Download A lamp rated at 12 V 60 W is connected to the secondary coil

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A lamp rated at 12 V 60 W is connected to the secondary coil of a step-down transformer and is
at full brightness. The primary coil is connected to a supply of 230 V. The transformer is 75%
efficient.
What is the current in the primary coil?
A
0.25 A
B
0.35 A
C
3.75 A
D
5.0 A
A transformer has 1150 turns on the primary coil and 500 turns on the secondary coil.
The primary coil draws a current of 0.26 A from a 230 V ac supply. The current in the secondary
coil is 0.50 A. What is the efficiency of the transformer?
A
42%
B
50%
C
84%
D
100%
(Total 1 mark)
Page 1 of 16
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The primary coil of a step-up transformer is connected to a source of alternating pd.
The secondary coil is connected to a lamp.
Which line, A to D, in the table correctly describes the flux linkage and current through the
secondary coil in relation to the primary coil?
A
>1
<1
B
<1
<1
C
>1
>1
D
<1
>1
(Total 1 mark)
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A transformer has 1200 turns on the primary coil and 500 turns on the secondary coil. The
primary coil draws a current of 0.25 A from a 240 V ac supply. If the efficiency of the transformer
is 83%, what is the current in the secondary coil?
A
0.10 A
B
0.21 A
C
0.50 A
D
0.60 A
Page 2 of 16
5
A hydroelectric power station has a power output of 2.0 MW when water passes through its
turbines at a rate of 1.4 m3 s–1 . The water is supplied from a reservoir which is 750 m above the
power station turbines, as shown in the diagram below.
density of water = 1000 kg m–3
(a)
Calculate
(i)
the mass of water passing through the turbines each second,
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(ii)
the loss of potential energy per second of the water flowing between the reservoir and
the power station turbines,
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(iii)
the efficiency of the power station.
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(6)
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(b)
The turbines drive generators that produce alternating current at an rms potential difference
of 25 kV which is then stepped up to an rms potential difference of 275 kV by means of a
transformer.
(i)
Calculate the rms current supplied by the generators to the transformer when the
power output of the generators is 2.0 MW.
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(ii)
The transformer has an efficiency of 95%. Calculate the output current of the
transformer.
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(4)
(Total 10 marks)
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A transformer is required to produce an r.m.s. output of 2.0 × 103 V when it is connected to the
230 V r.m.s. mains supply. The primary coil has 800 turns.
(a)
Calculate the number of turns required on the secondary coil, assuming the transformer is
ideal.
(2)
(b)
The transformer suffers from eddy current losses.
(i)
Explain how eddy currents arise.
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(4)
(ii)
State the feature of transformers designed to minimise eddy currents.
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(1)
(Total 7 marks)
Page 4 of 16
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(a)
Explain what is meant by the term magnetic flux linkage. State its unit.
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(2)
(b)
Explain, in terms of electromagnetic induction, how a transformer may be used to step
down voltage.
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(4)
(c)
A minidisc player is provided with a mains adapter. The adapter uses a transformer with a
turns ratio of 15:1 to step down the mains voltage from 230 V.
(i)
Calculate the output voltage of the transformer.
(2)
(ii)
State two reasons why the transformer may be less than 100% efficient.
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(2)
(Total 10 marks)
Page 5 of 16
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A 230 V, 60 W lamp is connected to the output terminals of a transformer which has a 200 turn
primary coil and a 2000 turn secondary coil. The primary coil is connected to an ac source with a
variable output pd. The lamp lights at its normal brightness when the primary coil is supplied with
an alternating current of 2.7 A.
What is the percentage efficiency of the transformer?
A
3%
B
10%
C
97%
D
100%
(Total 1 mark)
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(a)
Calculate the length of copper wire that has a diameter of 1.6 × 10–3 m and a resistance of
25 Ω.
resistivity of copper = 1.7 × 10–8 Ω m
Length of wire .......................................................
(3)
Page 6 of 16
(b)
The resistance of copper wire is not zero. Explain why this fact leads to the use of
alternating current rather than direct current when transmitting electrical energy.
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(3)
(Total 6 marks)
10
Which one of the following statements concerning power losses in a transformer is incorrect?
Power losses can be reduced by
A
laminating the core.
B
using high resistance windings.
C
using thick wire.
D
using a core made of special iron alloys which are easily magnetised.
(Total 1 mark)
Page 7 of 16
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A transformer with 3000 turns in its primary coil is used to change an alternating pd from an rms
value of 240 V to an rms value of 12 V.
When a 60 W, 12 V lamp is connected to the secondary coil, the lamp lights at normal brightness
and a rms current of 0.26 A passes through the primary coil.
Which line, A to D, in the table gives correct values for the number of turns on the secondary coil
and for the transformer efficiency?
number of turns on the
secondary coil
efficiency
A
150
96%
B
60 000
96%
C
150
90%
D
60 000
90%
(Total 1 mark)
12
Which one of the following would not reduce the energy losses in a transformer?
A
B
C
D
using thinner wire for the windings
using a laminated core instead of a solid core
using a core made from iron instead of steel
using a core that allows all the flux due to the primary coil to be linked to the secondary coil
(Total 1 mark)
Page 8 of 16
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Which one of the following is not a cause of energy loss in a transformer?
A
good insulation between the primary and secondary coil
B
induced currents in the soft iron core
C
reversal of magnetism in the soft iron core
D
resistances in the primary and secondary coil
(Total 1 mark)
Page 9 of 16
Mark schemes
1
2
3
4
5
B
[1]
C
[1]
A
[1]
C
[1]
(a)
(i)
mass per sec ( = density × vol per sec) = 1000 × 1.4 (1)
= 1400 kg (s–1)
(ii)
loss of Ep per sec
= 1400 × 9.8 × 750 (1)
= 1.0 × 107 J (s–1) (1) (1.03 × 107 J s–1)
(allow C.E. for value of mass per sec from (i))
(iii)
efficiency
= 0.2 (1)
(allow C.E. for value (ii))
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Page 10 of 16
(b)
(i)
(use of P = IV gives) Irms =
(1)
= 80 A (1)
(ii)
power output = (0.95 × power input) = 0.95 × 2.0 (MW) = 1.9 (MW) (1)
= 6.9 A (1)
[or I for 100% efficiency
= 7.3 (A) (1)
I for 95% efficiency = 95% of 7.3 = 6.9 A]
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[10]
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(a)
C1
7000 (6960)
A1
(2)
(b)
(i)
changing magnetic field
B1
emf or changing magnetic field is in the core
B1
e.m.f. induced (due to changing magnetic field) not back emf
B1
current flows as core is made from a conducting material
B1
(4)
(ii)
laminated core
B1
(1)
[7]
7
(a)
product of flux and number of turns
B1
Wb or equivalent
C1
(2)
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(b)
changing primary magnetic field due to alternating voltage
B1
(applied to primary)
varying flux links with secondary
B1
induced emf ∑ rate of change of flux linkage
B1
NS < NP so less voltage on secondary
C1
(4)
(c)
(i)
equation or correct substitution
C1
15.3 V
A1
(2)
(ii)
<100% flux linkage / flux leakage / copper losses / iron losses / hysterysis
losses not just “heating” or “heat loss”
B2
(2)
[10]
8
9
C
[1]
(a)
R = ρL/A
C1
A = 2.0 × 10–6 (m2) or π(0.8 × 10–3)2 seen in equation
(condone π(1.6 × 10–3)2 or 8.04 × 10–6 seen)
C1
L = 2900 m, 2940 m, 2960 or 3000 m
A1
3
Page 12 of 16
(b)
resistance leads to loss of heat/energy/power
or I2R loss or voltage drop (across cable)
B1
lower current lowers loss of heat/energy/power
or reduces voltage drop
B1
ac can be transformed (to lower transmission current)
B1
3
[6]
10
11
12
13
B
[1]
A
[1]
A
[1]
A
[1]
Page 13 of 16
Examiner reports
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This question was a transformer calculation that caused few problems. Its facility was 78% and it
discriminated very well.
It is possible that the values chosen for three of the distractors in this question, on transformer
efficiency, were so obviously wrong that there was no need for the students to perform any
calculation. However, only 54% gave the correct response when the question was pre-tested in
2009. The 2014 A level students made amends for this, because the facility of the question was
87% this time.
A fairly demanding test of candidates’ knowledge of transformers; slightly fewer than half of them
selected the correct answer. Among the incorrect responses, distractor C was a common choice
(23%), showing that the flux linkage ratio was better understood than the current ratio.
Transformers were also the subject under test, where 68% of the responses were correct. This
was a fairly straightforward calculation involving transformer efficiency.
The majority of candidates gained maximum marks in both parts (a)(i) and (a)(ii), but a few
candidates, who scored reasonably well elsewhere, lost marks in part (ii) as a result of
attempting unnecessarily complicated solutions involving expressions for loss of potential energy
and gain of kinetic energy. In part (iii), candidates often knew the correct equation for efficiency
but used incorrect data.
Candidates often gained both marks in part (b)(i) but gave confused and incorrect answers in
part (ii), with many of the weaker candidates attempting to convert from rms to peak values or
vice versa. Some candidates failed to recognise that the output current from the generators was
the input current to the transformer.
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(a)
Most candidates used the correct equation but a surprisingly large number rearranged it
incorrectly and gave an answer indicating that there would be fewer turns on the secondary
coil than on the primary coil. Significant figure penalties were common here.
(b)
(i)
This explanation was not well done. Few candidates made it clear that the induction
was taking place in the core and very few mentioned that the core was an electrical
conductor. Many candidates made weak references to back e.m.f’s and a large
number seemed to be trying to explain what happened in the secondary coil.
Questions of this type are often asked and candidates would be well advised to
practice making sequential explanations.
(ii)
Many candidates knew about laminations. A few forgot to specify that it was the core
that should be laminated. Some wanted to laminate the coils.
Page 14 of 16
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(a)
Few of the candidates were able to give a good explanation of magnetic flux linkage.
several confusing the term with flux density. Few correctly identified the Weber as the
correct unit.
(b)
Most candidates were unable to give a coherent explanation of how electromagnetic
induction underpinned the working of the step down transformer. The most common
observation gaining credit was that there are more turns on the primary than the secondary
coil.
(c)
(i)
Most candidates calculated the output voltage correctly.
(ii)
Few candidates’ answers gained credit for this part. It was common for candidates to
cite heating or eddy currents as the reason for less than 100% efficiency. Although
these answers are not incorrect they are too vague to allow credit at this level.
Knowledge of the efficiency of a transformer was tested in this question, which had a facility of
58%. The output power from the transformer must have been 60 W, because the lamp was lit at
its normal brightness. The turns ratio indicated that the primary voltage was 23 V, whilst the
question stated that the primary current was 2.7 A. Hence the input power could be found using
2.7 × 23 = 62.1 W. 25% of the candidates chose the incorrect distractor B.
(a)
There were many errors in recall of the resistivity formula and a sizeable proportion of
those who recalled it correctly simply substituted the value of the diameter for the area. Of
those who appreciated that a calculation of area was necessary, a disappointing number
used an incorrect formula or did not distinguish between diameter and radius.
(b)
This part was not done well. Candidates needed to explain that resistance leads to energy
loss in transmission, that lower current reduces this loss and that lower current can be
achieved using a transformer with ac. Many misconceptions were, however, apparent in
responses. For example, ‘using dc you lose current’; ‘ac goes both ways so you lose
energy when it goes one way and get it back when it goes the other’; ‘current makes
resistance so if current goes both ways the resistances cancel out’; ‘use a lower current
because if current is reduced then resistance is reduced’.
This qestion was about transformers. Causes of power loss were well known in this question,
where three quarters of the candidates evidently saw that using windings with higher resistance
would have a detrimental effect.
11
This question tested both the turns ratio equation and efficiency; again there were few problems
and the facility was 72%.
12
This question which were each correctly answered by just over three-quarters of the students,
respectively tested the rotating coil and energy losses in a transformer.
Page 15 of 16
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The facts about energy losses in transformers were well known in this question, where 73% of
candidates gave the correct answer. The strongest incorrect distractor was C, suggesting that the
energy losses caused by magnetic hysteresis are not always recognised.
Page 16 of 16