Colonel Frank Seely School
Exampro A-level Physics
(7407/7408)
Name:
Class:
3.5.1.4 Circuits
Author:
Date:
Time:
663
Marks:
580
Comments:
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Colonel Frank Seely School
Q1.(a)
Figure 1 and Figure 2 show two circuits in which a supply of e.m.f. 6.0 V and internal
resistance 5.0 Ω is delivering power to a pair of resistors.
Figure 1
Figure 2
When maximum power is dissipated in an external circuit, the resistance of the
external circuit is equal to the internal resistance of the supply.
(i)
For the circuit in Figure 1, determine the value of R which results in the
maximum power being delivered to the external circuit.
(3)
(ii)
Calculate the terminal potential difference when the supply is delivering
maximum power to the circuit in Figure 1.
(1)
(iii)
Calculate the power that will be dissipated by the 15 Ω resistor when the supply
is delivering maximum power to the external circuit.
(2)
(iv)
For the circuit in Figure 2, explain why the supply cannot deliver maximum
power in this circuit for any value of the resistor R.
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(2)
(b)
(i)
The 15 Ω resistor is made from wire of length 2.3 m. The wire has a diameter of
3.0 × 10–4 m. Calculate the resistivity of the material from which the wire is
made.
(3)
(ii)
Sketch below a graph showing how the resistance of 2.3 m of wire made from
this material varies with the diameter of the wire. The value for a wire of
diameter 3.0 × 10–4 m has been plotted for you.
(2)
(Total 13 marks)
Q2.The diagram shows a network of four 2 Ω resistors.
The effective resistance, in Ω, between X and Y is
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Colonel Frank Seely School
A
0.5
B
1.2
C
1.7
D
2.0
(Total 1 mark)
Q3.(a)
Figure 1 and Figure 2 show two circuits that may be used for controlling the voltage
across a 3.0 Ω resistor. In each circuit the supply has an e.m.f. E of 10 V and
negligible internal resistance.
Figure 1
(i)
Figure 2
Calculate the minimum voltage which can exist across the 3.0 Ω resistor using
the circuit shown in Figure 1.
(3)
(ii)
State one advantage of using the circuit shown in Figure 2 for controlling the
voltage across the 3.0 Ω resistor.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(iii)
The total resistance of the potentiometer wire in Figure 2 is 30 Ω. Explain why
the voltage across the 3.0 Ω resistor would not be half of the maximum when the
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Colonel Frank Seely School
slider of the potentiometer is half-way along the wire, as shown in Figure 2.
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(2)
(iv)
Label with a letter P the approximate position of the slider in Figure 2, when the
voltage across the 3.0 Ω resistor is about half the maximum possible.
(1)
(b)
The circuit in Figure 3 is used to balance the power dissipated by two components
that have different resistances. This is achieved by adjusting the position of S.
Figure 3
(i)
Show that for the power dissipation to be the same, the ratio V1/V2 = 3/2.
(2)
(ii)
Calculate the power dissipated by one of the components when they are
balanced.
(1)
(Total 10 marks)
Q4.The graph in Figure 1 shows how the resistance of a thermistor varies with temperature.
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Figure 1
(a)
Explain why the resistance decreases at higher temperatures.
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(2)
(b)
The thermistor is included in the circuit shown in Figure 2.
Figure 2
The thermistor has to be maintained at a temperature of 60°C.
Calculate:
(i)
the potential difference across the thermistor;
(3)
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(ii)
the power that has to be removed from the thermistor to maintain the
temperature at 60°C.
(2)
(c)
(i)
Sketch below a possible variation of resistance with temperature for a material
that becomes superconducting at a temperature of –80°C.
(1)
(ii)
State one application of superconductors and explain briefly the advantage of
superconductors over ordinary conductors in the application you have chosen.
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(2)
(Total 10 marks)
Q5.The 12 V battery in the circuit shown has negligible internal resistance. The diodes have 'ideal'
characteristics.
The current through the battery is approximately
A
0A
B
0.10 A
C
0.20 A
D
0.40 A
(Total 1 mark)
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Q6.Three identical resistors X, Y and Z are connected across a battery as shown.
The ratio
is
A
B
C
1
D
2
(Total 1 mark)
Q7.A student is provided with three resistors with the values of 1 Ω, 2 Ω and 3 Ω.
(a)
(i)
Show how all 3 resistors can be connected together to provide the greatest
value of resistance.
(1)
(ii)
Calculate this largest value of resistance.
(2)
(b)
(i)
Sketch the arrangement of the same resistors that will provide the smallest
value of resistance.
(1)
(ii)
Calculate this smallest value of resistance.
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(2)
(Total 6 marks)
Q8.A manufacturer asks you to design the heating element in a car rear-window de-mister. The
design brief calls for an output of 48 W at a potential difference of 12 V. The diagram below
shows where the eight elements will be on the car window before electrical connections are
made to them.
(a)
Calculate the current supplied by the power supply.
Current = ....................................................
(1)
(b)
One design possibility is for the eight elements to be connected in parallel.
(i)
Calculate the current in each element in this parallel arrangement.
Current = ....................................................
(1)
(ii)
Calculate the resistance required for each element.
Resistance = ....................................................
(2)
(c)
Another design possibility is to have the eight elements connected in series.
(i)
Calculate the current in each element in this series arrangement.
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Current = ....................................................
(1)
(ii)
Calculate the resistance required for each element.
Resistance = ....................................................
(2)
(d)
State one disadvantage of the series design compared to the parallel arrangement.
........................................................................................................................
........................................................................................................................
(1)
(e)
The series design is adopted. Each element is to have a rectangular cross-section of
0.12 mm by 3.0 mm. The length of each element is to be 0.75 m.
(i)
State the units of resistivity.
...............................................................................................................
(1)
(ii)
Calculate the resistivity of the material from which the element must be made.
Resistivity = ....................................................
(2)
(Total 11 marks)
Q9.Figure 1 shows several 12 V, 21 W lamps connected in parallel. The circuit is protected by a
fuse which melts if the current in the circuit exceeds 15 A.
Figure 1
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(a)
Determine the maximum number, n, of lamps that can be used without melting the
fuse.
n .......................................................................
(4)
(b)
Show that the working resistance of a single 12 V, 21 W lamp is 6.9 Ω.
(2)
(c)
Two of the 12 V, 21 W lamps are connected in parallel with a 12 V, 4.0 W lamp of
resistance 36 Ω as shown in Figure 2.
Figure 2
Calculate the resistance of the parallel combination of lamps, when they are working
normally.
Resistance .........................................................................
(3)
(Total 9 marks)
Q10.
The diagram below shows a potential divider consisting of a resistor in series with a
light dependent resistor. The voltmeter connected in parallel with the light dependent
resistor has an infinite resistance. The battery has an emf of 16V with a negligible internal
resistance.
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(a)
Calculate the reading on the voltmeter when the light dependent resistor has a
resistance of 1200 Ω.
Voltmeter reading .................................................
(2)
(b)
The light intensity in the room is increased. State and explain what happens to the
resistance of the LDR and the reading on the voltmeter.
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(3)
(Total 5 marks)
Q11.
The diagram shows a power supply connected to a car battery in order to charge the
battery. The terminals of the same polarity are connected together to achieve this.
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(a)
The power supply has an emf of 22 V and internal resistance 0.75 Ω. When charging
begins, the car battery has an emf of 10 V and internal resistance of 0.15 Ω. They are
connected together via a variable resistor R.
(i)
Calculate the total emf of the circuit when charging begins.
Total emf of circuit ...............................................
(1)
(ii)
The resistor R is adjusted to give an initial charging current of 0.25 A. Calculate
the value of R.
Resistance of R ....................................................
(3)
(b)
The car battery takes 8.0 hours to charge. Calculate the charge that flows through it in
this time assuming that the current remains at 0.25 A.
Charge ..................................................................
(1)
(Total 5 marks)
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Q12.
The circuit shown in the diagram below can be used as an electronic thermometer.
The battery has negligible internal resistance.
The reading on the digital voltmeter can be converted to give the temperature of the
thermistor T which is used as a temperature sensor.
(a)
Explain why the reading on the voltmeter increases as the temperature of the
thermistor increases.
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........................................................................................................................
(2)
(b)
When the thermistor is at 80.0 °C the voltmeter reading is 5.0 V. Show that the
resistance of the thermistor at this temperature is 4.0 Ω.
(1)
(c)
When the thermistor is at 20.0 °C its resistance is 24.5 Ω. Calculate the reading on
the voltmeter.
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Voltmeter reading ..............................
(2)
(d)
The battery is replaced with another having the same emf but an internal resistance
of 3.0 Ω.
(i)
Calculate the new voltmeter reading when the thermistor temperature is 80.0
°C.
Voltmeter reading ................................
(2)
(ii)
State and explain the effect, if any, on the measured temperature when the
thermistor is at 20.0 °C.
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(1)
(Total 8 marks)
Q13.
The diagram below shows two circuits X and Y that were used by a student to test a
battery of three identical cells. In circuit X there was no load resistor and in circuit Y a load
resistor was connected. You can assume that the meters in the circuits were ideal. Their
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readings are shown on each diagram.
(a)
(i)
Explain what is meant by the internal resistance of a battery and why this
explains the difference between the voltages recorded in the two circuits.
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(3)
(ii)
Calculate the internal resistance of a single cell.
Internal resistance ......................................................
(3)
(b)
One of the cells in the battery is reversed. Determine the new reading:
(i)
on the voltmeter in circuit X;
Voltmeter reading in X ......................................................
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(1)
(ii)
on the ammeter in circuit Y.
Ammeter reading in Y ......................................................
(2)
(Total 9 marks)
Q14.
The circuits in Figure 1 and Figure 2 both contain a 6.0 V supply of negligible internal
resistance. Each circuit is designed to operate a 2.5 V, 0.25 A filament lamp L.
The lamp works normally in both circuits.
Figure 1
(a)
Figure 2
Calculate the resistance of the filament lamp when working normally.
resistance .................................................
(2)
(b)
Calculate the resistance of the resistor that should be used for R in Figure 1.
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resistance .................................................
(2)
(c)
Calculate the total resistance of the circuit in Figure 2.
total resistance ................................................
(3)
(d)
Explain which circuit dissipates the lower total power.
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(3)
(Total 10 marks)
Q15.
(a)
Define the electromotive force (emf) of an electrical power supply.
........................................................................................................................
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........................................................................................................................
(2)
(b)
Explain why, when a battery is supplying a current to a circuit, the voltage measured
between its terminals is less than its emf.
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........................................................................................................................
........................................................................................................................
(2)
(c)
In the circuit shown in the figure below the voltmeter has a very high resistance and
the resistance of the ammeter is negligible. The motor M is being tested using a
battery with an emf of 9.00 V.
(i)
State the reading on the voltmeter when the switch S is open.
voltmeter reading .................................
(ii)
When S is closed and the motor is allowed to run freely the voltmeter reading is
8.41 V and the ammeter reads 0.82 A. Calculate the internal resistance of the
battery.
internal resistance .............................
(iii)
Explain why the ammeter reading is greater than 0.82 A when the motor does
work by lifting a load.
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(5)
(Total 9 marks)
Q16.
(a)
Define the electrical resistance of a component.
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........................................................................................................................
........................................................................................................................
(2)
(b)
Calculate the total resistance of the arrangement of resistors in the figure below.
total resistance ...............................
(3)
(c)
(i)
Calculate the current in the 3.0 Ω resistor in the figure above when the current
in the 9.0 Ω resistor is 2.4 A.
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current in the 3.0 Ω resistor ....................................
(ii)
Calculate the total power dissipated by the arrangement of resistors in the
figure above when the current in the 9.0 Ω resistor is 2.4 A.
total power ....................................
(4)
(Total 9 marks)
Q17.In each of the following circuits the battery has negligible internal resistance and the bulbs
are identical.
Figure 1
(a)
Figure 2
For the circuit shown in Figure 1 calculate
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(i)
the current flowing through each bulb,
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...............................................................................................................
(ii)
the power dissipated in each bulb.
...............................................................................................................
...............................................................................................................
(2)
(b)
In the circuit shown in Figure 2 calculate the current flowing through each bulb.
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........................................................................................................................
(3)
(c)
Explain how the brightness of the bulbs in Figure 1 compares with the brightness of
the bulbs in Figure 2.
........................................................................................................................
........................................................................................................................
(2)
(Total 7 marks)
Q18.A battery of e.m.f. 12 V and internal resistance r is connected in a circuit with three resistors
each having a resistance of 10 Ω as shown. A current of 0.50 A flows through the battery.
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Calculate
(a)
the potential difference between the points A and B in the circuit,
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(b)
the internal resistance of the battery,
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(c)
the total energy supplied by the battery in 2.0 s,
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(d)
the fraction of the energy supplied by the battery that is dissipated within the battery.
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(Total 7 marks)
Q19.(a)
Draw, on the axes below, the current/voltage characteristic for a filament lamp.
Do not insert any values for current or voltage.
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(3)
(b)
Explain why the characteristic has the shape you have drawn.
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........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The current/voltage characteristic of a filament lamp is to be determined using a
datalogger, the data then being fed into a computer to give a visual display of the
characteristic. Draw the circuit diagram required for such an experiment and state
what is varied so as to produce a range of values.
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(5)
(Total 11 marks)
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Q20.A battery of e.m.f. 12 V and negligible internal resistance is connected to a resistor network
as shown in the circuit diagram.
(a)
Calculate the total resistance of the circuit.
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........................................................................................................................
(3)
(b)
Calculate the current through the 50 Ω resistor.
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(1)
(Total 4 marks)
Q21.A heating element, as used on the rear window of a car, consists of three strips of a resistive
material, joined, as shown in the diagram, by strips of copper of negligible resistance. The
voltage applied to the unit is 12 V and heat is generated at a rate of 40 W.
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(a)
(i)
Calculate the total resistance of the element.
...............................................................................................................
...............................................................................................................
(ii)
Hence show that the resistance of a single strip is about 11 Ω.
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(5)
(b)
If each strip is 2.6 mm wide and 1.1 mm thick, determine the length of each strip.
resistivity of the resistive material = 4.0 × 10–5 Ω m
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(3)
(Total 8 marks)
Q22.(a)
A cell of emf ϵ and internal resistance r is connected in series to a resistor of
resistance R as shown. A current I flows in the circuit.
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(i)
State an expression which gives ϵ in terms of I, r and R.
...............................................................................................................
(ii)
Hence show how VR, the potential difference across the resistor, is related to ϵ, I
and r.
...............................................................................................................
(2)
(b)
A lamp, rated at 30 W, is connected to a 120 V supply.
(i)
Calculate the current in the lamp.
...............................................................................................................
...............................................................................................................
(ii)
If the resistor in part (a) is replaced by the lamp described in (b), determine how
many cells, each of emf 1.5 V and internal resistance 1.2 Ω, would have to be
connected in series so that the lamp would operate at its proper power.
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(5)
(Total 7 marks)
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Q23.In the circuit shown, the battery has negligible internal resistance.
Calculate the current in the ammeter when
(a)
the terminals X and Y are short-circuited i.e. connected together,
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(2)
(b)
the terminals X and Y are connected to a 30 Ω resistor.
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(4)
(Total 6 marks)
Q24.The circuit diagram shows a light-emitting diode connected in series with a resistor R and a
3.0 V battery of negligible internal resistance. The potential difference across the terminals
of the diode is 2.0 V and the current through it is 10 mA. The diode emits photons of
wavelength 635 nm.
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(a)
Calculate the resistance of R.
........................................................................................................................
........................................................................................................................
(b)
Calculate the electrical power supplied to the diode.
........................................................................................................................
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(c)
Calculate the energy of a photon of wavelength 635 nm.
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(d)
Estimate the number of photons emitted per second by the diode.
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(e)
State an assumption you made in your estimation in part (d).
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(Total 8 marks)
Q25.In the circuit shown, the battery has negligible internal resistance.
(a)
(i)
If the emf of the battery = 9.0 V, R1 = 120 Ω and R2 = 60 Ω, calculate the current
I flowing in the circuit.
...............................................................................................................
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(ii)
Calculate the voltage reading on the voltmeter.
...............................................................................................................
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(4)
(b)
The circuit shown in the diagram acts as a potential divider. The circuit is now
modified by replacing R1 with a temperature sensor, whose resistance decreases as
the temperature increases.
Explain whether the reading on the voltmeter increases or decreases as the
temperature increases from a low value.
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(3)
(Total 7 marks)
Q26.
A battery of emf 24 V and negligible intemal resistance is connected to a resistor
network as shown in the circuit diagram in the diagram below.
(a)
Show that the resistance of the single equivalent resistor that could replace the four
resistors between the points A and B is 50 Ω.
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(4)
(b)
If R1 is 50 Ω, calculate
(i)
the current in R1,
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(ii)
the current in the 60 Ω resistor.
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(4)
(Total 8 marks)
Q27.The circuit diagram shows a light emitting diode (LED) connected in series with a resistor, R,
and a 3.0 V battery of negligible internal resistance.
(a)
The LED lights normally when the forward voltage across it is 2.2 V and the current in
it is 35 mA.
Calculate
(i)
the resistance of R,
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(ii)
the number of electrons that pass through the LED each second.
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...............................................................................................................
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(4)
(b)
The LED emits light at a peak wavelength of 635 nm.
(i)
Calculate the energy of a photon of light of this wavelength.
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...............................................................................................................
(ii)
Estimate the number of photons emitted by the LED each second when the
current through it is 35 mA. Assume all the photons emitted by the LED are of
wavelength 635 nm and that all the electrical energy produces light.
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(4)
(Total 8 marks)
Q28.
(a)
X and Y are two lamps. X is rated at 12 V, 24 W and Y at 6.0 V, 18 W. Calculate
the current through each lamp when it operates at its rated voltage.
X: .................................................................................................................
Y: ..................................................................................................................
(2)
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(b)
The two lamps are connected in the circuit shown. The battery has an emf of 27 V and
negligible internal resistance. The resistors R1 and R2 are chosen so that the lamps
are operating at their rated voltage.
(i)
What is the reading on the voltmeter?
.............................................................................................................
(ii)
Calculate the resistance of R2.
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(iii)
Calculate the current through R1.
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(iv)
Calculate the voltage across R1.
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(v)
Calculate the resistance of R1.
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(7)
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(Total 9 marks)
Q29.
In the circuit shown, the battery has an emf of 12 V and an internal resistance of 2.0 Ω.
The resistors A and B each have resistance of 30 Ω.
Calculate
(i)
the total current in the circuit,
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.............................................................................................................
(ii)
the voltage between the points P and Q,
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(iii)
the power dissipated in resistor A,
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(iv)
the energy dissipated by resistor A in 20 s.
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(Total 8 marks)
Q30.
Four resistors, each having resistance of 50 Ω, are connected to form a square. A
resistance meter measured the resistance between different corners of the square.
Determine the resistance the meter records when connected between the following corners.
(a)
Between A and C, as in Figure 1.
Figure 1
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......................................................................................................................
(2)
(b)
Between A and B, as in Figure 2.
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Figure 2
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(3)
(Total 5 marks)
Q31.
In the circuit shown the battery has emf
(a)
(i)
and internal resistance r.
State what is meant by the emf of a battery.
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(ii)
When the switch S is open, the voltmeter, which has infinite resistance, reads
8.0 V. When the switch is closed, the voltmeter reads 6.0 V.
Determine the current in the circuit when the switch is closed.
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.............................................................................................................
(iii)
Show that r = 0.80 Ω.
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.............................................................................................................
(4)
(b)
The switch S remains closed. Calculate
(i)
the power dissipated in the 2.4 Ω resistor,
.............................................................................................................
(ii)
the total power dissipated in the circuit,
.............................................................................................................
.............................................................................................................
(iii)
the energy wasted in the battery in 2 minutes.
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(4)
(Total 8 marks)
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Q32.
(a)
In the circuit in Figure 1, the battery, of emf 15 V and the negligible internal
resistance, is connected in series with two lamps and a resistor. The three
components each have a resistance of 12 Ω.
Figure 1
(i)
What is the voltage across each lamp?
.............................................................................................................
(ii)
Calculate the current through the lamps.
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.............................................................................................................
(3)
(b)
The two lamps are now disconnected and reconnected in parallel as shown in Figure
2.
Figure 2
(i)
Show that the current supplied by the battery is 0.83 A.
.............................................................................................................
.............................................................................................................
Page 39
Colonel Frank Seely School
.............................................................................................................
.............................................................................................................
(ii)
Hence show that the current in each lamp is the same as the current in the
lamps in the circuit in Figure 1.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(c)
How does the brightness of the lamps in the circuit in Figure 1 compare with the
brightness of the lamps in the circuit in Figure 2?
Explain your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 8 marks)
Q33.The circuit in Figure 1 has a thermistor connected in series to a 200 Ω resistor and a 12 V
battery of negligible internal resistance. Figure 2 shows how the resistance, Rth, of the
thermistor varies with temperature.
Page 40
Colonel Frank Seely School
(a)
(i)
Calculate the current in the circuit when the temperature is 25°C.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the potential difference across the thermistor at 25°C.
...............................................................................................................
...............................................................................................................
(3)
(b)
Without further calculation, explain how you would expect the potential difference
across the thermistor to change as the temperature increases from 25°C.
You may be awarded marks for the quality of written communication in your answer.
Page 41
Colonel Frank Seely School
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The circuit in Figure 1 is modified by removing the 200 Ω resistance to give the circuit
in Figure 3.
The temperature of the thermistor is increased at a steady rate from 25°C to 45°C in
10 minutes.
(i)
Calculate the power dissipated in the thermistor at
25°C ......................................................................................................
...............................................................................................................
45°C ......................................................................................................
...............................................................................................................
(ii)
Use the mean value of the powers determined in part (c)(i) to calculate the
energy supplied by the battery during the period in which the temperature of the
thermistor increases.
Page 42
Colonel Frank Seely School
...............................................................................................................
...............................................................................................................
(iii)
State why the energy value, determined in part (c)(ii) is not an accurate value.
...............................................................................................................
...............................................................................................................
(6)
(Total 12 marks)
Q34.
In the circuit shown, a battery of emf
and internal resistance r is connected to a
variable resistor R. The current I and the voltage V are read by the ammeter and voltmeter
respectively.
(a)
The emf is related to V, I and r by the equation
= V + Ir
Rearrange the equation to give V in terms of
, I and r.
......................................................................................................................
(1)
(b)
In an experiment, the value of R is altered so that a series of values of V and the
corresponding values of I are obtained. Using the axes, sketch the graph you would
expect to obtain as R is changed.
Page 43
Colonel Frank Seely School
(2)
(c)
State how the values of
and r may be obtained from the graph.
..................................................................................................................
r ....................................................................................................................
(2)
(Total 5 marks)
Q35.
In the circuit shown, a battery of emf
and internal resistance r is connected to a
variable resistor R. The current I and the voltage V are read by the ammeter and voltmeter
respectively.
(a)
The emf is related to V, I and r by the equation
= V + Ir
Rearrange the equation to give V in terms of
, I and r.
......................................................................................................................
(1)
Page 44
Colonel Frank Seely School
(b)
In an experiment, the value of R is altered so that a series of values of V and the
corresponding values of I are obtained. Using the axes, sketch the graph you would
expect to obtain as R is changed.
(2)
(c)
State how the values of
and r may be obtained from the graph.
..................................................................................................................
r ....................................................................................................................
(2)
(Total 5 marks)
Page 45
Colonel Frank Seely School
M1.(a)
1 / R1 + 1 / R2 = 1 / RT or
(i)
= Rt
C1
correct substitution using total resistance of 5 Ω
C1
7.5 Ω
A1
(3)
(ii)
3 V or 3.0 V
(no significant figure penalty)
B1
(1)
(iii)
power in 15 Ω resistor = V2R or I2R or IV
C1
power = 0.6 or 0.60 W
allow e.c.f. from (ii)
A1
(no significant figure penalty)
(2)
(iv)
4 Ω resistor is too small (for maximum power to be delivered)
C1
placing a resistor in parallel will reduce the load resistance further
A1
leading to R = –20 Ω
or
B1
negative resistance needed which is impossible
B1
(2)
(b)
(i)
R=
or ρ =
or numerical evidence
C1
area = π(1.5 × 10–4)2 or
(3 × 10–4)2 or 7.1 × 10–8 (m2)
C1
Page 46
Colonel Frank Seely School
4.6 × 10–7 m
c.a.o.
A1
(3)
(ii)
correct curvature, convex to origin, not intersecting diameter axis
M1
inverse square law shown
A1
(check one point eg 6 × 10–4 m, 3.5 → 4.0 Ω; 4 × 10–4 m, 8.5 Ω)
(2)
[13]
M2.B
[1]
M3.(a)
(i)
current = emf / total resistance or
or equivalent
C1
current = 0.77 A
C1
V = 2.3(1) V
A1
(or alternative approach using potential divider formula)
(3)
(ii)
wider range of voltages available
from 0 to 10 V or since lower minimum voltage
B1
(1)
(iii)
resistance of lower part is reduced
(since 3 Ω resistor is in parallel with lower half of potentiometer)
or calculation of resistance = 2.5 Ω
B1
Page 47
Colonel Frank Seely School
so voltage across lower section is reduced or calculation of p.d. = 1.43 V
B1
allow 1 for calculation of p.d using {3 / (3 + 15) × 10} = 1.67 V
(2)
(iv)
point labelled at least
of the way toward the top of the potentiometer
(correct position is just over
of the way up)
B1
(2)
(b)
(i)
power = V2 / R
B1
V / 9 = V / 4 with correct clear manipulation
2
1
2
2
A1
or
calculates V1 (6 V) and V2 (4 V)
C1
calculates power in each resistor (4 W) showing them to be equal
A1
(2)
(ii)
4(.0) W (c.a.o.)
B1
NB not half of (102 / 13)
(1)
[10]
M4.(a)
at higher temperatures
more charge carriers / electrons are liberated
or more electrons moving in the material
or more electrons become mobile
not electrons move faster or have more energy
B1
(for a given p.d.)
more charge carriers flow increasing current
or use of I = nAvq to explain increased current
or there are more electrons in the conduction band
or this more than makes up for increased resistance due to lattice vibrations
B1
(2)
Page 48
Colonel Frank Seely School
(b)
(i)
at 60 °C, R = 340 Ω
(allow if 340 is seen used in equations)
B1
quotes and attempts to apply the potential divider formula
or tries to determine the circuit current using emf/ total resistance (0.011
A)
and hence the pd across the thermistor using V = IR
C1
3.7(4) V
allow ecf for incorrect R for the thermistor
320 Ω gives 3.6(3) V; 380 Ω gives 3.4 V
A1
(3)
(ii)
power generated = power removed or P = I2R; V2 / R; VI
C1
0.040 W
allow e.c.f. from (b)(i)
320 Ω 3.63 V gives 0.041W; 380 Ω, 3.4 V gives 0.038 W
A1
(2)
(c)
(i)
graph decreasing in any shape until –80 °C where it decreases rapidly to
zero
graph must show a transition temperature at –80 °C
B1
(1)
(ii)
acceptable application
as connectors in amplifiers / electronic circuits/computers
to produce higher memory capacity in computers
to produce strong magnets (condone maglev or MRI scanners)
to make high power motors
to transmit electrical power
in transformers
B1
explanation of advantage
because amplifiers would have low thermal noise
because large currents can be produced
because little or no (thermal) energy is generated
because no energy is transferred in conductors
B1
(2)
[10]
Page 49
Colonel Frank Seely School
M5.C
[1]
M6.A
[1]
M7.(a)
(i)
diagram showing all three resistors in series
B1
(1)
(ii)
R = 1 + 2 + 3 (allow e.c.f. from (a)(i))
C1
= 6 ohm [allow 6.0; 6.00]
A1
(2)
(b)
(i)
diagram showing all three resistors in parallel
B1
(1)
(ii)
use of formula correct
C1
= 6 / 11 = 0.54(5) ohm
A1
(2)
[6]
M8.(a)
I = P / V = 48 / 12 = 4.0 A
B1
(1)
Page 50
Colonel Frank Seely School
(b)
(i)
0.5 A
ecf their (a) / 8
B1
(1)
(ii)
R = V / I = 12 / 0.5
M1
= 24 ohm
A1
(2)
(c)
(i)
candidate (a) value
B1
(1)
(ii)
1.50 / 4
M1
= 0.37(5) ohm [ecf ]
A1
(2)
(d)
failure of one element breaks whole unit
B1
(1)
(e)
(i)
ohm-metre [allow correct symbols]
B1
(1)
(ii)
p = RA / I [= 0.375 (e.c.f.) × (3.0 × 10 ) × (0.12 × 10 ) / 0.75]
–3
–3
M1
= 1.8 × 10–7 [ Ω m]
A1
(2)
[11]
M9.(a)
P = VI OR P =
and V = IR OR P = I2 R and
V = IR
Page 51
Colonel Frank Seely School
C1
I = 21/12 or 1.75 (A)
C1
Divides 15 by their 1.75
C1
8
OR max power = 12 × 15
Max power = 180
Max n
= 180 / 21
8
A1
(b)
correct equation given in symbols or words
B1
correct substitution seen (look for 144 / 21 or 122 / 21
or 12 / 1.75)
B1
(c)
C1
(1 / RT) = 0.32 even if they forget to invert
C1
3.1 or 3.2 Ω
A1
[9]
M10.
(a)
[V1 = V × R1/(R1 + R2)]
C1
= 16 × 1200/2000 = 9.6 V
A1
2
(b)
LDR resistance drops
B1
Page 52
Colonel Frank Seely School
voltmeter reading decreases
B1
because more conduction electrons/charge
carriers released
B1
3
[5]
M11.
(a)
(i)
22-10 = 12 V
B1
1
(ii)
use of V = IR
C1
Rtotal = 12/0.25 = 48
C1
So R = 48 – 0.9 = 47.1 Ω
A1
3
(b)
Charge = It = 0.25 × 8 × 3600 = 7200 C
[cnao]
B1
1
[5]
M12.
(a)
as the temperature of T increases its resistance decreases
/more charge carriers are released
B1
increasing the current in the circuit
Page 53
Colonel Frank Seely School
/changing the ratio of resistance/reducing pd across T
B1
(so that so that the pd across the resister increases)
2
(b)
T/ 20.0 = 1.0/5.0 OR 5.0/6.0 = 20/(20+T) OR equivalent
(Therefore T = 4.0 ohms)
Note T = (1/5)20 just ok but T = 20/5 not enough
M1
1
(c)
Use of Vout = R1/(R1 + R2) × Vin OR I = 6/44.5 = 0.135 A
C1
V = 2.7 V
A1
2
(d)
(i)
V/6.0 = 20.0/(20.0+4.0+3.0) OR I = 0.222 A
C1
V = 4.4V
A1
2
(ii)
The measure temperature would be lower because
the pd across the resistor would be less (ie 2.53V)
B1
1
[8]
M13.
(a)
(i)
resistance inside the battery
or
resistance of component/chemicals in the battery
Page 54
Colonel Frank Seely School
or
resistance of the cell/battery
B1
some pd is used (lost) to pass current/charge through
the internal resistance
or internal resistance gives rise to lost volts
or terminal pd = emf – current x internal resistance
(lost volts) (no symbolic equations unless terms defined)
M1
in X there is no current /open circuit (so no pd is lost)
or in Y there is a current (so pd is lost)
A1
3
(ii)
V = E – Ir or 3.1 = 4.5 – 0.39 r
(n.b. 4.5 = 3.1 – 0.39 r gets 0)
C1
r =3.6 (3.59) Ω
A1
their r/3 (1.2 Ω if correct)
B1
3
(b)
(i)
Reading on voltmeter in X = 1.5 V
B1
1
(ii)
Resistance in circuit Y = 8.0 + 3.6 = 11.6Ω
C1
their (i) / (8 + their total internal r)
(1.5/11.6 = 0.13 A if correct)
A1
2
or
lost volts = 3.6/11.6 × 1.5 = 0.47 V (allow ecf for r)
terminal pd = 1.5 – 0.47 = 1.03 V (must see working for
this mark)
C1
Page 55
Colonel Frank Seely School
1.03/8 = 0.13 A if correct
A1
[9]
M14.
(a)
R = V/I
C1
10 Ω
A1
2
(b)
total resistance = 6/0.25 = 24 Ω
or correct substitution in potential divider formula
C1
14 Ω
A1
2
(c)
correct substitution in parallel resistance formula
or current through 15 Ω = 0.16 A
C1
or RΩ=
or
resistance of parallel combination = 6.0 Ω
or parallel combination resistance + 8.4
(allow in substitution form)
or total current = 0.41 A
14.4 Ω (cnao)
C1
Page 56
Colonel Frank Seely School
{
incorrect answer would get C1, C1
(may use incorrect (a) value here)
A1
3
(d)
identifies circuit with lower or higher total resistance correctly
(e.c.f.)
B1
power = V2/R or VI
B1
V is same for both so circuit with higher resistance
dissipates lower power or V is same for both and
circuit with lower current dissipates lower power
or reason by means of calculations
B1
3
[10]
M15.
(a)
the (total) energy transferred/work done when
one unit/coulomb of charge
B1
is moved around a circuit/provided by the supply
B1
2
(b)
work is done inside the battery/there is resistance
inside the battery
B1
so less energy is available for the external circuit/someoltage
is lost between the terminal/mention of lost volts
Page 57
Colonel Frank Seely School
B1
2
(c)
(i)
9.00 V
c.a.o.
B1
(ii)
lost voltage = E - V or E = I(R + r)
C1
0.82r = 0.59
C1
5
internal resistance = 0.720 Ω
A1
(iii)
because the battery has to provide more energy/power
B1
[9]
M16.
(a)
R = V/I
M1
with all three variables defined
accept voltage
A1
2
(b)
use of 1/R = 1/R1 + 1/R2
C1
Page 58
Colonel Frank Seely School
effective resistance of parallel resistors = 2
C1
3
total resistance = 11 Ω
A1
(c)
(i)
ratio 2/3 seen/ V= 4.8 V used
/clear attempt to find pd across parallel resistors
C1
current = 1.6 A
c.a.o.
A1
4
(ii)
use of P= I2R (= 2.42 × 11)
C1
total power = 63.4 W
allow e.c.f. from (b)
A1
[9]
M17.(a)
(i)
I=
(ii)
P = (0.80) × 5 = 3.2 W (1) (allow e.c.f. from (a)(i))
= 0.80 A (1)
2
(2)
Page 59
Colonel Frank Seely School
(b)
I =
(1) = 1.60 (A) (1)
tot
I=
= 0.80 (A) (1) (allow e.c.f. from Itot)
(3)
(c)
same brightness (1)
because same current (1)
[or an answer consistent with their current values]
(2)
[7]
M18.(a)
(b)
R = 5.0 (Ω) (1)
V (= 5.0 × 0.50) = 2.5 V (1)
AB
V = 12 – 2.5 + 5.0 (1) = 4.5 (V) (1)
r
= 9.0 Ω (1)
r=
(c)
W(= EIt) = 12 J (1)
(d)
W (= V It) = 4.5 (J) (1)
r
r
= 0.375 (1)
[Max 7]
Page 60
Colonel Frank Seely School
M19.(a)
correct curve in positive quadrant (1)
correct curve in negative quadrant (1)
passing through origin (1)
(3)
(b)
the current heats the filament (1)
(temperature rises) resistance increases (1)
pd. and current do not increase proportionally (1)
some reference to mirror image in negative quadrant (1)
The Quality of Written Communication marks were awarded primarily for the
quality of answers to this part
(max 3)
(c)
diagram to show:
battery, variable resistance (or variable supply) and filament (1)
current sensor in series circuit (1)
voltage sensor across filament (1)
the two sensor boxes connected to datalogger (1)
method:
variable resistor or variable supply altered
[or choose recording interval] (1)
thus changing both V and I (1)
The Quality of Written Communication marks were awarded primarily for the
quality of answers to this part
(max 5)
[11]
M20.(a)
(three parallel resistors) give
R = 10 (Ω) (1)
10 Ω and 50 Ω in series gives 60 Ω (1)
(allow e.c.f. from value of R)
(3)
Page 61
Colonel Frank Seely School
(b)
(V = IR gives) 12 = I × 60 and I = 0.2 A (1)
(allow e.c.f. from (a))
(1)
[4]
M21.(a)
(i)
P=
gives 40
(1)
R = 3.6 Ω (1)
[or P = VI to give I = 3.3 (A) (1) and R = P / I = 3.7 Ω (3.67 Ω) (1)]
T
2
(ii)
three resistors in parallel (1)
R = 3.6 × 3 = 10.8 (Ω) (1)
(allow C.E. for RT from (i))
5
(b)
(use of R =
= gives) 10.8 =
(1)
l=
(1)
(allow C.E for R from (a)(ii)
= 0.77 m (1)
3
[8]
M22.(a)
(i)
ϵ = I (R + r) (1)
(ii)
V = IR gives V = ϵ - I (1)
R
R
r
2
(b)
(i)
P = VI gives 30 = 120 I (1)
I = 0.25 A (1)
(ii)
I through lamp = 0.25 (A) and p.d. across it = 240 V (1)
Page 62
Colonel Frank Seely School
p.d. due to 1 cell = 1.5 – (0.25 × 1.2) = 1.2 (V) (1)
number of cells =
= 100 (1)
[or RL given by 30 = 0.252 RL and RL = 480 (Ω) (1)
1.5n = 0.25(480 + 1.2n) (1)
1.2n = 120 and n = 100 (1)]
[or ϵ = V + Ir gives 1.5n = 120 + 0.25 × 1.2n (2)
n = 100 (1)]
5
[7]
M23.(a)
only 30 Ω in the circuit (1)
(use of V = IR gives) 6 = I × 30 and I = 0.20 A (1)
2
(b)
two resistors in parallel gives
and R = 20 (Ω) (1)
total resistance = 20 + 30 = 50 (Ω) (1)
(allow C.E. for value of R)
I=
= 0.12 A (1) (allow C.E for total resistance)
4
[6]
M24.(a)
V = (3.0 – 2.0) = 1.(0) (V) (1)
R
R=
(b)
= 100 Ω (1)
(use of P = IV gives) Pdiode (= 10 × 10–3 × 2.0) = 0.02(0) W (1)
Page 63
Colonel Frank Seely School
(c)
(use of c = fλ gives)
) = (4.7 × 10 Hz) (1)
14
(use of E = hf gives) E (= 6.63 × 10–34 × 4.7 × 1014) = 3.1 × 10–19 J (1)
(allow C.E. for A.E. in value of f)
(d)
energy supplied in 1 sec = 0.02(0) (J) (1)
(allow C.E. for value of P from (ii))
number of photons emitted in 1 sec =
= 6.5 × 1016 (1)
(allow C.E. for value of E)
(e)
all the energy supplied converted to light energy [or 100% efficient]
[or monochromatic light]
[or all photons (emitted by LED) have the same energy] (1)
[8]
M25.(a)
(i)
(use of V = IR gives) V = I(R1 + R2) (1)
= 50 mA ✓
(ii)
V (= IR ) = 0.05 × 60 = 3 V (1)
(allow C.E. for value of I from (i))
out
2
4
(b)
(temperature increases, resistance decreases), total resistance decreases (1)
current increases (1)
voltage across R2 increases (1)
[or R2 has increased share of (total) resistance (1)
new current is same in both resistors (1) larger share of the 9 V (1)]
[or Vout = Vin
(1) R1 decreases (1) Vout decreases (1)]
3
[7]
Page 64
Colonel Frank Seely School
M26.
(a)
first pair in parallel
=
= gives R’ = 20 (Ω) (1)
(1)
gives R” = 30(Ω) (1)
second pair in parallel
resistance between A and B = 20 + 30 (1) (= 50 Ω)
(allow C.E. for values of R’ and R")
4
(b)
(i)
total resistance = 50 + 50 = 100 Ω (1)
(V = IR gives) 24 = I 100 and I = 0.24 A (1)
(ii)
current in 60 Ω =
I (1)
= 0.080 (A) (1)
[or alternative method]
(allow C.E. for value of I from (b)(i))
4
[8]
M27.(a)
(i)
pd across resistor (= 3.0 – 2.2) = 0.8 (V) (1)
= 23 Ω (1) (22.9 Ω)
(use of V = IR gives) R =
(ii)
charge flow in 1 s = 0.035 (C) (1)
= 2.2 × 1017 (1)
no. of electrons (in 1 s)
(2.19 × 1017)
4
Page 65
Colonel Frank Seely School
(b)
(i)
(use of E = hf =
gives) E =
(1)
= 3.1(3) × 10–19 J (1)
(ii)
(use of P = VI gives) P (= 2.2 × 0.035) = 0.077 (W)
= 63 (Ω)]
[or use of P = I2R with R
maximum no. of photons emitted per sec. =
= 2.5 × 1017 (1) (2.48 × 1017)
(allow C.E. for value of E from (i) and value of P from (ii))
4
[8]
M28.
(a)
(i)
for X: (P = VI gives) 24 = 12I and I = 2 A (1)
for Y 18 = 6I and I = 3 A (1)
2
(b)
(i)
(ii)
12 V (1)
voltage across R2 (= 12 – 6) = 6 (V) (1)
I = 3 (A) (1)
(V = IR gives) 6 = 3R2 and R2 = 2Ω (1)
(allow C.E. for I and V from (a) and (b)(i))
[or V = I(Ry + R2) (1) 12 = 3(2 + R2) (1)
R2 = 2Ω (1)]
(iii)
current = 2 (A) + 3 (A) = 5 A (1)
(allow C.E. for values of the currents)
(iv)
27 (V)– 12 (V) = 15 V across R1 (1)
(v)
for R1, 15 = 5 R1 and R1 = 3Ω (1)
(allow C.E. for values of I and V from (iii) and (iv)
7
[9]
Page 66
Colonel Frank Seely School
M29.
(i)
(V = IR gives)
I=
(ii)
12 = (30 + 30 + 2)I (1)
= 0.19 A (1)
(0.194 A)
VPQ = 12 – (0.19 × 2) (1)
= 11.6 V (1)
(allow C.E. for incorrect I in (i))
[or VPQ = 0.19 × 60 = 11.6 V]
[or VPQ = 12 ×
(iii)
(I = 0.194 A gives 11.6 V)
= 11.6 V
(PA = I2R gives) PA = (0.19)2 × 30 = 1.08 (1)
[or PA =
W (1)
]
(allow C.E. for incorrect I in (i) or incorrect V in (ii))
(iv)
(E = PAt gives) E = 1.08 × 20 (1)
= 21.6 J (1)
(allow C.E. for incorrect PA in (iii))
[8]
M30.
(a)
between A and C: (each) series resistance = 100Ω (1)
+ = gives RAC = 50Ω (1)
(parallel resistors give)
2
(allow C.E. for incorrect series resistance)
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Colonel Frank Seely School
(b)
between A and B: series resistance = 150Ω (1)
parallel =
(1)
(allow C.E. for series resistance)
RAB = 37.5Ω (1) (38Ω)
3
[5]
M31.
(a)
(i)
energy changed to electrical energy per unit charge/coulomb
passing through
[or electrical energy produced per coulomb or unit charge]
[or pd when no current passes through/or open circuit] (1)
(ii)
I=
(iii)
(use of
= 2.5 A (1)
= I(R + r) gives)
= V + Ir and 8 = 6 + Ir (1)
substitution gives 8 – 6 = 2.5r (1) (and r = 0.8 Ω)
4
(b)
(i)
(use of P = I2R gives) PR = 2.52 × 2.4 = 15 W
[or P = VI gives P = 6 × 2.5 = 15 W] (1)
(allow C.E. for value of I from (a))
(ii)
PT = 15 + (2.52 × 0.8) (1)
= 20 (W) (1)
(allow C.E. for values of PR and I)
(iii)
E = 5 × 2 × 60 = 600 J (1)
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(allow C.E. for value of P from (i) and PT from (ii))
4
[8]
M32.
(a)
(i)
5 V (1)
(ii)
RT = 36 (Ω)
(use of V = IR gives) 15 = I × 36 and I = 0.42 A (1)
3
(b)
(i)
equivalent resistance of the two lamps
RT = 6 + 12 = 18 (Ω) and 15 = I × 18 (1)
(ii)
(1)
(to give I = 0.83 A)
current divides equally between lamps (to give I = 0.42 A)
(or equivalent statement) (1)
3
(c)
same brightness (1)
(because) same current (1)
2
[8]
M33.(a)
(i)
at 25 (°C), total resistance = 300 + 200 = 500 (Ω) (1)
I=
= 24 mA (1)
(allow C.E. for value of total resistance)
(ii)
pd across thermistor = 24 × 10–3 × 300 = 7.2 V (1)
(allow C.E. for value of current from (i) and Rth from graph)
3
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Colonel Frank Seely School
(b)
as temperature increases, resistance (of thermistor) decreases (1)
total resistance decreases (1)
current in circuit increases (1)
pd across resistor increases (1)
(since battery remains at 12 V) pd across thermistor decreases (1)
[or Rth decreases (1)
potential divider situation (1)
V = 12 ×
(1)
th
denominator decreases less slowly than numerator (1)
Vth decreases (1)
or for last two marks, thermistor gets smaller share of voltage
explanation of this]
(max 3)
QWC 1
(c)
(i)
gives) at 25°C
(use of P =
P=
= 0.48 W (1)
at 45°C correct reading of R = 30 (Ω) (1)
P=
(ii)
= 4.8 (W) (1)
E = Pt = 2.64 × 10 × 60 (1)
= 1.6 × 103 J (1)
(allow C.E. from part (i)
(iii)
rate of decrease of resistance is not linear
[or resistance not directly proportional to temperature] (1)
6
[12]
M34.
(a)
V = –Ir +
(1)
1
(b)
straight line (within 1st quadrant) (1)
negative gradient (1)
2
(c)
: intercept on voltage axis (1)
r: gradient (1)
2
[5]
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Colonel Frank Seely School
M35.
(a)
V = –Ir +
(1)
1
(b)
straight line (within 1st quadrant) (1)
negative gradient (1)
2
(c)
: intercept on voltage axis (1)
r: gradient (1)
2
[5]
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Colonel Frank Seely School
E1.(a)
(b)
There were many correct responses but the equation 5 = 1 / 15 + 1 / R was often
seen. Some were unable to extract the important information that the external
resistance must equal 5 Ω from the context of the question.
(i)
(ii)
Many resorted to lengthy calculations having failed to spot that with an external
resistance of 5.0 Ω and an internal resistance of 5.0 Ω the e.m.f. would divide
equally giving a terminal p.d. of 3.0 V.
(iii)
A minority of the candidates obtained the correct answer. Most candidates were
able to quote a power formula but most were then unable to use appropriate
data to determine the power in the 15 Ω resistor. Many used the supply current
calculated in an earlier part (usually 0.6 A) and assumed that this was the
current in the 15 Ω resistor.
(iv)
Many performed calculations to show that the resistance would need to be –20
Ω and went on to state that this was not possible. Others pointed out
successfully that the resistance of a parallel combination of resistors is always
less than the smallest component and then drew an appropriate conclusion.
Weak answers included those who stated that the resistance would be negative
without offering any evidence and those who simply stated that the external
resistance could not equal 5.0 n without further discussion.
(i)
Although the majority was successful in this part many attempts failed because
candidates did not know the formula for the area of a circle.
(ii)
As with the previous graphical communication question many candidates were
careless, showing only the general trend of the resistance-diameter curve. It
was expected that the inverse square shape would be clearly communicated in
graphical form.
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Colonel Frank Seely School
E3.(a)
(i)
Many were unable to make any progress with this part. The only slight complication was the
requirement to understand that the voltage is a minimum when the rheostat is set to 10 Ω
but this usually seemed to be the least of the candidates’ concerns. The formula quoted was
often that for terminal p.d. that led nowhere. Some quoted the potential divider formula and
were then unable to apply it. Other candidates worked out the circuit current and gave this
as the minimum voltage.
(b)
(ii)
Most answers to this part were very vague and relatively few acknowledged the
fact that it was possible to vary the voltage from 0 V – 10 V. Some stated only
that it ‘increased the range’ without further qualification. Some focused attention
on the resistor giving ‘finer control’ but this would depend on the structure of the
potentiometer (e.g. the resistance per unit length).
(iii)
Although many stated that the 3 Ω resistor would be in parallel with 15 Ω of the
potentiometer relatively few continued the argument to a satisfactory
conclusion. Many assumed that the final situation was 3 Ω in series with 15
Ωand drew a conclusion from this which gained some credit. Although only a
correct qualitative argument was required for full marks, the best candidates
actually calculated the voltage correctly.
(iv)
Many placed the new position toward the bottom of the potentiometer even after
making progress to the correct explanation in part (iii). Generally this was not
well done.
(i)
Many assumed the currents in the two components to be equal. Some who
quoted the correct starting formula (V2 / R) tried to determine the power
dissipated and prove them to be the same. This was an acceptable approach
but candidates were expected to use the actual voltages in their explanations
rather than the ratio given.
(ii)
There was a good proportion of correct answers but often, even when part (i)
was correct, this part was wrong. A significant proportion calculated the total
power using 102 / 13 and then halved it.
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E4.(a)
(b)
(c)
This is a fairly standard question at this level. Most were able to gain some credit for
stating that more electrons become free at higher temperatures but far fewer went on
to explain that this led to a higher rate of flow of charge and hence a higher current. A
significant proportion thought that free electrons would be given more energy so that
they moved more quickly. Some discussed the increase in the amplitude of lattice
vibrations and hence an increase in resistance which was the opposite of what they
were asked to explain.
(i)
Incorrect graph reading (320 Ω instead of 340 Ω) was not uncommon but there
were many correct responses to this part. Another common error was
calculating current from 5 / 120 A and then using this to determine the voltage
across the thermistor.
(ii)
Most were able to quote a correct formula for power and there was a good
proportion of correct answers (allowing errors carried forward in many cases).
Assuming that the p.d. across the thermistor was 5 V was a not uncommon
error.
(i)
Few showed a clear transition temperature at which the resistance suddenly
dropped to zero at −80°C. A common response was a graph showing the
resistance falling continuously and reaching zero at this temperature.
(ii)
Good applications and reasons were rare but the use in transmission of
electrical power was a frequent correct response. ‘Use in power stations’
without reference to what part of the power station or ‘to produce
electromagnets’ without reference to the fact that the magnets could be made
very strong were common as was ‘in computers’ or in ‘electronic circuits’ without
mention of what part they would be useful for. That large currents could be
produced or that there would be little or no thermal energy generated in the
application were the anticipated advantages. Some thought that electrical
signals would travel more quickly through superconductors and some seemed
to be confusing superconductivity with semiconductors.
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Colonel Frank Seely School
E7.(a)
(i)
Most candidates were able to recognise that three resistors in series give a maximum additive
resistance.
(b)
(ii)
Most completed this successfully.
(i)
The majority of the candidates knew that three in parallel combine to give a
minimum value.
(ii)
Predictably, the parallel calculation gave more difficulty than that in (a) with only
about 50% of the answers correct. The commonest fault was that the
candidates forgot to form the reciprocal of the sum of reciprocals and quoted 1.8
Ω instead of 0.55 Ω. Another frequent error was to write Rt = 1 / R1+ 1 / R2 + 1 / R3
E8.This question revealed a number of misconceptions about the physics of electrical circuits.
Strong candidates frequently scored high or full marks on the question. Weaker candidates
were confused by the physics of parallel and series circuits and did not understand the
consequences of the type of circuit for the values of potential difference across, and current
in, the resistors.
(a)
The calculation of current, given the power and p.d., was routine and well done by
many.
(b)
(i)
There were many correct calculations of the current here.
(ii)
However, the p.d. drop across each resistor was not well understood and
consequently there were frequent errors.
(i)
A common error was the failure to recognise that the current in each element of
a series circuit is equal to the total current.
(ii)
Many candidates went on to use a value of 12V for the p.d. across each of the
eight elements.
(c)
(d)
Most were able to suggest that, in the series arrangement, the entire heater fails if a
single element breaks.
(e)
(i)
About half of the candidates correctly stated the unit of resistivity. Common
incorrect responses included Ω m-1 and Ω m-3.
(ii)
Although many were able to manipulate R = ρl / A well, they then failed to carry
the calculation through usually because conversions to metres in the wire
dimensions were poor.
E9.(a)
Many candidates got this section right and there many ways of tackling the problem.
Those who did not manage a full solution would have been more likely to get more
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Colonel Frank Seely School
credit had they set their work out better and made it more obvious what they were
trying to do.
(b)
This part was also well done and, once again, there were several different
approaches that were used successfully.
(c)
The majority of candidates managed to find the resistance of the parallel combination
correctly but there were some that could not remember how to do so or who forgot the
final invertion in their calculation.
E10.
(a)
(b)
E11.
(b)
E12.
This question was well answered by large numbers of the entry.
Thinking was confused in this question, however. About half the candidates could not
correctly state or explain how the resistance of a light-dependent resistor varies with
light intensity. Those who knew this often could not go on to state correctly how the
voltmeter reading varies. Only rarely could a candidate write fluently in terms of the
ratio of the resistance values in the potential divider. Explanations were usually on a
much lower level and rarely went beyond: ‘resistance goes down so the voltage
drops’.
(a)
(i)
Many candidates simply failed to read the question or, alternatively, to
study the diagram. The relative polarity of the cells was stated in both places.
Consequently there were very many incorrect answers of 32 V.
(ii)
Candidates failed to take into account the internal resistance of the supplies
and failed to gain full credit.
This was an easy mark often, but not universally, scored. Failures were amongst
those who forgot that in the calculation the time needs to be in seconds.
Another difficult question when judged by the candidates. responses: more
candidates failed to score on this question than any other. The calculations proved to be
beyond most candidates. abilities.
Many good answers to part (a) were seen, and the better candidates scored on part (b) as
well, although the idea of showing something mathematically is unfamiliar with many.
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Colonel Frank Seely School
It was quite clear that very few candidates understood what was being asked in part (d)(ii).
E13.
(b)
E14.
(a)
(i)
Very few were able to provide clear reasoned arguments for the
difference in readings between the two circuits. Most were able to give a
sensible statement about what was meant by internal resistance, but fewer
could take the argument further. Surprisingly, few even referred to the internal
resistance giving rise to ‘lost volts’. It was not uncommon for candidates to
suggest that there was no internal resistance in circuit X because there is no
current, whereas in electricity were evident. There was a not uncommon
misconception that resistance is caused Y the internal resistance has
increased. Misconceptions about current by a current and increases as current
increases.
(ii)
Although not being able to explain the operation of the circuit in words, more
were able to make progress with the calculation and many correct answers
were seen. Forgetting to divide by 3 led to a lost mark, but on the other hand
many who mixed up emf and terminal pd, or made other errors, gained the mark
for appreciating that a division by 3 was necessary.
(i)
Few could visualise the effect of reversing one cell and 3.0 V was a common
alternative to 1.5 V.
(ii)
Allowing ecf there were many correct answers, but many had difficulty in
identifying the correct total resistance in the circuit, seeming to think that
reversing a cell meant a reduction in total internal resistance as well as total
emf.
(a)
This was usually completed successfully.
(b)
Again, there were many correct answers. A sizeable proportion of the candidates
calculated the total resistance in the circuit (24Ω) but then did not subtract the filament
resistance.
(c)
This was done poorly. Correct analysis of this simple circuit proved too much for a
significant proportion of the candidates. Although identifying the formula, many could
not identify which components were in parallel and which were in series. Many were
inadequately skilled in the management of the reciprocals.
(d)
Most were able to identify the circuit which had the higher resistance correctly. The
next step was to appreciate that as the voltage is the same in both cases and, as P =
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Colonel Frank Seely School
V2/R, the circuit with the higher resistance dissipates lower power. A common error
was to choose, inappropriately, the equation P = 12R and then to assume that the
current in both circuits is the same which leads to the wrong conclusion.
E15.
E16.
A question which discriminated well between the good and the very good candidates,
while many of the less able candidates found it difficult to gain more than one or two marks
at best.
(a)
Many candidates were unable to give a good definition of electromotive force. The
examiners had hoped to see statements referring to total energy transfer per unit
charge as it is moved around a circuit. While a high impedance voltmeter can be used
to measure the emf of a supply, on an open circuit this does not define electromotive
force.
(b)
A few more candidates were able to give a satisfactory answer here than in part (a).
(c)
(i)
As the emf was given to 3 significant figures in the question, the reading was
expected to also be to 3 significant figures. This proved to be harder than
expected.
(ii)
Many candidates simply divided the voltage by the current.
(iii)
Few candidates were able to provide a satisfactory answer to this part. A
statement to the effect that the battery needs to provide more energy to do the
extra work was all that was required, but very many candidates wrote vaguely
about decreased resistance and/or greater voltage.
(a)
Fewer than half of the candidates were able to fully define electrical resistance.
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(b)
This was answered correctly by most. Those who did go wrong usually demonstrated
a weakness in mathematics rather than in physics.
(c)
This part was generally not answered well.
(i)
Only the more able candidates were able to correctly answer this part.
(ii)
Even with errors carried forward from part (i), only a few more candidates
gained full marks for this question.
E18.Most candidates could successfully analyse the circuit in this question. and the answers were
quite good on the whole. Even though some candidates had problems in calculating the
effective resistance of the parallel combination of resistors they were able to do the rest of
the question effectively.
E19.The characteristic of the filament lamp presented in part (a) was usually acceptable, although
there was a smattering of answers showing the characteristic curving in the opposite way.
Several candidates offered the characteristic of a semiconductor diode and gained no
credit.
The explanation for the shape of the characteristic in part (b) was not as complete and
logical as examiners had hoped for. The majority of candidates were aware that the filament
was non-ohmic but could not explain why the characteristic curved the way it did. Most
candidates realised that heat was generated but were vague as to what increased in
temperature and what caused it. Other candidates had the resistance of the filament
decreasing with increasing temperature in order to justify an incorrect curve.
In part (c) credit was usually gained both for the diagram and the account. Some of the
diagrams were very poor and did not show a data logger, but the most significant omission
was a variable resistor (or potentiometer) or variable power supply, in order to vary the
voltage across the filament. The general standard of the diagram and subsequent account
showed however that the majority of candidates were familiar with this type of experiment.
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Colonel Frank Seely School
E20.Part (a) was, in general, very well done with the mathematical manipulation of three resistors
in parallel posing a problem to only a handful of candidates. There were many attempts
however where the candidates did not seem sure of the expression and had the four
resistors in parallel, or, more frequently, considered the 50 Ω to be in series with 1 / Rt,
where Rt was the sum of the three parallel resistors.
In contrast to part (a), part (b) gave very poor results with some 50% of the candidates
failing to gain the allocated mark. The common error was giving the current as I = 12 / 50 i.e.
ignoring the equivalent resistance of the three parallel resistors. This showed a very poor
understanding of the dc circuit.
E21.Answers to this question realised quite high marks. In part (a) the calculation of the total
resistance was invariably correct. Most candidates then realised that the three strips were in
parallel and proceeded accordingly. A significant number of candidates however,
proceeded by assuming that the resistance of the strip was 11 Ω and worked backwards
through the calculation. This method was not acceptable, since it was felt that the
candidates had been set a task to calculate the actual resistance of each strip from the total
resistance of the element.
Very few difficulties arose in part (b), apart from several candidates taking the area of cross
section as circular. Many candidates used the approximate value of 11 Ω for the resistance
of each strip instead of 10.8 Ω, which they had obtained in part (a). This was not acceptable
and a penalty was applied.
E22.Part (a) was a very straightforward question for 2 marks. The equation in part (i) was usually
correct, which is not surprising since it is given in the data sheet. A surprising number of
candidates failed to rearrange this equation in terms of VR.
Calculating the current in part (b)(i) was also straightforward and the majority of candidates
gained all the marks allocated. The calculation of the number of cells in part (ii) was
considered by the examiners to be reasonably difficult and they were well pleased by the
number of candidates who arrived at the correct answer, through a variety of methods.
E23.Most candidates scored well on this question, although part (a) proved to be the most
troublesome. A considerable number of candidates seemed unfamiliar with the effect of
shorting out, or connecting terminals together and many assumed that doing so would not
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Colonel Frank Seely School
affect the effective overall resistance. In part (b) the large majority of candidates realised
that two of the resistors were in parallel and proceeded accordingly to obtain the correct
answer. There were very few errors in calculating the sum of the parallel resistors.
E24.In part (a), a significant number of candidates obtained a value of 200 Ω or 300 Ω for the
resistor through not using the correct pd across it. In part (b), the same candidates usually
proceeded with an incorrect calculation of power in the diode by using in the expression I2R
the resistance calculated in part (a). Some candidates were not aware of the correct value
of the prefix m in mA.
The energy of the photon was usually calculated correctly in part (c), but a small minority
wrongly considered 1 / λ as the frequency or used an incorrect equation. In part (d), most
candidates knew how to proceed and gave a correct calculation. In the final part most
candidates gave a correct assumption made in the previous estimation. Those candidates
who were not specific in stating the assumption were not awarded this mark.
E25.Part (a) provided the candidates with a reasonably easy four marks, and very few failed
completely on the calculation. Usual errors such as units and arithmetic errors occurred but,
in general, the candidates knew how to proceed with the calculation.
Part (b) required clear, logical thinking and sadly, the majority of candidates failed to gain
the full three marks. Having been told in the question that the resistance of the sensor
decreased with increasing temperature, many candidates simply wrote that the reading of
the voltmeter would increase. Such a statement, although in itself correct, without any
reasoning did not gain marks. Many candidate realised that the current in the circuit would
decrease, but failed to go any further. The best approach seemed to be using the potential
divider equation and candidates who tackled the question from this angle were usually
awarded two or three marks.
E26.
Part (a) was the calculation of the equivalent resistance of a network of resistors
consisting of resistors connected in series and in parallel. The majority of candidates gained
full marks on this section and were not troubled by the calculation. However, it is worth
pointing out that since the final answer of 50 Ω was given in the question, then in order to
gain full marks it was necessary to show that the two equivalent series resistors were being
added together.
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Part (b) did not prove to be as easy; the problem in (i) was that many candidates gave the
total resistance as 50 Ω rather than 100 Ω. No consequential error for calculating the current
was allowed and frequently no marks were awarded for this section. It was possible in part
(b)(ii) to gain the two marks even if the answer to (i) was incorrect, but very few candidates
managed to gain these marks. The usual error was giving the current in the circuit as 24/20,
i.e. ignoring the second batch of parallel resistors. Again, many candidates, having
calculated the total current correctly, assumed that 2/3 would pass through the 60 Ω
resistor, not realising that the greater the resistor, the lower the current for a given voltage.
E27.Part (a) proved to be very accessible and many candidates scored full marks. Most
candidates calculated the resistor pd as 0.8 V and then calculated the resistance, as
expected. Other candidates however, calculated the total circuit resistance, then the diode
resistance and obtained the required resistance by subtraction. In this particular problem
some candidates used an incorrect pd and were not awarded any credit. Many clear and
correct answers were seen in part (ii).
The energy of the photon was calculated correctly in part (b) by many candidates, but some
failed to score because the wavelength was taken as 1 / f or because the energy was taken
to be ½QV. The general principle behind the question in part (ii) was understood by most
candidates and many correct answers were seen. A small minority of candidates however,
calculated and used the power supplied to the resistor and not the diode.
E28.
This question involved the analysis of a relatively difficult circuit, which included two
lamps and two resistors. The question however, was so structured that the majority of
candidates were able to work through and gain full marks. Others, unfortunately, although
making a reasonable attempt, failed to gain many marks. In part (a), the majority of
candidates calculated the correct value of the currents passing through each lamp.
In part (b), obtaining the correct answers to parts (i) and (ii) depended to a large extent on
realising that the reading on the voltmeter equalled the voltage across lamp X. Many
candidates missed this point, but were still able to gain some marks. In part (ii) the error that
was committed regularly was determining the resistance of lamp Y instead of the resistance
of resistor R2. But at least, more candidates realised that the same current passed through
lamp Y and R2. Answers to parts (iii) and (iv) used the answer to part (a) as a starting point,
but many candidates failed to realise that the current through R1 was the sum of the current
through the two lamps. Considerable guesswork took over at this stage and although most
of it was wrong, candidates could still get a mark for part (v) by using the answers obtained
to parts(iii) and (iv).
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Colonel Frank Seely School
E29.
This question worked well and many candidates gained full marks. The majority of the
other candidates only failed to gain maximum marks because of a unit error or significant
figure error. Disappointingly, many answers were expressed as a fraction. It should be
noted that this practice is not acceptable and the first answer expressed as a fraction was
treated as a significant figure error.
In part (i) the error which occurred most frequently was ignoring the internal resistance of
the battery. The correct answer was 0.19 A, to two significant figures, but many candidates
rounded this down to 0.2 A, which apart from incurring a penalty, also, when carried forward
to part (ii), gave a voltage across the resistors of 12 V. This implied that there was no
voltage developed across the internal resistance of the battery. Although many candidates
produced such an answer no one noted that such a situation was not possible. Many
answers to part (ii), when carrying forward an incorrect value of the current from part (i),
gave an answer well in excess of 12 V. Again this did nor seem to worry the candidates.
In part (iii) many candidates made the error of calculating the power dissipated in the total
external resistance instead of in resistor A alone. The unit of power was usually correct as
was the unit of energy in part (iv). Many candidates arrived at the correct answer in part (iv).
Consequential errors were carried forward throughout the whole question. This gave many
candidates the chance to gain some marks even if their initial calculation and subsequent
answer was incorrect.
E30.
In this example of calculating equivalent resistance, the same resistor network was
used twice, the equivalent resistance being calculated between different terminals. The
majority of candidates had no difficulty with the calculations, but it was worrying to find many
answers where the candidates had attempted a solution, not by calculation, but with
phrases such as “electricity takes the path of least resistance and therefore the effective
resistance (in part (b)) is 50 Ω.”
It was surprising to find that a significant number of candidates obtained the correct result in
part (b) but failed on part (a), since part (b) was deemed to be the most difficult of the two.
Considerable arithmetical difficulty was encountered by many candidates with the
reciprocal of the resistance when calculating the resistance of parallel resistors.
E31.
Candidates found this question very accessible and many gained full marks. In part (a)
the meaning of emf seems to be reasonably well understood with most candidates opting
for the voltage when no current passed through the circuit. Others defined it correctly in
terms of energy per unit coulomb. There were, unfortunately, many candidates who,
apparently, had not encountered the definition of emf and merely quoted electromagnetic
force, or even tried to define it in terms of a force in the circuit. The calculation of the current
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Colonel Frank Seely School
in part (ii) was well done and in part (iii) correct substitution of values into the equation
V + Ir gave r = 0.80 Ω.
=
Part (b) was involved with calculation of power and energy and although the majority of
candidates obtained the correct answer for the power dissipated in the 2.4 Ω resistor, fewer
had the correct answer for the total power dissipated in the circuit and a disappointing
number had the correct value for the energy wasted the battery. The usual answer to the
last part was to give the energy in the complete circuit. Whether this was due to inaccurate
reading of the question or due to lack of understanding could not be decided.
E32.
The question involved straightforward calculations on voltage, resistance and current.
In part (a)(i) it was hoped that candidates would have spotted the correct voltage across
each lamp by inspection. Surprisingly, even those who managed to get the wrong answer in
part (i) nevertheless ignored their answer and proceeded from first principles to obtain the
correct answer to part (ii).
Part (b) involved the same circuit components as in part (a) but connected differently. The
majority of candidates showed that the current from the battery was the value given in the
question. Using this value they then proceeded to argue or calculate the current in each
lamp. Those candidates who merely halved the current value obtained in part (i) without any
reasoning did not gain the mark.
Although the question told the candidates that the current through each lamp was the same
in both circuits it was disappointing to find in part (c) how many candidates tried to argue
that the brightness of the bulbs in the 2nd circuit would be different to that in the first, the
main thrust of their argument being that the voltage across each bulb was different and
therefore that the brightness would be different.
E33.Part (a) provided three reasonably easy marks for the large majority of candidates. The most
common error was forgetting to add the resistance of the thermistor to that of the resistor.
Errors in part (ii) arose from not reading the question properly and many candidates
calculated the potential across the resistor instead of the thermistor.
To obtain full marks in part (b) required a logical answer, starting with a statement that the
resistance of the thermistor decreased with rising temperature. Almost all candidates
obtained credit for this. The next step was realising that this resulted in the total resistance
decreasing. Not many candidates mentioned this, but full marks could still be obtained if
they realised that the current in the circuit increased as the temperature fell. Many
candidates got to this point but then failed to go any further. A common answer was stating,
without any justification, that the voltage across the thermistor decreased because the rate
of increase of current was greater than the rate of decrease of resistance. In order to gain
the remaining marks, it was necessary to discuss the potential across the fixed resistance.
As an alternative approach, many candidates recognised the circuit as a potential divider,
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gave the potential divider equation and argued correctly from that point. Part (b) proved to
be a good discriminator.
Part (c) also earned high marks. The calculation of power dissipated in the thermistor was
usually correct for both temperatures and very few failed to read the graph correctly. The
usual error was using the same current as in part (a), not realising that the removal of the
fixed resistor increased the current. It was obvious from answers to part (ii) that a significant
number of candidates were not familiar with the term mean value. Several candidates
subtracted the two values before dividing by two, others just added them. Having arrived at
the mean power, most candidates were aware that to change to energy required multiplying
by the time, in seconds. Only the better candidates realised that the answer to part (iii) was
that the rate of decrease of resistance with temperature was not linear, or that the graph
was not a straight line.
E34.
The response to this question was very disappointing, especially in view of the fact
that this topic has been examined several times previously, including questions on the
graphical nature of the quantities involved. Rearranging the equation in part (a) was
intended as a guide to drawing the graph in part (b). The majority of candidates did
rearrange the equation correctly, but some candidates failed to do this and ended up with a
quotient.
Sketching the I-V graph in part (b) was, quite literally, a disaster area. The large majority of
candidates drew a straight line of positive gradient passing through the origin. Obviously
this was the easiest line to draw without applying any thought to the question. No marks
were awarded for such attempts. If a line of positive gradient was drawn, and did not pass
through the origin, then 1 mark was awarded. A large number of curved lines, some starting
at zero, others at a positive value of V and decreasing to zero, were also presented. Of the
candidates who drew a straight line with a negative gradient, many lost marks by either
extending the line into negative values of V, or negative values of I. It must be pointed out
that when carrying out an experiment to obtain this graph, it is not possible to obtain zero
values of V or I. However, since some textbooks do show the graph extending to the V axis,
this was accepted, but graphs extending to the I axis were not.
Most candidates gained at least one mark in part (c), but the impression gained was that
candidates had learned the answers parrot fashion with no reference to the graph. The
gradient of a curved graph was often given as the answer.
E35.
The response to this question was very disappointing, especially in view of the fact
that this topic has been examined several times previously, including questions on the
graphical nature of the quantities involved. Rearranging the equation in part (a) was
intended as a guide to drawing the graph in part (b). The majority of candidates did
rearrange the equation correctly, but some candidates failed to do this and ended up with a
quotient.
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Colonel Frank Seely School
Sketching the I-V graph in part (b) was, quite literally, a disaster area. The large majority of
candidates drew a straight line of positive gradient passing through the origin. Obviously
this was the easiest line to draw without applying any thought to the question. No marks
were awarded for such attempts. If a line of positive gradient was drawn, and did not pass
through the origin, then 1 mark was awarded. A large number of curved lines, some starting
at zero, others at a positive value of V and decreasing to zero, were also presented. Of the
candidates who drew a straight line with a negative gradient, many lost marks by either
extending the line into negative values of V, or negative values of I. It must be pointed out
that when carrying out an experiment to obtain this graph, it is not possible to obtain zero
values of V or I. However, since some textbooks do show the graph extending to the V axis,
this was accepted, but graphs extending to the I axis were not.
Most candidates gained at least one mark in part (c), but the impression gained was that
candidates had learned the answers parrot fashion with no reference to the graph. The
gradient of a curved graph was often given as the answer.
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