Download Transformers and transmission solutions

Chapter 3: Transformers and
transmission solutions
1.
[2 marks]
[3 marks]
This system would only have worked if
there were no power loss in the wires. The
voltage is dropped down at the first
transformer; this increases the current in the
wires.
The more current in the wires the more the
power loss there will be.
The relationship between power loss and
current is Ploss = I2 R , so if the current
increase by a factor of 2 the power loss
increases 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 that after it is
transformed back then it will not be as high
as 240volts.
If the voltage supplied to the light globe is
not 240 volts then the power dissipated in
the light globe will not be 120 Watts, it will
be lower, thus the light is not putting out as
much light energy per second, so it will not
work effectively.
3.
[2 marks]
(2001 Q5)
V2 N 2
=
V1 N 1
240
V2 =
×10
12
V2 = 200V (ANS)
4.
[2 marks]
(2001 Q6)
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.
12
240
= 0.415 A (ANS)
5.
(2001 Q4)
240
.
12
 currentout must be 8.3 
(2001 Q3)
P = VI
120 = 240  I
 I = 0.5 A (ANS)
2.
Vout = 10 
[2 marks]
(2001 Q7)
Determine the voltage loss because of the
wires.
The step down transformer supplies 12V
but there is 10 volts at the step up
transformer.
Therefore there is 2 volts lost across the
wires.
Using Vloss = IR , the current through the
wires was 8.3A.
V
Using R = loss
I

6.
2
8.3
= 0.24Ω (ANS)
R=
[2 marks]
(2001 Q8)
If the resistance of the wires 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.
B and C (ANS)
7.
[2 marks]
(2001 Q12)
The total resistance is
V2
R total =
P
2402
R total =
960
R total = 60Ω
The two (identical) heating elements are
joined in series. Therefore each has a
resistance of 30 ohm.
 30  (ANS)
8.
[2 marks]
(2001 Q13)
When the resistors are connected in parallel
their total resistance becomes
1
1
1
=
+
R total R 1 R 2
1
1
1
=
+
R total 30 30
1
=
R total 15
R total = 15Ω
V2
R
R
PSeries
V2
=
× Parallel
PParallel R Series
V2
Use P =
R
PSeries
= Parallel
PParallel
R Series
PSeries
15
=
PParallel 60
1
= (ANS)
4
[3 marks]
P=
V
R
1202
60
= 240Ω (ANS)
R=
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
11.
[1 mark]
(2002 Q3)
This is a step down transformer with a ratio
220000
= 22 (ANS)
of
10000
12.
[3 marks]
(2002 Q4)
The resistance of the wires needs to include
the supply and return lines, over the
distance of 2000m. Therefore a total length
of 4000m
 Voltage drop 20 x 1.6 = 32V
and a supply voltage of 240 - 32
= 208V (ANS)
This is a voltage divider question. To have
120 V across the light bulb you need to
halve the supply voltage. This is done by
placing the same resistance in series with it;
hence the new resistor must have the same
resistance as the bulb.
From
(2002 Q2)
 R is 0.0004 x 4000 = 1.6Ω.
(2002 Q1)
2
[3 marks]
i.e.: there will be 22 times more turns on the
primary coil compared with the secondary
coil
1
9.
10.
13.
[2 marks]
(2002 Q5)
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 be
240 – 160 = 80V
If each bulb in group Q has a voltage drop
across it of 10V there will be
80/10 = 8 bulbs in the group.
Group R will also have 8 bulbs (ANS)
14.
[2 marks]
(2002 Q6)
The string of lights is the total combination
of P + Q + R.
The current through group P is 0.50 A.
This current then splits for Q and R
current in P must be 500 + 200 (= 700), and
the current in Q must be just 500 A.
 P = 700A
 Q = 500A
 R = 0A
(ANS)
The supplied current = 0.50 A (ANS)
18.
15.
[3 marks]
(2002 Q7)
The current through P is 0.50A and as Q
and R are identical they will each have
0.25A.
Each bulb has a voltage drop of 10V across
it.
The power dissipated by each bulb in
groups Q and R will be
P = VI
= 10 x 0.25
= 2.5W (ANS)
16.
[3 marks]
(2002 Q8)
If one of the globes in group Q burns out,
then the current in this path will be zero.
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.
This will result in a reduction of the current
in group P.
Group R however will have an increase in
the current.
Hence
Group
ON/OFF
Brightness
P
ON
Dimmer
Q
OFF
R
ON
17.
[3 marks]
Brighter
(2003 Q1)
This question is pretty basic, but only 20%
of students got it right.
Since R is not connected, there cannot be a
current in it.
This is basically a parallel circuit, using
Kirchhoff’s conservation of current, the
[2 marks]
(2003 Q2)
Power = VI
= 540  500
= 2.7  105W (ANS)
19.
[3 marks]
(2003 Q3)
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.
(Where R is the resistance of 1.0 km of
overhead wire)
The voltage drop along Q is 500R.
This means that 60V = 700R + 500R
60 = 1200R
R = 0.05 (ANS)
20.
[3 marks]
(2003 Q4)
The voltage at tram 1 must be
600 – 700R
= 600 – 700  0.05
= 600 – 35
= 565V (ANS)
21.
[3 marks]
(2003 Q5)
Since the elements in the circuit are in
series, the current through both will be the
same.
Power in globe = I2R,
 9 = I2  8
 I = 1.06.
The voltage drop across the globe will be
V = IR
 V = 1.06  8
= 8.49V.
This means that the voltage drop across the
variable resistor is
12 – 8.49 = 3.51.
 Vresistor = IR
 3.52 = 1.06  Rresistor
 R = 3.31
 R = 3.3 (to 2 sig figs.) (ANS)
25.
[3 marks]
(2003 Q6)
The fuse burns out when the current is 2A.
Using V = iR for the entire circuit, the
effective resistance of the two elements in
parallel must be 6.
1 1 1
= +
6 8 R
1 1 1
= 

R 6 8
1
2
=

R 48
 R = 24 (ANS)
23.
[3 marks]
(2003 Q7)
When justifying your answer it is preferable
if you can prove your case and disprove the
others.
In circuit A, when altering the variable
resistor, the voltage drop (and current) of
the globe will change, this will cause the
brightness to change.
In Circuit B, altering the variable resistor
will not change the voltage across the globe,
it will always be 12V. The resistance of the
globe does not change.
V2
 The power (intensity) of the globe (
)
R
will be constant.
 A (ANS)
24.
[2 marks]
P = VI
 48 = 240  I,
 I = 0.2A (ANS)
(2004 Q1)
(2004 Q2)
The three sets of globes are in parallel,
therefore the current of 0.2A will be shared
between the three sets.
 0.067A (ANS)
26.
22.
[2 marks]
[2 marks]
(2004 Q3)
There are 12 globes in each set. The
potential drop across each set is 240V,
 the voltage drop across each globe is
240
= 20V (ANS)
12
27.
[2 marks]
(2004 Q4)
If the power rating of the 36 globes is 48W,
then each globe has a power rating of
48
= 1.33W (ANS)
36
28.
[2 marks]
(2004 Q5)
If the globe (circled) is removed, then the
middle set will not operate at all.
The voltage across the two other sets
remains at 240V.
Their resistance is constant, so they will be
at the same brightness.
 B (ANS)
29.
[2 marks]
(2004 Q6)
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 (ANS)
Both B and D needed to be given to score
the 2 marks for this question. Any other
answer scored zero.
30.
[2 marks]
(2004 Q7)
The turns ratio can be calculated via the
formula
Ns Ip
=
Np Is
5
.
80
= 0.063 (ANS)
As an example, if the transmission voltage
was reduced to one-tenth (1/10) of the
original
voltage,
and
the
power
consumption maintained, the line losses
would [ INCREASE / DECREASE ] by a
factor of [ 1 / 10 / 100 / 1000 ].
=
35.
[2 marks]
(2004 Sample Q13)
This is a step down transformer, so the ratio
of
Vin:Vout is 1200:30
30
 Vout = 240 
1200
= 6V AC (RMS)
The voltage across each lamp is given by
V = 12 – IRwires.
The resistance of the leads reduces the
voltage drop across the lamp.
For lamp 2 this drop is 1V.
In the other part of the circuit there are two
lamps that need to operate. Combined they
will draw more current from the battery
than lamp 2.
This means that the voltage drop in the lead
to lamp 1 will be greater than lamp 2.
Lamp 3 will have the smallest potential
difference across it because there are more
resistance losses before the lamp.
 C (ANS)
33.
36.
31.
[2 marks]
(2004 Q8)
The voltage drop across the wires can be
determined by using Ohm’s law V = IRwires.
 V = 5.0 x 0.32
= 1.6 V (ANS)
32.
[2 marks]
[3 marks]
(2004 Sample Q4)
(2004 Sample Q5)
To get a 24V AC (RMS) output you need to
have 120 turns in the secondary coil.
This means that you need to use all the
turns on the secondary.
So connect R and S together.
This give 120 turns on the secondary.
Then connect Q to U, and T to V to provide
the input for the stereo.
34.
[4 marks]
(2004 Sample Q12)
Almost all electricity sold today is in the
form of [ DC / AC ], mainly because of the
ease with which it can be [TRANSFORMED
/ RECTIFIED ] from one [ CURRENT /
VOLTAGE ] to another.
Large [VOLTAGES / CURRENTS] in wires
produce heat and therefore energy loss
because of the resistance in the wires. To
overcome this loss, power is transmitted
over large distances at [ LOW / HIGH ]
voltages and correspondingly [ LOW /
HIGH] currents.
[3 marks]
(2004 Sample Q14)
The voltage across lamp 2 is given by
V = 12 – IRwires.
The lamp will not be operating at 12W, but
the resistance of the lamp will remain
constant.
The resistance of the lamp is given by
V2
P=
R
12 2
R=
12
= 12.
The potential drop across lamp 2 is
11V = IRlamp
11 = I  12
11
 I=
12
This is the current through the wires and so
11
Vwires =
 Rwires
12
12
Rwires =
11
= 1.09 (ANS)
37.
[2 marks]
(2004 Sample Q15)
39.
a period. The frequency of the power source
is 50 Hz, so the period is 0.02 sec.
 34  0.02 = 0.015 sec (ANS)
The resistance of each lead is 1.09. so the
circuit can be drawn as below.
Itotal
1.09
40.
I3
12
12V
12
I1
The effective resistance of this circuit is
given by 1.09 + the resistance of the parallel
component of the circuit.
The resistance of the parallel part is
1
1
1
=
+
1.09 + 12
R total
12
 Rtotal = 6.26
 Resistance of the circuit is
1.09 + 6.26 = 7.35
12
 Itotal =
7.35
= 1.63 amp.
 Voltage drop across the first 1.09
resistor is 1.78V.
Voltage across the parallel section of the
circuit is 12 – 1.78 = 10.22V
 Current through lamp 3 is given by
10.22 = I  13.09.
 I = 0.78A
 Vlamp 3 = 0.78  12
= 9.41V (ANS)
38.
[4 marks]
(2004 Sample Q16)
The voltage output of the AC power
adapter is 12 VRMS.
 the peak value = 12  2
= 17V (ANS)
(2004 Sample Q17)
The time difference between 0 and Q is 34 of
This is an extremely difficult question, I
think that it is beyond the scope of the
course.
1.09
[2 marks]
[3 marks]
(2004 Pilot Q1)
If a current-carrying wire is placed in a
magnetic field, the wire experiences a force that
is maximum when the wire and magnetic field
are [parallel / perpendicular] to each other.
In a similar way, when a magnet is passed
repeatedly through a coil of wire, the changing
[electric / magnetic] field induces voltage
across the coil. This is called electromagnetic
induction. Increasing the speed at which the
magnet passes through the coil [increases /
reduces / does not change] the amount of
electromagnetic induction.
41.
[2 marks]
(2004 Pilot Q5)
Closing the switch caused an increasing
current which led to an increasing magnetic
field.
The resultant change in flux induced 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.
42.
[2 marks]
(2004 Pilot Q6)
Opposite flux induced, (whilst the flux is
changing) but the change is only
momentary because the current has
stopped.
 C (ANS)
43.
[2 marks]
(2005 Q9)
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 A (ANS)
44.
[2 marks]
(2005 Q10)
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 (ANS)
45.
[2 marks]
(2005 Q11)
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.
180 V (ANS)
46.
[3 marks]
(2005 Q12)
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 ),
If the current is (
11000
then the power loss in the lines will be
0.652  2 = 0.86W (ANS)
(compared with 1800W initially)
47.
[2 marks]
(2005 Q13)
To convert VRMS to Vpeak to peak
multiply VRMS by 2 2
 Vpeak to peak = 2 2  11 000
= 31113V
= 3.1  104V (ANS)
48.
[2 marks]
(2005 Q14)
The definition of an ideal transformer is
when Pin = Pout
 VinIin = VoutIout
From the data
240  IPrimary = 11.3  2.2
 IPrimary = 0.10 A (ANS)
49.
[3 marks]
(2005 Q15)
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
 B
=– n
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”. This is the ‘minus’ sign in the
equation
50.
[2 marks]
(2005 Q16)
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.
 C (ANS)
51.
[2 marks]
(2006 Q9)
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 (ANS)
52.
[2 marks]
(2006 Q10)
Assume that the transformer is ideal,
 PowerIN = PowerOUT
 (VI)IN = (VI)OUT
 22 000  I = 10  105
1× 105
I=
22000
 I = 4.55 A (ANS)
(rounded off to 2 decimal places)
53.
[3 marks]
(2006 Q11)
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 
 P = 4.552  2
= 41 W (ANS)
(Remember to include the unit in your
answer)
54.
[2 marks]
An increase in V allows a decrease in I,
which means that losses related to I2R are
lowered.
55.
[2 marks]
(2006 Q13)
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.
To step the voltage up from 250 to 22 000,
you need the number of turns in the
22000
secondary to be
times the number of
250
turns in the primary.
22000
= 88.
250
 500  88 = 44 000.
 C (ANS)
56.
[3 marks]
(2006 Q14)
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.
(2006 Q12)
The power being delivered is given by P =
VI. The power loss in the cables is given by
P = I2R.
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.
The magnitude of the induced EMF is given
by Faraday’s Law
ΔBA
EMFAVE = -n
Δt
57.
[2 marks]
(2006 Q15)
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.
58.
[2 marks]
[2 marks]
(2006 Q17)
The alternator produces 250VRMS at 50 Hz.
The peak voltage will be 250  2 = 354 V.
The 50 Hz means that the period is 0.02 sec.
 B (ANS)
60.
[3 marks]
 A2 = 2.5 A
V1 = 10V
V2 = 2V
(ANS)
(2006 Q16)
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.
 B (ANS)
59.
The transformer ratio is 5:1, so the
10
secondary voltage will be
= 2V
5
The transformer is ideal, so
PowerIN = PowerOUT
 (VI)IN = (VI)OUT
 10  0.5 = 2  Isecondary
 Isecondary = 2.5 A
(2007 Q11)
This circuit can be considered as a 12VRMS
AC supply in series with three elements.
Element 1 is a 2 resistor, element 2 is the
transformer, and element 3 is the second 2
resistor.
With a current of 0.50 A, each of the
resistors will lose 2  0.5 = 1 V across them.
The circuit will have a 12 V drop along it, so
the potential difference across the
transformer needs to be 10V.
61.
[3 marks]
(2007 Q12)
Power Supply
Poutput = VI
= 12  0.5
= 6 W (ANS)
Primary coil
P = VI
= 10  0.5
= 5 W (ANS)
Globe
P = VI
= 2  2.5
= 5 W (ANS)
62.
[3 marks]
(2007 Q13)
The globe will not glow.
A transformer requires a changing current
in the primary coil to produce a changing
magnetic field. The resulting changing flux
in the secondary coil induces a voltage.
Since the 12V battery supplies a constant
voltage to the input of the transformer. The
magnetic field in the transformer core,
remains constant and results in a constant
magnetic flux in the secondary coil.
According to Faraday’s law of
electromagnetic induction
ΔBA
EMFAVE = -n
Δt
the output voltage of the transformer is zero
because ΔBA = 0 .
63.
[3 marks]
(VCE 2008 Q11)
240
18
 N1 = 400 (ANS)
N1  30 
Use:
V2
R
12 2
P
3.0
 P = 48W (ANS)
P
64.
[3 marks]
67.
(VCE 2008 Q12)
Combine the resistance of the wires to be 1.0
Ω. The total resistance of the circuit would
be 4.0 Ω.
Use:
R
V1  VS 1
RT
3
V1  12 
4
 V1 = 9.0V (ANS)
65.
[3 marks]
(VCE 2008 Q13)
First find the effective resistance of the two
light globes.
1
1
1


RT R1 R2
1
1 1
 
RT 3 3
R T  1.5
Those light globes are connected in series
with the 1.0 Ω of the wires. So the effective
resistance of the circuit is 2.5 Ω.
The current at point A will be found by
using:
V
I
R
12
I
2.5
 I = 4.8A (ANS)
66.
[2 marks]
Use:
V1 N1

V2 N2
240 N1

18
30
(VCE 2008 Q14)
[3 marks]
(VCE 2008 Q15)
Since the power input is the same as the
power output. The input power must also
be 40 W. The primary voltage is 240 V.
Using:
P  VI
P
I
V
40
I
240
 I = 0.17A (ANS)