Download Chapter 5 - Vicphysics

Chapter 4
Transformers and transmission.
Question 128
The number of turns is directly proportional to the voltage output.
So to step the voltage up from 215 to 240, the number of turns in the secondary coil needs to increase in the
same ratio.
 the new number of turns is 500 × 240/215 = 558 turns.
Question 129
E
This is a step down transformer, because the output voltage is less than the input voltage. This means that the
number of turns in the secondary must be less than in the primary, but in the same ratio as the voltages.
Ratio = 240/19.8 = 12.1
Question 130
Power = V × I Pin = 240 × 1.2 = 288W.
Pout = 19.8 × 12 =237.6
 Ratio of power out/power in = 237.6/288 = 0.825 = 0.83 (correct to 2 sig. figs.)
Question 131
You would always expect the actual value of the ratio power out/power in to be less than 1. It can never be
greater than 1, because this would mean that you were getting more power out than you were putting in. So the
transformer must have been gaining energy from somewhere. This is impossible.
So the actual ratio will always be less than 1, on the assumption that the transformer is losing (using) energy.
If you touch any transformer, eg. The little transformer to my computer, you will immediately notice that it is
warm. This heat energy is being generated because the transformer is not 100% efficient. Energy is being lost to
heat. The heat is produced by stray currents in the iron core, that are not contributing 100% to the field being set
up.
Question 132
You need to transform the voltage from 240 V down to 12 V. This requires a step-down transformer with a ratio
of 240/12 which is 20/1 in the primary / secondary.
 20
Question 133
The transformer works on the principle that an AC current in the primary coil will induce a changing magnetic
field in the iron of the transformer (because the field associated with a current carrying wire varies with the
current in the wire). The changing magnetic flux in the iron core will induce an EMF in the secondary coil.
If the current in the primary coil is constant i.e. DC, then the field in the iron core is constant, so there is not an
induced EMF (and current) in the secondary coil.
Question 134
The ratio of the number of turns is the same as the ratio of the input and output voltages.
V
V
n
n1
= 1 which becomes 2 = 2 for this problem
V2
V1
n2
n1
n2
330kV
=
= 22
15kV
n1
You need to realise that this has to be a step-up transformer, so the secondary coil has to have more turns than
the primary coil.

Question 135
If the transformer is ideal (100% efficient) then the power in = power out.
 VI in = VI out.
This means that if the voltage steps down by a factor of 9 000 to 110, then the current must step up the same
amount.
9000
The voltage steps down by
= 81.82
110
So the output current will be 81.82 times the input current.
 81.82 × 1.5 = 122.7 Amp.
= 123 Amp
Question 136
There is always a voltage drop across a resistor, (assuming that there is a current). For most situations the
resistance of the wires in the circuit are so small in comparison to the resistance of the appliance that it is
appropriate to assume that the voltage drop across the wires is zero.
When we have very long cables this assumption is not reasonable. There will be a noticeable potential drop
along the cable. In this case the drop is 5 000V. It is given by V = iR. Where I is the current and R the
resistance of the wires.
Question 137
The active and the neutral should always carry the same current, because all the current that flows into a device
should flow out of it. There is no build up or loss of electrons in any part of the circuit.
If the currents are not equal, i.e. the current in the active is greater than the current in the neutral, then some of
the current has found an alternative path to earth. This should not happen, it might be through the person using
the device or through the casing of the device. This would make the casing 'live' to touch.
This may also mean that the insulation protecting the user has broken down.
Question 138
With the wires around the core as shown the magnetic flux changes due to each should be equal and opposite,
so they will cancel each other out. Magnetic fields are vectors.
Question 139
C
If the current in the active wire was greater than the current in the neutral wire, then the flux induced in the coil
from the active wire would be greater than that produced from the neutral wire. Because the current is AC, then
the flux through the coil will be changing
Question 140
There is always a voltage drop across a resistor, (assuming that there is a current). For most situations the
resistance of the wires in the circuit are so small in comparison to the resistance of the appliance that it is
appropriate to assume that the voltage drop across the wires is zero.
When we have very long cables this assumption is not reasonable. There will be a noticeable potential drop
along the cable. It is given by V = iR. Where I is the current and R the resistance of the wires.
As more appliances are being used then the current being drawn increases. This means that the potential drop
along the cable increases, so the voltage supplied to the house is decreased.
Question 141
If V = iR, then the voltage drop along the cable is given by V = 240 - 225 = 15 volt.
So if the current is 45 amp, then the resistance comes from
V = iR
 R = V/i = 15/45 = 0.33 
Question 142
The voltage drops along the line because the electrons lose energy travelling down the line. The definition for a
volt is the energy in Joule the each Coulomb of charge contains. The electron must do work to travel down the
lines so they lose energy, hence potential (Voltage)
The loss is calculate from V = I x R.
(I is the current and R is the resistance of the wire)
Question 143
AD
The electrons lose energy, but we don't lose electrons, so the current stays the same, but there is less energy
being transmitted. This means that the rate of energy being delivered is less, so the power is less.
Question 144
Power loss = V × I
The change in potential is from 66 kV to 65.8 kV.  V = 200 V
The current can be found from the power being put into the circuit.
At the Terminal station T, the power = 10 MW (107 W).
P=VxI
So the input current can be found. This current will be the same throughout this circuit.
 107 = 66 × 103 × I
 I = 107  66 × 103
 I = 151.5 amp
 Power loss
= V × I
= 200 × 151.5
= 30303.0 W
= 3.0(3) × 104 Watt.
Question 145
The V will equal the product of the current and the resistance.
 V = I × R
 200 = 151.5 × R
 R = 200  151.5
 R = 1.3(2) 
Question 146
The resistance of the cable will be different.
The new resistance
= 0.24 /km × 80 km (From question stem)
= 19.2 
 power loss = i2 × R.
from the station hasn't altered.)
 Power loss = 151.52 × 19.2
440683.2 W
0.44 MW
(we know that the current will be the same, because the power input
Question 147
The power being produced is 10 kW at 200 Volt.
At the generator, the power = 10 kW (104 W).
P=VxI
So the input current can be found. This current will be the same throughout this circuit.
 104 = 200 × I
 I = 50 amp.
Question 148
If the voltage drops by 30 volt across the lines then
The V will equal the product of the current and the resistance.
 V = I × R
 30 = 50 × R
 R = 0.6
Question 149
Power loss in cables
= i2 R
= 50 × 50 × 0.6
= 1500 W
Question 150
The idea of installing the transformers is to cut down on this energy loss. To do that you need to minimise I, so
we use a step-up transformer at the 'generator end' and a step-down transformer at the 'load end'.
If the transformers have a ratio of 1:20, then we will step the voltage up to 200 × 20 = 4 000 V and then transmit
at this voltage.
Generator
10 kW
Step up
transformer
Step down
transformer
Load
Question 151
This will mean that to get the equivalent power down the line we now only need 1/20 th of the current. So the new
current is 50/20 = 2.5 amp.
Question 152
The voltage loss across the cables will now be
where I = 2.5 amp and R = 0.6
So V = 2.5 × 0.6
= 1.5 Volt
 V = I × R
Because the farmer had used a transformer, the initial voltage was 4 000 V. !.5 was lost along the lines so the
input voltage to the step-down transformer at the load was 4 000 - 1.5 = 3998.5 V.
The step-down transformer will then divide this by 20, so the output voltage (of the step-down transformer) is
3998.5  20 = 199.9 Volt
Question 153
The resistance of the wires will stay the same, so the power loss in the cables is now given by
P = i2 × R
= 2.52 × 0.6 
= 3.75 Watt
=3.8 Watt
(correct to 2 sig. figs.)
Question 154
power loss with transforme rs
power loss without transforme rs
If the ratio of Vp:Vs = 1:20, then the ratio of Ip:Is = 20:1
As the resistance of the lines will stay constant then the ratio of the power loss will be the square of the current
1
i2r
with transformers compared with the current without transformers.
=
2
400
ir
Question 155
The ratio of the turns is the same as the ratio of the voltages.
The input voltage is 240 with an output of 6, so there must be (240/6) times as many turns on the primary side of
the coil.
240  6 = 40.
So the number of turns on the output (secondary) side of the coil is 960  40 = 24 turns.
(Remember: The more turns the greater the voltage.)
Question 156
C
The supply is AC, (otherwise the transformers wouldn't work) So with an RMS current of 3.5 Amp, this means
that the peak value of the current will be 3.5 × 2 = 4.95 amp ~ 5 amp
The output voltage will look a little like this.
EMF (volts)
This means that the average value of this graph is the middle.  the average is zero.
Question 157
The voltage loss across the cables will now be
where I = 3.5 amp and V = 1.0 Volt
 R = V  I
= 0.288
 R = 0.29
 V = I × R
Question 158
Power loss in cables
= V × I
= 1.0 × 3.5
= 3.5 Watt
This answer could also be found by using power loss in cables
= i2 × R
= 3.52 × 0.29
= 3.5(3) W
Question 159
E
Power = V × I = I2 × R = V2/R
15 kW = V2/0.8 (Don't forget the kW)
15 000 × 0.8 = V2
 V2 = 12 000
 V = 109.5
 V = 110 Volt
120 Volt is a better answer than 100V because the motor needs at least 15 kW.
Question 160
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 'A' so that the current in the lines is minimal. We then need to use a step-down
transformer at the pump, to reduce the voltage to 240 V.
Question 161
The power being used by the pump is P = V × I
= 240 × 20
= 4 800 W
So this must be the power being supplied to transformer B
So the V × I (in the wires ) = 4 800
V = 4 800 / 0.8
= 6 000V
Question 162
Assume that the voltage VPQ = 6 000 V. (On the exam, if you weren't able to calculate this, then you are to
substitute in any number and explain it) 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'.
VP
N
= P
NS
VS
' x'
240
=
6000
100
240  100
x=
6000
 x = 4 turns.

Question 163
If the Power loss is given by i2R
 P = 0.82 × 4
= 2.56 W
= 2.6 W
(correct to 2 sig. figs)
Question 164
If the Power loss is given by i2R
 P = 102 × 4
= 400 W
= 4.0 × 102 W
Question 165
Use
P=V×I
P = 330 kV x 200
P
= 330  103  200
= 66  106 W
= 66 MW
(correct to 2 sig. figs)
Question 166
As the transformer is ‘ideal’ this means that PIN = POUT
 VINIIN = VOUTIOUT
 15  103  IIN = 330  103  200
 IIN = 66  106  15  103
= 4 400 Amp
Question 167
VLOSS = IRTRANSMISSION LINES
 VLOSS = 200  5
= 1000 V
Question 168
If the transformer is ideal (100% efficient) then the power in = power out.
 VI in = VI out. This means that if the voltage steps down by a factor of 20, then the current must step up the
same amount. So the output current will be 20 times the input current.
 10 × 20 = 200 Amp.
Question 169
Since V = IR, if you use a step-down transformer, you have lowered V to 12 volts. This makes its very difficult to
get a large current in any circuit, because you need R to be small, to allow I to be large.
Question 170
P = VI
120 = 240  I
I = 0.5 A
Question 171
This system would only have worked if there were no power loss in the wires. The voltage is dropped down on
the first transformer; this increases the current that is travelling through the wires. The more current that is
flowing through the wires the more the power loss there will be. The relationship between power loss and
current is Ploss  I 2 R , so if the current goes up by 2 the power loss goes up by 4.
There is also the effect on the voltage loss in the wires Vloss  IR this means that at the second transformer there
is not a full 12V, the value will be lower so the after it is transformed back then it will not be as high as 240volts.
If the voltage supplied to the light globe is not 240volts then the power dissipated in the light globe will not be
120Watts, it will of course be lower, thus the light is not putting out as much light energy per second, so it will not
work effectively.
Question 172
V2 N2

V1
N1
240
V2 
 10
12
V2  200V
Question 173
As the transformers are ideal we know that Pin = Pout
If the voltage increases by a factor of 20, to maintain the same power the current must decrease by a factor of
20.
240
Vout = 10 
.
12
12
 currentout must be 8.3 
= 0.415 amp.
240
Question 174
Determine the voltage loss because of the wires.
There is 10 volts for the transformer; therefore there is 2 volts lost in the wires. Using Vloss  IR , we know the
current through the wires was 8.3A. Manipulate the formula to get R by itself, R 
R
2
 0.24
8.3
V
and you get the
I
Question 175
B and C,
If the resistance of the wire was lower, then the voltage drop across them would decrease, so the voltage
across the globe would increase.
If the transformer was 240:24, then the voltage on the output side of the step down transformer
would be 24 volts. To supply the power needed the current would be smaller than if the transformer was
240:6.  the smaller current means less power loss due to i2r in the wires, a higher voltage at the
second transformer.
Question 176
Power loss in transmission lines is given by P = I2R 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.
Since P  I2, this corresponds to a power saving factor of about 106
Question 177
220k
 22
10k
i.e.: there will be 22 times more turns on the primary coil compared with the secondary coil
This is a step down transformer with a ratio of
Question 178
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)
Question 179
This question is pretty basic, but only 20% of students got it right. Since R does not connected, there cannot be
a current in it.
This is basically a parallel circuit, using Kirchhoff’s conservation of current, the current in P must be 500 + 200 (=
700), and the current in Q must be just 500 A.
 P = 700A,
Q = 500A
R = 0A
Question 180
Energy = VI = 540  500 = 2.7  105W.
Question 181
If you find this question very difficult, then you are in excellent company. Only 3% of students got this right on
the exam.
If the voltage at tram 2 is 540V, then 60V has been lost along P + Q. Using V = IR, The voltage drop along P is
700R. The voltage drop along Q is 500R.
This means that 60V = 700R + 500R
60 = 1200R
R = 0.05
This is the resistance of 1.0km of wire.
Question 182
The voltage at tram 1 must be 600 – 700R = 600 – 700  0.05 = 600 – 35 = 565V.
Question 183
B&D
The power transmitted is given by P = VI. High transmission voltages are used so as to reduce the line
current.
Power loss in the wires can be calculated using the formula P = I2Rwires. To minimise power loss need to have
I as a minimum.
Hence, statements B and D were the expected answers.
Both B and D needed to be given to score the 2 marks for this question. Any other answer scored zero.
Question 184
The turns ratio can be calculated via the formula
N s Ip
=
Np Is
5
.
80
= 0.063
=
Question 185
The voltage drop across the wires can be determined by using Ohm’s law V = IR wires.
Substitution into this equation gives V = 5.0 x 0.32 = 1.6 V.
Question 186
Closing the switch caused an increasing current which led to an increasing magnetic field.
The resultant change in flux produced a current in the secondary.
Once the supply current reached a constant value, the magnetic field was constant and thus there was no
change in flux.
Therefore, there was no more current induced in the secondary and the meter returned to zero.
Question 187
C
Opposite flux induced, but the change is only momentary because the current is now off.
Question 188
As there is a 1 resistance in the lead, there will be a ΔV along its length
•which will reduce the ΔV across the lamp,
•and hence the current and power.
Therefore Lamp 1 is brighter than Lamp 2
Question 189
32 W
Using Ohm’s Law, the voltage loss in the extension lead was 4 V.
So the voltage across lamp two was 12 – 4 = 8 V.
Therefore the power dissipated in lamp two was P = V I = 8 x 4 = 32 W.
Question 190
D
The resistance in the branch with lamp one is less than that in the branch with lamp two. Therefore the current
through lamp one is greater than 4 A. Accordingly the combined current is greater than 8 A.
With calculations to support your argument, the voltage across Lamp 2 is 8V, because the voltage drop due to
the resistance of the wire is 4A  1 = 4V. So 12 – 4 = 8V.
 the resistance of the lamp is given by V = iR
 R = 8  4 = 2
This means that the current flowing through Lamp 1 is given by 12  2 = 6A.
 the current supplied by the generator must be 6 + 4 = 10A
Question 191
Transformer symbol or diagram showing/labelling primary, secondary and core with a statement that the primary
has 20 times as many turns as the secondary.
Soft iron
core
Ratio of turns Primary:Secondary = 20:1
Question 192
The current in the extension lead is only 1/20th of that in 12 V supply and •hence there is much less ΔV along the
lead, with less power loss. •
1
P=I²R so power loss only
.
400th
Question 193
We are given the power and the voltage (RMS indicates that we are given the average voltage)
 Power = VI
 1200 = 240  I
 I = 1200  240
= 5 amp
Question 194
The power loss in the transmission lines is given by P = i2R. where ‘i’ is the current in the lines and ‘R’ is the
total resistance of the lines.
 Power loss = 302  2
= 1800W
Question 195
The initial supply voltage is 240V. There is a potential drop across the 1 km of line.
This potential drop is given by V = IR
 V = 30  2
= 60V.
If the initial supply voltage is 240V, then after a loss of 60V there is only 180V at the house.
This question is poorly worded. The question should have asked for the ‘voltage drop’ at the house.
Question 196
Since the power loss in the lines is given by Ploss = i2R, if we decrease the current we will minimise the power
loss in the cables.
The current that is flowing in the lines depends on the demand. Assuming maximum demand then the power
being supplied is given by Psupplied = VI.
The power required is the same, so if V is 11 000V (compared with 240V), then the current will be much less.
240
 30  0.65amp ), then the power loss in the lines will be 0.652  2 = 0.86W (compared
11000
with 1800W initially)
If the current is (
Question 197
To convert VRMS to Vpeak to peak you must multiply VRMS by 2 2
 Vpeak to peak = 2 2  11 000
= 31113V
= 3.1  104V
Question 198
The definition of an ideal transformer is when Pin = Pout
 VinIin = VoutIout
From the data
240  IPrimary = 11.3  2.2
 IPrimary = 0.10 amp
Question 199
Faraday’s law gives the size of the induced EMF, while Lenz’s law gives the direction of the induced EMF and
the direction of the induced current.
Faraday’s law states that “The magnitude of the induced emf is directly proportional to the rate of change of
magnetic flux”. Faraday’s law is summarised as  = – n
 B
t
Lenz’s Law states that “the direction of the induced EMF is the same as that of a current whose magnetic action
would oppose the flux change”.
Question 200
C is the best answer. B is fairly close but if the power point is wired incorrectly it will not be safe. The question
specifically asks for the best precaution. The other two options are trivial solutions.
Question 201
The power being used is 10  105 W. This power is supplied at a voltage of 250 V.
Using P = VI
Gives 10  105 = 250  I
1 105
I=
250
= 400 A
Question 202
Assume that the transformer is ideal,
 PowerIN = PowerOUT
 (VI)IN = (VI)OUT
 22 000  I = 10  105
1 105
I=
22000
 I = 4.55 Amp (rounded off to 2 decimal places)
Question 203
When 100kW of power is being used in the village the current in the high voltage lines is 4.55 Amp.
The power loss in the lines = i2R
Where R = 2 
= 4.552  2
= 41 W
Question 204
The power being delivered is given by P = VI. The power loss in the cables is given by P = I 2R.
To minimise power loss in the cables either ‘I’ or ‘R’ need to be kept to a minimum. The resistance of the cables
is difficult to control or influence, so it is best to minimise ‘I’.
As the power being delivered is given by P = VI, then to keep the power delivered constant, if ‘I’ decreases then
‘V’ needs to increase by the same factor.
An increase in V allows a decrease in I, which means that losses related to I 2R are lowered.
Question 205
Transformer T1 is a step up transformer.  The number of turns in the secondary needs to be greater than the
number of turns in the primary.
22000
To step the voltage up from 250 to 22 000, you need the number of turns in the secondary to be
times
250
the number of turns in the primary.
22000
= 88.
250
 500  88 = 44 000.
C
Question 206
A changing current at the primary coil produces an alternating B inside the soft iron core. The secondary coil is
linked to the primary through the soft iron core. The changing B in the soft iron core results in a changing flux in
the secondary coil. This changing flux induces an EMF in the secondary coil.
According to Faraday’s Law
EMFAVE = -n
Δφ
Δt
Question 207
The transformers in the transmission system require AC. The alternator provides the AC, changing the magnetic
flux to induce an EMF in the secondary. If the voltage produced is constant (from a DC generator), then the
output of the transformer would be zero. Therefore an alternator should be used rather than a DC generator.
Question 208
B
The power demand has increased but the voltage at the alternator (P) remained the same.
Since Psupplied = VI, the current supplied had to increase.
This would lead to an increased current in the power lines.
This would mean an increased voltage drop along the lines, (Vdrop = IR).
Therefore a slightly lower voltage at the primary of transformer 2.
Since the transformer has not changed, with a lower voltage across the primary there will be a lower voltage
across the secondary.
Question 209
B
The alternator produces 250VRMS at 50 Hz. The peak voltage will be 250 
the period is 0.02 sec.
2 = 354 V. The 50 Hz means that