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MAGNETIC FIELDS
Question 1
SMF 1 to MFP c
SMF 2 to MFP a
SMF 3 to MFP b
Diagram 1 is of a coil, we use coils to become 'bar' magnets, so the field around a coil must look like the field
around a bar magnet. Mfp 3 is the best option.
The next easiest one to deal with is '3'. Don't feel constrained to do the questions in the order that they are
written, sometimes it is easier to do them in another order.
So '3' is a straight current carrying conductor, so the filed is given by the direction that the fingers curl if the
thumb is in the direction of the current. This means that they are concentric circles, the direction of the field
is controlled by the direction of the current (which is not shown)
That means that '2' must behave like two versions of '3'. The field lines will be in the opposite direction if the
direction of the current was given, then you could show this. The non-concentric circles are a result of the
interaction between the two fields.
FORCES ON WIRES
Question 1
Force
= BIl
= 0.5  3.0  2.0
= 3.0 N
Question 2 ANS A
The current is flowing from left to right. The field is into the page. The Right Hand Slap Rule shows that the
force is up the page.
Question 3
A
The right hand grip rule specifies the direction of the field. This is covered better in dot point 3.
DC MOTOR
Question 1
F = nBIl = 50 x 0.005 x 3.0 x 0.05 = 0.038 N (remember to convert cm to m)
Question 2
The direction of the force is always perpendicular to both P and the magnetic field, hence response B
Question 3
The commutator needs to be free to rotate since this is the way that the connection is made between the
external current source and the coil itself via the brushes rubbing on the commutator.
He motor rotates since the force on the sides of the coil that are perpendicular to the magnetic applies a
torque to the coil which is down on the side nearest the south pole and up on the side nearest the north pole,
given the direction of current flow shown on the diagram. However, when the coil rotates into the vertical
position, the force on side P at the bottom needs to reverse in order for the torque to be applied in the
clockwise direction. For this to happen, the direction of the current needs to reverse and it is the role of the
split-ring commutator to do this. The split in the ring is such that as the coil gets to the vertical position, the
direction of the current flow through the coil reverses and the direction of the torque applied to the coil
remains in the same direction.
Question 4 B
The magnetic field points from left to right (north to south). Therefore, the current must flow in at Z and out at
X (right hand grip rule)
Question 5
Magnitude of the force
= B  I  l = 0.10  1.5  0.12 = 0.018 N
Question 6 A
Using the right hand slap rule (fingers – field, palm – push, thumb – current), the armature will start to rotate
anticlockwise
(side JK moves down)
Question 7
H
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is
pointing to the top of the page, so the palm is facing upwards. Therefore the force is out of the page.
Question 8
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point y?
Question 8
I
The force at the point y is zero, because the direction of the current is parallel to the field here. Therefore
the component of the current perpendicular to the field is zero.
Question 9
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point z?
Question 9
G
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is
pointing to the bottom of the page (in general), so the palm is facing downwards. Therefore the force is into
the page.
Question 10
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point
w?
Question 10
H
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is
pointing to the top of the page, so the palm is facing upwards. Therefore the force is out of the page.
In order to convert the arrangement in Figure 5 into a motor, Thomas provides an axis for rotation. He
realises that there must be current flowing through the coil when it is rotating, so he attaches a set of slip
rings that rotate with the coil as illustrated in Figure 5.
The coil is initially set with its plane parallel to the magnetic field as shown, and the switch is open so that no
current flows.
Question 11
The switch is closed. Which one of the statements (A-D) best describes the situation after current begins to
How?
A. The coil begins to rotate, but stops after turning through 90°.
B. The coil begins to rotate and, after rotating one half turn, rotates back to its original position. It then
continues to oscillate in this way.
C. The coil does not move. D. The coil rotates continuously.
Question 11
A (or B)
The best answer is A, because when the switch is closed the current will flow around the loop. The magnetic
field will exert a force on this current and cause the loop to rotate. When the loop has rotated through 90 0 ,
the direction of the force is now non acting to rotate the loop, but to dilate it. If we assume that there is some
frictional forces, then the loop will stop. If we assume that there are no frictional forces then the loop will
oscillate. Eventually the loop would come to rest, because there will be some friction acting.
Question 12
If the slip rings used by Thomas in the circuit of Figure 5 were replaced with a commutator, which one of the
statements (A-D) best describes the situation after the switch is closed, and the current begins to flow?
A. The coil begins to rotate, but stops after turning through 90°.
B. The coil begins to rotate and, after rotating one half turn, rotates back to its original position. It then
continues to oscillate in this way.
C. The coil does not move. D. The coil rotates continuously.
Question 12
D
The best answer is D, because the function of the commutator is to reverse the direction of the current every
half cycle. This will mean that the direction of the force changes every half cycle. This will cause the coil to
rotate continuously.
Two physics students set up a 50-turn coil (JKLM), which is free to rotate about the axis shown as the
dashed line below. The loop is placed between the poles of a magnet, in a uniform magnetic field of 0.040 T.
The current in the coil is 1.5 A.
Question 13
What is the magnitude of the magnetic force on side JK (length = 0.050 m) of the 50-turn coil, when oriented
as shown above?
Question 13
Note that there are now 50 turns in the coil, so the force will be given by F = nBiL.
 F = nBil
 F = 50 × 0.040 × 1.5 × 0.050
 F = 0.15 N
Question 14
What is the magnitude of the magnetic force on side KL (length = 0.040 m) of the 50-turn coil, when oriented
as shown above?
Question 14
Zero
Because the wire KL is parallel to the field, so there isn't a force acting on it. A force only acts when the wire
is perpendicular (or has a component that is perpendicular) to the field. As side KL is oriented parallel to the
magnetic field it experiences zero force.
The students now turn off the current and set the coil at rest, oriented as shown below. They then turn the
current on again.
Question 15
What happens to the coil after the current is turned on? Explain your answer.
Question 15
If the coil is at rest in this position when the current is turned on, the coil will not rotate.
Consider JK. The field (your fingers) is from left to right. The current (your thumb) is pointing into the page.
This means that your palm (the force) is downwards.
If you consider ML, then the field is still from left to right, but the current is out of the page. This means that
the force on ML is upwards.
Consider KL. Field to the right, current down the page, so the force is out of page.
Consider JM. Field to the right, current up the page, so the force is into the page.
The effect is to compress all the sides, because there are forces acting 'inwards' on all sides. There is no
torque, so the coil will not spin. In a simple motor like this, it will not start from this position, but if it wasn't
'oriented' exactly in this position then there would be a torque acting and the coil would begin the spin. It
would usually have enough momentum to get past this point and it the current changed it would continue to
spin.
A single loop of wire in a uniform magnetic field is shown below. The loop can rotate, and is shown at
three different orientations. In each case there is a current flowing around the coil from W to X to Y to Z.
Question 16
The magnetic field is 0.10 T, and the current in the loop is 0.30 A. With the loop in orientation (a), what is the
magnitude of the force acting on side WX of the coil? The length of side WX is 0.030 m. Show your working.
Question 16
F = nBIL
F = 1 × 0.010 × 0.30 × 0.30
 F = 0.0009 N
= 9 × 10-4 N
Below, the arrows indicate possible directions of the force on side WX of the loop in the three orientations
(a), (b) and (c). The arrows in each orientation are in a plane perpendicular to the axis of rotation of the
loop.
Question 17
For each orientation below, circle the head of the arrow which best represents the direction of the
magnetic force on side WX of the coil. If there is no force on the side, write NF under the diagram.
Question 17
a) pointing to the left
b) pointing to the left
c) pointing to the left
(a)
The field is down, the current is into the page, so the thumb points into the page with the fingers
pointing down. This makes the palm of your hand face left, so the force is in the direction
pointing to the left
(b)
The field is down, the current is into the page, so the thumb points into the page with the fingers
pointing down. This makes the palm of your hand face left, so the force is in the direction
pointing to the left
(c)
The field is down, the current is into the page, so the thumb points into the page with the fingers
pointing down. This makes the palm of your hand face left, so the force is in the direction
pointing to the left
There isn't any change in either 3 orientations, the field was constant, and the wire had a current going into
the page in these three situations, so each answer is identical. Don't get fussed that on the exam, they
actually gave you such a simple question. You need to solve every problem on its merits.
In Figure 6 below, the arrows indicate possible directions of the force on side XY for the loop in orientations
(a) and (c) shown in Figure 4.
Question 18
For each of the two orientations in Figure 6, circle the head of the arrow which best represents the direction
of the magnetic force on side XY of the coil. If there is no force on the side, write NF under the diagram.
Question 18
a) NF
b) Into the plane of the page
(a)
So the section of wire XY is not at right angles to the field. It is parallel to the field, so this
means that the force acting on it will be zero.
(b)
Now the wire is at right angles to the field, so there will be a force acting on it. The field is down
the page (direction of your fingers) the current is from left to right across the page (direction of
your thumb). So your palm must be facing into the page (you need a flexible wrist for this). So
the force is into the plane of the page.
Figure 5 below shows four positions (A, B, C, D) of the coil of a DC motor. The coil can be assumed to be a
single wire which is in a uniform magnetic field parallel to the coil when in the orientation shown in diagram
A.
It rotates in the direction indicated, about the axis which passes through the middle of sides LM and NK. The
coil is attached to a commutator, to which current is passed by brushes (not shown in the figure).
Figure 5
Question 19
For the coil as shown in orientation A of Figure 5, in which direction is the current flowing in the side KL?
Explain your answer.
Question 19
Place your fingers in the direction of the field, from left to right. For the loop to rotate clockwise the force on
KL must be upwards. So have your palm facing up. This means that your thumb points from K - L.
So the current flows from K to L.
Side KL of the coil is 0.10 m long, and a magnetic force of 0.60 N acts on it.
Question 20
If the magnetic field has a magnitude of 1.5 T, what is the magnitude of the current in the coil?
Question 20
F = nBIL
0.60 = 1 × 1.5 × I × 0.1
I = 4 amp
Question 21
Consider two cases:
a. The coil is at rest with the orientation shown in diagram A of Figure 5.
b. The coil is at rest with the orientation shown in diagram B of Figure 5.
In the answer book explain what would happen, in each case, if current is allowed to flow in the coil. Your
answer should discuss the forces on each side of the coil, and their net effect.
Question 21
a.
If the coil starts in this position, the forces acting on KL and MN will spin the loop in the direction
shown, because of the torques acting See answer 234.
b.
If the coil starts in this position the forces acting on KL and MN will tend to try to squeeze or
expand the loop, but not spin it.
EMF
A 2.0 m long wire is moved down the page out of the magnetic field (of strength 0.5T ) at a velocity of
-1
2.0 ms as shown in the diagram below.
2.0 m
X
X
X
0.5 T
X X X
Z
Y
X
X
X
X X X
2.0 ms-1
Question 1
Calculate the emf generated in the wire.
Question 1
EMF = vBl = 2.0  0.5  2.0 = 2.0 V
Question 2
Describe two different ways in which the emf generated could be doubled.
Question 2
Either double the velocity or double the magnetic flux intensity or double the length of the wire. (any two of
these)
Question 3
Do induced currents in a circuit oppose or magnify the change in magnetic flux that induces them? Fully
explain your answer in detail.
Question 3
This is Lenz’s Law. The induced current flows in a direction such that the magnetic flux produced opposes
the change in magnetic flux that produced the current. This is conservation of energy. If the reverse were
true, energy would be created out of nothing.
A square loop of wire of side 0.01 m is moved through a magnetic field.
-4
The maximum magnetic flux generated is 5.0 x 10 Wb.
Question 4
Calculate the magnetic flux density of the magnetic field.
Question 4
Maximum magnetic flux = BA
BA  5.0  10 4
B
5.0  10 4
5 T
0.01  0.01
The square loop is moved at a constant speed through the magnetic field in one direction and then back
again through the magnetic field in the reverse direction.
The graph showing the magnetic flux through the coil as a function of time is shown below.
Magnetic
-4
flux through 5.0 x 10
coil (Wb)
Time (s)
Question 5
Draw a graph which shows the variation of emf through the loop against time. Explain the main features of
your graph.
Question 5
The induced emf is the gradient of the magnetic flux-time graph. The gradient changes from constant
positive to constant negative and then again to constant positive to constant negative.
The graph is shown below. (5 marks)
EMF (V)
Time (s)
Jackie and Jim are studying electromagnetic induction. They have a small permanent magnet and a coil of
wire wound around a hollow cylinder as shown below.
Jackie moves the magnet through the coil in the direction shown at constant speed.
Question 6
Indicate on the diagram the direction of the induced current that flows in the resistor. Explain the physics
reason for your choice.
Question 6
Use Lenz’s Law and RH grip rule
From Lenz’s Law the induced field produced by the coil must oppose the increasing external magnetic field
caused by the magnet moving closer. The induced field must therefore flow to the right towards the magnet.
From the RH grip rule, to get a magnetic field that points out of the centre of the coil to the right will require
the current to flow anti-clockwise when viewed from the magnet towards the end of the coil.
The current through the external resistor will thus flow from left to right
They next decide to move the magnet, at a constant speed, all the way through the coil and out the other
side.
Question 7
Which one of the diagrams (A–D) below best shows how the current through the coil varies with time?
Question 7
As the magnet approaches the induced current will increase to a maximum as the magnet enters the coil, it
will then be constant while the magnet moves through the coil (it will not be zero as the magnet is still
moving). As the magnet exits the coil and moves away this will result in a reduction of the field within the coil
and the induced current will flow in the direction needed to try and maintain the field ie it will reverse.
Diagram D is the only one that fits this reasoning
The on-off action of a lightning stroke produces an electromagnetic field surrounding the stroke. It is this field
that causes the crackling in your radio or TV during a thunderstorm.
The magnetic field, B, produced by a lightning stroke varies with time as shown in Figure 1.
A small coil is placed perpendicular to the magnetic field, and the induced emf is monitored on an
oscilloscope.
Question 8
Which one of the graphs (A-D) best shows the variation of the emf with time?
Question 8
D
It must be the negative gradient of the B vs t graph. Initially the graph had a constant positive gradient,
(which will be shown as a negative straight line) followed by zero gradient (shown as zero) and then a
constant negative gradient (shown as a constant positive line).
Sofia and Max are investigating electromagnetic induction using a square coil. They place the coil between
the poles of a magnet as shown in Figure 5. The sides of the coil are 0.020 m long. The uniform magnetic
–2 T, and elsewhere in air it is assumed to be zero.
Figure 5
Question 9
Calculate the magnetic flux through the coil when it is entirely within the magnetic field, as shown in Figure 5.
Question 9
 = BA
  = 5.0 × 10-2 × (0.020 × 0.020)
= 0.00002 Wb
= 2 × 10-5 Wb
Sofia and Max now move the coil from the position shown in Figure 5 (entirely inside the magnetic field) to
the position shown in Figure 6 below (entirely outside the magnetic field). It takes 0.040 s to move the coil
from the position in Figure 5 to the position in Figure 6.
Figure 6
Question 10
What is the average emf induced in the coil as it moves from the position in Figure 5 to the position in Figure
6?
Question 10
The induced EMF is the rate of change of flux. In this case the flux changes from 2 × 10-5 Wb to zero in
0.040 secs.

 = -n
t
n= 1, so  = 2 × 10-5/ 0.040
= 0.0005 Volt
= 5 × 10-4 V
Sofia and Max now move the coil through the magnetic field, as shown in Figure 7.
The magnetic flux through the coil as it moves from position (a), through position (b), to position (c) is shown
in Figure 8.
Question 11
Circle the letter (A–D) of the graph below that best shows the emf induced in the coil as a function of time.
2 marks
Figure 7
Question 11
B
The induced EMF is the negative gradient of the magnetic flux vs time graph.
Between t = 0 and t = 1, the negative of the gradient of the line is zero
Between t= 1 and t = 2
the negative of the gradient of the line is a negative constant
Between t= 2 and t = 3
the negative of the gradient of the line is a positive constant
Between t = 3 and t = 4, the negative of the gradient of the line is zero
Because the slope of the flux graph has the same gradient (magnitude), the two 'pulses' need to also have
the same magnitude.
So the graph that best fits is B
Gary and Kate are investigating electromagnetic induction. They have a single wire loop of dimensions
0.030 m long by 0.020 m wide which is placed in a uniform magnetic field. The loop can be rotated by hand
about an axis as shown below in Figure 7. The ends of the loop slide within slip rings so that a measurement
of the emf between the ends of the loop can be made between terminals A and B.
The uniform magnetic field is 0.12 T.
Question 12
What is the value of the magnetic flux through the loop when it is oriented as in Figure 7?
Question 12
 = BA
  = 0.12 × (0.030 × 0.020)
= 0.000072 Wb
= 7.2 × 10-5 Wb
The first question in a set on an exam is usually a very straight forward question. Either, a simple concept
or substitution into a basic formula. The next couple are generally more complex
Gary now rotates the coil at a constant rate of 5 rotations per second. Using an oscilloscope they observe
that the voltage between points A and B varies with time as shown in Figure 8.
Figure 8
Kate decides to double the rate of rotation to 10 rotations per second.
Question 13
On Figure 8 above, sketch a graph which shows the variation with time of the voltage between points A and
B at this faster rate of rotation.
Question 13
Voltage (mV)
0.4
original
new
0.2
0.2
0.4
time (secs)
The graph remains a sine curve, but the amplitude doubles ,

the induced EMF is given by
 = -n
, if t is halved then  is doubled and the period is halved,
t
because it is rotating twice as fast.
A student investigates electromagnetic induction using a single loop coil and an electromagnet as shown
below in Figure 6. The loop is placed between the poles of the electromagnet, perpendicular to the magnetic
field, and connected to an oscilloscope so that any voltage induced in the loop can be measured.
Figure 6
The current in the coils of the electromagnet is reduced to zero and then reversed so that the magnetic field,
B, changes as shown below in Figure 7.
Figure 7
Question 14
Which one of the options (A–D) shown below best represents the induced voltage measured on the
oscilloscope?
Question 14
B.
The induced voltage is the negative gradient of the B vs t graph. This would give you a positive constant,
but we do not know how the CRO has been wired. If we reversed the wiring on the CRO we would get a
positive constant.
B is by far the best answer.
Question 15
Justify your answer to Question 12, referring to Figures 6 and 7.
Question 15
The induced EMF will depend on the change of flux. The area of the loop remains constant so the change in
flux is related to the change in the magnetic field.
The gradient of the magnetic field is constant,  the induced EMF is constant from time 0 to t= t 2.
The only graph to show this is graph B. You might have expected the graph to have a positive value, but this
depends entirely on the connections to the CRO. If you swapped the red and the black connections you
would get a positive output from the CRO. The most important concept is the gradient of the flux vs time
graph.
Below is a diagram of a model generator which a student used to investigate electromagnetic induction. The
student moved the coil at a slow constant speed and observed the current using a galvanometer as shown.
The galvanometer has a central zero setting.
One complete cycle for the coil, viewed from position Z, is shown below.
Question 16
In which one or more positions would the magnitude of the magnetic flux through the coil have
been a maximum? (one or more answers)
Question 17
Which one of the following diagrams best shows how the current through the galvanometer varied with time?
The student then modified the generator by removing the galvanometer and slip rings, and including a splitring commutator instead, as shown in Figure 3 below. The way the voltage across the bnrshes varied with
time was observed using a cathode ray oscilloscope (C.R.O.).
Question 18
Which one of the waveforms below best shows what the student observed on the C.R.O., when the coil was
rotated as before?
The student disconnected the C.R.O. from the modified generator as shown in below, in order to operate it
as a DC motor.
To make the DC motor, the student connected a battery between.x and y. The coil now rotated
continuously in the same direction.
Question 19
Explain how the split-ring commutator enabled the coil to rotate always in the same direction
Question 20
How should the terminals of the battery be connected to x and y to make the coil rotate clockwise as viewed
from Z?
RMS and PEAK, TRANSFORMERS, POWER LOSS
A house on a rural block of land 10 km from the main road has been obtaining its electrical power from a
diesel generator. The opportunity has arisen to connect to the state electrical grid.
The owner of the house has two options:
Option 1
Run a 11kV RMS line from the main road to the house and then use a step down
transformer to produce power at 240 V RMS
Option 2
Use a step down transformer at the main road from 11 kV RMS to 240 V RMS and then run
the power to the house.
For both options, the wires connecting from the main road to the house have a total resistance
of 0.050 ohm.
Question 1
What is the peak voltage across the input terminals of the transformer?
Question 1
Vpeak  VRMS 
2 11 
2 15.6 kV
Question 2
In Option 2, if there are 10,000 turns in the primary winding of the transformer, how many turns are there in
the secondary winding?
Question 2
Ns
V
 s
N p Vp
Ns
240


10000 11000
2400
 Ns 
 218 turns
11
Question 3
When 25 kW of electric power is being drawn from the output terminals of the ideal transformer, calculate,
for Option 1, the RMS current flowing in the transmission wires
Question 3
Since the transformer is ideal, 25 kW must be coming into the transformer.
Current flowing in the transmission wires =
P 25  103

 2.27A
V
11000
Question 4
When 25 kW of electric power is being drawn from the output terminals of the ideal transformer, calculate,
for Option 2, the RMS current flowing in the transmission wires
Question 4
Since the transformer is ideal, 25 kW must be coming into the transformer.
P 25  103

 104A
Current flowing in the transmission wires =
V
240
Question 5
When 25 kW of electric power is being drawn from the output terminals of the ideal transformer, calculate,
for Option 1, the power loss in the transmission wires.
Question 5
Power loss =
2
I R  2.27  2.27  0.05  0.26 W
Question 6
When 25 kW of electric power is being drawn from the output terminals of the ideal transformer, calculate,
for Option 2, the power loss in the transmission wires.
Question 6
Power loss =
2
I R 104 104  0.05  5408 W
The Smith family and the Jones family are farmers near Warragul. Their electricity supply comes more than
100 km, from a power station in the LaTrobe valley. It is carried by transmission lines, at a voltage of 220
kVRMS. Near the town, a switchyard-transformer (T1) steps the voltage down to 10 kVRMS for the local area. A
10-kVRMS line runs to the Jones’ farm, where there is a transformer (T 2) that provides 240 VRMS for the farms.
A 240-VRMS line then runs 2 km to the Smith’s farm. A sketch of the situation is shown below. The
transformers can be considered to be ideal.
Question 7
Explain why the supply from the power station to the local area is chosen to be 220 kV RMS rather than at 240
VRMS. Use numerical estimates to support your answer.
Question 7
This question was clearly written by a city person, Warragul is approximately 40km from the Latrobe Valley.
Ignoring this.
Power loss in transmission lines is given by P = I 2R and thus reduce power loss most by reducing the
current in the wire.
This is done by increasing the voltage using a transformer.
Stepping up the voltage from 240V to 220kV represents an approximate change of 1000 which will result in a
current change of approximately 1/1000.
2,
this corresponds to a power saving factor of about 106
Assume that the input voltage to transformer T 1 is 220 kVRMS, and the output is 10 kVRMS.
Question 8
What is the value of the ratio:
number of turns on the primary coil
number of turns on the secondary coil
Question 8
This is a step down transformer with a ratio of
220k
 22
10k
i.e.: there will be 22 times more turns on the primary coil compared with the secondary coil
The supply and the return lines between transformer T2 and the Smith’s farm have a total resistance of
0.0004 ohm m–1.
At a particular time, 20 A of current is being supplied to the Smith’s farm. Assume that the potential at the
secondary for transformer T2 is 240 VRMS.
Question 9
What is the voltage at the Smith’s farm?
Question 9
The answer depends upon the interpretation given to the term ‘total resistance’ for the supply and return
lines;.
If you interpret this as the combined resistance over the distance of 2000m, then the resistance will be
0.0004 x 2000 = 0.8Ω. This gives a voltage drop of 20 x 0.8 = 16V and a supply voltage of 240 - 16 = 224V
If you interpret this as a total line length of 4000m then the resistance will be doubled giving a voltage drop of
32V and a supplied voltage of 208V (Correct response)
It is common practice for the wires in the cables associated with garden lights to carry only low-voltages
(often 12RMS). However it is more efficient to use 240-volt globes in the lights. In order to achieve this, the
circuit shown in Figure 2 is used. At the 240-V supply, the voltage is stepped down using a 240-V to 12-V
transformer, and at the light it is stepped up using a 12-V to 240-V transformer. The wires joining the two
transformers are each many metres long. The transformers can be assumed to be ideal.
Question 10
The light globe is rated at 120 W when connected to a 240-VRMS supply. What current should flow through it
under this condition?
Question 10
P=VI
120 = 240  I
I = 0.5 A
When the system shown in Figure 2 was tested, it was clear that the globe was not operating at the rated
120 W.
Question 11
Explain the reason for this.
Question 11
There is a significa
voltage input at the second transformer is less than 12 V. As a result, the voltage output is less than 240.
When the garden light is operating, the voltage across the input to the transformer that supplies the globe is
10RMS.
Question 12
What is the voltage across the globe?
Question 12
The transformer is working on a 20 to 1 ratio. Hence,
the output voltage (voltage across the globe)
= 10  20 = 200 V
RMS
Under these conditions the current flowing through the long wires is 8.3 A.
Question 13
What current is flowing through the globe?
Question 13
Transformer is ideal. Therefore, power rating of globe = 10  8.3 = 83 W. Hence, current in the globe =
83
 0.415A
200
Question 14
What is the total resistance of the two wires? Remember that the transformers are ideal.
Question 14
Power loss
I 2R = (12 x 8.3) – 10 x 8.3) = 16.6 W
16.6
 0.241 ohm
R
8.3  8.3
=
Question 15
Which one or more of the following changes would increase the voltage across the globe?
A. use wires of higher resistance
B. use wires of lower resistance
C. use transformers with ratios of 240:24 and 24:240
D. use transformers with ratios of 240:6 and 6:240
Question 15 B and C
By using wires of lower resistance, there will be less power loss and, therefore, a higher input voltage to the
globe transformer. This will give a higher voltage across the globe. B
By using transformers with a 10 to 1 ratio, there will be a higher transmission voltage along the wires. This
will give a smaller current and less power loss and, therefore, a higher input voltage to the globe transformer.
This will give a higher voltage across the globe. C
Andreas is having trouble with the reading lamp on his desk. Globes rated at 60 W and designed for an
RMS voltage of 240 V keep burning out. He seeks help from his friend Emma who is a qualified electrician.
They measure the RMS supply voltage and find it to be 264 V.
Question 16
What would the RMS current in one of these 60 W globes be if it was connected to an RMS voltage of 240
V?
Question 16
Use
P=V×I
60 = 240 x I
I
= 60/240
= 0.25 Amp
Emma and Andreas have different solutions to the problem of the globes burning out. Emma suggests to
solve the problem by stepping down the supply voltage from 264 V to 240 V, using a transformer with 1000
turns in the primary coil as shown in Figure 1.
Figure 1
Question 17
How many turns would there be in the secondary coil, assuming the transformer is ideal?
Question 17
The ratio of the number of turns in the primary coil : secondary coil is the same as the input voltage : output
voltage. Remember always think about the 'voltages'.
264
1000

=
240
' x'
240  1000
x=
264
 x = 909 turns.
Andreas suggests the problem could be solved by connecting an appropriate second globe (globe X in
Figure 2) in series with the 60 W globe.
Figure 2
Question 18
What would be the power used in globe X?
Question 18
The voltage across globe 'x' needs to be 24 V.
The current in both globes will be the same, because they are connected in series.
The current in the 240 V globe is found from P = V × I
60 = 240 × I
I = 0.25 Amp
 the power in the 24 V globe
P=V×I
= 24 × 0.25
= 6 Watt
Joe and Jan are installing two low-voltage lights in their garden. The lights are supplied from a transformer
that has an output RMS voltage of 12 V, and is connected to the 240 V household supply.
Question 19
What is the value of the ratio number of turns on the primary coil
number of turns on the secondary coil
Question 19
20 (or 20:1)
N
V1
240
= 1 =
= 20.
N2
12
V2
Each light is designed to operate at an RMS voltage of 12 V, and has a resistance of 18 ohm, which does
not depend on temperature.
Question 20
What is the power dissipated in such a light when operated at an RMS voltage of 12 V?
Question 20
8.0 Watt
P = VI =
V2
122
=
= 8.0
18
R
Joe and Jan now connect light 1 to the transformer using two wires, each 16.0 m long, as shown in Figure 2.
Each wire has a resistance of 0.050 ohm per metre.
Question 21
What is the RMS voltage across light I? Show your working.
Question 21
11 Volt
The resistance of the wire is given by 16.0  2  0.050 = 1.6  (You need to include both wires, hence the
length is 16.0  2)
 Rtotal = 18.0 + 1.6 = 19.6  .
V
12
I=
=
= 0.6122
R
19.6
V = I  R = 0.6122  18 = 11.0 Volt
They now connect light 2 directly across the secondary of the transformer as shown in Figure 3.
Joe and Jan thought that with the circuit in Figure 3, the two lights would be equally bright. In fact light 2 is
brighter than light 1.
Question 22
In the space below explain why this is so. Your answer should also include
• the value of the current flowing through light 2
• the value of the current flowing through light 1.
Question 22
Light 1:
I1 = 0.6122 A (as per question 6)
12
Light 2: I2 =
= 0.6667 A
18
 current through light 2 > current through light 1
 Light 2 brighter.
Light 1 is now unplugged.
Question 23
Which of the statements below (A-D) best describes the change in the brightness of light 2?
A. light 2 gets less bright
B. light 2 gets brighter
C. light 2 gets much brighter and bums out
D. light 2 does not change in brightness
Question 23
B or D
The better answer is D, because assuming that the transformer is ideal, then there will not be any change to
either V so the current is constant, therefore the globe will be of the same brightness. If the transformer was
not ideal, then you could assume that as the current went down, then there would ave been a very slight
increase in the voltage available across light 2. This would have resulted in an increase in the current in light
2, therefore it would have been brighter. (This answer needed to be explained).
A power station in the Latrobe Valley generates electric power at an RMS voltage of 20 kV (20 000 V). The
switchyard transformer steps up the voltage to an RMS value of 500 kV (500 000 V) for transmission to
other parts of Victoria. On the outskirts of large towns a local-area transformer steps down the voltage for
local transmission. The circuit in Figure 9 shows how the power station and the two transformers are
connected.
Figure 9
The RMS current in the secondary of the switchyard transformer is 300 A. Assume the two transformers are
ideal.
Question 24
What is the RMS current in the primary of the switchyard transformer?
Question 24
If the transformer is ideal then the power out = power in
 500 × 103 × 300 = 20 × 103 × I
  I = (500 × 103 × 300)/20 × 103
  I = 7500 Amp
The total resistance
Question 25
How much power is lost in the transmission lines?
Question 25
Power loss in transmission lines is given by i2R
= 3002 × 8
= 720000
= 7.2 × 105 W
Question 26
What is the RMS voltage across the primary of the local-area transformer? Give your answer to three
significant figures and make sure you show your working clearly.
Question 26
The voltage drop will be given by P = V × I
7.2 × 105 = V × 300
V = 7.2 × 105  300
V = 2400 V
 the Voltage at the local area transformer = 500 000 - 2400 = 4.98 × 105 V (Correct to 3 sig. figs.)
A 4.0-kW electric generator is used to supply power to an electric motor which is located some distance
away.
The RMS voltage at the output of the generator is 400 V, and the RMS current supplied to the motor is 10 A.
The total resistance of the two cables, AC and BD, is 3.0 W. The situation is shown below in Figure 2.
Question 27
How much electric power is dissipated in the cables?
Question 27
If the Power loss is given by i2R
 P = 102 × 3
= 300 W
= 3.0 × 102 W
(correct to 2 sig. figs)
An engineer suggests that by connecting a transformer (T1 ) at the generator output, and another (T2 ) at the
motor input, a larger fraction of the generated power could be transferred to the motor. The plan is shown
below in Figure 3.
Question 28
Explain why this procedure increases the power transferred to the motor. Your answer should clearly show
the physics involved. Indicate what type of transformer (step-up or step-down) should be used in each
location.
Question 28
Power losses due to heating in transmission lines are given by P = i2R.
To minimise this we need to minimise I. Since the power input is P = V × I, this means we need to increase
V as much as possible.
We use a step-up transformer at 'T1'. The step-up transformer steps up the voltage but steps down the
current, resulting in a lower electric current in the cables and reduced power loss within the cables.
We then need to use a step-down transformer 'T2' at the motor, to reduce the voltage to the original value,
400 V.
After installation of the transformers, the RMS current in the transmission wires is 0.5 A. Assume the
transformers are ideal.
Question 29
What is the RMS voltage between the terminals A and B in Figure 3 now?
Question 29
If the initial current was 10A, and the new current is 0.5A, then the current has decreased by a factor of 20.
This means that the potential difference between A and B must increase by a factor of 20. This is because
in ideal transformers the power in = power out. So VIin = VIout.
So the new voltage across AB will be 400 × 20 = 8,000 V
= 8.0 × 103V
Question 30
How many turns of wire are in the primary of transformer T 1 if the secondary consists of 5000 turns? Show
your working.
Question 30
The voltage across AB increases by a factor of 20. So it must be a step-up transformer. This means that the
number of turns in the primary will be 5000/20 = 250 turns.
An electrician has imported an electric light globe that is designed to operate with an RMS voltage of 110 V.
When operating at this voltage it has a resistance of 55 ohm. The globe has a power rating of 220 W. Two
methods are considered that will allow the light globe to be used with a mains supply with an RMS voltage of
240 V.
In method 1 a resistor of 65 ohm is placed as shown below in Figure 1, so that when used with an RMS
supply voltage of 240 V, the RMS voltage across the light globe is 110 V.
Figure 1
In method 2 a transformer is used to convert the RMS voltage of 240 V to an RMS voltage of 110 V, as
shown below in Figure 2. Assume the transformer is ideal.
Figure 2
There are 1440 turns on the primary coil of the transformer.
Question 31
How many turns are on the secondary coil of the transformer?
Question 31
The ratio of the number of turns in the primary coil : secondary coil is the same as the input voltage : output
voltage. Remember always think about the 'voltages'.
240
1440

=
' x'
110
1440  110
240
 x = 660 turns.
x=
Check that your answer makes sense by asking the following questions. Is it a 'step-down' or 'step-up'
transformer? In this case it is a step-down transformer because the output voltage is less than the input
voltage. This then means that the number of turns in the secondary coil must be less than the primary coil.
Question 32
What is the RMS current in the primary coil of the transformer?
Question 32
We are told that the transformer is 'ideal'. This means that the power in = power out.
So (V × I)in = (V × I)out. The power out is the rating of the globe. It is rated as 220W, so the power in must
equal 220W.
So Pin = 220 = V × I
= 240 × I
I
= 220/240
= 0.92 Amp
Question 33
What is the power supplied from the mains for the two different methods?
Question 33
In Method 1, some power is used in the resistor, this is given by V2/R.
V = 240 - 110
= 130 Volt (sum of voltages drops around the circuit = 240 V)
So power used in resistor = 1302/ 65 = 260W.
 total power used in circuit
= 260 + 220
= 480W
In method 2, the power loss is 220W. Because the globe is the only appliance that is using power.
A generator at a farmhouse supplies electricity to two motors in a workshop, which is several kilometres from
the house. The RMS voltage at the generator is 230 V. When the two motors are operating, the RMS voltage
at the workshop is 220 V. When operating, the RMS current in motor 1 is 13 A and that in motor 2 is 10 A.
Figure 3 below shows the generator connected to the workshop, but the wiring to the motors has been
omitted.
Figure 3
Question 34
On the copy of Figure 3 in the answer book, show the wiring to the two motors.
Question 34
From reading the stem of the question very carefully, we glean that the currents in the two motors are
different, so they must be connected in parallel, not in series.
Both will have 220 V across the motor.
13 A
1
10 A
2
230 V
You need to show that the generator supplied 230 V, and that motor 1 used 13 A, with 220 V across it, and
motor 2 used 10 A with 220 V across it.
You will also need to consider that the charge loses 10 V of PD along the lines.
Question 35
How much power is supplied by the generator when the two motors are running?
Question 35
The total power = VI, where V is the voltage that the power is supplied at (230 V), and I is the total current
supplied, (23 amp)
P = 230 × 23
= 5290 W
= 5.29 × 103 W
Question 36
How much electric power is lost in the cables which connect the generator and the workshop? Show your
working.
Question 36
The total power supplied is 5290 W.
Motor 1 uses P = VI
= 220 × 13
= 2860 W
Motor 2 uses P = VI
= 220 × 10
= 2200 W
So total power used by both motors is 2860 + 2200 = 5060 W
 5290 - 5060 = 230W, must be the power being lost in the lines.
The other option was to consider that there was a voltage drop of 10 V across the lines, so if the lines were
carrying a current of 23 Amp, then the power loss in the lines
=V×I
= 10 × 23
= 230 Watt.
ELECTRIC LAWS
The 60-watt light bulb in Sam’s desk lamp has burnt out. There is no spare replacement that operates on
the 240-V supply. However, a 60-watt light bulb is found that operates at 120 V. Sam suggests building the
following circuit in order to use the 120-V light bulb.
Figure 1
Question 1
What is the value of the resistor R that will allow the 120-V light bulb to operate correctly?
Question 1
This is a voltage divider question. To have 120v across the light bulb you need to halve the supply voltage
and you do this by placing the same resistance in series with it; hence the resistor must have the same
resistance as the bulb.
From
V2
we have
P  VI 
R
1202
R
 240
60
Sometimes strings of Christmas-tree lights consist of three groups of globes that are connected as shown in
Figure 3.
There are 16 globes in group P. Each of the globes has a voltage of 10 V across it and a current of 0.50 A
flowing through it when the string of lights is operating as designed. The globes in groups Q and R have a
different power rating to those in group P.
The number of globes in groups Q and R is equal. Although they have a different power rating from the
globes in group P, the potential difference across each globe is still 10 V when operating.
Question 2
How many globes are there in group Q?
Question 2
The voltage drop across group P will be 16 x 10 = 160V
The voltage drop across both group Q and R (the groups are in parallel) will thus be 240 – 160 = 80V
If each bulb in group Q has a voltage drop across it of 10V there will thus be 80/10 = 8 bulbs in the group.
Group R will also have 8 bulbs
Question 3
What current is being supplied from the electricity supply to the string of lights?
Question 3
The string of lights is the total combination of P + Q + R and this is supplied with 0.50A of current
Assume that the power rating of all globes in Q and R are identical.
Question 4
How much power is dissipated by each globe in groups Q and R?
Question 4
From Q6 the current thru P is 0.50A and as Q and R are identical they will each have 0.25A flowing through
them.
Each bulb has a voltage drop of 10V across it and hence the power dissipated by each bulb in groups Q and
R will be P = VI = 10 x 0.25 = 2.5W
One of the globes in group Q burns out.
Question 5
Indicate in the box beside each group whether the globes in that group are on or off. If the group is on,
indicate whether the globe is brighter or dimmer compared to when the system is operating correctly.
Question 5
Since Q and R are in parallel, if group Q is ‘removed’ from the circuit this will result in doubling the resistance
of this part of the circuit which will result in a reduction of the current flowing through part P. Part R however
will have an increase in the current that flows through it. Hence
Group
ON/OFF
Brightness
P
ON
Dimmer
Q
OFF
R
ON
Brighter
Question 6
What is the resistance of the thick copper wire?
Question 6
R
V 1  10 4

 0.001  1.0  10 3 ohm
I
0.1
Question 7
How many electrons enter the copper wire each second? An electron has a charge of 1.60 x 10 -19 C.
Question 7
Current = coulombs per second
0.1 A
 0.1 Cs

1
0.1
electrons per second
1.6  10 19
17
 6.25  10
electrons per second
A portable electric heater has two settings, 'high' and 'low’. These heating levels are obtained by connecting
two heating elements either in series or in parallel, across the 240- VRMS mains supply. Each element has the
same electrical resistance.
When the heating elements are connected in series, the total power dissipated in them is 960 W.
Question 8
What is the resistance of each element?
Question 8
P  VI
 960  240  I
960
I
 4A
(
240
V 240
 Rtotal  
 60
I
4
 Each element  30
Question 9
What is the value of the ratio
total power dissipated in heating elements in series
total power dissipated in heating elements in parallel
Question 9
In parallel,
1
1
1
2
1




RT 30 30 30 15
 RT  15
V  IR
240
I
 16A
15
Power Loss  I 2 R  16  16  15  3840W
Ratio 
960
 0.25
3840
A student is provided with several lengths of wire, plus:
a 6-V battery
a 6-W light globe
a switch
a 2-ohm resistor
Question 10
In the space provided draw a circuit that could be connected using all the components, so that the globe
lights up all the time, but becomes brighter when the switch is closed.
Question 10
The globe must light up all of the time and so the globe must be connected to the battery independent of the
switch. When the switch is closed, the total resistance of the circuit decreases (since the resistor and the
globe are in parallel) and so the current through the lamp increases. The globe becomes brighter when the
switch is closed.
A lightning discharge between the ground and a cloud is composed of a series of strokes. The duration of
each lightning stroke is typically 30 microseconds. The power per stroke is 1.0 x 10 12 W.
Question 11
If the voltage between the cloud and the ground is 4.0 x 106 volt, how much charge is transferred in a single
stroke?
Question 11
energy
Power =
time
U
1.0  1012 =
30  10 6
 U = 3.0  107 J
U = qV
 3.0  107 = q  4.0  106
 q = 7.5 C
An alternative way was P = VI
I=
Then using I =
1.0  1012
P
=
= 2.5  105 Amp.
V
4.0  10 6
Q
 Q = 2.5  105  30  10-6 = 7.5 C.
t
ENERGY AND POWER
Figure 9 below shows a graph of the average daily usage of electricity in a house. The vertical scale of the
graph
is in units of kilowatt-hour (kW h).
Figure 9
The unit kW h used on the graph is rarely used in physics.
Question 1
Circle the letter (A–D) that corresponds to the unit that could be used, with a different scale, on the vertical
axis instead of kW h.
A. J
B. J s
C. J s-1
D. W
Question 1
A
The unit kWh is a measure of the power × time, so it is a measure of ENERGY. The units of energy are
more commonly Joule
The average daily usage for the 3-month period September–October–November 1996 is 16 kW h. All this
electricity was used to heat the house with an electric heater with a power consumption of 2000 W.
Question 2
How many hours each day, on average, was the heater used?
Question 2
It took 16 kWh to run the heater each day.
So 16 000W = 2 000 W × time (in hours)
 t = 8 hours.
You must ask yourself is your answer reasonable? This one appears to be a reasonable estimate of how
long the heaters were on in the house on average over winter.
The electricity bill for a household usually shows the average daily usage of electric energy for one year as
shown in Figure 4 below.
Figure 4
Question 3
Estimate the total electric energy (in joule) used by the household in the three months of June, July and
August 1996.
Question 3
You need to read the axes very carefully. It is giving us the average daily usage, so to find the total electrical
energy we need to multiply this by the number of days. There are 92 days in these 3 months.
 Energy used = 30 kWh/day × 92
= 2760 kWh's.
To convert this to joule you need to multiply by 103 to include the 'kiloWatt' and by 60 × 60 to convert
hours to seconds.
= 9.936 × 109
= 9.94 × 109 J
The cost of supply is 12 cents per kWh.
Question 4
How much does it cost to operate a 2.4-kW heater for 8 hours?
Question 4
2.4 kW × 8 hours = 19.2 kWh.
The cost for 1 kWh = 12 cents.
 the cost = 19.2 × 12 = 230.4 c
= $2.30.
Note that the answer can't be $2.304, because you cant get 0.4 of a cent.
It is now possible to buy an electrical safety unit that tums off the elecaical current in an appliance if a current
flows in the earth wire. The power lead of the appliance plugs into a box containing the unit. The box in turn
plugs into a power point.
Figure 5 above shows a clothes dryer connected to the mains supply via a safety unit. The iron ring is an
essential part of the safety unit. The active and neutral wires run through the ring. The clothes dryer has an
electrical fault, a connection between the active wire and the earthed metal case of the dryer.
Question 5
The magnitude of the current in the earth wire is 2 A, and the magnitude of the current in the neutral wire is
5 A (see Figure 5). What is the magnitude of the current in the active wire?
Question 6
After the clothes dryer is repaired, there is no current in the earth wire when the dryer is operating correctly.
In this case, which one of the following statements best describes the situation in the iron ring?
A. There is an alternating magnetic flux and an alternating current in the ring.
B. There is an altemating magnetic flux and zero current in the ring.
C. There is zero magnetic flux and zero current in the ring.
D. There is zero magnetic flux and an alternating current in the ring.
E. There is a constant non-zero magnetic flux and a constant non-zero current in the ring.
Figure 6 is a sketch of a single phase electricity supply that the SEC installed for a farmer whose farmhouse
was off the main road.
Before the electrician connected the secondary of the transformer to the house circuit, the RMS voltage
across the secondary coil was rFasured to be 240 V. The transformer, which we can assume to be 100
percent efficient, had a ratio of primary tums to secondary turns equal to 45.0. When the installation was
complete, the electrician conducted a load test at the maximum designed load for the supply. This involved
tuming on most of the electrical appliances in the house. During this load test, the electrician measured an
RMS voltage of 230 V across the secondary coil of the transformer, and an RMS current of 52.5 A in the
secondary coil.
Question 7
During the load test, estimate the power use (in W) in the house.
Question 8
During the load test described above, what would be the RMS voltage (in V) across the primary coil?
Question 9
During the load test described on page 19, what would be the RMS current (in A) through the primary coil?
Question 10
During the load test described on page I 9, what was the power loss (in W) in the wires from the I I 000 V
(RMS) supply at the main road to the transforrner near the house?