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GCSE level physics:
National Grid electricity supply:
Part 6.4
More detailed notes on the theory of transformers, their structure & how they work? How to do transformer
calculations?
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Dr WP Brown PhD:
Doc Brown's physics exam revision notes suitable for students studying UK
IGCSE & GCSE level physics courses, ~ US grades 9-10 physics, page updated
Feb 16th 2026 *]
This page contains some online questions
with worked out answers for you to check.
Jot down
your answers and check them against the worked out answers at the end of
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INDEX for physics notes on
National Grid power supply, use of transformers-calculations and
environmental issues
6.4
(A)
More detailed notes explaining the theory and
structure of transformers
See Part 6.6 for
Examples of transformer calculations
A transformer is a device that can
change the potential difference (p.d.) of an alternating current (a.c.)
It is another example of electromagnetic induction - the effect of magnetic fields inducing
an ac current in a coil.
Transformers are nearly 100%
efficient - important in the context of the National Grid
electricity supply.
A basic transformer
consists of two coils of insulated wire, a primary coil and a secondary coil
independently wound on an iron
core - see the diagram further down.
The wire is usually made of
copper with a thin coating of insulation material.
Iron is used as because it is easily
temporarily magnetised.
Both coils MUST consist of complete,
but
separate circuits - no electrical connection between them.
When an a.c. p.d. is applied across
the primary coil, the iron core magnetises and demagnetises
quickly due to the nature of the alternating current.
Therefore an
alternating current (a.c.) in the primary coil (p) of a transformer
automatically produces a
alternating (changing) magnetic field in the iron core and hence in the secondary
coil (s) by induction.
This induces an alternating potential difference across the
ends of the secondary coil (Vs).
If the secondary coil is part of a
complete circuit, an induced current will flow in the secondary
coil.
Remember, the current MUST BE
ac to get an alternating magnetic field that 'cuts'
through the iron core', otherwise continuous induction will NOT take
place.
Using a d.c. current is no
good - you don't get a constantly changing magnetic field if the
current only flows one way!
KEY to transformer diagram:
Vp
= p.d.
across the input primary coil
Vs
= induced p.d. generated across the output secondary coil
np = number of
wire turns on the primary input coil
ns = number of
wire turns on the secondary output coil
Ip = input
current flowing through the primary coil
Is = induced output
current flowing out through the secondary coil
A step-up transformer increases
the p.d. and has a greater number of coils in the secondary coil
than the primary.
A step-down transformer reduces
the p.d. and has a smaller number of coils in the secondary coil
than the primary.
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For example, the images above
illustrate the local electricity supply to some farms in a rural area
using small-scale step-down transformers.
If you know the input p.d. and the
number of turns on each coil, you can calculate the output p.d. using
the equation below.
Subscript p
is form the primary coil and subscript
s for
the secondary coil.
The transformer formula for calculations
The ratio of the potential
differences across the input primary coil and the output secondary coil of a transformer,
Vp and Vs, depends on the ratio of the number
of turns of wire on each coil, np and ns
and is given by a simple ratio equation. The data is 'pictured' on the
diagram above.
vp / versus
= np /
ns The p.d. and coils transformer equation
the ratios are the same i.e. input p.d. / output p.d = turns on primary coil / turns on secondary
coil
and the potential differences,
Vp and Vs in volts, V
In a step-up transformer
Vs > Vp
In a step-down transformer
Vs < Vp
See the diagram above for a
visual appreciation of the equation.
If transformers were 100% efficient, the
electrical power output would equal the electrical power input.
Vs × Is = Vp × Ip The
transformer power equation
(remember power P = I x V, P in watts, W and
I current in amps, A)
where
Vs × Is
equals the power output
from the secondary coil (never 100% efficient)
and
Vp × Ip
equals the power input
to the primary coil
However, no transformer is ever 100%
efficient, so the power output never equals the power input,
because there are always energy losses eg as thermal (heat energy) -
remember that the two coils of wire in a transformer are still acting as
resistances, so there will always be a little energy change of
electrical energy to thermal energy.
See Part 6.6 for
Examples of transformer calculations
INDEX of notes on
National Grid power supply & use of transformers
6.4 (B) Examples of transformer calculation
practice questions and worked out answers
KEY to transformer diagram
and transformer formula for calculations
Everything you need to know is on
the transformer diagram above and take note of the abbreviations
below!
To save repetition in the questions
PLEASE note the following abbreviations which I will use for problem
solving:
P = power (W) = I (A) x V
(V), P = IV = I2R
Vp
= p.d.
across the input primary coil
Vs
= induced p.d. generated across the output secondary coil
np = number of
wire turns on the primary input coil
ns = number of
wire turns on the secondary output coil
Ip = input
current flowing through the primary coil
Is = induced output
current flowing out through the secondary coil
The ratio of the potential
differences across the coil equals the ratio of the turns on each
coil.
Vp / Vs =
np / ns The p.d. and
coils transformer equation
Vpns
= Vsnp
VsIs = VpIp
The transformer power equation
Since P = IV, it means
power input = power output,
assuming 100 % efficiency
(never this in reality)
Rearrangement: Vs / Vp
= Ip / Is
Q1 A transformer has
200 turns on the primary coil and 10 turns on the secondary coil.
If the output p.d. required is
12.0 V from the secondary coil, what p.d. must be put across the
primary coil?
ANSWERS
Q2 The p.d. across
the primary coil of a transformer was 12000 V and the current flowing
through it was 20 A.
If the current flowing through
the secondary coil was 1000 A, what is the p.d. across the secondary
coil?
ANSWERS
Q3 A power station
step-up transformer has 500 turns of wire on the primary coil and 8000
turns of wire on the secondary coil.
(a) If the generator output is 25
000 V, what is the output p.d. in kV for the transmission lines of
the National Grid?
(b) What is the power input, in
MW, from the generator to the transformer, if the input current to
the primary coil is 20 A? (assume 100% efficiency)
(c) If the power output
from the transformer to the transmission line is 0.48 MW, what is
the % efficiency of the energy transfer and what has become of the
lost energy?
(d) How much energy is wasted
from the transformer every minute?
ANSWERS
Q4 The p.d. across
the primary coil of a transformer is 240 V carrying a current of 5.0
A.
If the p.d. across the secondary
coil is 12.0 V, what current is flowing in secondary coil?
ANSWERS
Q5 A transformer has
20 turns of wire on the secondary coil and a p.d. across it is 3.0 V.
How many turns must be on
the primary coil if the primary coil p.d. is 230 V?
ANSWERS
Q6 The output p.d.
across the secondary coil of a transformer is 6.0 V and a current
flowing through it of 0.50 A.
ANSWERS
Q7 The secondary coil
of a transformer has a p.d. output of 120 V.
ANSWERS
Q8 The p.d. across
the secondary coil of a transformer is 12.0 V and 0.30 A flows through
it.
ANSWERS
Q9 A laptop charger
works off the 230 V AC mains supply.
Inside the adapter, the
transformer produces an a d.c. supply current of 3.0 A at a p.d. of
19 V.
(a) What current did the charger
draw from the mains supply?
(b) What assumptions has been
made? and how can you tell from your everyday experience that the
assumption is incorrect?
ANSWERS
Keywords, phrases and learning objectives for
transformer calculations
Be able to do exam questions based on the formula
for
transformers e.g. calculations involving ratio of turns of wire in primary
coil and secondary coil, input voltage and output voltage and the input
and output
current.
Appreciate the calculations assume there is no loss
of power i.e. power input = power output, even though there is some
wasted thermal energy.
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Phil Brown 2000+.
All copyrights reserved on Doc Brown's physics revision notes, images,
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INDEX of notes on
National Grid power supply & use of transformers
|
Keywords, phrases and learning objectives for National Grid electricity supply
Be able to describe how a transformers works.
Be able to describe the structure and function of
transformers.
Describe the design and construction of
transformers and what materials are they made of.
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environmental issues
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Website content © Dr
Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries and references to GCSE science course specifications
are unofficial.
INDEX of notes on
National Grid power supply & use of transformers
|
6.4 (B) ANSWERS to the transformer
questions
Q1 A transformer has
200 turns on the primary coil and 10 turns on the secondary coil.
If the output p.d. required is
12.0 V from the secondary coil, what p.d. must be put across the
primary coil?
Vp / Vs =
np / ns
Vpns =
Vsnp
Vp = Vsnp
/ ns
Vp = 12 x
200/10 =
240 V
Q2 The p.d. across
the primary coil of a transformer was 12000 V and the current flowing
through it was 20 A.
If the current flowing through
the secondary coil was 1000 A, what is the p.d. across the secondary
coil?
VsIs = VpIp
(assuming 100 % efficiency)
Vs = VpIp
/ Is
= 12000 x 20/1000 =
240 V
Q3 A power station
step-up transformer has 500 turns of wire on the primary coil and 8000
turns of wire on the secondary coil.
(a) If the generator output is 25
000 V, what is the output p.d. in kV for the transmission lines of
the National Grid?
Vp / Vs =
np / ns
Vpns =
Vsnp
p.d. output = Vs =
Vpns / np = (25 000 x
8000)/500 = 400 000 V =
400 kV
(b) What is the power input, in
MW, from the generator to the transformer, if the input current to
the primary coil is 20 A? (assume 100% efficiency)
Pinput = IpVp
= 20 x 25 000 = 500 000 W = 500 kW =
0.50 MW
(Note: In reality, a
power station will usually have several generators and
step-up transformers running at the same time.)
(c) If the power output
from the transformer to the transmission line is 0.48 MW, what is
the % efficiency of the energy transfer and what has become of the
lost energy?
% efficiency = 100 x useful
power output / total power input = 100 x 0.48/0.50 =
96%
There are always thermal
energy losses
from the transformer to the surrounding thermal energy store.
(d) How much energy is wasted
from the transformer every minute?
The power wastage is equal to
0.50 - 0.48 = 0.02 MW
0.02 MW = 20 kW = 20 000 W
This equals a heat loss of 20
000 J/s
Therefore in one minute 20
000 x 60 =
1 200 000 J is lost (1.2 MJ/min)
Q4 The p.d. across
the primary coil of a transformer is 240 V and carries a current of 5.0
A.
If the p.d. across the secondary
coil is 12.0 V, what current is flowing in secondary coil?
VsIs =
VpIp (assuming 100 % efficiency)
Is = VpIp
/ Vs
= 240 x 5.0 / 12.0 =
100 A
Q5 A transformer has
20 turns of wire on the secondary coil and the p.d. across it is 3.0 V.
How many turns must there be on
the primary coil if the primary coil p.d. is 230 V?
Vp / Vs =
np / ns
Vpns = Vsnp
np = Vpns
/ Vs
= 230 x 20/3.0 = ~1533
turns of wire.
Q6 The output p.d.
across the secondary coil of a transformer is 6.0 V and a current
flowing through it of 0.50 A.
What p.d. must be applied across
the primary coil to give a current of 0.0125 A to flow through it?
VsIs = VpIp
(assuming 100 % efficiency)
Vp = VsIs
/ Ip
= 6.0 x 0.50/0.0125 = 240
V
Q7 The secondary coil
of a transformer has a p.d. output of 120 V.
If the primary coil has 300
turns, how many turns must there be on the secondary coil if the
primary coil input p.d. is 3.0 V?
Vp / Vs =
np / ns
Vpns = Vsnp
ns = Vsnp
/ Vp
= 120 x 300/6.0 = 6000
turns
Q8 The p.d. across
the secondary coil of a transformer is 12.0 V and 0.30 A flows through
it.
If the p.d. across the primary
coil is 240 V, what current is flowing through the primary coil?
VsIs = VpIp
(assuming 100 % efficiency)
Ip = VsIs
/ Vp
= 12 x 0.3 / 240 =
0.015 A
Q9 A laptop charger
works off the 230 V AC mains supply.
Inside the adapter, the
transformer produces an a d.c. supply current of 3.0 A at a p.d. of
19 V.
(a) What current did the charger
draw from the mains supply?
power input (mains) = power
output (adapter)
P = IV, power output = 3.0 x
19 = 57 W
power input = 57 W, therefore
since current drawn I = P / V, I = 57 / 230 =
0.25 A (2 sf)
(b) What assumptions has been
made? and how can you tell from your everyday experience that the
assumption is incorrect?
The calculation assumes
there is no energy loss in the transformer process.
The adapter feels 'warm'
due to electrical energy ==> thermal energy by the resistances
of the wire causing a rise in temperature - this loss increases
the thermal energy store of the surrounding air.
INDEX of notes on
National Grid power supply & use of transformers
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