

53.

a polythene rod was negatively charged by rubbing it with a cloth. the rod was then stroked several times across the metal cap of a meter used for measuring charge. C h a rg e d p o ly th e n e ro d

M e te r the initial reading on the meter was zero. –8

after 3.8 s the final reading was –6.4 × 10 c. calculate the number of electrons that were transferred to the metal cap. ……….……………………………………………………………………………………………. ……….……………………………………………………………………………………………. number of electrons = ………………………… (3) –1

calculate the average rate in c s at which charge was transferred to the metal cap. ……….……………………………………………………………………………………………. ……….……………………………………………………………………………………………. rate = …………………………………… c s

–1

(2)

state the base unit for the rate of flow of charge. ……….……………………………………………………………………………………………. (1) (total 6 marks)

54.

number of electrons –9

–19

11

(–64 × 10 c) / (–1.6 × 10 c) = 4.0 × 10 electrons use of n = q/e (1) –19 seeing 1.6 × 10 c (1) 11 answer of 4.0 × 10 (electrons) (1)

3

[use of a unit is a ue] [–ve answer: 2/3]

ealing, hammersmith and west london college

48

rate of flow –8

–1

–9

–1

(6.4 × 10 c)/3.8 s = 16.8/17 [nc s ] or 16.8/17 × 10 [c s ] (6.4) / 3.8 s i.e. use of i = q/t [ignore powers of 10] (1) –8

–8

correct answer [no e.c.f.] [1.7 or 1.68 x 10 or 1.6 × 10 ] (1)

2

unit amp(ere)/a (1)

1 [6]

55.

a 12 v car battery is recharged by passing a current of 5.0 a through it in the reverse direction using a 15 v battery charger. the internal resistance of the charger and the battery are 0.56 ω and 0.04 ω respectively. the circuit used is shown below. C h a rg in g cu rren t 5 .0 A

15 V

12 V

0 .5 6 Ω

0 .0 4 Ω

V

B a tte ry c h a rg e r

B a tte ry 5 .0 A

the terminal p.d. across the battery charger is found by solving the following equation: terminal p.d. = 15 v – (5.0 a × 0.56 ω) determine the reading on the voltmeter. ……………………………………………………………………………………………. (1)

write an equivalent equation for the terminal p.d. across the battery. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. (3)

ealing, hammersmith and west london college

49

calculate the rate at which energy is being wasted during the recharging process. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. rate = .................................................. (2)

hence determine the efficiency of the recharging process. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. efficiency = .......................................... (2)

the internal resistance of this car battery is 0.04 Ω. explain why a car battery is designed to have a very low internal resistance. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. …………………………………………………………………………………………….…. (2) (total 10 marks)

56.

car battery voltmeter reading: 12.2 (v) (1)

1

equation terminal p.d. = 12 v + (5.0 a × 0.04 ω) see 12v (1) see 5.0 a × 0.04 ω (1) addition of terms (1)

3

wasted power see 0.04 Ω + 0.56 Ω or 2.8 v + 0.2 v or 5 x (15 – 12) w (1) power = 15 w (1)

2

efficiency (same current) 12 v / 15 v or pout/pin = 60 w/75 w (1) efficiency = 0.8/80%

efficiency = 0.8/80% (1)

ealing, hammersmith and west london college

2

50

explanation any two from: • starter motor / to start car needs (very) large current

E R+r • i= • (e and r fixed) rmin ⇒ imax (1) (1) (1)

2 [10]

57.

the circuit diagram shows a 12 v d.c. supply of negligible internal resistance connected to an arrangement of resistors. the current at three places in the circuit and the resistance of two of the resistors are given on the diagram. 12 V

R

2 .0 Ω

1

1 .5 A

4 .0 Ω

2 .0 A

2 .0 A

R (a)

2

calculate the potential difference across the 4.0 Ω resistor. ...................................................................................................................................... ...................................................................................................................................... potential difference = .................................. (1)

(b)

calculate the resistance of resistor r2. ...................................................................................................................................... ...................................................................................................................................... ...................................................................................................................................... resistance of r2 = ...................................... (2)

ealing, hammersmith and west london college

51

(c)

calculate the resistance of resistor r1. ...................................................................................................................................... ...................................................................................................................................... ...................................................................................................................................... ...................................................................................................................................... ...................................................................................................................................... resistance of r1 = ...................................... (3) (total 6 marks)

58.

(a)

p.d. across 4 Ω resistor

1.5 (a) × 4 (Ω) = 6 v (1) (b)

1

resistance r2 current through r2 = 0.5 a (1) 6 (V) r = 0.5(A) 2

r2 = 12 Ω (1)

2

[allow ecf their pd across 4 Ω] (c)

resistance r1 p.d. across r1 = 12 − 6 − 4 = 2 v (1) current through r1 = 2 a (1) 2(V) r = 2(A) = 1Ω (1) 1

[allow ecf of pd from (a) if less than 12 v] alternative method parallel combination = 3Ω (1) circuit resistance = 12(v)/2 (a) = 6Ω (1) r1 = 6 – (3 + 2) = 1 Ω (1)

3

[allow ecf of pd from (a) and r from (b)] [6]

ealing, hammersmith and west london college

52

59.

(a) a student sets up a circuit and accidentally uses two voltmeters v1 and v2 instead of an ammeter and a voltmeter. the circuit is shown below.

V

1

9 .0 V

100 Ω

(i)

V

2

circle the voltmeter which should be an ammeter. (1)

(ii)

both voltmeters have a resistance of 10 mΩ. the student sees that the reading on v2 is 0 v. explain why the potential difference across the 100 Ω resistor is effectively zero. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2)

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(b)

the student replaces the 100 Ω resistor with another resistor of resistance r. the reading on v2 then becomes 3.0 v. (i)

complete the circuit diagram below to show the equivalent resistor network following this change. label the resistor r.

9 .0 V

(2)

(ii)

calculate the value of r. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... r = .............................................................. (3) (total 8 marks)

60.

(a) v1 (1)

(i)

replacement 1

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(ii)

explanation [one pair of marks] resistance: resistance of v1 [not just the voltmeter] is much larger than 100 Ω or combined resistance of parallel combination is (1) approximately 100 Ω voltage: p.d. across v1 is much greater than p.d. across 100 Ω or (1) all 9 v is across v1 or current: no current is flowing in the circuit / very small current (1) resistance: because v1 has infinite/very large resistance (1) or (correct current calculation 0.9 x 10 –6 90 x 10 a (1) this is a very small/negligible pd (1)

(b)

–6

a and) correct pd calculation 2

circuit diagram (i)

or equivalent resistor symbol labelled 10 mΩ (1) or equivalent resistor symbol labelled 10 mΩ (1)

2

[they must be shown in a correct arrangement with r] (ii)

value of r 6 (v): 3 (v) = 10 (mΩ): 5 (mΩ) / rtotal of parallel combination is 5 (1) mΩ 1/5 (mΩ) = 1/10 (mΩ) + 1/r or some equivalent correct (1) substitution to show working r = 10 mΩ (1)

3 [8]

61.

a battery of e.m.f. 6.0 v is connected to a 10 ω resistor as shown in the circuit diagram. V 6 .0 V

10 Ω

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(a)

define the e.m.f. of the battery. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2)

(b)

when the switch is open the voltmeter reading is 6.0 v. the internal resistance of the battery is 0.40 ω. calculate the reading on the voltmeter when the switch is closed. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... voltmeter reading = ……………………………… (3)

(c)

a second identical battery is connected in parallel with the first one. describe and explain qualitatively what would happen to the voltmeter reading if the switch remains closed. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (total 8 marks)

62.

(a)

definition of e.m.f.

energy (conversion) or work done (1) per unit charge (1) or e = w/q (1) symbols defined (1) [e = 1j/c scores 1] or e = p/i (1) symbols defined (1) [terminal pd when no current drawn or open circuit scores max 1]

ealing, hammersmith and west london college

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(b)

voltmeter calculation any attempt to find any current (1) attempt to calculate pd across 10 ω resistor (1) 5.77 v

2

or potential divider method; ratio of resistors with 10.4 ω on the bottom (1) multiplied by 6.0 v (1) 5.77 v (1) 3 [for either method, an answer of 0.23 v scores max 1] (c)

second battery added voltmeter reading increased (1) any two of: emf unchanged total resistance reduced current increases or “lost volts” decreases (2)

3 [8]

63.

a freezer contains a heat pump which pumps energy from the inside of the freezer to the outside. the diagram shows the energy flow for one day of use.

4 .5 M J ta k e n f r o m in s id e fre e z e r

3 .0 M J f r o m e le c tric ity s u p p ly

H EAT PU M P

e n e rg y tra n s fe rre d in to ro o m

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(a)

(i)

how much energy is transferred into the room?

........................................................................................................................... energy = ...................................................... what principle have you applied to do this calculation? ........................................................................................................................... (2)

(ii)

why do you need an energy source to pump energy from the inside of the freezer to the outside? ........................................................................................................................... ........................................................................................................................... (1)

(iii)

assuming the inside of the freezer remains at a constant temperature, calculate the rate at which energy is flowing in through the walls of the freezer. ........................................................................................................................... ........................................................................................................................... rate of flow of energy = ............................. (2)

(b)

an ice cube tray is filled with 0.35 kg of water at 20 °c and placed in the freezer. the 3 –1 –1 specific heat capacity of water is 4.2 × 10 j kg k and the specific latent heat of fusion 5 –1 of water is 3.3 × 10 j kg . calculate the minimum amount of additional energy the heat pump has to pump out of the freezer in order to freeze the water. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... energy = ...................................................... (3)

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(c)

inside the freezer there are cooling fins towards the top but not at the bottom. explain how these fins cool the air in the freezer and why there are no fins at the bottom. you may be awarded a mark for the clarity of your answer. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4) (total 12 marks)

64.

(a) (i) energy = 7.5 mj (1) conservation of energy (1)

(b)

(ii)

energy source needed because (thermal) energy is moving (1) from cold to hot or moving up a (temperature) gradient or moving against the (temperature) gradient [do not penalise heat used for energy]

(iii)

recall of p = e/t (1) –1 –1 power = 52 w // 187500 j h // 3125 j min (1)

use of e = m c δθ and correct δθ (20 k) (1) use of e = m l (1) 5 5 answer = 1.4 × 10 j[1.449 × 10 j] (1)

2 1

2

3

example of answer: 3 –1 –1 energy temp drop = 0.35 kg × 4.2 × 10 j kg k 20 k = 29400 j 5 –1 energy for change of state = 0.35 kg × 3.3 × 10 j kg = 115 500 j total energy = 29400 + 115500 = 144900 j (c)

• • • •

qowc (1) fins remove thermal energy / cool the air (1) hot air rises or cold air sinks or hot air less dense or cold air more dense (1) fins at top cause convection, fins at bottom do not cause convection (1)

4 [12]

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