Physics
for
CAPE®
Unit
2
Physics
for
Terry
Joyce
CAPE®
David
Crichlow
Dwight
Carlos
de
Freitas
Hunte
Unit
2
3
Great
Clarendon
Oxford
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Acknowledgements
Cover
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Mark
Lyndersay,
Lyndersay
Digital,
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Illustrations:
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in
all
at
of
Contents
Introduction
Module
1
Chapter
Electricity
and
magnetism
Chapter
1
Electrostatics
1.2
Applications of
6.
1
Magnetic fields
6.2
Force on a current-carrying conductor
50
6.3
Force on
52
6.4
Measuring
6.5
Force
electrostatics
moving
charge
magnetic flux density
between
current-carrying
conductors
56
The
58
electromagnet
Electrical quantities
2.
1
Electric current and potential difference
6
2.2
Drift velocity
8
2.3
Resistance
and
power
7
Electromagnetic
7
.
1
Faraday’s
and
7
.2
Motors
7
.3
Examples on
and
Lenz’s
induction
laws
generators
and
64
electromagnetic
internal
induction
resistance
68
12
Revision questions
3
d.c.
Module
circuits
3.
1
Kirchhoff ’s
3.2
Resistors
3.3
Worked
3.4
Potential dividers
laws
in
series
and
in
parallel
examples on d.c.
The Wheatstone
circuits
4
Electric fields
4.2
Electric field
74
16
Module
2
a.c. theory
18
electronics
and
8
Alternating
currents
22
1
8.
1
Alternating
currents
78
8.2
The transformer
80
8.3
Semiconductors
82
8.4
Rectification
84
24
28
strength
and
Revision questions
electric
potential
4
86
30
Chapter
4.3
Relationship
4.4
Worked
5
exam questions
Electric fields
4.
1
Chapter
Practice
70
20
bridge
Revision questions
Chapter
1
3
14
Chapter
3.5
60
10
Electromotive force
Chapter
54
4
Chapter
2.4
a
48
2
6.6
2
Magnetic fields
Electrostatics
1.
1
Chapter
6
between
examples on
E
and
V
electric fields
9
Analogue
electronics
32
9.
1
Transducers
88
9.2
Operational
9.3
Inverting and non-inverting amplifiers
9.4
The
34
amplifiers
90
Capacitance
5.
1
Capacitance
5.2
Charging
5.3
Capacitors
36
and discharging
in
series
Revision questions
and
2
a
in
capacitor
parallel
44
amplifier
and the
voltage follower
38
42
summing
9.5
Analysing op-amp
92
94
circuits
98
Contents
Chapter
10
Digital
electronics
Chapter
12
Atomic
structure
and
radioactivity
10.
1
Logic
gates
100
10.2
Equivalent
10.3
Logic
10.4
Applications of
10.5
Sequential
logic
gates
gates
and timing diagrams
logic
12.
1
Atomic
structure
12.2
Nuclear
12.3
Binding
12.4
Calculating
142
102
gates
reactions
146
104
energy
148
106
circuits
energy
Revision questions
Revision questions
Module
Module
2
Practice
3 Atomic
changes
150
108
5
exam questions
and
7
152
112
12.5
Radioactivity
154
12.6
Types of
radiation
156
12.7
Radioactive decay
160
12.8
Measuring
164
12.9
Uses of
116
nuclear
half-life
physics
Chapter
11
The
particulate
electromagnetic
nature of
radioisotopes
Revision questions
11.
1
The
11.2
Investigating the photoelectric effect
124
photoelectric
effect
11.3
Examples on the photoelectric effect
126
Revision questions
128
11.4
Millikan’s oil drop
11.5
Emission
experiment
11.6
Examples of
11.7
Wave–particle duality
136
11.8
X-rays
138
13
spectra
exam questions
spectra
Analysis
172
and
Analysis
and
interpretation
Analysis
and
interpretation:
176
130
Practice
line
Practice
interpretation
13.
1
6
absorption
3
120
Chapter
exam questions
180
132
134
List of
iv
170
radiation
Module
and
8
168
physical
constants
182
Glossary
183
Index
186
Introduction
This
Study
Guide
has
been
developed
exclusively
with
the
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Council
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Proficiency
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The
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in
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CAPE
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to
syllabus,
support
learning
®
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providing
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features
and
for
guidance
this
activities
On
ᔢ
of
Y
ourself
you
problem
ᔢ
to
This
an
in
an
for
Do
examiner
will
show
the
key
to
your
includes
and
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syllabus
format!
electronic
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your
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refer
Physics
examination
example
examination
examination
inside
to
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sample
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CAPE
master
and
CD
good
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in
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activities
you
best
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answering
sections
are
questions
as
is
in
your
and
guide
so
to
provide
helpful
that
experience
feedback
you
can
will
revise
areas.
Answers
work
syllabus.
activities
multiple-choice
refer
easier
course
type
from
achieve
it
activities
These
confidence
T
est
the
the
essay
feedback
you
make
you
Marks
and
improved.
and
help
Guide
assist
Y
our
and
of
on
Study
to
answer
to
included
requirements
full
Inside
ᔢ
tools
you
unique
examination
included
that
on
require
the
CD
for
calculations,
multiple-choice
so
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can
questions
check
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and
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Physics.
1
1
Electrostatics
1.
1
Electrostatics
Learning outcomes
Charging objects
Lightning
On
completion
of
this
section,
charged
should
be
able
state
that
there
are
two
types
charges
first
describe
and
explain
charging
to
other
or
tall
ᔢ
show
and
that
unlike
like
charges
charges
distinguish
of
an
effect
through
the
of
static
electricity.
atmosphere.
This
Clouds
charge
is
become
able
clouds
or
to
objects
on
the
surface
of
the
Earth
to
such
buildings.
how
In
order
objects
to
explain
become
this
phenomenon,
you
as
must
charged.
by
repel
and
A
by friction
polythene
insulating
attract
between
rod
is
thread
with
a
rubbed
so
piece
that
of
with
it
is
wool
a
piece
free
and
to
of
wool
move.
placed
and
suspended
Another
close
to
the
from
polythene
suspended
rod
an
is
rod.
The
conductors
suspended
and
move
induction
rubbed
ᔢ
example
understand
Charging
friction
an
they
of
trees
ᔢ
as
to:
move
ᔢ
is
you
rod
is
repelled.
insulators.
When
a
perspex
suspended
rod
is
polythene
Experiments
ᔢ
there
are
ᔢ
similarly
ᔢ
objects
of
this
two
rubbed
nature
types
charged
with
a
piece
of
wool
it
attracts
the
rod.
of
have
shown
that:
charge
objects
repel
each
other
repulsion
When
ebonite
that
(a
fur
.
Repulsion of two charged
polythene rods
It
is
were
the
T
o
attraction
to
rod.
some
the
acquires
hard
a
attract
were
positive
rubber)
first
made
a
up
other
.
conducted,
charge
acquires
was
each
when
negative
long
note
The
that
rod
from
was
the
of
In
this
wool.
rubbed
with
wool.
rod
of
not
rubbed
before
the
wool
acquired
process,
an
a
glass
the
the
of
the
silk
and
rubbed
existence
of
acquire
a
a
+
+
+
+
polythene and a perspex rod
+
+
woollen
cloth
polythene
Electrons
the
woollen
are
transferred from
cloth
Explaining charging by friction
to
the
action.
electrons
of
the
charge .
At
charge ,
negative
positive
+
gained
rubbing
the
because
conserved.
electrons
2
with
wool,
positive
lost
Figure 1.1.3
arbitrarily
when
surface
negative
is
acquire
by
piece
to
equal
charge
ebonite
and
created
with
the
acquired
and
Perspex
is
rubbed
surface
wool
polythene
with
charge
polythene
piece
rubbed
Attraction between a
was
charge
electrons
Figure 1.1.2
it
known.
electrons.
summarise,
charges
experiments
polythene
time
unlike
convention
transferred
polythene
same
of
became
important
When
lost
glass
type
This
electrons
Figure 1.1.1
have
electrostatic
decided
with
that
polythene.
charge
charge
when
when
it
Chapter
Charging
An
by
in
the
then
A
metal
sphere
has
by
metal
briefly
then
sphere
can
be
charged
negatively
sphere.
uncharged
earthed
metal
object
induction.
uncharged
Electrostatics
induction
uncharged
called
1
The
a
to
touching
move
it
and
positively
charged
rod
negatively
sphere
become
by
charged
is
charged
to
one
the
object
brought
rod
side.
rod
is
by
a
close
causes
The
taken
process
to
an
the
metal
away.
electrons
sphere
The
is
metal
charged.
negatively
negatively
charged
charged
rod
wire
connected
rod
+
+
+
to
+
+
earth
+
+
+
A
+
insulated
support
1
2
Figure 1.1.4
are
electrons
the
4
Charging by induction
Conductors
Metals
3
very
that
and
good
orbit
innermost
insulators
conductors
the
shells
nucleus
are
held
of
are
electricity.
situated
tightly
by
the
In
in
a
metal
‘shells’.
nucleus.
atom,
The
In
the
the
nucleus
electrons
in
outermost
electrons
shells,
are
the
able
electrons
to
escape
are
the
held
less
tightly
electrostatic
force
by
of
the
nucleus.
attraction
of
These
the
electrons
nucleus.
outer
These
electrons
are
free
to
move
throughout
the
metal
structure.
shell
Since
electron
these
electrons
electron
an
has
electric
electrons
and
iron
called
the
that
are
good
that
structure
Plastic
in
the
is
and
Materials
It
a
are
is
on
are
of
of
a
types
to
good
neither
said
high
be
of
delocalised.
charge
of
of
free,
electricity.
Graphite
is
Each
shells
constitutes
an
mobile
Copper
,
example
Figure 1.1.5
Simple structure of an atom
silver
of
a
electricity.
of
that
of
electricity.
materials
of
the
allow
to
concentration
conductors
conductors
from
are
movement
electricity.
of
throughout
electrons
are
they
The
excellent
these
different
are
it.
because
poor
In
rubber
that
move,
conductor
bonds
mobile
to
conductors
is
insulators.
free
free
charge
metals
materials
tightly
no
are
tiny
current.
non-metal
Some
a
a
the
metal.
structure.
for
electrical
These
type
All
the
This
of
materials
bonding
electrons
means
conduction
that
to
are
found
are
there
take
in
held
are
place.
insulators.
good
conductors
nor
good
insulators
are
called
semiconductors .
Key points
ᔢ
There
are
two
ᔢ
Objects
ᔢ
Charge
ᔢ
Like
ᔢ
Unlike
ᔢ
Materials
that
conduct
ᔢ
Materials
that
are
can
is
types
be
charge:
charged
always
charges
of
positive
by friction
conserved
when
or
an
and
by
negative
charge.
induction.
object
is
charged.
repel.
charges
attract.
poor
electricity
are
conductors
of
called
conductors.
electricity
are
called
insulators.
3
1.2
Applications
Learning outcomes
of
electrostatics
Practical
applications of
Agricultural
On
completion
should
be
able
of
this
section,
describe
to:
There
practical
applications
are
many
One
such
appreciate
the
potential
pest
with
control.
cultivating
Pests
can
be
agricultural
very
of
the
plants.
They
feed
on
the
leaves
of
harmful
plants,
to
usually
with
on
the
underside
of
the
leaves.
This
makes
spraying
with
pesticides
hazards
a
associated
associated
is
phenomena
living
ᔢ
problems
problem
of
development
electrostatic
spraying
you
land.
ᔢ
electrostatics
charging
tricky
process.
A
technique
has
been
developed
where
the
nozzle
of
by
the
spray
is
connected
to
a
high
voltage
supply.
This
high
voltage
supply
friction.
has
two
effects.
opposite
droplets
charge
to
prevents
Dust
Dust
It
be
charges
on
the
spray
ground
attracted
wastage
the
to
during
both
the
and
droplets
the
sides
plants.
of
spraying
positively
the
This
leaves.
and
causes
This
induces
the
an
spray
technique
also
process.
extraction
particles
chimneys,
may
using
be
the
extracted
electric
from
field
the
that
flue
exists
gases
released
between
a
wire
by
industrial
and
a
metal
cylinder
.
flue
gas
dust
positively
without
particles
charged
cylinder
collecting
plate
chimney
negatively
charged
flue
dust
Figure 1.2.1
The
A
inner
series
cylinder
.
The
of
away.
of
wall
of
wires
A
large
The
Photocopiers
A
and
use
light-sensitive
grid.
An
dr um.
4
then
image
The
the
dust
with
particles
areas
the
on
in
fall
the
ions
laser
the
air
are
to
air
to
at
wires
of
the
the
and
the
the
are
to
cylinder
bottom
of
of
metal.
the
cylinder
.
molecules,
to
the
to
thus
accelerated
attached
attracted
the
piece
inside
air
wire
become
then
strike
traps
the
some
close
the
cylindrical
mounted
the
inner
dislodge
the
dust
walls
dust
chimney.
printers
principle
of
electrostatics
dr um
document
the
from
in
used
a
are
between
particles
into
with
voltages
exists
are
cylindrical
of
fitted
electrons
negative
Hammers
which
Photocopiers
field
electrons
and
is
negative
removes
charged
cylinder
.
particles,
chimney
high
electric
Free
Electrons
the
the
at
field
ions.
particles.
gas
Electrostatic dust extraction
electric
forming
wall
grid
dr um
is
charged
being
exposed
copied
to
to
copy
positively
is
light
documents.
by
projected
lose
their
a
charged
on
to
the
positive
charge.
Chapter
A
negatively
The
toner
then
A
is
The
paper
a
of
drum.
attracts
image
Electrostatic
Most
car
paint
leaves
Since
the
all
sticks
of
be
the
droplets
with
to
the
an
surface
use
of
to
equal
the
by
dr um
is
it
dusted
image.
over
the
and
war ming
to
on
be
the
written
on
over
A
grid.
an
the
principle.
document
focus
is
charged
passes
similar
printed
sprayer
,
the
form
and
of
electrostatics
the
have
out
to
toner,
sheet
The
image
fi nal
When
is
drum
of
paper
positively
for med
on
a
printer
laser
light-sensitive
the
dr um.
product.
the
printed,
the
Electrostatics
by
uses
a
cylindrical
the
laser
.
spraying
nozzle
spreads
a
the
lenses
paint
the
it
from
using
manufacturers
paint
charged
to
as
per manent
version
and
the
positively
toner
made
digital
the
called
charge
operates
mirrors
The
to
positive
is
printer
receives
series
a
image
laser
powder,
attracted
receives
charged
it.
charged
1
same
a
the
car
to
even
they
cloud.
charge.
tightly
paint
their
droplets
charge,
large
opposite
the
and
The
less
are
repel
The
a
each
body
result
paint
vehicles.
given
is
is
As
the
charge.
other
of
so
the
that
the
wasted
in
that
car
is
paint
the
process.
+
Hazards of
static
+
+
electricity
–
–
– –
Static
electricity
electric
shock
charged
to
flow
by
to
from
the
you
touch
great
fuel
no
In
there
care
is
The
flow
of
shock.
handle.
the
always
must
causing
a
the
fuel
spark
be
There
For
a
charge
It
is
is
always
example,
metal
door
it
is
our
a
in
to
an
see
electric
to
a
lightning
become
the
rod
charge
current
hear
be
of
may
causes
and
possible
danger
bodies
handle
results
possible
Sometimes,
danger
aircraft
and
are
of
when
dangerous
spark
charged
up
and
as
to
a
spark
explosion.
the
at
fuel
the
being
refuelling
A
A
conducting
tanker
.
same
produced
aircraft.
This
electric
by
spark
cable
ensures
potential.
is
electrostatics,
can
ignite
the
connected
that
the
This
aircraft
ensures
Figure 1.2.2
Lightning rod
that
arises.
thunderstorms,
When
a
taken
tanker
atmosphere.
been
is
clouds
These
charge
lightning
has
objects.
touching
electric
door
hazardous.
kilovolts.
between
and
the
an
very
Then
handle.
experience
Since
be
charged
friction.
you
many
can
flows
seen.
known
large
start
charged
have
between
V
ery
to
are
clouds
huge
clouds
or
currents
fires,
as
they
move
amounts
of
towards
flow
damage
tall
during
buildings
through
charge
the
stored
buildings
this
and
process.
even
in
and
kill
them.
trees,
Lightning
people
by
electrocution.
A
commonly
use
copper
the
used
lightning
strip
fixed
building.
technique
rods.
A
to
The
to
lightning
an
part
outside
of
the
protect
rod
wall
rod
buildings
takes
the
reaching
above
the
from
form
of
above
highest
a
lightning
thick
the
highest
part
of
is
piece
the
to
of
part
of
building
Key points
consists
buried
of
in
several
the
sharp
earth
spikes.
below.
The
When
other
lightning
end
of
the
strikes,
it
copper
usually
strip
is
strikes
the
ᔢ
highest
least
to
point
of
resistance
the
ground
a
building
towards
it
is
and
the
the
earth.
effectively
current
Since
earthed.
travels
the
through
lightning
When
a
the
path
conductor
charged
cloud
is
Electrostatics
crop
the
a
large
lightning
electric
to
ionise.
produced.
electric
the
Electrons
The
current
earth,
rod.
field
air
This
is
are
set
up
between
without
electric
stripped
breaks
flows
large
down
from
and
harmlessly
damaging
the
field
the
the
air
begins
through
causes
cloud
the
and
molecules
to
the
conduct
air
the
by
electrostatic
paint
spraying,
dust
the
photocopiers
and
spikes
printers.
molecules
and
ions
electricity.
lightning
in
spraying,
fixed
passes
laser
of
used
electrostatic
extraction,
building,
is
of
are
The
conductor
towards
ᔢ
Charging
by friction
problems,
reduce
but
the
there
can
are
create
ways
to
effects.
building.
5
2
Electrical
2.
1
Electric
quantities
current
Learning outcomes
Electric
Metals
On
completion
of
this
section,
ᔢ
be
able
ᔢ
define
ᔢ
recall
ᔢ
define
is
that
an
the flow
charge
and
are
good
present.
use
and
Q
=
tightly
bound
and
charge
conductors
The
outer
of
electricity.
shell
electrons
In
of
a
metal
the
there
metal
are
atoms
many
are
carries
a
potential
and
are
free
to
move
throughout
the
lattice.
not
Each
electron
electric
of
charge
the
move
tiny
in
a
amount
particular
of
charge.
direction,
electric
current
is
the
rate
of
electric
current
is
the
ampere
When
an
flow
these
electric
of
charged
cur rent
charged
is
particles
(electrons)
produced.
particles.
The
SI
An
unit
of
coulomb
(A).
It
When
difference
a
salt
such
as
sodium
chloride
is
dissolved
in
water
,
sodium
and
and
chloride
the
current
difference
to:
understand
current
potential
you
atoms
should
and
ions
are
produced.
The
ions
present
in
the
solution
are
charged.
volt
Therefore,
a
solution
of
sodium
chloride
will
allow
an
electric
current
to
W
ᔢ
recall
and
use
V
flow
=
through
it.
Q
Charged
faster
The
particles
the
SI
can
charged
unit
of
have
either
particles
charge
is
move
the
a
negative
the
or
greater
coulomb
(C).
a
the
An
positive
electric
electron
charge.
The
current.
has
a
charge
of
Definition
−19
1.6
An
of
electric
current
is
the
rate
of flow
charge.
In
×
10
order
simple
move
the
C.
for
the
circuit,
the
free
chemical
electrons
an
in
electric
electrons
reactions
a
cell
in
metal
can
the
taking
to
be
metal.
place
flow,
used
The
=
the
produces
be
supplied.
energy
energy
In
needed
because
a
to
of
it.
cell
+
Q
must
provide
cell
inside
Equation
energy
to
–
It
actual
Q
–
charge/C
I
–
current/A
t
–
time/s
direction
of
electron flow
metal
free
direction
electrons
of
conventional
electrical
Figure 2.1.1
metal
ions
current
Diagram showing the direction of electron flow in a metal
Definition
In
Figure
left.
1
coulomb
is
the
quantity
of
The
passes
through
any
a
conductor
in
1
second
when
of
1
ampere
direction.
flowing
is
=
from
important
flowing
current,
through
however
,
the
is
metal
defined
from
as
right
flowing
to
in
In
an
electrical
circuit
we
think
of
the
electric
the
current
the
to
positive
terminal
remember
that
if
a
of
the
cell
stream
of
to
the
negative
electrons
is
terminal.
flowing
in
one
is flowing.
direction,
1 C
are
electric
a
It
current
electrons
section
as
of
the
charge
opposite
that
2.1.1
conventional
1 A s
it
can
the
opposite
the
electric
Electrical
the
is
direction.
currents
as
If
will
are
current
connected
the
thought
current
electric
with
be
a
be
as
flowing
measured
shown
in
An
the
stream
flowing
component.
of
of
in
Figure
positive
that
using
through
ideal
conventional
a
ions
same
an
electric
is
2.1.2.
flowing
ammeter.
The
in
In
a
ammeter
has
in
flowing
one
in
direction,
direction.
component
ammeter
current
zero
order
to
circuit,
is
measure
an
ammeter
connected
in
series
resistance.
Example
I
In
a
the
ammeter
tube,
there
cathode.
the
time
taken
for
a
Measuring an electric current
cathode
6
ray
and
is
a
current
of
160 µA
in
the
vacuum
between
Calculate:
component
a
Figure 2.1.2
cathode
anode
and
anode
charge
of
2.5 C
to
be
transferred
between
the
Chapter
b
the
number
of
electrons
emitted
per
second
from
the
2
Electrical
quantities
cathode.
A
F
–19
(Charge
on
one
electron
e
=
–1.6
×
10
C)
1.56
×
10
s
×
10
I
Q
2.5
R
4
a
t
=
=
=
R
1
2
–6
I
160
×
10
B
–19
b
Charge
on
one
electron
Charge
flowing
per
=
e
second
=
=
–1.6
C
D
E
C
Figure 2.1.3
160 µC
–6
160
×
10
15
Number
of
electrons
emitted
per
second
=
=
1
×
10
–19
1.6
×
10
Definition
Potential difference
The
Whenever
Energy
is
usually
energy
a
the
term
required
required
into
circuit,
unit
to
to
move
power
electrical
energy
is
positive
charge
charge
an
energy.
around
electrical
When
converted
is
from
used,
an
The
passes
electrical
refers
electrical
circuit.
charge
it
to
to
circuit.
cell
A
converts
through
other
+1 C
of
charge.
cell
of
is
chemical
energy.
is
at
defined
moving
components
forms
potential
field
infinity
unit
to
a
as
point
the
positive
that
in
an
work
electric
done
in
charge from
point.
in
In
a
potential / V
filament
In
the
An
in
lamp,
case
electric
In
field
infinity
the
is
to
circuit
through
resistor
,
in
from
between
as
point.
Figure
R
around
.
energy
the
The
R
converted
one
the
2.1.3
and
1
potential
is
electrical
defined
that
resistors
energy
flows
potential
electric
from
a
current
electric
an
of
electrical
a
point
two
cell
is
Figure
to
another
done
unit
of
light
converted
points.
work
SI
is
into
The
in
being
into
used
thermal
potential
moving
shows
thermal
because
potential
2.1.4
and
is
to
of
at
unit
the
a
energy.
difference
point
positive
volt
provide
the
a
energy.
a
in
charge
(V).
current
variation
of
I
electric
2
the
circuit
relative
to
point
F
.
A
Electric
potential
cannot
be
measured
directly.
The
difference
in
B
C
Figure 2.1.4
potential
can,
however
,
be
measured.
A
voltmeter
is
used
to
D
E
F
electric
Variation of electric
measure
potential around the circuit
potential
difference .
component
whose
A
voltmeter
potential
is
connected
difference
is
being
in
parallel
with
measured.
An
the
ideal
Definition
voltmeter
As
to
charge
some
thermal
of
a
has
an
passes
other
between
form.
energy
.
filament
infinite
In
The
lamp,
resistance.
the
the
two
case
resistor
of
gets
electrical
points,
a
resistor
,
warm
energy
energy
is
as
is
converted
electrical
charge
converted
energy
flows
into
from
is
converted
through
light
and
electrical
it.
In
the
thermal
into
case
energy
.
The
two
unit
of
potential
difference
is
the
volt
points
done
in
(energy
electrical
of
The
potential
volt
is
circuit
flows
defined
when
1
between
as
the
joule
the
of
two
potential
energy
is
circuit
in
is
V
between
the
work
converted from
energy
energy)
to
other forms
moving
unit
positive
(V).
charge from
1
difference
a
difference
converted
between
when
1
two
points
coulomb
of
in
one
point
to
the
other.
a
charge
points.
−1
1 V
=
1 J C
resistor
Key points
ᔢ
An
ᔢ
1
electric
current
is
the
rate
of flow
of
charge.
voltmeter
coulomb
is
the
quantity
of
charge
that
passes
through
any
section
of
a
Figure 2.1.5
conductor
in
1
second
when
a
current
of
1
ampere
Measuring potential
is flowing.
difference
ᔢ
The
potential
charge from
ᔢ
The
at
a
point
infinity
potential
to
is
defined
that
difference
V
as
the
work
done
in
moving
unit
positive
point.
Equation
between
two
points
in
a
circuit
is
the
work
done
W
V
ᔢ
(energy
converted from
moving
unit
1
defined
volt
when
is
1
joule
between
the
positive
of
as
charge from
the
energy
two
electrical
potential
is
energy
one
to
point
difference
converted
when
other forms
to
the
energy)
=
in
Q
other.
between
1
of
two
coulomb
of
points
in
a
circuit
charge flows
V
–
potential
W
–
work
Q
–
charge/C
difference/V
done/J
points.
7
2.2
Drift velocity
Learning outcomes
and
Drift velocity
When
On
completion
of
this
section,
be
able
there
rapidly
explain
the
term
derive
an
current
a
flowing
range
of
in
a
speeds,
metal,
in
the
random
electrons
are
directions.
moving
When
difference
is
applied
across
a
metal
an
electric
field
is
a
set
up.
drift velocity
The
ᔢ
no
with
to:
potential
ᔢ
is
you
about
should
power
expression for
drift
As
free
the
This
electrons
free
begin
electrons
movement
of
moving
accelerate
the
under
they
electrons
is
the
influence
collide
of
the
continuously
superimposed
on
the
electric
with
field.
metal
random
ions.
motion.
velocity
The
drift
velocity
is
the
mean
value
of
the
velocity
of
the
electrons
in
a
2
ᔢ
recall
and
use
P = IV,
P = I
R and
conductor
when
an
electric
field
is
applied.
2
P = V
/R
Consider
ᔢ
recall
the
symbols for
area
used
circuit
a
section
of
a
metallic
conductor
of
length
L
and
cross-sectional
commonly
A.
Let:
components.
I
=
current
flowing
in
the
conductor/A
3
n
=
the
number
e
=
charge
v
=
the
on
of
free
each
electrons
per
unit
volume/m
electron/C
L
−1
V
olume
A
mean
of
Number
T
otal
current I
electrons
drift
section
of
velocity
=
electrons
amount
of
of
the
electrons/m s
AL
in
charge
the
section
=
nAL
next
=
flowing
metallic
Time
conductor
taken
for
electrons
to
travel
L
from
one
end
of
the
section
to
the
v
Figure 2.2.1
Deriving an expression for
Q
nALe
drift velocity
Electric
current
I
=
=
t
I
=
L/v
nevA
Example
A
potential
difference
is
applied
−6
sectional
area
of
1.3
×
across
a
piece
of
copper
wire
m
.
The
current
flowing
through
28
The
the
concentration
drift
velocity
of
of
free
the
electrons
free
with
cross-
2
10
in
electrons
copper
in
the
is
8.7
×
10
it
is
1.2 mA.
−3
m
.
Calculate
wire.
–3
I
Drift
Exam tip
velocity
=
v
1.2
=
8.7
×
10
–6
×
1.3
−8
=
is
that
small,
yet
instantly
drift
a
10
28
nAe
Notice
×
=
velocity
light
when
a
bulb
light
in
a
×
10
10
–19
×
1.6
×
10
−1
m s
metal
turns
switch
6.6
×
on
is
In
a
metal,
conduction
solution
of
sodium
negative
ions
is
due
chloride)
to
free
the
electrons.
mobile
In
charge
an
electrolyte
carriers
are
(e.g.
positive
a
and
+
turned
on.
charge
Free
electrons
conducting
are
present
wires. As
soon
in
as
(Na
carriers
and
are
Cl
in
electrons
this
and
case).
In
a
semiconductor
,
the
mobile
‘holes’.
the
the
Example
circuit
is
closed
moving. A
a
electrons
movement
current. Therefore
of
the
start
charge
light
is
Suppose
a
solution
(electrolyte).
on
carried
A
tube
of
current
cross-sectional
I
flows
through
area
the
A
contains
solution.
a
The
salt
equally
by
positive
and
negative
ions.
The
charges
on
current
the
instantly.
positive
each
and
ion
current
charge
8
glass
bulb
is
turns
uniform
I
negative
species
per
flowing
carriers
in
ions
unit
through
the
salt
are
+2 e
volume
the
is
and
n.
solution
solution.
−2e
W
rite
in
respectively.
down
terms
of
an
the
The
number
expression
drift
for
velocity
of
the
of
Chapter
The
the
positive
positive
and
negative
ions
flow
ions
with
a
flow
drift
in
opposite
velocity
v,
directions.
the
negative
Therefore,
ions
will
2
Electrical
quantities
if
flow
cell
with
a
drift
velocity
−v
+
Current
I
=
n(+2e)(v)(A)
+
n(−2e)(−v)(A)
=
battery
–
of
4nevA
cells/d.c.
power
supply
Energy
and
power
resistor
Consider
of
time
t.
a
steady
As
current
current
energy
dissipated
moves
through
is
the
I,
flows
equal
flowing
through
to
the
potential
through
the
a
resistor
resistor
potential
difference
it
energy
between
R
for
dissipates
lost
by
a
energy.
the
terminals
duration
charge
of
the
The
as
filament
it
lamp
resistor
.
ammeter
From
the
definition
of
potential
difference
W
V
voltmeter
=
Q
switch
where
W
Q
is
is
the
the
energy
charge
dissipated
that
in
flowed
a
time
during
variable
t
a
time
resistor
t
earth
V
is
The
the
potential
charge
that
difference
flows
across
during
the
time
t
is
resistor
Q
=
alternating
It
signal
∴
W
Power
is
defined
as
=
the
ItV
rate
at
which
energy
is
capacitor
converted.
W
P
=
t
thermistor
ItV
∴
P
=
t
P
=
light-dependent
IV
resistor
The
SI
From
unit
the
of
power
definition
is
of
the
watt
resistance
V
=
IR
P
=
I(IR)
(LDR)
(W).
(see
2.3)
light-emitting
2
∴
=
I
R
diode
(LED)
2
V
Also,
P
=
(
R
V
)
V
=
R
semiconductor
diode
Commonly
used
circuit
symbols
Figure 2.2.2
Figure
2.2.2
shows
a
list
of
commonly
used
electrical
circuit
symbols
Commonly used circuit
that
symbols
will
be
encountered
in
the
chapters
that
follow.
Key points
ᔢ
The
drift
Definition
velocity
conductor
when
is
an
the
mean
value
electric field
is
of
the
velocity
applied.
of
the
electrons
in
a
1
watt
energy
is
a
of
rate
1
of
joule
conversion
per
of
second.
−1
ᔢ
Power
is
the
rate
at
which
energy
is
converted.
1 W
=
1 J s
9
2.3
Resistance
Learning outcomes
Resistance
In
On
completion
of
this
section,
a
metal,
electrons
should
be
able
define
resistance
and
the
recall
and
use
I–V
are
free
to
free
electrons
move
within
throughout
the
metal.
the
structure.
When
a
cell
is
These
connected
the
ᔢ
sketch
ᔢ
state Ohm’s
ᔢ
define
ends
of
a
piece
of
metal,
the
electrons
begin
moving.
The
cell
ohm
provides
ᔢ
are
to:
across
ᔢ
there
you
the
electrons
V = IR
necessary
move
metal
ions.
of
metal
energy
through
These
the
to
allow
metal
collisions
the
they
restrict
electrons
collide
the
flow
with
of
to
move.
each
As
other
electrons.
the
and
This
the
property
characteristics
the
is
known
as
electrical
resistance .
law
The
resistivity.
circuit
in
component
reading
gives
gives
the
Figure
X.
The
the
potential
current
difference
to
the
2.3.1
is
used
component
flowing
current
to
in
difference
through
flowing
determine
this
case
across
the
is
a
the
resistance
resistor
.
resistor
.
through
the
resistor
.
The
The
ratio
component
X
The
is
of
of
the
voltmeter
ammeter
the
its
reading
potential
resistance.
Definition
X
Resistance
conductor
(R)
to
is
defined
the
as
current
the
ratio
(I) flowing
of
the
potential
through
difference
(V)
across
the
it.
component
Equation
Figure 2.3.1
Measuring electrical
V
resistance
R
=
I
Definition
1
ohm
is
the
conductor
of
through
1 A flows
difference
resistance
when
of
1V
of
a
which
there
across
is
a
current
a
potential
R
–
resistance/Ω
V
–
potential
I
–
current/A
The
SI
unit
difference/V
of
resistance
is
the
ohm
(Ω).
it.
−1
1 Ω
=
1V A
Current–voltage
graphs
In
2.3.2
the
circuit
in
Figure
a
variable
resistor
is
used
to
adjust
the
variable
current
flowing
through
the
component
X.
resistor
The
variable
resistor
corresponding
plotted
X
Figure
to
The
obtain
2.3.3
constant
the
shows
is
I
is
is
adjusted
of
I
I–V
the
temperature.
current
relationship
Figure 2.3.2
values
and
several
recorded.
characteristic
I–V
The
directly
known
are
of
characteristic
graph
is
a
proportional
as
A
Ohm’s
values
graph
the
for
the
a
law
Circuit used to obtain data
I/A
Definition
law
states
that
flowing
through
directly
proportional
difference
there
is
no
conditions
a
across
of
the
current
conductor
it
change
the
to
the
the
V/V
potential
provided
in
is
that
physical
conductor.
Figure 2.3.3
10
of
V
and
against
metallic
line
the
V
is
then
I–V characteristic for an ohmic conductor
conductor
through
potential
to plot an I–V curve
Ohm’s
I
component.
straight
to
of
the
at
origin.
difference
V.
This
Chapter
A
conductor
that
obeys
Ohm’s
law
is
called
2
Electrical
quantities
‘
ohmic’.
I/A
The
physical
Figure
The
2.3.4
graph
conditions
shows
for
the
the
could
I–V
be
temperature
characteristic
filament
lamp
does
for
not
or
a
mechanical
filament
obey
Ohm’s
strain.
lamp.
law.
The
resistance
V/V
of
the
From
filament
the
I–V
resistance
flows
of
lamp
is
not
constant.
characteristics
the
through
lamp
the
of
the
increases
lamp
the
It
varies
filament
as
the
with
lamp
voltage
temperature
of
current.
it
can
be
increases.
the
filament
seen
As
that
more
the
current
increases.
The
Figure 2.3.4
kinetic
about
energy
their
collide
the
resistance
Figure
the
mean
with
through
of
atoms
positions.
these
vibrating
filament
of
2.3.5
the
inside
lamp
filament
shows
the
The
it
moving
atoms.
As
becomes
lamp
I–V
increases.
a
The
electrons
result,
in
the
restricted.
atoms
the
vibrate
filament
movement
This
rapidly
explains
of
I–V characteristic for a
filament lamp
lamp
I/A
electrons
why
the
increases.
characteristic
for
a
semiconductor
diode.
V/V
The
I–V
When
(see
graph
the
8.3)
when
it
the
semiconductor
semiconductor
it
is
negative
for
does
not
diode
allow
any
reverse-biased.
voltages.
If
the
is
diode
connected
current
This
is
does
the
to
reason
semiconductor
so
flow.
not
that
The
why
diode
obey
is
it
is
Ohm’s
reverse-biased
resistance
the
now
law.
current
is
is
infinite
zero
connected
so
for
that
it
Figure 2.3.5
I–V characteristic for a
semiconductor diode
is
forward-biased
voltage
is
(see
8.3),
approximately
it
begins
to
allow
a
current
to
flow
when
the
0.6 V
.
resistance / Ω
It
may
One
is
be
desirable
such
one
device
type
coefficient.
at
a
is
whose
temperature.
a
These
particular
the
resistance
thermistor
.
resistance
There
temperature
for
another
temperature.
There
decreases
thermistors
is
of
are
kind
a
are
two
to
vary
kinds
exponentially
said
to
whose
These
device
have
a
of
are
temperature.
thermistor
.
with
temperature
increases
said
to
There
increasing
negative
resistance
thermistors
with
suddenly
have
a
positive
coefficient.
temperature / °C
Figure
2.3.6
shows
temperature
how
coefficient
the
resistance
varies
with
of
a
thermistor
having
a
negative
temperature.
Figure 2.3.6
Thermistors
are
used
as
temperature
sensors
in
many
electrical
devices.
Variation of resistance for
a thermistor with negative temperature
coefficient
They
are
used
in
aircraft
wings
to
monitor
external
temperature.
Definition
Resistivity
Resistance
the
depends
material,
material
and
the
the
on
several
length
of
factors.
the
temperature
It
material,
of
the
depends
the
on
the
resistivity
cross-sectional
area
of
of
the
material.
The following
define
the
The
unit
SI
equation
resistivity
of
of
resistivity
is
a
used
to
material.
is
Ω m.
−8
The
resistivities
of
silver
and
copper
are
1.6
×
10
Ω m
and
−8
1.7
×
10
Ω m
conductors
and
respectively.
therefore
Copper
have
low
and
silver
are
good
Equation
electrical
resistivities.
ρ
R
L
=
A
Key points
ᔢ
Resistance
the
ᔢ
1
conductor
ohm
of
(R)
is
is
to
defined
1 A flows
defined
the
as
when
as
the
current
the
there
ratio
a
Resistivity
(ρ)
is
defined
by
of
a
potential
ρ
ᔢ
R
the
(I) flowing
resistance
is
of
potential
through
(V)
–
resistance/Ω
ρ
–
resistivity/Ω m
across
L
–
length/m
A
–
cross-sectional
it.
conductor
difference
difference
R
2
through
of
1V
which
across
a
area/m
current
it.
L
=
A
11
2.4
Electromotive force
Electromotive force
Learning outcomes
In
On
completion
of
this
section,
an
electrical
be
able
current
understand
between
a
(e.m.f.)
source
and
provides
resistance
internal
the
energy
resistance
in
the
circuit.
The
source
has
a
required
positive
and
to
a
drive
an
negative
to:
terminal.
ᔢ
circuit,
internal
you
electric
should
and
the
difference
potential
circuit.
difference
and
Electrons
Examples
are
of
forced
sources
out
of
the
include
negative
cells,
terminal
batteries,
solar
in
a
cells
closed
and
dynamos.
e.m.f.
In
ᔢ
understand
internal
the
concept
of
a
cell,
chemical
electromotive
resistance.
energy
to
potential
to
other
energy
force
electrical
energy
difference
forms
of
is
converted
(e.m.f.)
was
of
per
a
unit
defined
energy
per
cell
as
unit
into
is
electrical
the
energy
charge
flowing
energy
converted
charge
flowing
energy.
The
converted
through
from
between
from
the
chemical
cell.
In
electrical
two
2.1,
energy
points.
Definition
Equation
The
as
e.m.f.
the
of
a
energy
source
is
W
converted from
V
chemical
(or
Definition
defined
mechanical)
=
The
potential
two
into
electrical
energy
per
unit
charge
V
–
electromotive force,
points
through
it.
W
–
work
Q
–
charge/C
in
a
circuit
between
is
the
energy
e.m.f./V
converted
flowing
difference
Q
energy
from
electrical
energy
done/J
to
other
per
unit
two
forms
of
charge
energy
fl owing
(e.g.
heat)
between
the
points.
Exam tip
1
e.m.f.
is
devices
associated
(e.g.
produce
with
batteries)
electrical
active
that
power.
The
chemicals
The
resistance
A
cell
having
represented
2
inside
that
an
as
a
is
cell
provide
internal
e.m.f.
shown
of
in
E
to
and
Figure
a
the
an
resistance
cell
is
to
called
internal
the
the
flow
of
internal
resistance
of
r
current.
resistance .
can
be
2.4.1.
Potential difference is associated
with passive devices (e.g. resistors).
Effect of
3
Potential
difference
is
Suppose
associated
with
internal
resistance
a
cell
is
connected
to
an
external
load
R
and
supplies
a
current
I
electric fields
in
(see
increasing
also
the
circuit
as
shown
in
Figure
2.4.2.
4.
1).
The
e.m.f.
can
be
written
E
cell
=
as
I
follows:
×
(R
+
r)
E
r
where
(R
+
Therefore,
internal
load
and
r)
E
Ir
is
=
is
the
IR
the
+
total
Ir,
resistance
where
potential
IR
is
in
the
difference
the
circuit.
potential
across
the
difference
internal
across
the
resistance
of
external
the
resistance
The
Figure 2.4.1
potential
difference
across
the
external
load
is
V.
e.m.f. and internal resistance
∴
From
E
the
last
equation,
=
V
+
when
Ir
I
is
equal
to
zero,
E
=
V
(the
potential
E
r
A
B
difference
ter minal
The
between
potential
power
A
and
B).
Therefore,
difference
dissipated
in
the
is
equal
external
when
to
the
resistor
,
I
is
equal
e.m.f.
P
,
is
of
2
=
I
R
R
V
The
total
power
generated
by
the
cell
P
is
T
2
R
P
=
T
Figure 2.4.2
12
I
(R
+
r)
given
the
given
R
I
P
to
by:
zero,
cell.
by:
the
cell.
Chapter
The
fraction
of
the
total
power
dissipated
in
the
external
resistor
is
given
2
Electrical
quantities
by:
E
P
I
r
A
2
B
R
R
=
2
P
I
(R
+
r)
T
P
R
R
I
=
P
(R
+
r)
T
R
After
prolonged
increased
can
use,
internal
supply.
This
the
internal
resistance
reduces
resistance
reduces
the
total
the
of
a
cell
may
maximum
power
P
.
This
increase.
current
reduces
that
the
The
the
cell
fraction
of
T
the
to
total
R
is
power
supplied
to
the
resistor
R.
As
a
result,
the
power
supplied
reduced.
Measuring
e.m.f.
and
internal
resistance
Figure 2.4.3
Measuring e.m.f. and
internal resistance
The
circuit
measure
diagram
the
e.m.f.
in
Figure
and
2.4.3
internal
shows
the
resistance
of
circuit
a
cell
that
can
be
used
to
experimentally.
V /V
The
resistance
voltmeter
V
R
and
is
adjusted
the
so
that
corresponding
a
series
of
readings
readings
( I)
of
the
( V)
of
the
ammeter
A
are
gradient
e.m.f.
recorded.
The
A
e.m.f.
resistance
The
graph
can
in
equation
difference
be
the
of
V
against
written
as
I
E
is
=
then
I(R
=
– internal
resistance
plotted.
+
r),
where
(R
+
r)
is
the
total
circuit.
can
across
be
written
as
the
resistor
R
E
=
and
IR
Ir
+
is
Ir,
the
where
IR
potential
is
the
potential
difference
across
the
I / A
internal
resistance
of
the
cell.
Figure 2.4.4
The
potential
difference
across
the
resistor
R
is
Determining e.m.f. and
V
internal resistance graphically
∴
E
=
V
+
Ir
∴
V
=
E
−
Ir
1.32 V
By
comparing
can
be
seen
the
that
equation
plotting
a
above
graph
with
of
V
y
=
mx
against
I
+
c
for
gives
a
a
straight
straight
line,
line
it
where
r
the
y-intercept
=
e.m.f.
the
gradient
negative
of
the
cell
I =
=
the
internal
resistance
of
the
0.65 A
cell.
V
Example
A
cell
has
an
e.m.f.
connected
across
provides
current
of
the
1.32 V
and
terminals
internal
of
a
resistance
resistor
of
of
r.
resistance
The
1.20
cell
Ω.
1.20 Ω
is
The
cell
Figure 2.4.5
a
of
0.65 A.
(See
Figure
2.4.5.)
Calculate:
a
the
total
b
the
internal
c
the
potential
resistance
in
the
circuit
Key points
resistance
r
ᔢ
difference
across
the
terminals
of
the
The
as
a
The
e.m.f.
of
the
cell
is
given
by
E
=
I(R
+
r)
total
resistance
in
the
circuit
=
R
+
r
the
=
of
a
energy
chemical
E
The
e.m.f.
source
(or
mechanical)
=
total
resistance
r
=
=
R
+
r
2.03
=
−
=
into
2.03 Ω
electrical
energy
2.03
=
0.83 Ω
−
per
through
unit
it.
2.03
ᔢ
R
The
to
=
energy
0.65
charge flowing
The
defined
converted from
1.32
I
b
is
cell.
a
resistance
cell
is
that
called
is
the
internal
internal
1.20
resistance.
ᔢ
Therefore
c
Potential
the
internal
difference
resistance
across
the
r
=
The
e.m.f.
of
cell
and
internal
resistance
0.83 Ω
terminals
of
the
a
can
be
determined
cell
experimentally.
V
=
IR
=
0.65
×
1.20
=
0.78 V
13
3
d.c.
3.
1
Kirchhoff ’s
circuits
Learning outcomes
On
completion
should
be
able
of
this
laws
Kirchhoff ’s
section,
Kirchhoff ’s first law
you
to:
Kirchhoff ’s first
ᔢ
state
ᔢ
apply
laws
Kirchhoff ’s
law
states
that
the
sum
of
the
currents flowing
into
any
point
laws
in
Kirchhoff ’s
a
circuit
is
equal
to
the
sum
of
the
currents flowing
out
of
that
point.
laws.
Consider
a
current
I
flowing
into
a
junction
J.
I
1
the
junction
and
I
2
are
flowing
out
of
3
J.
I
2
According
to
Kirchhoff ’s
I
first
=
I
1
I
law :
+
I
2
3
1
The
total
charge
flowing
into
the
junction
must
equal
the
total
charge
J
leaving
based
the
on
junction
the
in
a
given
conservation
of
time.
Therefore,
Kirchhoff ’s
first
law
is
charge.
I
3
Figure 3.1.1
Kirchhoff ’s first law
Kirchhoff ’s
second
In
in
a
the
cell
circuit
of
Figure
negligible
potential
law
3.1.2
internal
differences
two
resistors
resistance.
across
resistors
are
The
R
and
R
1
E
connected
e.m.f.
of
are
the
V
2
in
series
cell
and
is
V
1
E.
with
The
respectively.
2
Kirchhoff ’s second law
I
I
Kirchhoff ’s
any
R
loop
in
second
a
circuit
law
is
states
equal
the
the
algebraic
algebraic
sum
sum
of
of
the
the
p.d.s
e.m.f.s
around
around
the
loop.
R
1
2
This
can
be
stated
as
E
=
V
+
V
1
2
V
V
2
1
If
1
coulomb
through
Figure 3.1.2
that
to
the
of
charge
cell
and
flows
loses
around
energy
as
a
loop,
it
it
passes
gains
energy
through
as
each
it
passes
resistor
.
If
the
Kirchhoff ’s second law
charge
it
moves
started,
it
Therefore,
equal
to
around
will
the
the
Kirchhoff ’s
have
energy
energy
second
the
loop
the
same
gained
is
the
by
based
ends
energy
by
dissipated
law
and
at
at
the
charge
the
on
up
end
same
as
passing
charge
the
the
of
the
at
which
beginning.
through
passing
principle
at
point
the
through
cell
is
resistors.
conservation
of
energy.
Example
For
9V
the
circuit
in
Figure
3.1.3,
calculate
the
value
of
the
current
I
3V
Using
Kirchhoff ’s
sum
of
second
e.m.f.s
in
law
loop
=
sum
=
20 I
of
p.d.s
in
loop
I
20 Ω
9
60 Ω
Note
Figure 3.1.3
3 V
the
cell
negative
is
−
sign
opposite
in
3
in
front
of
direction
80 I
=
I
=
to
+
60I
the
3.
that
This
of
the
6
6
=
80
14
0.075 A
is
because,
9 V
cell.
the
e.m.f.
of
the
Chapter
Example
the
circuit
resistances
in
and
Figure
the
3.1.4,
the
resistance
of
batteries
the
have
voltmeter
negligible
is
infinite.
d.c.
circuits
A
I
In
3
internal
Determine
the
12 V
reading
Using
on
the
Kirchhoff ’s
sum
of
9V
voltmeter
.
e.m.f.s
second
in
law,
loop
=
sum
9
=
5I
7I
=
3
I
=
of
p.d.s
in
loop
2 Ω
12
−
+
5 Ω
2I
3
B
=
0.429 A
7
Figure 3.1.4
Current
The
always
potential
flows
from
difference
a
higher
across
the
potential
2 Ω
to
resistor
a
=
lower
potential.
−(0.429
×
2)
I
I
1
=
The
voltmeter
reading
is
therefore
=
12
−
2
I
−0.858 V
0.858
=
3
11.1 V
2 Ω
20 Ω
3 Ω
Example
Calculate
the
currents
in
the
circuit
shown
in
Figure
3.1.5.
12 V
Using
Kirchhoff ’s
first
18 V
law,
loop
I
=
I
3
Using
Kirchhoff ’s
second
12
+
I
1
law
=
on
2I
+
loop
=
loop
2
1,
Figure 3.1.5
20I
1
12
1
(1)
2
3
2I
+
20(I
1
+
I
1
)
2
Exam tip
12
=
22 I
+
20I
1
(2)
2
Practise
Using
Kirchhoff ’s
second
law
on
loop
many
Kirchhoff ’s
18
=
3I
+
=
3
3I
+
=
+
(2)
and
Multiplying
(3)
can
20 I
equation
be
(2)
I
1
+
to
on
ensure
that
you
the
(3)
the
concepts.
direction
of
the
Pay
attention
currents
and
to
the
e.m.f.s.
2
solved
by
)
2
23I
1
Equations
grasp
20(I
2
18
laws
20I
2
18
questions
2,
simultaneously
as
follows:
20,
Key points
240
=
440I
+
400I
1
Multiplying
equation
(3)
by
(4)
2
22,
ᔢ
Kirchhoff ’s first
the
396
=
440I
+
506I
1
sum
(5)
−
Equation
law
states
that
currents flowing
(5)
any
point
in
a
circuit
is
(4)
equal
156
the
2
into
Equation
of
=
106 I
to
flowing
the
out
sum
of
of
that
the
currents
point.
2
156
ᔢ
I
=
=
Kirchhoff ’s first
law
is
based
on
1.47 A
2
106
the
Substituting
the
value
of
I
into
Equation
conservation
of
charge.
(2)
2
ᔢ
12
=
22 I
+
20
×
1.47
22 I
=
12
−
Kirchhoff ’s
the
1
around
29.4
second
algebraic
any
sum
loop
law
of
in
states
the
a
that
e.m.f.s
circuit
is
1
equal
22 I
=
to
the
algebraic
sum
of
the
−17.4
1
p.d.s
around
the
loop.
17.4
I
=
=
−0.791 A
1
ᔢ
22
Kirchhoff ’s
on
Substituting
I
and
1
I
into
Equation
the
second
law
conservation
of
is
based
energy.
(1)
2
I
=
3
I
+
1
I
=
−0.791
+
1.47
=
0.679 A
2
15
3.2
Resistors
Learning outcomes
On
completion
should
be
able
of
this
in
series
Resistors
section,
Consider
two
a
I
an
parallel
series
resistors
flowing
R
and
R
connected
them.
The
series
potential
(Figure
differences
3.2.1),
across
with
R
1
R
are
V
expression for
and
V
1
respectively.
The
potential
difference
across
the
two
2
resistors
is
V
.
series
R
R
R
1
2
derive
an
expression for
T
I
I
I
ᔢ
in
2
through
to:
resistors
in
in
1
current
2
derive
in
you
and
ᔢ
and
resistors
≡
in
parallel.
V
V
2
1
V
V
Figure 3.2.1
From
to
Kirchhoff ’s
the
flows
Also,
Resistors in series
current
I
through
the
first
each
sum
law,
leaving
of
of
the
∴
on
the
the
current
the
right.
=
V
differences
+
the
definition
of
means
from
that
the
the
left
is
same
equal
current
across
each
resistor
is
equal
V
1
From
entering
resistors.
potential
V
I
This
=
(1)
2
IR
1
V
=
V
(2)
1
IR
2
(3)
2
=
IR
(4)
T
R
is
the
combined
resistance
of
the
two
resistors
T
Substituting
Equations
IR
(2),
=
(3)
IR
T
Dividing
by
summing
The
=
the
(4)
into
Equation
(1),
IR
2
R
the
resistance
expression
can
be
R
2
resistance
of
each
extended
=
T
+
1
combined
R
16
+
1
T
Therefore,
and
I,
R
V
resistance,
V
where
to
R
+
1
of
resistors
in
series
is
found
by
resistor
.
to
R
two
+
2
or
R
+
3
more
…
resistors
in
series:
Chapter
Resistors
Consider
in
two
d.c.
circuits
parallel
resistors
R
and
R
1
potential
3
difference
connected
in
parallel
(Figure
3.2.2).
The
2
across
R
and
R
1
is
V
2
R
1
I
1
R
I
T
I
I
≡
V
I
2
R
2
V
Figure 3.2.2
Resistors in parallel
Exam tip
From
Kirchhoff ’s
first
law,
In
I
=
I
+
I
1
to
Since
the
two
resistors
are
the
examination,
in
parallel,
they
each
have
a
derive
an
of
V
across
you
are
expression for
asked
the
potential
combined
difference
if
(1)
2
resistance for
resistors
in
them.
parallel:
From
the
definition
of
resistance,
1
Sketch
a
simple
diagram
to
V
I
=
show
(2)
the
resistors,
currents
and
1
R
1
potential
differences.
V
I
=
(3)
2
R
2
2
Write
out
the
steps
in
the
V
I
=
(4)
derivation
as
clearly
as
possible.
R
T
where
R
is
the
combined
resistance
of
the
two
resistors
T
Substituting
Equations
(2),
V
(3)
V
=
(1),
R
1
1
2
1
=
the
R
1
combined
1
1
=
be
of
resistors
in
parallel
is
found
by
1
+
R
T
can
2
resistance
R
expression
1
+
R
T
The
Equation
V,
R
Therefore,
into
V
R
T
by
(4)
+
R
Dividing
and
R
1
2
extended
1
1
=
to
+
R
R
T
two
1
or
+
R
1
more
resistors
in
parallel:
1
+
…
R
2
3
Key points
ᔢ
The
combined
R
=
T
ᔢ
The
R
+
1
+
2
1
two
or
more
resistors
in
series
is
two
or
more
resistors
in
parallel
given
by
…
of
is
given
by
1
+
R
1
+
of
3
+
R
R
resistance
1
=
T
R
1
combined
R
resistance
+
…
R
2
3
17
3.3
Worked
examples
Calculate
completion
should
ᔢ
be
able
analyse
of
circuits
section,
the
resistance
of
the
network
shown
in
Figure
3.3.1:
you
a
between
A
and
B
b
between
A
and
C.
to:
and
involving
this
d.c.
Example
Learning outcomes
On
on
solve
d.c.
problems
circuits.
3 Ω
5 Ω
4 Ω
2 Ω
Exam tip
Figure 3.3.1
In
an
examination
you
may
be
a
given
a
diagram
of
a
Between
The
combination
that
A
and
B:
resistance
first
thing
to
do
is
to
redraw
the
circuit
as
shown
in
Figure
3.3.2.
appears
3 Ω
complicated. Take
break
up
parallel
the
your
diagram
time
into
and
series
and
resistors.
D
C
5 Ω
2 Ω
4 Ω
Figure 3.3.2
The
combined
because
This
they
resistance
are
combined
in
of
the
resistance
1
is
in
1
∴
5 Ω,
2 Ω
and
4 Ω
resistors
is
11 Ω
series.
=
R
parallel
with
the
3
Ω
resistor
.
1
+
3
11
AB
1
=
0.424
R
AB
1
R
=
=
2.36 Ω
AB
0.424
b
3 Ω
Between
A
The
first
The
3 Ω
thing
C:
to
do
is
to
redraw
the
circuit
resistor
resistance
The
5 Ω
D
This
is
is
shown
in
Figure
3.3.3.
is
and
the
4 Ω
resistor
are
in
series,
so
their
combined
and
the
2 Ω
resistor
are
in
series,
so
their
combined
7 Ω
resistor
resistance
7 Ω
equivalent
to
two
resistors
of
value
2 Ω
other
.
Figure 3.3.3
1
∴
1
=
R
1
+
7
7
AC
1
=
0.285
R
AC
1
R
=
=
AC
0.285
18
as
4 Ω
B
5 Ω
and
3.5 Ω
7 Ω
in
parallel
with
each
Chapter
3
d.c.
circuits
Example
A
battery
The
is
power
connected
dissipated
to
in
three
the
resistors
150 Ω
as
shown
resistor
is
in
Figure
3.3.4.
51.8 mW
.
Calculate:
25 Ω
a
the
current
b
the
combined
c
the
e.m.f.
d
the
potential
e
the
power
a
current
flowing
of
in
the
resistance
the
circuit
in
the
150 Ω
circuit
battery
20 Ω
difference
across
the
25 Ω
resistor
Figure 3.3.4
dissipated
flowing
in
in
the
the
20 Ω
circuit
=
resistor
.
current
flowing
through
the
150
Ω
resistor
2
P
=
I
−3
51.8
×
R
2
10
=
I
(150)
–3
51.8
×
10
2
I
=
150
2
−4
I
=
3.45
×
10
−4
I
b
The
25 Ω
resistor
3.45
=
and
√
the
1
×
20 Ω
1
−2
=
resistor
1.86
are
in
×
10
A
parallel
with
each
other
.
1
=
R
10
+
25
20
1
=
0.09
=
11.1 Ω
R
R
This
resistance
Therefore,
the
is
in
series
combined
R
=
with
the
resistance
150
+
150 Ω
in
11.1
the
=
resistor
.
circuit
is:
161.1 Ω
T
c
e.m.f.
of
battery
=
current
flowing
in
circuit
×
combined
resistance
−2
d
Potential
=
=
1.86
=
3.00 V
difference
e.m.f.
of
×
across
battery
10
the
−
×
161.1
25 Ω
resistor
potential
difference
across
150
Ω
resistor
−2
e
=
3.00
=
0.21 V
Potential
−
(1.86
difference
=
potential
=
0.21 V
×
10
across
×
the
difference
150)
20 Ω
across
resistor
the
25 Ω
resistor
2
2
V
(0.21)
−3
Power
dissipated
in
20 Ω
resistor
=
=
R
=
2.21
×
10
W
20
19
3.4
Potential
Learning outcomes
dividers
Potential dividers
A
On
completion
of
this
section,
potential
from
should
be
able
divider
circuit
is
used
to
produce
a
small
a
larger
potential
difference.
The
larger
potential
across
two
resistors
in
series,
R
and
R
1
describe
how
to
use
a
as
variable
a
source
potential
, as
shown
in
V
is
Figure
2
potential
3.4.1.
divider
difference
difference
to:
connected
ᔢ
potential
you
of fixed
or
This
provides
difference.
circuit
a
circuit
greater
can
be
is
very
p.d.
useful
than
supplied
when
that
with
the
required
the
correct
only
by
p.d.
available
some
V
by
power
electrical
supply
circuit.
connecting
it
The
across
the
1
terminals
The
total
A
and
B.
resistance
in
the
circuit
is
I
R
=
R
T
+
R
1
(1)
2
R
2
From
the
definition
of
resistance,
V
I
A
V
=
=
(2)
R
V
R
T
+
R
1
2
V
1
But,
I
=
(3)
R
1
R
V
1
1
Equation
(2)
is
equal
to
(3)
V
V
1
∴
=
R
+
R
1
R
2
1
B
V
(R
1
Figure 3.4.1
+
R
1
)
=
VR
2
1
A potential divider circuit
R
1
∴
V
=
1
(
R
+
Apart
I
from
variable
X.
This
can
supplying
voltage
by
resistor
supply
any
is
a
fixed
replacing
called
voltage
a
)
R
1
V
2
voltage,
the
the
circuit
resistors
with
potentiometer
.
between
zero
and
a
can
be
single
Therefore,
made
to
variable
the
supply
a
resistor
potential
divider
V
A
V
X
Example
Suppose
requires
R
a
student
3 V
has
and
a
he
12 V
has
d.c.
no
supply.
other
He
power
wants
supply
to
power
available.
a
circuit
How
1
accomplish
He
can
Figure
B
V
A potential divider providing
this?
achieve
this
by
using
a
potential
divider
circuit
as
shown
in
3.4.3.
=
12 V
and
V
=
3 V
1
a variable p.d.
He
can
choose
any
combination
V
of
R
1
3
V
=
R
+
R
1
If
2
he
chooses
R
=
100 kΩ,
2
then:
1
12 V
=
100
+
R
4
2
100
+
R
=
400
=
400
=
300 kΩ
2
R
−
100
2
R
1
V
=
3V
1
R
2
Figure 3.4.3
20
1
=
12
1
100
V =
provided
1
=
I
R
resistors,
4
that
the
that
can
V
1
Figure 3.4.2
d.c.
ratio:
he
Chapter
3
d.c.
circuits
Example
A
thermistor
950 Ω
room
a
at
as
T
ii
in
of
resistance
thermistor
the
that
resistance
a
The
shown
Assuming
i
has
25 °C.
the
the
of
circuit
in
battery
value
the
temperature
R
such
the
reading
on
is
the
Figure
is
that
2500 Ω
used
has
voltmeter
the
of
is
a
to
0 °C
and
a
the
resistance
of
temperature
of
T
a
3.4.4.
negligible
innite,
the
at
monitor
internal
resistance
and
the
3V
calculate:
reading
on
the
voltmeter
is
1.8 V
when
R
0 °C
voltmeter
when
the
temperature
in
the
room
is
25 °C.
b
The
voltmeter
is
replaced
with
one
having
a
resistance
of
9 k Ω.
Figure 3.4.4
Calculate
the
new
voltmeter
reading
at
25 °C.
R
1
a
i
Using
the
potential
divider
equation
V
(
=
1
R
+
)
R
1
V
2
R
×
R
+
3
=
1.8
2500
R
1.8
=
R
+
R
+
2500
3
R
=
0.6( R
+
0.6 R
2500)
+
R
0.6
2500
−
=
R
1500
=
R
0.6R
=
1500
0.4 R
=
1500
R
=
1500
=
3750 Ω
0.4
ii
At
25 °C
the
potential
resistance
divider
of
equation
the
thermistor
is
950 Ω.
Using
the
again
R
1
V
=
1
(
R
+
)
R
1
V
2
3750
V
=
×
3
1
3750
V
=
+
950
2.39 V
1
Therefore,
the
reading
on
the
voltmeter
is
V
=
2.39 V
.
1
b
Since
The
the
voltmeter
combined
has
a
resistance
resistance
of
the
it
needs
voltmeter
1
1
and
be
the
taken
into
resistor
R
is
account.
given
by:
1
=
R
to
+
3750
9000
T
1
−4
=
3.78
×
10
R
T
1
R
=
=
2647 Ω
−4
T
3.78
×
10
R
1
V
oltmeter
reading
V
=
1
(
R
+
)
R
1
V
2
2647
V
=
×
3
=
2.21 V
1
2647
+
950
Key point

A
potential
divider
can
be
used
as
a
source
of xed
or
variable
potential
difference.
21
3.5
The Wheatstone
Learning outcomes
The Wheatstone
A
On
completion
of
this
section,
Wheatstone
be
able
describe
component
how
to
use
circuits.
Wheatstone
ᔢ
bridge
to
compare
an
resistances
understand
used
to
determine
It
is
the
resistance
essentially
made
up
of
an
of
two
unknown
technique
resistance.
unknown
are
that
This
uses
a
null
method
for
potential
determining
the
a
unknown
two
is
accurately.
to:
divider
ᔢ
bridge
bridge
you
resistive
should
bridge
known.
The
resistance,
The
circuit
it
can
relationship
P
a Wheatstone
diagram
be
is
shown
determined
between
P,
Q,
if
R
in
the
and
Figure
3.5.1.
resistances
S
is
as
P,
If
Q
R
is
and
S
follows:
R
=
bridge
is
a
double
Q
potential
divider.
In
practice,
sectional
The
a
fixed
area,
slider
D
voltage
using
is
S
a
is
applied
standard
adjusted
until
cell
the
across
(e.g.
a
a
wire
AC
Leclanché
galvanometer
of
uniform
cell)
reading
(Figure
is
zero.
cross-
3.5.2).
The
B
distances
AD
uniform,
P
the
and
DC
are
resistance
measured
of
the
wire
using
is
a
metre
proportional
rule.
to
Since
the
the
wire
is
length.
Q
AD
R
∴
=
I
DC
1
S
G
When
the
galvanometer
reading
is
zero,
the
potential
difference
across
I
2
the
R
points
B
and
D
is
zero.
S
∴
Potential
The
difference
current
flowing
across
through
AB
P
=
and
potential
Q
is
difference
across
AD
I
D
1
The
current
flowing
through
R
and
S
is
I
2
I
P
=
I
1
Figure 3.5.1
R
(1)
2
The principle of the
I
Wheatstone bridge
Q
=
I
1
Dividing
the
two
equations
I
thick
copper
S
strip
(2)
2
P
we
I
1
get:
R
2
=
I
Q
I
1
S
2
P
R
B
=
Q
S
G
Example
A
A
slide-wire
Wheatstone
bridge
is
balanced
when
the
uniform
wire
is
C
R
D
S
divided
resistor
as
shown
in
Figure
3.5.3.
Calculate
the
value
of
the
unknown
R.
R
20
=
400
Figure 3.5.2
80
Practical setup of a
Wheatstone bridge
20
R
×
400
=
=
100 Ω
80
The
20 cm
80 cm
A
potentiometer
potentiometer
potential
differences.
cell
400 Ω
used
applied
resistivity
is
22
called
across
connected
Figure 3.5.3
a
difference.
uniform
R
is
potential
at
the
the
device
It
uses
A
and
the
potential
driver
X.
X
The
measure
of
difference
an
the
is
The
Y
.
unknown
The
jockey
J
point
(a
unknown
potential
applied
cross-sectional
cell.
and
to
principle
uniform
points
point
used
the
area
metal
across
(Figure
potential
with
e.m.f.
divider
the
a
a
measure
wire
3.5.4).
difference
higher
contact)
or
to
is
AB
is
potential
moved
of
The
is
along
AB
Chapter
until
the
deflection
on
the
sensitive
galvanometer
is
zero.
This
3
d.c.
circuits
technique
E
is
called
the
no
a
deflection
galvanometer
current
measured
A
null
flows
using
is
zero
method.
is
through
a
potentiometer
metre
is
called
the
The
the
point
at
balance
galvanometer
.
which
point.
The
the
At
deflection
the
length
of
balance
the
wire
driver
on
point,
AJ
is
rule.
used
cell
L
J
to:
wire
ᔢ
compare
of
uniform
e.m.f.s
resistivity
ᔢ
compare
resistances
ᔢ
measure
currents.
X
uniform
Y
sectional
Advantages
ᔢ
No
A
is
current
Since
a
be
on
from
coil
from
with
the
results
potentiometer
the
a
the
p.d.
calibration
are
is
degree
of
dependent
the
on
measure
being
has
being
method
high
to
p.d.
voltmeter
deflection
found
depend
The
null
a
drawn
moving
some
can
ᔢ
using
current
point.
ᔢ
of
a
potential
measured
resistance
at
and
Figure 3.5.4
differences:
the
and
cross-
area
A potentiometer
balance
therefore
draws
measured.
used
of
to
find
accuracy.
sensitive
the
The
balance
method
point,
does
this
not
galvanometer
.
lengths.
E
1
Disadvantages
ᔢ
The
ᔢ
It
is
of
using
resistivity
difficult
and
to
a
potentiometer
cross
section
measure
of
changing
to
measure
the
wire
potential
potential
must
be
differences:
uniform.
differences,
since
it
takes
l
2
time
to
find
the
balance
point
and
measure
lengths.
l
1
A
Comparing
A
e.m.f.s
potentiometer
how
this
E
initially
is
is
can
done.
A
be
s
used
to
standard
connected
B
E
as
compare
cell
(a
shown.
cell
The
two
e.m.f.s.
with
an
jockey
Figure
accurately
is
adjusted
3.5.5
known
until
a
shows
e.m.f.)
balance
s
point
l
is
found.
1
E
is
replaced
by
a
second
cell
E
s
(of
unknown
e.m.f.).
2
E
2
The
jockey
is
adjusted
until
a
balance
point
l
is
found.
2
Figure 3.5.5
E
Comparing e.m.f.s
E
s
2
Therefore,
=
l
l
1
2
E
1
Comparing
A
resistances
potentiometer
can
be
used
to
compare
two
resistances.
Suppose
l
2
a
circuit
consists
of
a
cell
E
and
two
resistors
R
2
and
R
1
.
The
2
l
1
potentiometer
is
initially
connected
across
R
and
the
balance
point
1
found.
The
potentiometer
is
then
connected
l
is
1
across
R
and
the
balance
2
point
l
is
found.
2
R
R
1
2
Therefore,
=
l
R
l
1
R
1
2
2
E
2
Key points
Figure 3.5.6
ᔢ
A Wheatstone
bridge
is
used
to
determine
the
value
of
an
Comparing resistances
unknown
resistance.
ᔢ
A
potentiometer
potential
ᔢ
A
is
a
device
used
to
measure
an
unknown
e.m.f.
or
a
difference.
potentiometer
can
be
used
to
compare
two
e.m.f.s
and
compare
two
resistances.
23
Revision
Answers
found
1
a
to
on
questions
the
that
require
questions
calculation
can
be
1
7
a
Define
Describe,
in
terms
of
an
electron
model,
terms:
difference
between
an
electrical
i
resistance
ii
resistivity.
conductor
A
potential
[2]
difference
of
3V
insulator.
of
copper
sectional
Give
an
i
An
ii
A
iii
example
each
of
wire
of
area
2.2
×
length
a
0.65 m
and
cross-
2
10
m
i
the
resistance
of
the
. Calculate:
ii
the
current flowing
iii
the
power
copper
wire
[3]
[1]
that
is
non-metal
a
conductor
that
is
a
a
Discuss
two
hazards
of
b
Discuss
two
applications
conductor
dissipated
in
the
[1]
electrostatics.
the
wire
[2]
wire.
[2]
−8
of
copper
=
1.7
×
10
Ω m)
[4]
8
of
through
[1]
(Resistivity
2
across
the following:
insulator
metal
A
of
applied
[2]
−9
b
is
and
piece
electrical
[2]
the
b
an
the
accompanying CD.
electrostatics.
a
Explain
what
is
meant
by
the
term
drift velocity
[6]
[2]
3
A
1.5 V
cell
delivers
a
constant
charge
of
420 C for
b
a
4
period
of
2.8
×
10
s
in
a
Derive
the
expression for
conductor
circuit.
in
terms
electrons flowing
of
in
the
the
the
current
drift
in
velocity
a
of
the
conductor.
[4]
Calculate:
c
a
the
current flowing
through
the
circuit
Explain
of
b
the
resistance
in
the
why
circuit
the
charge
the
total
number
of
a
semiconductor,
carriers
is
much
the
larger
drift velocity
than
in
a
[2]
metallic
c
in
[3]
electrons flowing
during
conductor
of
the
same
dimensions
with
the
the
same
current flowing
in
it.
[3]
4
period
of
2.8
×
10
s.
[2]
9
4
a
Explain
what
is
meant
by
the
terms
a
Use
energy
considerations
electromotive force
current
and
potential difference.
A
student
connects
his
cellular
phone
to
a
period
of
2
minutes for
a
quick
delivers
a
constant
between
potential
current
(p.d.).
[3]
Explain
what
is
meant
by
the
internal
resistance
charge. The
of
charger
distinguish
and
charger
b
for
(e.m.f.)
[4]
difference
b
to
electric
of
a
cell.
[1]
300 mA.
c
Explain
why
the
potential
difference
across
the
Calculate:
terminals
i
the
charge flowing
ii
the
number
of
during
the
period
electrons flowing
[2]
during
the
period.
5
battery’s
d
[2]
a
Define
potential
b
Define
the
c
Using
difference.
show
the
of
potential
definition
that
the
of
power
a
battery
is
normally
lower
than
the
e.m.f.
Under
what
across
a
[2]
condition
battery’s
is
the
terminal
potential
equal
to
difference
its
e.m.f.?
[1]
[2]
10
unit
of
difference.
potential
dissipated
[2]
difference,
in
a
Two
are
resistors
having
connected
and
negligible
in
resistances
series
internal
with
a
of
1.5 kΩ
battery
resistance
of
shown
and
4.5 kΩ
e.m.f.
9.0 V
below.
resistor
a
Calculate
the
potential
difference
across
each
of
2
V
of
resistance
R
is
given
by
P
=
,
where
the
V
resistors.
[4]
R
is
the
potential
difference
across
the
resistor.
[3]
9.0 V
18
6
A
potential
electrons
difference
to flow
of
12 V
through
1.2
minutes. Calculate:
a
the
charge
b
the
electric
a
causes
metallic
2.4
×
10
conductor
in
1.5 kΩ
that flowed
through
current flowing
through
conductor
c
24
the
resistance
the
conductor
[2]
the
[2]
of
the
conductor.
[2]
4.5 kΩ
Revision questions
b
A
voltmeter
4.5 kΩ
i
ii
of
resistance
resistor
Calculate
What
and
the
will
reads
R
is
placed
across
the
15
a
Define
electrical
b
Sketch
the
c
State
resistance
R
the voltmeter
the
1.5 kΩ
of
the
A
household
switched
bulb
on for
4
is
read
when
ohm.
[2]
I–V
what
graph for
happens
to
a filament
the
lamp.
[2]
resistance
of
the
placed
resistor?
marked
the
voltmeter. [4]
lamp
as
V
increases.
[4]
Suggest
11
and
5.92 V
filament
across
resistance
1
125 V,
60 W. The
bulb
the
is
I–V
an
explanation for
graph for
the
a filament
shape
of
lamp.
[3]
hours.
16
A
car
battery
has
an
e.m.f. of
12 V
and
internal
Calculate:
resistance of
a
the
current flowing
b
the
charge
that
through
passes
the
through
bulb
the
[2]
0.
10 Ω. When the
battery delivers
a
current of
ignition
is turned, the
80 A to the
starter
motor.
bulb
Calculate:
for
c
the
the
4
hours
energy
[2]
supplied
the
4-hour
the
working
to
the
bulb
a
the
resistance
b
the
potential
of
the
starter
motor
[3]
during
period
difference
across
the
starter
[2]
motor
d
12
a
Explain
resistance
what
is
meant
of
by
the
the
bulb.
resistivity
of
a
material.
Show
c
the
power
dissipated
in
the
starter
that
d
the
power
dissipated
in
the
battery.
the
unit
in
which
resistivity
A
battery
is
Ω m.
The
resistance
and
cross-sectional
of
a
piece
of
wire
of
length
of
is
and
e.m.f.
3.0 V
and
negligible
internal
area
of
2.0
×
10
a
connected
in
series
the
resistivity
of
as
with
shown
a
in
resistor
the
of
diagram
room
temperature,
the
resistance
of
the
2
m
is
thermistor
Determine
thermistor T
12 cm
below. At
−8
this
wire.
is
2.0 kΩ.
[3]
a
13
[2]
[2]
1.5 kΩ
3.66 Ω.
[2]
is
resistance
measured
c
motor
[2]
17
b
[2]
[2]
a
Derive
an
expression for
two
resistors
in
series. [4]
b
Derive
an
expression for
two
resistors
in
parallel.
Calculate
the
thermistor
at
potential
room
difference
across
the
temperature.
[2]
C
[4]
c
Calculate
the
effective
resistance
between
the
1.5 kΩ
points X
and Y.
[3]
3.0 V
10 Ω
B
X
T
10 Ω
10 Ω
A
b
Y
A
uniform
is
connected
thermistor
10 Ω
is
14
For
the
circuit
a
the
value
b
the
power
below
of
resistance
calculate:
I
as
connected
contact
M
in
wire
parallel
shown
the
of
the
below. A
between
on
PQ
with
the
length
1.00 m
resistor
and
the
sensitive voltmeter
point
B
and
a
movable
wire.
[4]
P
C
dissipated
in
the
4 Ω
resistor.
[3]
12 V
1.5 kΩ
B
1.00 m
4 Ω
M
I
5 Ω
A
Q
2 Ω
25
i
Explain
wire,
two
why, for
the
points
distance
a
constant
potential
on
the
wire
between
current
difference
the
is
in
the
between
proportional
two
20
any
to
In
the
and
the
points.
diagram
negligible
e.m.f.
of
6V
below,
a
internal
and
its
battery
P
has
resistance.
internal
an
e.m.f.
Battery Q
resistance
is
of
has
9V
an
1.2 Ω.
[2]
P
ii
The
contact
M
is
moved
along
PQ
until
the
I
1
voltmeter
1
State
reading
the
contact
2
is
zero.
potential
at
Calculate
M
and
the
difference
the
length
between
the
point Q.
of
the
9V
[1]
wire
MQ.
3 Ω
[2]
2 Ω
iii
The
thermistor
explain
the
between
at
18
a
zero
Explain
M
with
of
a
effect
cooled
on
slightly. State
the
and Q for
length
of
the
the voltmeter
to
and
wire
remain
deflection.
the
potentiometer
e.m.f.
is
aid
[2]
of
could
a
be
diagram
used
to
how
a
3 Ω
slide-wire
measure
6 V,
1.2 Ω
the
cell.
A
[5]
B
I
2
b
Explain
a
more
why
the
slide-wire
accurate
potentiometer
reading for
the
e.m.f.
of
Q
gives
the
cell
a
Calculate
the
values
of
I
and
I
1
than
a
moving
coil
galvanometer.
b
19
a
State
Kirchhoff ’s
principle
upon
laws
which
and
each
state
is
the
Using
Kirchhoff ’s
laws,
based.
calculate
the
Determine
points A
physical
[6]
the
and
potential
difference
between
the
B.
[2]
[6]
21
b
.
2
[3]
An
electric
heater
is
made
using
nichrome
wire.
magnitude
−6
Nichrome
of
the
currents
I
,
A
I
and
B
I
has
a
resistivity
of
1.0
×
10
Ω m
at
the
[8]
C
operating
heater
is
temperature
designed
to
of
the
have
a
heater. The
power
electric
dissipation
of
1.
1 Ω
75 W
I
when
across
A
When
the
the
a
potential
terminals
heater
is
of
difference
the
operating
a
the
current flowing
b
the
resistance
of
18 V
is
applied
heater.
at
through
75 W,
calculate:
it
[2]
0.012 Ω
+
–
I
of
the
nichrome
wire
inside
the
heater
B
[2]
9.0 V
c
0.
15 Ω
+
the
length
0.75 mm
I
C
12.0 V
26
of
nichrome
wire
of
diameter
of
–
required
to
make
the
heater.
[3]
Revision questions
22
The
diagram
below
shows
a
network
of
resistors
and
24
Use
Kirchhoff ’s
and
I
laws
to
determine
the
currents
I
1
,
1
batteries.
Each
battery
has
an
internal
resistance
of
in
the
circuit
below.
I
2
[6]
3
0.
1 Ω.
20 Ω
9V
0.
1 Ω
I
I
1
2
3.0V
4.5V
10 Ω
1 Ω
1 Ω
12 V
0.
1 Ω
P
Q
I
3
2 Ω
2 Ω
6V
25
The following
circuit
is
used
to
compare
two
e.m.f.s.
0.
1 Ω
E
1
1.5 V
a
Calculate
the
current flowing
in
the
9V
0.40 Ω
battery.
[3]
b
A
thick
wire
connect
the
of
negligible
points
current flowing
P
resistance
is
and Q. Calculate
through
the
9V
used
the
to
I
R
new
battery.
[6]
0.85 m
X
23
For
the
circuit
below
Y
calculate:
J
E
r
12 V
uniform
length
resistance
wire
1.00 m
E
2
The
uniform
1.00 m
2 Ω
and
resistance
radius
wire XY
has
0.55 mm. The
a
length
resistivity
of
of
the
−6
1 Ω
material
of
the
wire
is
1.
1
×
10
Ω m.
E
has
an
e.m.f.
1
of
1.5 V
and
through
E
internal
is
resistance
of
0.40 Ω. The
current
I.
1
E
3 Ω
4 Ω
has
an
e.m.f.
of
E
and
internal
resistance
of
r.
2
The
movable
flowing
contact J
through
E
is
is
adjusted
zero. When
so
that
this
the
occurs,
current
the
2
a
the
total
resistance
in
the
circuit
[4]
length XJ
b
the
current flowing
through
the
12 V
d.c.
supply
c
the
is
resistance
0.85 m
of
and
the variable
resistor
has
a
1.8 Ω.
[2]
potential
difference
d
the
potential
difference
e
the
current flowing
f
the
power
across
across
through
dissipated
in
the
the
the
the
3 Ω
2 Ω
1 Ω
2 Ω
resistor
resistor
resistor
resistor.
a
Calculate
the
resistance
b
Calculate
the
current
c
Calculate
the
potential
of
wire XY.
[3]
[2]
I.
[2]
[2]
[2]
[2]
length
d
State
of
the
wire XJ.
value
of
difference
across
the
[2]
E.
[1]
27
4
Electric fields
4.
1
Electric fields
Learning outcomes
Electric fields
A
On
completion
of
this
section,
charged
charged
should
be
able
body
explain
what
is
meant
by
in
ᔢ
define
ᔢ
state
an
electric
inside
the
force
the
field.
For
example,
electric
electric field
this
field
electric
around
field
it
itself.
When
experiences
a
another
is
determined
by
the
type
of
charge
force.
being
The
placed
field
around
within
that
it.
an
isolated
When
field
it
an
will
positive
object
charge
having
experience
a
a
+ Q
will
positive
repulsive
have
charge
force.
The
an
+ q
is
direction
lines
of
electric field
and
placed
of
placed
sketch
produces
an
electric field
ᔢ
is
to:
direction
ᔢ
body
you
the
field
is
therefore
pointing
away
from
the
positive
charge
+ Q
strength
apply Coulomb’s
law.
Definition
An
a
Definition
electric field
charged
body
is
a
region
where
around
a force
is
experienced.
The
direction
any
point
on
a
is
small
that
of
the
an
electric field
direction
positive
of
charge
at
the force
placed
at
point.
+
Figures
An
the
Figure 4.1.1
4.1.1–5
isolated
show
positive
isolated
various
charge
–
positive
charge.
negative
charge
electric
the
fields
field
is
and
radial
their
and
directions.
pointing
away
from
Electric field due to an
An
isolated
–
the
field
is
radial
and
pointing
towards
the
isolated positive charge
isolated
T
wo
negative
like
charge.
charges
T
wo
unlike
charges
–
+
+
Figure 4.1.2
+
–
Electric field due to an
isolated negative charge
Figure 4.1.3
Figure 4.1.4
Electric field due to two like
Electric field due to two
unlike charges
charges
+
A
pair
ends
of
of
parallel
the
plates
plates
Electric field
the
–
the
field
field
between
becomes
strength
the
plates
is
uniform.
At
the
curved.
E
Definition
Equation
–
F
The
electric field
strength
E
at
a
E
=
Q
Figure 4.1.5
Electric field due to a pair of
point
in
an
electric field
is
the force
−1
parallel plates
acting
per
unit
positive
E
–
electric field
F
–
force/N
Q
–
charge/C
strength/N C
charge.
−1
The
SI
unit
Electric
28
of
field
electric
strength
field
is
a
strength
vector
is
the
quantity.
newton
per
coulomb
(N C
).
Chapter
Coulomb’s
The
force
4
Electric fields
law
experienced
when
two
charged
bodies
are
brought
close
to
each
d
other
is
dependent
between
is
them.
inversely
Coulomb’s
on
The
the
force
proportional
law.
Figure
magnitude
is
to
of
their
proportional
square
4.1.6
of
to
the
illustrates
charges
the
product
distance
two
and
the
of
between
bodies
having
separation
their
charges
them.
This
charges
Q
them
F
and
is
Q
and
Q
1
2
1
Q
.
They
are
separated
by
a
distance
d.
The
force
between
is
2
Figure 4.1.6
The
equation
charged
below
is
used
to
calculate
the
force
between
the
two
bodies.
Equation
Q
Q
1
F
2
=
2
4πε
d
0
F
–
Q
force/N
–
charge/C
–
charge/C
1
Q
2
d
–
separation
ε
–
permittivity
between
charges/m
8 cm
−12
of free
space,
8.85
×
−1
10
A
F m
B
0
Q
2
ε
is
a
constant
and
is
called
the
per mittivity
of
free
space .
Its
–
Q
–
3
magnitude
0
has
for
been
an
determined
electric
positive
and
field
experimentally.
to
be
negative,
It
transmitted
the
force
is
a
measure
through
calculated
space.
would
be
of
If
how
the
easy
it
charges
negative.
A
is
were
negative
10 cm
value
for
F
indicates
indicates
that
the
that
force
the
is
force
is
repulsive.
attractive.
Remember
,
A
positive
like
value
charges
for
repel
F
and
Q
+
1
unlike
charges
attract
each
other
.
C
Figure 4.1.7
Example
Figure
4.1.7
shows
three
point
charges
located
at
the
corners
of
a
right-
Q
2
F
angled
triangle
BAC
2
where:
–3 μC
θ
Q
=
+4 μC,
Q
1
=
−3 μC
and
Q
2
=
−4 μC
3
Calculate:
F
R
a
the
magnitude
of
the
force
F
acting
on
Q
1
b
the
magnitude
of
the
force
c
the
magnitude
of
the
resultant
F
,
due
to
Q
2
acting
on
Q
2
F
alone
1
1
,
due
to
Q
2
alone
3
Figure 4.1.8
force
acting
on
Q
,
due
to
2
Q
Q
1
a
F
–6
2
4πε
1
(+4
2
=
×
10
×
10
=
×
Q
3
Key points
)
–12
4π(8.85
and
1
–6
)(–3
=
d
0
Q
10
–2
)(6
×
10
−30.0 N
2
)
ᔢ
Q
Q
2
b
F
–6
3
=
2
2
4πε
(–3
×
10
×
10
)
=
=
d
–12
4π(8.85
0
×
10
–2
)(8
×
10
Resultant
force
acting
on
Q
=
2
angle
the
resultant
force
is
a
region
(–30.0)
√
around
a
force
experienced.
makes
with
charged
body
where
a
16.9 N
2
is
2
+
(16.9)
=
34.4 N
ᔢ
The
a
The
electric field
)
2
c
An
–6
)(–4
the
electric field
point
in
an
strength
electric field
E
is
at
the
horizontal
force
acting
per
unit
positive
30.0
−1
=
tan
=
60.6°
charge.
16.9
(See
Figure
4.1.8.)
ᔢ
Coulomb’s
between
to
the
and
to
is
the
law
states
charges
product
of
inversely
square
between
is
the force
proportional
their
charges
proportional
of
the
distance
them.
29
4.2
Electric field
Learning outcomes
strength
Electric field
Consider
On
completion
of
this
section,
be
able
electric
strength due to
isolated
point
charge
+ Q.
a
point
The
field
potential
charge
lines
around
the
point
charge
you
are
should
an
and
shown
in
Figure
4.2.1.
A
small
positive
charge
+q
is
moved
from
an
to:
infinite
distance
to
a
point
P
,
which
is
at
a
distance
r
from
the
point
charge.
Q
ᔢ
recall
and
use
E
=
2
4πε
The
r
electric
field
radiates
outwards
from
the
positive
charge
+ Q.
The
0
ᔢ
define
electric
direction
of
the
field
would
repelled.
is
outward,
because
the
small
positive
charge
+ q
potential
be
Q
ᔢ
recall
and
use
V
=
4πε
r
The
force
exerted
on
the
small
positive
charge
at
P
can
be
found
using
0
Coulomb’s
law:
Qq
F
=
2
+Q
4πε
r
0
P
F
Electric
field
strength
E
is
defined
as
the
force
per
unit
positive
charge:
E
F
+q
E
=
Q
Therefore
the
electric
field
strength
at
the
point
P
is
given
by
2
Qq/4πε
r
Q
0
E
Figure 4.2.1
r
=
=
2
4πε
q
Electric field strength due to
r
0
a point charge
Electric
field
strength
is
a
vector
quantity.
Equation
Electric
Electric field
strength
due
to
a
Consider
charge
is
given
potential due to
a
point
charge
point
an
isolated
point
charge
+ Q.
The
field
lines
around
the
point
charge
by
are
shown
in
Figure
4.2.2.
A
small
positive
charge
+q
is
moved
from
an
Q
E
infinite
=
distance
to
a
point
P
,
which
is
at
a
distance
r
from
the
point
charge.
2
4πε
r
0
−1
E
–
electric field
Q
–
point
ε
–
permittivity
strength/N C
+Q
charge/C
P
W
of free
space,
0
−12
8.85
r
–
×
−1
10
distance from
∞
+q
F m
point
charge/m
r
Definition
Figure 4.2.2
Electric potential due to a point charge
W
The electric potential at a point in an
The
electric
potential
V
at
the
point
P
is
defined
as
V
=
,
where
W
is
Q
electric field is numerically equal to
the work done in bringing unit positive
the
P
.
work
The
done
SI
unit
in
of
bringing
electric
unit
positive
potential
is
charge
the
volt
from
infinity
to
the
point
(V).
charge from infinity to that point.
Electric
Equation
at
a
potential
point
potentials
Electric
potential
at
a
point
due
to
charge
is
given
to
due
to
is
a
scalar
several
each
quantity.
point
This
charges
is
means
equal
that
potential.
line
are
equidistant
from
a
point
charge
A
(Figure
line
drawn
through
these
points
is
4.2.3).
r
0
V
–
electric
Q
–
point
ε
–
permittivity
potential/V
Example
charge/C
Figure
of free
4.2.4
shows
two
point
charges
A
and
space,
0
−12
8.85
r
–
×
10
30
Calculate:
−1
F m
distance from
the
potential
algebraic
sum
of
are
at
the
same
electric
by
=
4πε
that
the
charge.
Q
V
to
a
Points
point
due
point
charge/m
a
the
force
between
the
two
point
charges
B.
called
an
equipotential
Chapter
b
the
c
the
d
the
e
the
electric
field
strength
at
the
point
X
due
to
the
point
charge
A
f
the
electric
field
strength
at
the
point
X
due
to
the
point
charge
B
g
the
electric
field
strength
at
the
point
X
due
to
the
two
electric
electric
potential
at
the
point
X
due
to
the
point
charge
A
potential
at
the
point
X
due
to
the
point
charge
B
4
Electric fields
only
only
+Q
electric
potential
at
the
point
X
due
to
the
two
point
charges
A
and
B
only
only
equipotential
point
charges
lines
A
and
B.
Figure 4.2.3
Q
a
Force
between
A
and
B
Equipotential lines
Q
1
2
=
2
4πε
d
0
–6
(20
×
10
–6
)(–30
×
10
)
A
=
=
–12
4π(8.85
×
10
–2
)(10
×
10
B
X
−540 N
2
)
10 cm
The
negative
sign
indicates
that
the
force
is
attractive.
+20 μC
b
Electric
potential
at
the
point
X
due
to
the
point
charge
20
×
–30 μC
A
Figure 4.2.4
–6
Q
10 cm
10
5
=
=
=
4πε
–12
r
4π(8.85
0
c
Electric
potential
at
×
10
the
8.99
×
10
V
–2
)(20
point
X
×
10
due
)
to
the
point
charge
B
–6
Q
–30
×
10
6
=
=
4πε
=
–12
r
4π(8.85
0
d
Electric
at
the
sum
potential
point
of
the
electric
X
is
due
×
a
the
10
quantity.
two
point
potentials
potential
×
the
point
electric
due
Therefore,
charges
to
each
field
V
potential
is
the
equal
electric
to
the
potential
algebraic
charge.
the
point
X
due
at
to
Electric
A
the
5
Electric
10
potential
8.99
strength
at
×
X
due
Q
+
point
(−2.70
X
due
×
to
20
×
10
6
)
=
charge
−1.80
×
10
Electric
=
–12
field
direction,
repelled
at
a
by
the
Electric
4π(8.85
×
10
strength
is
a
small
the
X
strength
to
at
×
A.
is
In
placed
charge
point
X
4.50
×10
−1
N C
2
)
Therefore
point
the
10
quantity.
charge
charge
due
–2
)(20
vector
positive
point
point
field
the
the
A
due
order
at
is
to
to
determine
point
direction
to
the
X.
of
It
will
the
Q
30
×
right.
charge
B
7
Electric
=
–12
field
direction,
attracted
at
be
electric
10
=
2
r
0
field
its
–6
=
4πε
V
A
6
f
B
10
=
2
r
0
field
to
6
10
the
point
–6
=
4πε
at
+
X
=
e
×
)
=
at
−2.70
–2
)(10
scalar
to
electric
10
a
×
10
strength
is
a
small
to
the
4π(8.85
the
vector
positive
point
point
X
to
×
B.
10
is
charge
A
In
placed
Therefore
point
−2.70
×
10
−1
N C
2
)
quantity.
charge
charge
due
–2
)(10
B
order
at
the
the
is
to
determine
point
X.
direction
to
the
of
It
will
the
its
Key points
be
electric
ᔢ
left.
B
per
unit
strength
is
the force
charge.
X
ᔢ
E
The
in
E
B
+20 μC
Electric field
electric
an
potential
electric field
is
at
a
point
numerically
A
–30 μC
equal
to
the
work
done
in
Figure 4.2.5
bringing
g
Since,
due
electric
to
both
field
point
strength
charges
is
can
a
vector
be
quantity,
found
by
the
finding
combined
the
vector
from
effect
sum
individual
field
infinity
positive
to
that
charge
point.
of
ᔢ
the
unit
Electric
potential
is
a
scalar
strengths.
quantity.
The
electric
field
strength
at
the
point
X
due
to
both
point
charges
is
ᔢ
7
=
2.70
×
Electric field
strength
is
a
vector
6
10
−
4.50
×
10
quantity.
7
=
2.25
×
10
−1
N C
ᔢ
The
electric
field
strength
at
the
point
X
due
to
B
is
larger
than
An
equipotential
through
electric
field
strength
at
the
point
X
due
to
A.
Therefore,
the
is
a
line
drawn
the
points
having
equal
direction
potentials.
of
the
resultant
electric
field
strength
at
the
point
X
is
acting
to
the
left.
31
4.3
Relationship
Learning outcomes
On
completion
should
be
able
of
this
between
The
section,
Consider
you
The
to:
recall
that
the field
strength
A
point
equal
in
to
a field
the
is
ᔢ
B
in
the
field
is
E.
and
between
Consider
uniform
V
a
a
electric
pair
of
charge
field
parallel
+Q
by
a
metal
being
force
plates.
moved
F,
as
from
shown
in
4.3.1.
gradient
at
The
distance
assumed
between
to
be
A
and
constant.
the formula for finding
electric field
strength
B
( Δx)
The
is
very
work
small
done
by
so
the
that
force
the
force
F
can
is
charged
compare
parallel
motion
in
with
=
FΔx
(1)
by
F
=
−EQ
(2)
definition,
plates
an
The
electric field
ΔW
between
But,
ᔢ
point
electric
strength
E
point
recall
two
to
field
between
V
numerically
potential
be
that
uniform
and
at
Figure
a
a
electric
point
ᔢ
relationship
E
motion
in
minus
sign
is
required
because
work
has
to
be
done
against
the
a
electric
field
in
order
to
move
the
charge
from
A
to
B.
gravitational field.
Substituting
Equation
∴
(1)
ΔW
into
=
Equation
(2)
−EQΔx
(3)
E
The
Let
point
the
B
is
at
potential
a
higher
electric
difference
potential
between
A
and
than
B
be
the
point
A.
ΔV
F
∴
+
ΔW
=
QΔV
(4)
–
Equating
Equation
(3)
and
Equation
− EQΔx
=
QΔV
E
=
−
(4)
ΔV
∴
(5)
Δx
ΔV
Figure 4.3.1
is
called
the
potential
gradient.
Since
this
expression
can
be
used
to
Δx
find
the
electric
field
strength,
10 mm
another
unit
that
can
be
used
for
electric
−1
field
The
strength
electric
is
the
field
V m
strength
numerically
equal
to
Figure
shows
the
at
a
point
potential
in
an
electric
gradient
at
field
that
is
therefore
point.
C
+ V
4.3.2
two
parallel
plates
separated
by
a
distance
of
60 mm.
0V
The
potential
represent
difference
between
equipotentials.
Points
the
plates
along
any
is
V.
The
dashed
dashed
line
have
lines
the
same
A
B
potential.
Therefore,
the
potential
difference
between
A
and
C
is
zero.
V
The
potential
difference
between
each
broken
line
is
volts.
6
60 mm
V
Therefore,
the
potential
difference
between
A
and
B
is
V
×
3
=
volts.
6
2
Figure 4.3.2
V
The
potential
difference
between
C
and
B
is
also
volts.
2
+
The
d
electric field
Consider
the
electric
between two
field
produced
parallel
between
two
charged
parallel
plates
plates
as
shown
V
in
Figure
The
–
separation
across
the
unifor m.
Figure 4.3.3
32
4.3.3.
between
plates
At
the
is
V.
ends
the
plates
Along
of
the
the
is
d
and
middle
plate
the
of
the
the
electric
potential
plates
field
the
difference
electric
becomes
field
is
non-unifor m.
Chapter
The
electric
voltage
The
field
across
the
following
between
two
in
the
middle
plates
equation
parallel
V
is
and
the
used
plate
section
to
of
the
plate
separation
determine
of
the
is
the
dependent
plates,
electric
on
4
Electric fields
the
d
field
strength
conductors.
Equation
V
E
=
–
d
Exam tip
−1
E
–
electric field
V
–
potential
strength/V m
difference
across
The
plates/V
in
d
–
separation
between
minus
minus
force
F
are
Suppose
metal
it
sign
an
with
a
accelerate
cur ved
the
in
of
a
as
it
Once
the
the
velocity
path
projected
v.
is
v
straight
force
is
before
V
and
the
plate.
through
with
is
keep
in
ignored
mind
electric
the
the
the
potential
electric force
difference
act
in
and
opposite
the
the
plates
is
zero.
the
the
electric
the
velocity
parallel
entering
(It
it
was
electron
is
fi eld.
projected
causes
not
The
it
cur ved
fi eld
possessed
it
to
of
acting
electron
The
electric
fi eld
the
component
force
unchanged.
outside
of
electric
horizontal
because
pair
the
on
The
a
enters
velocity
acting
therefore
electron
line
difference
between
electron
Just
unaffected
passes
the
sometimes
directions.
the
electron’s
positive
and
potential
horizontally
As
force.
downward
direction
parabolic.
is
the
directions.
velocity
downward
The
electron’s
moving
a
towards
horizontal
because
opposite
component
horizontally.)
is
in
electron
experiences
the
necessary
acting
plates
vertical
a
is
is
but
plates/m
that
The
sign
calculations,
in
a
follows
path
continues
coming
out
of
fi eld.
v
m
h
–
Electron
travels
in
v
a
the
electron
path
remains
h
v
parabolic
=
0
v
inside
constant
electric field.
v
increases
v
parabolic
path
+
Electron
straight
exits
travels
line
the
in
when
a
it
electric field.
Figure 4.3.5
Figure 4.3.4
The
path
taken
by
a
charged
particle
as
it
travels
A mass travelling in a
gravitational field
A charged particle travelling in an electric field
in
a
uniform
electric
Key points
field
can
be
compared
gravitational
field.
to
the
Suppose
a
path
taken
mass
m
is
by
a
mass
projected
travelling
in
horizontally
a
with
a
ᔢ
velocity
v
.
The
initial
vertical
component
of
its
velocity
v
h
force
of
force
causes
is
zero.
The
electric field
at
point
v
gravity
acts
vertically.
It
exerts
a
force
of
mg
on
the
mass.
the
mass
to
accelerate
vertically.
The
vertical
a
component
v
of
the
mass
therefore
increases.
The
horizontal
an
electric field
numerically
equal
to
the
of
potential
velocity
in
This
is
the
strength
The
gradient
at
that
point.
component
v
of
its
velocity
v
remains
constant
(assuming
air
resistance
is
negligible)
h
since
the
mass
is
force
of
gravity
does
not
act
horizontally.
The
path
taken
by
ᔢ
the
The
path
particle
taken
by
travelling
a
in
charged
a
uniform
parabolic.
electric field
ᔢ
The
path
travelling
is
parabolic.
taken
in
a
by
a
mass
uniform
gravitational field
is
parabolic.
33
4.4
Worked
Learning outcomes
examples
completion
of
this
section,
flat
be
able
metal
solve
of
involving
by
a
distance
of
1.2 cm
as
shown
in
charge
the
electric
field
strength
between
the
plates.
the
particles
in
Potential
uniform
separated
to:
problems
motion
are
4.4.1.
Calculate
ᔢ
plates
you
Figure
should
electric fields
Example
T
wo
On
on
difference
between
the
plates
is
100
−
(−100)
=
200 V
electric fields.
(Note
that
the
potential
difference
is
not
0 V
.)
V
Electric
field
strength
E
=
–
d
200
+100 V
=
–
–2
1.2
×
10
4
=
−1.67
×10
−1
V m
1.2 cm
Example
–100 V
T
wo
flat
1.5 cm
metal
as
plates,
shown
in
each
Figure
of
length
4.4.2.
A
4.0 cm,
are
potential
separated
difference
of
by
a
distance
110 V
of
between
Figure 4.4.1
the
plates
provides
a
uniform
electric
field
7
plates.
to
the
the
Electrons
field.
field
of
speed
Assume
outside
the
4.0
that
the
plates
×
10
space
is
in
the
region
between
the
−1
m s
enter
between
the
this
region
plates
is
a
at
right
vacuum
angles
and
zero.
+
7
–1
m
4 × 10
s
1.5 cm
110 V
electrons
–
4.0 cm
Figure 4.4.2
Calculate:
a
the
electric
b
the
force
c
the
acceleration
d
the
time
e
the
speed
motion
f
the
field
on
an
taken
of
as
velocity
electron
of
for
the
it
strength
the
an
of
due
an
to
electron
electron
electron
leaves
between
the
at
electron
the
as
plates
electric
along
to
right
region
the
the
travel
it
leaves
direction
between
angles
between
field
to
its
the
the
on
an
electron
=
−1.6
×
10
original
region
of
an
electron
=
9.11
×
10
C,
kg)
V
a
Electric
field
strength
E
=
–
d
110
=
–
–2
1.5
×
10
3
=
34
−7.33
×
10
electric
direction
between
−31
mass
the
field
plates
of
plates
−19
(Charge
of
the
−1
V m
the
plates.
Chapter
b
Force
on
an
electron
F
=
EQ
=
−7.33
=
1.17
3
4
Electric fields
−19
×10
×
−1.6
×
10
−15
c
Using
Newton’s
second
law
F
×
=
10
N
ma
F
Acceleration
along
the
direction
of
the
electric
field
a
=
m
–15
1.17
×
10
9.11
×
10
=
–31
15
=
1.28
×
10
−2
m s
7
d
The
horizontal
There
is
no
electrons
velocity
component
force
in
this
acting
in
direction
remains
of
the
this
is
electron’s
direction,
zero.
The
velocity
hence
the
horizontal
is
4.0
×
−1
10
m s
acceleration
component
of
of
.
the
the
constant.
1
2
Using
s
=
ut
+
at
2
s
=
ut,
since
a
=
0
–2
s
4
×
10
−9
t
=
=
=
1
×
10
s
7
u
4
×
10
−9
Time
e
The
taken
to
electron
travel
is
between
accelerating
the
plates
vertically
is
at
a
1
×
10
s.
constant
rate.
The
initial
6
1.28 × 10
–1
m
s
v
vertical
component
Using
of
v
the
electron’s
=
u
+
at
=
0
+
(1.28
velocity
is
zero.
15
×
6
=
f
The
vertical
component
6
is
1.28
×
10
Magnitude
1.28
of
×
×1
×
10
)
−1
10
the
−9
10
m s
electron’s
velocity
as
it
leaves
the
plates
−1
m s
of
θ
electron’s
velocity
as
it
leaves
the
plates
7
4 × 10
6
(1.28
=
√
×
10
2
7
)
+
Angle
the
velocity
×
10
s
)
Figure 4.4.3
7
=
(4
–1
m
2
4.00
makes
×
−1
10
with
m s
the
horizontal
6
1.29
–1
=
tan
10
(
7
4
=
×
×
)
10
1.85°
Example
An
isolated
Sketch
a
electric
Inside
zero
to
potential
the
of
from
a
For
<
the
point
R,
E
V
there
inside
charge
0
the
with
sphere
=
sphere
show
sphere,
everywhere
surface
r
conducting
graph
and
are
the
no
is
r
centre
a
r
>
R,
E
πε
If
of
the
The
the
the
constant.
has
a
centre
if
lines
they
charge
strength
of
electric
field
as
positive
field
the
field
are
were
E
R
+ Q.
and
sphere.
strength
drawn
being
E
from
is
the
E
produced
sphere.
0
r
Q
=
4
R
electric
from
appear
Q
For
of
charges.
sphere.
would
the
V
radius
distance
they
at
of
variation
2
r
0
and
V
=
4
πε
r
0
V
0
r
Figure 4.4.4
The variation of E and V
from the centre of an isolated charged
hollow sphere
35
5
Capacitance
5.
1
Capacitance
Learning outcomes
Capacitors
Capacitors
On
completion
of
this
section,
They
should
be
able
consist
state
the function
of
a
define
ᔢ
recall
capacitance
solve
called
and
the farad
problems
paper
.
5.1.1).
are
and
two
components
sheets
of
a
found
conducting
in
many
material
electrical
separated
devices.
by
a
dielectric.
Commonly
used
dielectrics
are
air
,
an
oil
capacitor
and
ᔢ
electrical
of
to:
insulator
,
ᔢ
are
you
A
also
The
two
capacitor
used
in
conducting
is
a
device
electrical
sheets
that
circuits
are
stores
to
attached
electric
prevent
the
to
leads
charge.
flow
of
(Figure
Capacitors
direct
currents
using
(d.c.).
Capacitors
are
widely
used
in
digital
cameras.
A
capacitor
inside
Q
C
=
the
camera
is
charged
up
using
a
battery.
Once
charged,
it
is
made
to
V
discharge
through
Electronic
metal
plates
following
‘Danger:
back
lead
of
a
bulb
equipment
warning
Even
this
and
such
as
message
when
a
bright
flash
cameras
at
the
switched
and
back
off,
it
is
of
may
produced.
television
sets
have
the
them:
be
dangerous
to
remove
the
equipment’.
lead
Inside
store
these
large
equipment,
switched
in
types
these
off,
them.
of
amounts
the
electrical
of
capacitors
capacitors
Opening
equipment,
electric
the
charge.
become
may
back
of
charged.
still
the
some
During
have
the
Even
some
equipment
of
normal
if
capacitors
operation
the
electric
exposes
of
the
equipment
charge
the
user
is
stored
to
the
dielectric
capacitors.
Figure 5.1.1
The parts of a capacitor
capacitor
The
In
to
circuit
order
across
for
it.
discharge
symbol
a
is
the
applied
touching
through
for
capacitor
Consider
difference
Figure 5.1.2
Accidentally
a
the
the
capacitor
to
store
circuit
across
a
leads
user ’s
is
a
in
capacitor
the
body,
shown
charge,
shown
of
in
causing
Figure
potential
Figure
an
capacitor
cause
electric
the
shock.
5.1.2.
difference
5.1.3.
electric
an
will
When
current
must
a
be
applied
potential
flows
momentarily
The symbol for a capacitor
in
the
from
circuit.
the
charge
Electrons
opposite
that
flows
flow
plate.
to
the
towards
Charge
plate
is
on
one
plate,
conserved
the
right
is
in
while
this
− Q,
electrons
process.
then
the
If
flow
the
charge
away
amount
that
of
flows
electron
to
V
+
the
left
plate
is
+Q.
The
total
charge
In
order
called
more
to
measure
capacitance
charge
capacitor
the
can
the
is
ability
used.
capacitor
store
of
The
is
depends
a
in
capacitor
greater
capable
on
the
of
the
the
to
capacitor
store
value
storing.
voltage
Definition
+Q
stored
is
Q
flow
–
of
The
applied
charge,
the
a
quantity
capacitance,
amount
across
of
the
charge
a
it.
Equation
Q
electron
The
capacitance
of
a
capacitor
Q
is
C
flow
the
Figure 5.1.3
charge
stored
per
unit
unit
1
farad
1
volt
if
is
of
capacitance
the
charge
applied
is
the
stored
across
is
farad
1
it.
36
=
C
–
Q
–
charge/C
V
–
potential
(F).
coulomb
−1
1 F
V
potential
difference.
The
=
1 C V
A
capacitance/F
capacitor
when
a
difference/V
has
a
potential
capacitance
difference
of
of
Chapter
The
of
farad
is
a
large
microfarads
T
ypically,
of
maximum
the
capacitor
.
charge
Factors
or
are
the
the
allow
( μF)
there
capacitance
unit
If
typical
picofarads
two
values
capacitor
.
allowable
to
this
flow
have
printed
The
is
values
on
other
the
a
capacitors
capacitor
.
value
difference
exceeded,
between
of
are
of
the
Capacitance
order
(pF).
potential
value
affecting
Experiments
and
5
the
is
a
One
voltage.
that
can
dielectric
be
value
This
is
applied
will
the
voltage
break
is
across
down
and
plates.
capacitance
shown
that
the
capacitance
of
a
capacitor
depends
on
d
the
cross-sectional
the
material
quantities
used
are
area
as
of
the
related
the
plates,
dielectric.
to
the
The
separation
equation
of
the
below
plates
shows
cross-sectional
and
how
area
A
these
capacitance.
Equation
ε
C
A
=
d
C
–
capacitance/F
ε
–
permittivity
A
–
cross-sectional
dielectric
−1
of
the
material/F m
2
Figure 5.1.4
area/m
Parameters affecting
capacitance
d
–
The
separation
between
permittivity
electric
field.
of
The
a
plates/m
material
relative
is
a
measure
permittivity
of
a
of
its
ability
material
ε
to
transmit
(also
known
an
as
its
r
dielectric
constant)
permittivity
constant
T
able
of
of
is
free
the
space.
vacuum
is
ratio
It
1.0.
is
of
a
the
permittivity
dimensionless
Some
common
of
the
material
quantity.
dielectrics
The
are
to
the
dielectric
shown
in
Table 5.1.1
5.1.1.
Values of some dielectric
constants
Equation
Dielectric
Dielectric
Air
1.0006
Paper
2–5
Glass
4.5–8
Oil
2.5
Ceramic
45–6000
constant
ε
ε
=
ε
r
0
−1
ε
–
permittivity
of
the
ε
–
permittivity
of free
–
relative
material/F m
−1
space/F m
0
ε
permittivity
of
the
material
(no
units)
r
When
a
dielectric
is
molecules
inside
it
across
plates
to
the
placed
between
become
the
polarised.
decrease.
As
a
plates
This
result,
of
an
causes
the
isolated
the
capacitor
,
electric
capacitance
of
the
potential
the
capacitor
1
increases
with
the
presence
of
a
dielectric
(C
∝
).
V
Key points
ᔢ
A
capacitor
ᔢ
The
ᔢ
A
capacitor
The
a
device
capacitance
when
ᔢ
is
a
has
a
potential
capacitance
plates,
of
the
a
that
stores
capacitor
capacitance
difference
of
a
of
of
capacitor
separation
of
the
is
1
electric
the
charge
1 farad
volt
is
if
stored
the
and
on
the
per
charge
applied
depends
plates
charge.
the
across
unit
voltage.
stored
is
cross-sectional
material
1
coulomb
it.
used
as
area
the
of
the
dielectric.
37
5.2
Charging
Learning outcomes
and
Charging
A
On
completion
of
this
section,
capacitor
be
able
recall
for
and
capacitor
capacitor through
can
be
charged
by
a
connecting
resistor
a
resistor
as
shown
in
Figure
5.2.1.
it
to
a
When
power
the
source
switch
S
in
is
series
at
position
to:
A,
ᔢ
a
a
you
with
should
discharging
use
charging
the
and
a
current
begins
to
flow
and
the
capacitor
C
begins
to
charge.
equations
discharging
A
a
B
capacitor
S
ᔢ
understand
constant
and
the
during
discharging
through
a
concept
the
of
of
time
charging
a
R
capacitor
V
R
resistor.
V
S
V
C
Figure 5.2.1
V
is
the
C
Charging and discharging a capacitor through a resistor
e.m.f.
of
the
d.c.
power
supply.
V
s
is
the
potential
difference
C
across
the
capacitor
C
and
V
is
the
potential
difference
across
the
R
resistor
varies
R.
Figure
with
time
5.2.2
as
the
illustrates
capacitor
how
the
charge
stored
in
the
capacitor
charges.
charge/
C
current/
A
V
/ V
C
Q
0
I
0
V
S
t
CR
Q = Q
(1– e
)
t
0
t
CR
V
C
CR
I = I
= V
(1– e
)
S
e
0
time/
s
time/
s
time/
s
Figure 5.2.2
Charge Q against time
Figure 5.2.3
I against time
Figure 5.2.4
V
against time
C
V
/ V
R
V
0
Figure
5.2.3
period
of
Figure
5.2.4
the
illustrates
time.
The
illustrates
capacitor
as
it
what
current
happens
to
approaches
what
the
current
in
the
circuit
over
zero.
happens
to
the
potential
difference
across
happens
to
the
potential
difference
across
charges.
t
–
CR
V
R
= V
e
0
Figure
the
time/
s
Figure 5.2.5
V
R
38
against time
5.2.5
resistor
illustrates
as
the
what
capacitor
is
being
charged.
a
Chapter
Discharging
When
B.
the
The
how
with
capacitor through
capacitor
capacitor
the
Figure
a
charge
5.2.7
C
will
in
fully
charged,
discharge
the
shows
is
through
capacitor
how
the
the
varies
current
a
the
flowing
Capacitance
resistor
switch
with
5
S
is
resistor
time
moved
R.
as
it
through
Figure
to
position
5.2.6
shows
discharges.
the
resistor
varies
time.
charge/C
current /
A
time/
s
Q
0
0
t
CR
Q = Q
e
0
I
0
time/
s
Figure 5.2.6
The
graph
process.
opposite
Time
The
=
to
shows
This
is
the
to
Figure 5.2.7
current
because
direction
the
that
of
as
being
current
the
negative
now
charging
I against time
during
flows
the
through
discharge
the
resistor
in
the
current.
constant
time
capacitor
τ
Charge against time
CR.
constant
to
fall
The
The
of
1/e
larger
discharge.
discharging
to
a
circuit
(0.368)
the
time
value
is
of
of
the
its
R,
constant
time
initial
the
can
taken
value.
longer
be
it
for
the
The
takes
determined
charge
time
for
the
using
on
a
constant
a
is
capacitor
charging
or
curve.
t/CR
A
charging
curve
is
of
the
form
x
=
x
(1
–
e
)
0
where
x
represents
charge
or
potential
difference.
t/CR
A
discharge
curve
is
of
the
form
x
=
x
e
0
where
Figure
from
a
x
represents
5.2.8
and
charging
charge,
Figure
or
current
5.2.9
show
discharging
or
potential
how
the
difference.
time
constant
is
determined
curve.
charge
initial
discharge
rate
Q
0
time
constant
0.368 Q
0
0
Figure 5.2.8
τ = CR
time
Graphical determination of the time constant using a discharging curve
39
Chapter
5
Capacitance
charge
Q
0
0.63 Q
initial
charging
rate
0
time
0
Figure 5.2.9
In
The
respect
time
experiments,
to
time
constant
illustrated
τ = CR
time
Graphical determination of the time constant using a charging curve
laboratory
with
constant
in
currents
during
can
Figures
be
the
measured
5.2.8
and
and
charging
voltages
or
are
usually
discharging
graphically
using
of
the
a
measured
capacitor
.
same
technique
5.2.9.
Example
A
resistor
of
capacitance
resistance
resistance
12.0 μF
(Figure
2.2 k Ω
and
a
is
connected
battery
of
e.m.f.
in
series
20.0 V
with
with
a
capacitor
negligible
of
internal
5.2.10).
12.0 μF
2.2 kΩ
20.0 V
Figure 5.2.10
The
a
b
capacitor
initially
uncharged.
Calculate:
i
the
current
ii
the
final
charge
iii
the
time
constant
The
with
battery
time
[Assume
i
=
at
circuit
on
the
of
switch
in
time
the
t
immediately
after
the
switch
is
closed
capacitor
the
are
circuit.
replaced
Figure
=0,
by
a
voltage
source
that
varies
5.2.11.
the
capacitor
is
uncharged.]
potential
difference
across
the
capacitor
when
potential
difference
across
the
capacitor
when
0.04 s.
the
0.08 s.
Draw
the
the
shown
that
Calculate
t
iii
=
in
and
as
Calculate
t
ii
40
is
a
sketch
capacitor
to
show
with
the
time.
variation
of
potential
difference
across
Chapter
5
Capacitance
potential
difference/ V
20
10
time /s
0
0
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
Figure 5.2.11
V
a
i
I
20.0
=
=
=
9.1 mA
3
R
2.2
×
10
−6
ii
Q
=
CV
=
12.0
×
10
×
20
=
0.24 mC
−6
iii
τ
=
CR
=
12
×
3
10
×
2.2
t/CR
b
i
V
=
V
(1
–
e
×
10
=
0.026 s
–0.02/0.026
)
=
20(1
–
e
)
=
10.7 V
0
t/CR
ii
V
=
V
–0.04/0.026
e
=
10.7e
=
2.30 V
0
iii
Between
t
potential
Between
t
potential
Between
=
0 s
and
difference
=
0.02 s
t
=
0.04 s
=
and
difference
potential
t
0.02 s,
across
t
=
across
and
difference
it
t
capacitor
the
reaches
0.08 s,
it
is
uncharged
and
the
zero.
0.04 s,
it
=
across
the
is
the
reaches
capacitor
charges
up
and
the
10.7 V
.
capacitor
discharges
and
the
2.30 V
.
Between
t
=
0.08 s
and
t
=
0.14 s,
the
capacitor
charges
Between
t
=
0.14 s
and
t
=
0.16 s,
the
capacitor
discharges.
up.
potential
difference/ V
20
10
time /s
0
0
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
potential
difference/ V
time /s
Figure 5.2.12
Key points
ᔢ
A
capacitor
can
be
charged
or
discharged
through
a
resistor.
–t/CR
ᔢ
A
charging
curve
is
of
the form
x
=
x
(1
–
e
)
where
x
represents
charge
or
0
potential
difference.
–t/CR
ᔢ
A
discharging
curve
is
of
the form
x
=
x
e
where
x
represents
charge
or
0
potential
difference.
ᔢ
The
time
constant
of
ᔢ
The
time
constant
τ
capacitor
to fall
to
a CR
of
1/e
a
circuit
circuit
(0.368)
is
of
is
τ
the
its
=
CR.
time
initial
taken for
the
charge
on
a
value.
41
5.3
Capacitors
Learning outcomes
in
series
Capacitors
Consider
On
completion
of
this
section,
be
able
and
of
V
parallel
(Figure
C
5.3.1).
,
C
and
C
2
,
in
series
with
a
d.c.
supply
having
3
The
potential
differences
across
the
capacitors
to:
V
,
V
1
derive
capacitors,
1
e.m.f.
are
ᔢ
in
series
you
an
should
three
in
and
use
an
and
V
2
respectively.
Electrons
from
the
d.c.
supply
flow
towards
3
expression for
the
right-hand
the
left-hand
plate
of
capacitor
C
,
while
electrons
flow
away
from
3
capacitors
in
series
derive
use
hand
ᔢ
and
an
plate
in
derive
the
and
capacitor
C
same
,
time,
while
electrons
electrons
an
flow
flow
The
same
process
occurs
for
capacitor
C
parallel
use
energy
the
towards
away
from
the
the
right-
left-hand
2
.
In
this
process,
charge
is
1
conserved.
ᔢ
At
expression for
plate.
capacitors
of
plate.
Therefore,
the
charge
stored
in
each
capacitor
is
the
same.
expression for
stored
in
a
The
charge
The
three
stored
in
each
capacitor
is
Q
capacitor.
capacitors
capacitance
of
C
can
and
be
a
replaced
charge
of
by
a
single
capacitor
having
a
Q
total
total
V
Using
+
Kirchhoff ’s
second
law
we
can
write
–
V
=
V
+
V
1
+
V
2
(1)
3
Q
C
C
1
But
C
2
Q
=
V
,
V
1
3
and
–
+
–
+
V
=
3
C
1
+
Q
=
2
C
C
2
3
–
Q
Q
total
Therefore,
V
1
Q
+
+
C
total
V
Q
=
C
(2)
C
1
C
2
3
V
2
3
where
Q
=
Q
total
Figure 5.3.1
Capacitors in series
Dividing
Equation
(2)
by
Q,
1
Equation
1
1
=
capacitors
1
1
1
+
C
total
C
1
C
2
3
series:
1
=
C
in
+
C
total
For
1
+
C
Capacitors
in
parallel
+
C
1
C
2
3
Consider
three
capacitors,
C
,
C
1
an
The
V
+
e.m.f.
of
V.
charges
The
potential
stored
in
C
,
C
1
and
C
2
,
in
parallel
with
a
d.c.
supply
having
3
difference
and
C
2
across
are
Q
3
,
each
Q
1
capacitor
and
Q
2
is
therefore
V.
respectively
.
3
–
The
three
capacitors
capacitance
of
C
can
and
be
a
replaced
charge
of
by
a
single
capacitor
having
a
Q
total
total
C
1
Q
=
C
1
+
V
Q
1
=
C
2
V
Q
2
=
C
3
V
3
–
The
total
charge
Q
=
Q
total
+
Q
1
+
Q
2
3
V
1
Q
=
C
total
V
+
C
1
V
+
C
2
V
=
V(C
3
+
C
1
+
C
2
)
3
C
2
Q
total
+
–
Therefore,
C
=
=
total
C
+
C
1
V
+
C
2
3
V
2
Energy
stored
in
a
capacitor
C
3
+
–
When
a
charge.
capacitor
Electrons
is
connected
flow
towards
to
a
one
power
plate,
supply,
while
it
begins
electrons
to
accumulate
flow
away
from
V
3
Figure 5.3.2
Capacitors in parallel
the
opposite
work
on
has
the
to
plate.
be
plate.
As
done
As
more
electrons
against
more
the
electrons
move
repulsive
move
towards
forces
away
of
from
the
the
the
negative
electrons
positive
plate,
already
plate,
work
Equation
has
on
For
capacitors
in
to
the
=
total
C
+
1
C
+
2
plate.
shows
difference
C
against
The
the
work
attractive
how
across
the
done
it.
charge
is
forces
stored
The
stored
energy,
W,
3
equal
42
done
as
of
the
electric
positive
potential
charges
energy.
already
Figure
parallel:
5.3.3
C
be
to
the
area
under
the
graph.
in
a
capacitor
stored
in
the
varies
with
capacitor
is
potential
numerically
Chapter
5
Capacitance
1
Energy
stored
in
a
capacitor
W
QV
=
Q/C
2
But,
Q
=
CV
,
so
1
1
energy
2
W
=
(CV)V
=
stored
in
CV
2
2
the
capacitor
Q
Also,
V
=
,
so
C
2
1
W
=
Q
Q
2
(
1
)
C
Q
0
=
2
V
C
S
V/ V
Figure 5.3.3
Example
An
isolated
25 V
capacitor
across
of
capacitance
250 μF
has
a
potential
difference
of
it.
Equation
a
Calculate:
1
i
the
charge
stored
in
the
W
capacitor
QV
=
2
ii
b
the
An
energy
stored
uncharged
across
the
by
two
the
in
the
capacitor
charged
of
250 μF
capacitors
capacitor
.
capacitance
capacitor
.
after
they
150 μF
is
Calculate
have
been
then
the
total
connected
W
–
energy
stored
in
capacitor/J
Q
–
charge
stored
in
capacitor/C
V
–
potential
connected
energy
and
the
stored
difference
across
potential
capacitor/V
difference
a
i
across
Charge
them.
stored
in
capacitor
−6
Q
=
CV
=
250
×
10
−3
×
25
=
6.25
1
ii
Energy
stored
in
the
capacitor
=
×
10
C
1
2
CV
=
−6
(250
2
×
10
2
)(25)
2
−2
=
b
During
the
process
This
means
total
charge
are
now
that
after
Effective
will
be
charge
connection
in
the
connecting
total
the
connected
capacitor
of
the
parallel,
the
capacitors,
before
is
the
7.81
Also,
potential
10
charge
connection
made.
×
is
since
J
is
conserved.
equal
the
difference
to
the
capacitors
across
each
250 μF
same.
capacitance
after
connection
C
=
C
total
+
150 μF
C
1
=
V
2
250
+
150
=
400
μF
−3
T
otal
charged
stored
after
the
connection
=
6.25
×
10
C
Figure 5.3.4
2
1
–3
Q
1
(6.25
×
10
2
)
−2
T
otal
energy
stored
=
=
×
=
4.88
×
10
J
–6
2
C
2
400
×
10
total
Potential
difference
across
the
capacitors
–3
Q
V
=
6.25
×
10
=
=
15.6 V
–6
C
400
×
10
total
Key points
ᔢ
The
equivalent
1
1
+
C
total
ᔢ
The
+
1
=
total
C
several
capacitors
in
series
is
of
several
capacitors
in
parallel
given
by:
…
C
2
equivalent
C
of
1
=
C
capacitance
+
1
C
capacitance
+
is
given
by:
…
2
1
ᔢ
The
energy
stored
in
a
capacitor
is
given
by:
W
=
QV
2
43
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
2
be
iii
the
accompanying CD.
to
iv
1
a
Explain
what
is
meant
by
an
b
Explain
what
is
meant
by
electric field
electric field.
electric field
B
the
c
electric
State
the
potential
E
[2]
between
the
two
quantities.
Sketch
the
point C
due
strength
at
the
point C
due
and
B
[3]
v
the
electric
potential
at C
due
to A
alone
[2]
vi
the
electric
potential
at C
due
to
alone
[2]
vii
the
electric
potential
at C
due
to A
and
[2]
B
B.
[2]
4
d
the
[1]
strength
V
relationship
at
[2]
electric field
to A
and
strength
alone
a
Two
charges A
and
B
are
separated
by
a
distance
electric field:
of
i
around
an
isolated
positive
ii
around
an
isolated
negative
iii
between
a
positive
and
point
charge
point
negative
point
in
a vacuum
in
the figure
shown
−19
[2]
charge
0.80 nm
below. A
has
a
charge
of
+
4.8
×
10
C
and
B
−19
[2]
has
charge
a
charge
exerted
of
by A
−1.6
on
×
10
C. Calculate
the force
B.
[3]
[2]
iv
between
metal
2
two
oppositely
charged
parallel
A
plates.
a
State Coulomb’s
b
The
diagram
B
[2]
law.
below
P
[3]
shows
three
point
charges.
b
Sketch
the
4 cm
the field
direction
of
lines
between A
and
B,
including
the field.
[3]
+2 μC
c
Make
+4 μC
a
arrow
sketch
to
of
the
points A,
B
and
P. Use
an
show:
2 cm
i
the
electric field
the
charge A
the
electric field
strength
at
the
point
P
due
only
to
[1]
–5 μC
ii
charge
B
strength
at
the
point
P
due
P
due
only
to
[1]
Calculate:
iii
i
the force
acting
on
the
+2 μC
charge
due
the
resultant
the
iii
+4 μC
the force
charge
acting
alone
on
the
−5 μC
the
resultant force
due
to
charge
the
+4 μC
the
strength
at
to
to
both
ii
electric field
charges.
[2]
[2]
+2 μC
charge
due
alone
[2]
acting
charge
5
to
on
and
the
the
+2 μC
−5 μC
a
Define
b
A
charge
the
capacitor
charged
charge.
term
plates
with
until
is
capacitance.
a
the
5.0 V.
[1]
capacitance
potential
of
2200 μF
difference
is
between
its
Determine:
[2]
3
a
Explain
and
what
is
meant
by
the
terms
electric field
electric field strength
Sketch
the field
charges
c
a
around
distance
The
diagram
and
B.
d
below
two
similar
positive
charge
the
energy
[2]
two
point
A
capacitor
is
220 μF. There
apart.
shows
the
ii
on
one
stored
of
in
the
the
plates
[2]
capacitor.
[2]
[2]
6
b
i
charges A
of
a
marked
is
as
another
having
a
marking
capacitance
indicating
of
a voltage
18 V.
Explain
what
is
meant
by
a capacitance of 220 μF.
[1]
b
B
A
4 mm
+6 μC
Calculate
the
charged
stored
on
the
capacitor
C
4 mm
when
c
–8 μC
State
a
p.d.
the
stored
by
of
18 V
is
maximum
this
applied
charge
across
that
can
it.
be
[1]
safely
capacitor.
[1]
Calculate:
i
the
magnitude
and
state
of
the force
whether
between A
the force
is
and
attractive
repulsive
ii
the
electric field
to A
44
alone
B
or
at
the
point C
Calculate
e
State
one
marked
[3]
strength
d
due
[2]
the
energy
reason
on
a
stored
why
a
capacitor.
in
the
capacitor.
maximum voltage
[2]
is
[1]
Revision questions
7
a
b
Define
A
the
terms
capacitor
from
i
a
of
20 V
capacitance
d.c.
Calculate
capacitance
and
12 μF
farad.
is fully
[2]
11
charged
a
State Coulomb’s
b
Define
charge
stored
by
the
c
capacitor.
Write
Calculate
the
energy
delivered
by
the
d
d.c.
supply.
Two
Calculate
the
energy
iv
Account for
the
answers for
ii
stored
in
the
capacitor.
c
A
12 μF
supply
capacitor
through
i
Calculate
ii
Given
is
a
The
the
difference
[1]
charged from
time
the
of
time
a
constant for
is
this
the
capacitor
i
[1]
the
ii
reach
iii
a
Derive
an
to
[3]
expression for
the
total
iv
similar
capacitors
connected
A
parallel-plate,
air-filled
in
series.
capacitor
consists
each
having
an
area
separation
p.d.
across
the
of
the
plates
plates
of
the
is
of
4.5
×
10
9
a
capacitance
the
energy
A
parallel
of
stored
plate
the
in
capacitor
is
strength
of
E
between
charge
midpoint
of
a
in
a
has
of
a
−15 μC.
line
joining
strength
at
the
point X
due
strength
at
the
point X
due
[2]
[2]
strength
at
the
point X
due
and Q
[3]
potential
at
the
point X
due
to
[2]
electric
potential
at
the
point X
due
to
potential
at
the
point X
due
to
only
the
[2]
electric
and Q.
12
Two
horizontal
[2]
[3]
constant
the
plates
are
situated
between
the
1.5 cm
plates
is
−1
2.5
×
N C
. Calculate:
a
the
potential
b
the
acceleration
[3]
capacitor.
a
10
metal
electric field
difference
of
an
between
electron
the
plates
between
[2]
the
electric
plates,
field
a
P
.
450 V.
capacitor
has
the
has
apart.
1.2 mm. The
capacitor
the
on
situated
only
4
the
and Q
are
18 cm
2
m
Calculate:
ii
lies
electric
the
apart. The
i
+45 μC
are
only
P
The
and Q
of
−3
plates,
relationship
[4]
vi
two
the
capacitance for
Q
b
P
charges
electric field
P
the
v
two
show
[2]
electric field
the
P
8
to
V.
only
to Q
95%
its final value.
[2]
electric field
P
the
uncharged,
potential
to
plates. The
area
assuming
there
is
a vacuum
between
of
them.
each
is
plate
is A
and
the
separation
of
the
13
an
expression for:
An
isolated
All
the
as
i
the
capacitance
C
of
the
capacitor
in
terms
though
d
metal
the
it
on
sphere
the
were
has
energy
per
unit volume
P
of
the
concentrated
sphere. The
air
to
conduct
terms
of
E.
greater
plate
be
of
20 cm.
considered
the
the
centre
metal
of
sphere
of
a
than
2
×
electricity
10
when
the
electric field
−1
V cm
. Suppose
that
a
spark
is
[3]
about
Each
at
around
4
b
radius
may
capacitor
is
in
a
surface
[1]
begins
ii
metal
charge
of
the
A and
[4]
plates
d.
Write
[3]
and
Calculate:
d.c.
circuit.
E
V.
charges
point X
to
initially
taken for
the
20 V
of
charges.
and Q.
1.8 kΩ.
capacitor
across
P
your
iii.
resistor
the
that
calculate
of
and
between
and
point
charge
[2]
difference
E
electrostatic
electric field strength
equation
vacuum. The
[1]
iii
an
between
[2]
ii
law for
terms
electric potential
supply.
the
the
2
parallel-plate
capacitor
has
to
leave
the
sphere.
an
2
area
of
plates
18.0 cm
is
. The
separation
4.0 mm. The
between
electric field
strength
6
between
the
plates
is
i
the
capacitance
ii
the
energy
2.5
of
×
the
10
Calculate:
the
10
A
6.0 μF
capacitor
is
capacitor
unit volume
charged
by
a
the
charge
b
the
potential
a
Define
b
Two flat
is
then
discharged
through
a
P.
Calculate
the
charge
on
the
electric field
12.0 V
d.c.
1.8 MΩ
parallel
being
are
c
After
6.0 s,
i
the
charge
capacitor
just
ii
the
p.d.
constant for
the
circuit.
[1]
electron
the
across
a
line
current
in
a
of
distance
between
travels
midway
speed
of
4.8
length
of
the
1.2 cm.
plates
parallel
to
the
is
plates
the
the
between
×
10
the
plates
with
a
−1
m s
.
Calculate:
capacitor
[3]
capacitor
the
magnitude
circuit?
of
the
electric field
strength
[3]
between
the
by
difference
7
i
iii
each
before
is:
on
[1]
plates,
[2]
time
what
strength.
metal
separated
potential
along
the
[3]
resistor.
discharged.
Calculate
[3]
sphere.
supply
200 V. An
b
sphere
the
[3]
The
a
the
of
[2]
10.0 cm,
and
on
. Calculate:
14
per
a
−1
V m
the
plates
[2]
[3]
ii
the
magnitude
of
the
acceleration
[4]
45
iii
the
time
length
taken for
of
the
the
plate
electron
to
travel
the
(10.0 cm).
17
a
[2]
Two flat
metal
plates,
each
of
length
4.5 cm,
A
by
a
distance
of
1.2 cm. A
a
uniform
of
100 V
between
electric field
in
the
the
region
space
Electrons
at
right
the
speed
angles
between
outside
of
plates
region
the
plates
to
×
is
in
simple
is
charged
to
a
potential
difference
of
and
then
connected
resistor
of
resistance
in
series
ammeter
as
shown
10 kΩ
and
with
a
a
switch,
sensitive
provides
between
below.
the
−1
10
m s
enter
the field. Assume
plates
is
3.0
capacitor
potential
7
plates.
a
are
a
difference
of
[1]
capacitor
12 V
separated
one function
circuits.
b
15
State
a vacuum
this
that
and
the
the field
10 kΩ
zero.
Calculate:
a
the
b
the force
electric field
c
the
on
an
strength
electron
between
due
to
the
the
plates
[2]
electric field
[2]
The
acceleration
direction
d
of
the
time
the
plates
the
speed
the
of
the
electron
along
the
of
electric field
taken for
an
switch
the
is
closed
current
I
in
and
the
the variation
circuit
is
shown
with
time
t
below.
[2]
electron
to
travel
I /mA
between
[2]
1.2
e
of
the
electrons
at
right
angles
to
its
1.0
original
direction
between
f
the
the velocity
of
motion
as
it
leaves
the
region
plates
of
an
[2]
electron
as
it
leaves
the
0.8
region
0.6
between
the
plates.
[2]
0.4
−19
(Charge
on
an
electron
=
−1.6
×
10
C,
mass
of
an
−31
electron
=
9.
11
×
10
0.2
kg)
0
16
a
Determine
the
equivalent
capacitance
of
the
0
network
of
capacitors
in
the
circuit
shown
[4]
i
State
a
14 μF
5
10
15
t /s
below.
the
circuit
relationship
and
the
between
charge
that
the
passes
current
a
point
ii
6 μF
The
circuit.
area
charge
12 μF
9 μF
[1]
under
charge. Use
the
stored
iii
Calculate
iv
What
is
the
the
the
graph
graph
in
the
to
above
estimate
the
initial
capacitor.
capacitance
time
represents
of
[4]
the
constant for
capacitor.
the
v
If
the
of
9 V
its
the
total
charge
stored
by
the
network
capacitors?
of
[2]
c
What
is
the
charge
stored
d
What
is
the
potential
in
the
6 μF
capacitor?
[2]
difference
across
the
6 μF
capacitor?
e
What
is
the
[2]
charge
stored
in
the
14 μF
capacitor?
[2]
f
What
is
the
capacitors?
46
total
energy
stored
by
the
network
of
[2]
[2]
capacitor
initial
difference
is
[2]
discharge
circuit?
What
in
8 μF
the
b
in
had
energy,
across
discharged
calculate
the
to
the
capacitor.
one
half
potential
[3]
Revision questions
18
An
isolated
charge
point
conducting
+Q. This
charge
shown
charge
situated
sphere
may
at
the
be
of
radius
r
assumed
centre
of
is
to
the
given
act
a
as
sphere
20
a
Electrons
with
as
in
accelerated
below.
an
a
cathode
negligible
electron
to
in
a
the
gain
b
the
loss
c
the
gain
d
the
speed
speed
an
in
a
anode
this
in
ray
at
tube,
at
potential
kinetic
leave
a
the
of
cathode
−8 kV
potential
of
and
−300 V.
are
For
calculate:
electrical
in
tube
potential
2
potential
[1]
energy
[1]
energy
[2]
+Q
on
reaching
the
anode.
[2]
−19
21
An
oil
droplet
situated
Sketch
a
graph
to
show
the variation
x from
the
centre
of
the
4.5
×
10
the
oil
V
due
to
the
charge
sphere
of
State
the
strength
c
Sketch
a
distance
charge
relationship
and
+Q.
of
to
the
the variation
electric field
An
with
strength
E
due
the
+Q.
electron
negative
The
is
[3]
cathode
potential
cathode
accelerated from
is
and
a
positive
difference
1.5 kV
and
rest
their
between
anode
between
electric field
10
of
C.
It
is
strength
−1
. Calculate
the force
experienced
by
droplet.
Two
capacitors
in
initially
[2]
of
capacitances
30 μF
and
50 μF
are
series
with
a
12 V
supply. The
capacitors
uncharged. Calculate:
a
the
total
b
the
charge
capacitance
delivered
c
the
charge
stored
d
the
potential
capacitor.
19
×
electric field
[1]
show
4.8
[2]
potential.
graph
x
between
of
the
are
b
charge
uniform
V m
connected
potential
a
with
22
distance
a
5
r
a
in
has
the
in
is
the
the
each
difference
circuit
power
[2]
supply
capacitor
across
the
[2]
[2]
30 μF
[2]
a
a vacuum.
anode
separation
in
in
by
and
the
80 cm.
Calculate:
a
the
and
b
c
electric field
strength
E
between
the
anode
cathode
the
kinetic
the
anode
the
speed
anode.
[2]
energy
of
the
electron
when
it
reaches
[2]
of
the
electron
when
it
reaches
the
[2]
47
6
Magnetic fields
6.
1
Magnetic fields
Learning outcomes
Magnetic flux
Materials
On
completion
of
this
section,
materials
should
be
able
understand
the
concept
of
of
at
the
a
distance.
two
types
of
Experiments
poles
define
iron
and
magnetic
steel
are
said
properties.
to
A
be
fer romagnetic .
magnet
is
able
to
These
attract
direction
of
Experiments
have
shown
that
a
bar
magnet
a
paper
consists
a
magnetic field
ᔢ
as
exhibit
to:
clip
ᔢ
such
you
attract
magnetic
have
also
each
poles .
shown
There
that
is
like
a
north
poles
pole
repel
and
each
a
south
other
and
pole.
unlike
other
.
a
magnetic field
The
ᔢ
sketch
long
the flux
straight
patterns
wire,
due
a flat
to
a
a
and
a
long
solenoid.
a
will
region
magnetic
These
the
lines
field
pole
S
behaves
it
like
align
a
weak
itself
magnet.
with
the
If
a
Earth’s
bar
magnet
magnetic
is
suspended
from
field.
circular
The
coil
Earth
string,
around
field.
are
line
placed
at
at
a
magnet
Lines
often
a
where
drawn
referred
point
that
are
in
point.
to
the
as
field
The
a
magnetic
around
lines
is
closer
the
of
the
the
force
is
magnet
magnetic
direction
field
lines
of
experienced
to
represent
flux.
the
are,
The
force
the
is
called
the
field.
direction
on
a
of
north
stronger
the
field.
N
S
magnetic
Figure 6.1.1
(a)
Poles of a magnet
S
Figure 6.1.4
poles
N
Weak magnetic field
Figure 6.1.2
(b)
Stronger magnetic field
Magnetic flux
N
N
Figure 6.1.3
Field around a bar magnet
S
Field between two north poles
Figure 6.1.5
Magnetic field
Field between a north pole and a south pole
produced
by
a
current-carrying
conductor
An
electric
current,
straight
The
be
ᔢ
wire
Maxwell’s
on
moving
lines
is
in
wire.
the
by
also
the
a
field
the
is
magnetic
shown
produced
rule
bottle.
direction
direction
grip
Y
our
a
produced.
by
The
in
the
field.
field
Figure
current
The
larger
around
a
the
long
6.1.6.
flowing
in
the
wire
can
using:
wine
the
produce
field
current
corkscrew
a
remaining
48
of
Right-hand
of
is
carrying
direction
cork
can
stronger
determined
a
ᔢ
current
the
rule
thumb
fingers
of
–
–
imagine
The
of
the
the
turning
point
in
in
the
corkscrew
of
the
corkscrew.
imagine
points
a
direction
your
the
The
action
right
of
direction
the
of
of
the
is
driven
into
imagined
direction
hand
direction
being
current
of
the
as
field
corkscrew.
gripping
the
a
length
current.
field
(Figure
The
6.1.7).
Chapter
right
6
Magnetic fields
hand
current
current
(a)
Current flowing into
the plane of the paper
Figure 6.1.6
The field around a wire
Figure 6.1.7
Fleming’s right-hand grip rule
carrying a current
Figure
6.1.8
shows
the
effect
of
changing
the
direction
of
the
current
on
(b)
the
magnetic
field.
the
plane
also
are
The
symbols
for
representing
current
into
and
out
of
Current flowing out of
the plane of the paper
shown.
Figure 6.1.8
The effect of changing the
direction of current flow
Magnetic field
Figure
6.1.9
shows
coil
with
produced
the
magnetic
current flowing
by
a flat
field
in
circular
produced
by
a
flat
coil
circular
coil.
it
field
lines
Key points
ᔢ
A
magnetic field
by
Figure 6.1.9
permanent
be
produced
or
Magnetic field produced by a flat circular coil
current-carrying
Magnetic field
produced
Figure
the
by
a
long
solenoid
ᔢ
The
region
where
6.1.10
shows
magnetic
field
produced
by
a
long
solenoid.
a
rule
is
used
to
determine
the
direction
of
the
field.
The
conductors.
around
a
magnet
magnetic force
is
The
experienced
right-hand
can
magnets
is
called
a
magnetic
right
field.
hand
is
used
direction
of
points
the
in
to
the
grip
the
current
direction
solenoid
flowing
of
the
such
in
the
that
your
solenoid.
magnetic
field
four
The
fingers
point
direction
produced
by
the
of
in
the
the
thumb
ᔢ
solenoid.
to
ᔢ
field
lines
Magnetic field
as
The
lines
are
referred
magnetic flux.
direction
of
the field
line
at
point
of
in
ᔢ
the field
the force
placed
at
The field
wire
is
on
that
is
a
the
direction
north
pole
point.
around
a
long
represented
as
straight
concentric
circles.
ᔢ
The
to
current
right-hand
determine
grip
the
rule
is
direction
used
of
a
direction
magnetic field.
Figure 6.1.10
a
solenoid
Magnetic field produced by a long solenoid
49
6.2
Force
on
a
Learning outcomes
current-carrying
Force
in
On
completion
should
be
able
of
this
section,
a
appreciate
to:
on
a
a force
current-carrying
when
placed
recall
and
in
a
might
use
understand
current-carrying
experience
hand
a
Consider
force.
define
the
force.
conductor
The
is
placed
magnetic
field
in
a
magnetic
produced
by
field
the
it
the
wire
interacts
with
the
external
magnetic
current
field
a
current-carrying
conductor
placed
at
to
right
produce
angles
magnetic field
to
use
a
magnetic
produces
a
field
as
magnetic
magnetic
field
as
shown
field.
in
The
shown
in
Figure
6.2.1.
The
current-carrying
the
diagram.
permanent
conductor
These
two
magnet
also
fields
produces
interact
to
Fleming’s
a
downward
force
F
acting
on
the
wire.
The
direction
of
this
force
rule
can
ᔢ
a
in
produce
left
placed
conductor
F = BIl sin θ
how
a
flowing
a
ᔢ
conductor
act
to
ᔢ
current-carrying
magnetic field
When
that
a
you
may
ᔢ
acting on
conductor
magnetic flux
density
be
predicted
using
Fleming’s
left
hand
r ule .
Using
the
left
hand:
and
ᔢ
the
index
finger
ᔢ
the
second
points
in
the
direction
of
the
magnetic
field
tesla.
finger
points
in
the
direction
of
the
current
flowing
in
the
conductor
ᔢ
the
thumb
magnetic field
by
permanent
points
in
the
direction
of
the
force.
produced
magnet
stronger field
magnetic field
by
produced
current-carrying
conductor
weaker field
current flowing
plane
of
the
into
the
paper
force
(a)
on
wire
F
(b)
Figure 6.2.1
The force acting on a current-carrying conductor placed in a magnetic field
A
magnetic
field
is
represented
by
a
vector
quantity
B.
This
is
called
the
thrust
magnetic
field
The
density
magnitude
placed
current
flux
ᔢ
the
ᔢ
ᔢ
ᔢ
in
a
of
the
magnetic
magnitude
of
the
magnitude
of
the
length
the
angle
magnetic
of
θ
the
and
its
force
field
SI
acting
depends
the
current
the
flux
I
between
l
is
on
a
on
B
tesla
(T).
current-carrying
the
inside
the
the
of
in
the
the
the
current
conductor
I
conductor
external
external
and
the
magnetic
magnetic
direction
field
field
of
the
field.
Equation
Figure 6.2.2
Fleming’s left hand rule
F
F
Exam tip
The force
plane
F
acts
at
containing
I
right
and
B
angles
to
the
=
–
BIl sin θ
force
on
B
–
magnetic flux
I
–
current flowing
l
–
length
θ
–
angle
Figure
of
is
conductor/N
density/T
in
conductor/A
conductor
made
6.2.3
conductor
50
acting
in
between
shows
how
affected
by
magnetic field/m
the
the
its
current
and
magnitude
orientation
magnetic field
of
in
when
following:
flowing
density
conductor
made
unit
the
the
force
acting
magnetic
on
field.
the
Chapter
6
Magnetic fields
force
I
Example
on
conductor
A
conductor
of
length
1.5 m,
carrying
a
current
of
8.0 A,
is
placed
in
a
F
magnetic
wire
field
when
a
at
b
along
c
at
it
right
of
is
flux
density
0.12 T
.
Calculate
the
force
acting
on
placed:
angles
to
=
BIl
the
B
the
magnetic
field
(top
an
the
direction
angle
of
30°
of
to
the
the
view)
field
(a) Conductor is at right angles
field.
to the magnetic field
a
F
=
BIl sin θ
=
0.12
×
8.0
×
1.5 sin 90°
=
1.44 N
b
F
=
BIl sin θ
=
0.12
×
8.0
×
1.5 sin 0°
=
0 N
c
F
force
I
on
conductor
=
BIl sin θ
=
0.12
×
8.0
×
1.5 sin 30°
=
F
0.72 N
=
B I sinθ l
θ
B
Magnetic flux density
(top
Magnetic
flux
a
conductor
straight
density
is
numerically
carrying
unit
equal
current
to
the
force
normal
to
per
the
unit
length
view)
on
field.
(b) Conductor makes an angle
of
1
tesla
metre
is
the
on
a
magnetic
wire
flux
carrying
a
density
current
of
of
a
field
1 A
producing
normal
to
the
a
force
of
1 N
θ with the magnetic field
per
field.
I
force
on
conductor
Equation
F
=
0
B
The
magnetic flux
current
I
is
given
density
at
a
distance
r from
a
straight
conductor
carrying
a
by:
(top
μ
view)
I
0
B
=
(c) Conductor is parallel to the
2πr
direction of the magnetic field
−1
μ
–
permeability
–
current/A
of free
space/H m
0
I
Figure 6.2.3
The effect of F when θ is
changed
r
–
In
perpendicular
the
equations
distance from
shown,
the
conductor/m
constant
µ
is
called
the
permeability
of
Exam tip
free
0
space.
It
is
a
measure
of
the
ability
of
a
medium
to
transmit
a
magnetic
field.
If
−7
Its
value
is
4π
×
10
you
cannot
remember
the
−1
H m
.
The
unit
is
the
henry
per
metre.
definition for
use
the
magnetic flux
equation
F = BIl
density,
and
state
the
Equation
meaning
The
magnetic flux
carrying
a
current
density
I
is
at
given
the
centre
of
a flat
circular
coil
of
radius
of
the
symbols.
r,
by:
Key points
μ
NI
0
B
=
2r
ᔢ
−1
μ
–
permeability
of free
A
current-carrying
may
space/H m
experience
conductor
a force
when
0
N
–
number
I
–
current/A
of
r
–
radius
turns
in
the
placed
coil
ᔢ
of
in
a
Fleming’s
magnetic field.
left
hand
rule
is
used
to
coil/m
predict
ᔢ
the
direction
Magnetic flux
of
density
the force.
is
Equation
numerically
per
The
magnetic flux
density
at
the
centre
of
a
long
solenoid
having
n
turns
unit
length
and
carrying
a
current
I
is
given
=
μ
to
a
the force
straight
carrying
unit
current
by:
normal
B
on
per
conductor
unit
equal
length
to
the field.
nI
0
ᔢ
1 tesla
is the
magnetic flux density
−1
μ
–
permeability
of free
space/H m
0
of
a field
producing
a force of
1 N
−1
n
–
number
of
turns
per
unit
length/m
per
I
–
metre on
a wire
carrying
a
current/A
current of
1A
normal to the field.
51
6.3
Force
on
Learning outcomes
a
moving
Force
An
On
completion
of
this
section,
be
able
acting on
electric
is
predict
the
direction
of
a
defined
moving
as
the
flow
in
of
a
magnetic field
in
motion,
it
constitutes
an
charge.
electric
When
current.
If
a
a
charged
positively
charge
moving
in
particle
is
moving
from
left
to
right,
the
conventional
current
also
the force
flows
on
was
charge
to:
charged
ᔢ
current
a
you
particle
should
charge
a
magnetic
from
from
left
left
to
to
right.
right,
the
If,
however
,
a
conventional
negatively
current
charged
flows
from
particle
right
to
is
moving
left.
field
An
ᔢ
recall
and
use
a
ᔢ
solve
electric
current
problems
charged
particle
involving
particle
is
particles
in
electric
v
Figures
a
at
BQv
–
force
field.
force
on
–
magnetic flux
Q
–
charge/C
no
density/T
–
work
using
speed/m s
In
a
magnetic
fields.
angles
path
outside
magnetic field
to
A
a
field.
field,
field
it
particle
magnetic
F
a
is
done
it
to
of
of
a
is
Therefore,
around
the
itself.
experiences
with
field
charge
a
If
motion
a
force
Q,
experiences
of
charged
because
travelling
of
with
the
a
a
force
F
=
BQv
a
plane
when
because
means
field.
field
particle,
F
Since
that
The
it
no
v
is
the
and
enters
does
F
on
path
of
respectively.
containing
particle
particle.
This
magnetic
not
at
act
right
energy
direction
B.
the
is
of
The
in
the
angles
gained
F
is
force
magnetic
to
by
v,
the
predicted
rule.
the
is
particle,
F
the
magnetic
6.3.2
magnitude
of
charged
unchanged
magnetic
F
to
charged
force.
the
force
effect
charged
hand
and
the
v
the
the
left
the
field,
path
by
the
of
of
the
negatively
angles
of
enters
6.3.1
angles
a
right
motion
when
force
at
show
and
direction
Fleming’s
the
6.3.2
magnitude
magnetic
to
and
acts
of
Figures
right
linear
F
The
charge
−1
v
in
two
charged
the
direction
charge/N
B
the
right
6.3.1
changes
=
of
positively
The
Equation
F
magnetic
magnetic
and
magnetic fields.
F
a
a
mutually
speed
perpendicular
produces
moving
interaction
charged
produces
F = BQv sin θ
charged
field
B.
particle
As
soon
experienced.
causing
acts
the
The
with
as
force
particle
continuously
on
the
to
the
a
speed
particle
acts
at
change
particle,
v
travels
enters
right
angles
direction.
the
at
the
As
particle
F
F
magnetic field
v
plane
of
into
follows
a
circular
paper
curved
path.
motion.
The
The
force
motion
F
of
provides
the
particle
the
is
necessary
similar
to
that
centripetal
of
force
+Q
needed
circular
path
to
maintain
the
circular
motion
of
the
charged
particle.
inside
Equation
magnetic field
Figure 6.3.1
Force acting on a positive
charge travelling in a magnetic field
circular
path
A
particle
with
charge
magnetic field
Q,
travelling
experiences
a force
with
F
a
given
speed
v
at
an
angle
of
θ
to
the
by:
inside
F
=
BQv sin θ
magnetic field
v
magnetic field
plane
of
into
F
–
force
B
–
magnetic flux
on
charge/N
Q
–
charge/C
v
–
speed/m s
θ
–
angle
paper
density/T
−1
Q
F
F
between
v
and
B
F
linear
path
outside
The
path
of
the
particle
is
helical
(Figure
6.3.3).
magnetic field
Charged
Figure 6.3.2
particles
from
the
Sun,
on
approaching
the
Earth,
may
become
Force acting on a negative
trapped
in
the
Earth’s
magnetic
field
near
the
poles.
The
trapped
charged
charge travelling in a magnetic field
particles
The
cause
V
an
the
Allen
sky
to
glow
.
radiation
belt
The
is
a
phenomenon
torus
of
is
highly
called
the
charged
aurora
borealis.
particles
B
around
by
the
the
Earth’s
electrons,
Figure 6.3.3
52
Helical path
Earth.
radiation
magnetic
while
belts
These
the
pose
charged
field.
outer
a
belt
particles
The
is
challenge
inner
made
for
are
belt
up
of
orbiting
held
is
in
made
two
up
energetic
satellites.
distinct
of
belts
protons
electrons.
and
These
Chapter
The
effect of
particle
The
force
with
a
in
a
following
v
curvature of
a
charged
region
Magnetic fields
of
uniform
magnetic
magnetic field
acting
speed
speed on the
6
on
at
a
particle
right
angles
of
mass
to
a
m,
having
magnetic
field
a
charge
B
is
Q,
given
travelling
by
the
equation:
F
=
BQv
electrons
This
a
force
curved
on
the
provides
path.
If
particle
the
the
is
centripetal
radius
given
by
of
force
curvature
the
following
required
is
r,
the
for
the
particle
centripetal
to
force
follow
acting
equation:
2
mv
F
slow
=
moving
fast
moving
r
electrons
electrons
2
mv
Therefore,
=
BQv
Figure 6.3.4
r
Effect of speed on curvature
mv
r
=
BQ
The
the
radius
of
charged
curvature
particle
is
is
directly
travelling.
proportional
Consider
a
to
the
stream
speed
of
with
electrons
which
of
+
varying
speeds
travelling
at
right
angles
to
a
uniform
magnetic
field.
The
F =
magnetic
field
acts
into
the
plane
of
the
paper
as
shown
in
Figure
EQ
6.3.4.
charged
E
Mutually
Consider
through
a
a
perpendicular
beam
region
of
of
particles
space
magnetic
having
where
the
a
charge
electric
and
−Q
field
particle
electric fields
travelling
strength
with
is
E.
a
speed
The
v
F =
BQv
force
–
acting
on
the
experience
charged
an
particles
upward
plates
force
is
given
(Figure
by
F
=
EQ.
The
charged
particles
6.3.5).
B
+
producing
Figure 6.3.7
source
charged
of
particles
Key points
–
Figure 6.3.5
Electric force
ᔢ
Suppose
right
a
uniform
angles
to
the
magnetic
electric
field
field.
of
magnetic
flux
The
magnetic
field
density
is
in
B
the
is
applied
plane
of
at
A
charged
at
right
The
force
acting
on
the
charged
particles
is
given
by
F
=
BQv
downwards.
In
order
for
the
charged
particles
to
pass
through
undeviated,
the
electric
force
and
the
magnetic
force
must
travelling
a
magnetic
a force
at
right
and
to
its
motion.
both
ᔢ
fields
to
experiences
angles
acts
particle
angles
the
field
page.
Electric force and magnetic
force are equal and opposite
electric field
be
The
path
taken
by
the
particle
is
equal
circular.
(Figures
6.3.6
and
6.3.7).
ᔢ
EQ
=
The
radius
of
curvature
is
BQv
proportional
to
the
speed
of
the
E
Therefore,
=
v
particle.
B
Mutually
velocity
plates
perpendicular
selection
in
a
electric
mass
producing
and
magnetic
fields
can
used
for
ᔢ
The speed of the particle does not
change when in the magnetic field.
spectrometer
.
path
electric field
be
of
ᔢ
charged
particles
is
undeviated
If
the
that
particle
it
makes
is
travelling
an
angle
θ
such
with
the
+
when
they
mutually
pass
through
magnetic field
and
in
a
path.
Mutually
and
charged
travels
magnetic fields
ᔢ
source
it
perpendicular
helical
electric
B,
perpendicular
electric fields
can
magnetic
be
used
of
electric field
particles
for
velocity
selection
of
charged
–
particles.
Figure 6.3.6
Electric field and magnetic field at right angles to each other
53
6.4
Measuring
Learning outcomes
magnetic flux
The
Hall
Consider
On
completion
of
this
section,
be
able
in
explain
the
use
a
Hall
Hall
to
B
measure
acts
at
velocity
the
The
are
of
principle
a
a
small
metal
current
conductor
I
is
there
flowing
are
free
as
mobile
v.
from
current
angles
to
Consider
left
the
the
to
flows
face
right.
A
to
the
left.
PQRS.
force
acting
magnetic
This
The
on
a
field
means
electrons
single
of
flux
that
the
have
a
electron.
density
mean
drift
According
Fleming’s
left
hand
rule,
the
electron
will
experience
a
force
F
acting
of
current
balance
magnetic flux
This
force
is
at
right
angles
to
the
plane
containing
v
and
B.
to
The
measure
conventional
flowing
right
of
downwards.
operation
which
the
density
to
describe
in
Inside
effect
probe
magnetic flux
ᔢ
conductor
6.4.1.
to:
electrons
ᔢ
metal
Figure
electrons.
ᔢ
effect
you
shown
should
a
density
force
F
is
given
by:
density.
F
where
and
v
B
is
is
the
the
=
magnetic
mean
drift
Bev
flux
density
velocity
on
and
an
e
is
the
charge
on
an
electron
electron.
Definition
Electrons
Hall
therefore
A
begin
negatively
effect
potential difference
is
set
collecting
charged
set
up
with
along
respect
between
sides
the
to
RS
side
RS,
which
the
side
PQ.
An
and
PQ.
This
therefore
electric
creates
a
becomes
field
is
potential
up
difference
V
called
Hall
between
the
sides
RS
and
PQ.
The
potential
difference
is
H
transversely
across
a
current-carrying
conductor when
a
magnetic field
applied.
is
the
voltage
perpendicular
B
I
A
mV
d
v
H
F
t
R
S
I
Figure 6.4.1
Explaining the Hall effect
V
H
The
electric
field
strength
is
given
by
E
=
.
This
causes
an
upward
d
force
side
Ee
RS,
to
act
the
on
the
electric
electron.
field
As
strength
more
E
and
more
increases.
electrons
The
voltage
collect
V
on
the
becomes
H
steady
when
the
Therefore,
two
forces
Bev
=
E
=
acting
on
the
electrons
each
other
.
Ee
V
V
H
But
cancel
H
,
so
Bev
=
e
and
V
=
Bvd
H
d
But
I
=
number
nevA,
of
where
electrons
the
mean
drift
the
conductor
.
I
is
per
velocity
d
the
current
unit
of
flowing
volume,
electrons,
e
is
and
in
the
A
is
the
conductor
,
charge
the
on
an
cross-sectional
I
So,
v
=
neA
BId
Therefore,
the
Hall
voltage
is
given
by
V
=
H
neA
The
cross-sectional
area
of
the
BI
∴
V
=
H
net
54
conductor
is
n
given
by
A
=
is
electron,
dt
area
v
is
of
Chapter
The
A
Hall
device
6
Magnetic fields
probe
called
a
Hall
probe
The
Hall
probe
makes
The
Hall
probe
is
use
is
of
used
the
to
Hall
measure
magnetic
flux
density.
effect.
Hall
made
of
a
thin
slice
of
a
probe
semiconductor
.
N
Semiconductors
larger
drift
velocities,
are
used
velocities
the
because
than
measured
those
Hall
the
in
charge
pure
voltage
is
carriers
metals.
larger
inside
Since
than
they
that
of
them
have
a
have
larger
pure
metal.
mV
H
In
order
the
to
measure
semiconductor
current
I
is
passed
magnetic
material
through
flux
is
at
the
density
,
right
probe
the
angles
and
Hall
to
the
probe
the
Hall
is
placed
magnetic
voltage
is
field.
so
A
that
S
small
measured
constant
(Figure
net.
6.4.2).
The
The
magnetic
manufacturer
flux
density
V
B
of
the
is
probe
normally
calculated
using
supplies
the
the
following
value
equation:
Figure 6.4.2
that
known
The
the
current
A
current
It
consists
current
side
I
measure B
be
is
at
of
(Figure
wire
enters
the
calibrated
6.4.3)
frame
one
end
magnetic
measured.
right
direction
so
a
is
by
measuring
the
Hall
voltages
in
balance
balance
of
probe
fields.
is
ABCD
of
the
used
to
pivoted
frame
measure
about
through
a
magnetic
horizontal
side
AD
and
flux
axis
density.
(PQ).
leaves
A
through
BC.
Suppose
to
Hall
magnetic
Using a Hall probe to
=
I
Note
I
net
H
B
current
of
that
of
a
mark
is
the
mg
is
verifi ed
on
to
the
side
the
I
of
the
F
acts
so
to
the
centre
frame
fi eld
the
the
see
if
AB
the
side
frame
the
of
is
produced
through
on
that
checking
at
wire
fl owing
force
adjusted
by
density
magnetic
cur rent
downward
weight
This
The
angles
fl ux
a
placed
by
the
frame
AB.
A
AB CD
pointer
solenoid
such
that
solenoid.
AB CD
small
up
is
mass
remains
lines
needs
it
The
adjusted
(rider)
horizontal.
with
the
zero
scale.
y
x
y
x
scale
F
D
P
0
I
rider
T
B
Q
mg
rider
B
C
mg
F
pivot
Figure 6.4.3
The
A simple current balance
principle
of
moments
F
×
x
=
F
=
is
Figure 6.4.4
then
mg
×
applied
to
determine
the
magnitude
of
Using a current balance to measure B
F
Key points
y
mgy
ᔢ
In
the
Hall
effect
a
potential
x
difference
If
the
length
of
the
side
AB
is
l,
then
the
magnetic
flux
density
B
across
determined
as
a
acting
on
side
AB
is
F
=
magnetic
flux
density
B
=
shows
how
a
current
balance
flux
density
inside
a
is
a
perpendicular
applied.
ᔢ
A
calibrated
Hall
probe
and
a
xIl
is
used
to
measure
balance
can
be
used
to
the
measure
magnetic
transversely
mgy
=
current
6.4.4
when
magnetic field
BIl
Il
Figure
up
follows:
F
Therefore,
set
current-carrying
conductor
Force
is
is
magnetic flux
density.
solenoid.
55
6.5
Force
between
Learning outcomes
The force
When
On
completion
of
this
section,
be
able
explain
the forces
current-carrying
predict
forces
the
current-carrying
conductors
conductors
other
a
force
is
experienced
between
are
them.
If
placed
the
parallel
current
to
flows
the
conductors
direction
between
same
direction
in
the
two
conductors,
the
force
is
attractive.
If
between
the
ᔢ
straight
current-carrying
to:
in
ᔢ
two
between
conductors
you
each
should
current-carrying
of
currents
in
the
repulsive.
Figures
through
thin
a
conductors
6.5.1
sheet
and
of
flow
6.5.2
in
opposite
show
aluminium
the
directions,
effect
of
the
passing
a
force
is
current
foil.
the
current-carrying
conductors.
I
I
2
1
F
F
aluminium foil
Figure 6.5.1
Currents flowing in the same direction
I
I
2
1
F
F
aluminium foil
Figure 6.5.2
Currents flowing in opposite directions
Explaining the
experiment
Currents flowing
Figure
wires
carrying
a
current
plane
of
the
shows
two
same direction
wires
X
and
Y
that
are
the
paper
.
They
carry
currents
I
paper
and
I
1
they
are
Since
separated
wire
X
is
by
a
distance
carrying
an
r
right
the
angles
plane
of
to
the
the
plane
paper
and
2
as
electric
into
shown.
current
I
,
a
magnetic
field
is
created
1
r
around
the
Y
X
F
at
into
of
the
6.5.3
in the
it.
wire.
wire
Y
is
This
The
magnetic
second
carrying
a
field
wire
Y
current
is
I
,
is
represented
situated
it
in
by
this
experiences
concentric
magnetic
a
force
F
circles
field.
as
around
Since
shown.
The
2
F
direction
of
Newton’s
F
can
third
be
law,
determined
an
equal
and
by
Fleming’s
opposite
left
force
hand
will
act
μ
B
rule.
on
field
produced
by
I
around
wire
X,
1
B
wire
X.
I
0
Magnetic
According
1
=
1
2πr
Figure 6.5.3
μ
Two currents acting in the
I
0
The
force
exerted
on
a
section
l
of
the
wire
Y
,
F
=
BIl
1
=
same direction
I
l
2
2πr
μ
F
The
force
per
unit
length
of
wire
which
I
causes
on
wire
Y
=
I
0
I
1
=
1
l
56
2πr
2
to
Chapter
6
Magnetic fields
Currents flowing opposite directions
Figure
plane
6.5.4
of
the
shows
paper
,
two
wires
carrying
X
and
Y
currents
which
I
and
I
1
of
the
paper
and
I
is
flowing
out
of
are
.
the
at
I
2
is
right
angles
flowing
to
into
the
the
plane
1
plane
of
the
paper
.
2
X
– Current
plane
Y
is flowing
of
– Current
the
the
is flowing
plane
into
the
paper.
of
the
out
of
paper.
r
F
F
Y
X
B
Figure 6.5.4
Since
wire
Two currents acting in opposite directions
X
is
carrying
a
current
I
,
a
magnetic
field
is
created
around
1
it.
This
wire.
Y
is
magnetic
The
second
carrying
a
field
is
wire
represented
Y
current
I
is
,
situated
it
by
in
concentric
this
experiences
a
circles
magnetic
force
F
around
field.
whose
Since
the
wire
direction,
by
2
Fleming’s
an
equal
left
and
hand
rule,
opposite
is
as
force
shown.
will
act
According
on
wire
to
Newton’s
I
0
field
produced
by
I
around
wire
law,
X.
μ
Magnetic
third
X,
B
1
1
=
1
2πr
μ
I
0
The
force
exerted
on
a
section
l
of
the
wire
Y
,
F
=
BIl
1
=
I
l
2
2πr
μ
F
The
force
per
unit
length
of
wire
which
I
causes
on
wire
Y
I
0
=
I
1
2
=
1
l
2πr
Example
T
wo
long
2.8 A
wires
flows
length
of
one
Magnetic
are
mounted
through
of
flux
the
each
vertically
wire.
and
Calculate
are
the
4 cm
force
apart.
acting
A
on
current
a
of
20 cm
wires.
density
produced
μ
one
of
the
wires:
–7
I
0
B
by
4π
1
×
10
×
2.8
−5
=
=
=
1.4
×10
T
–2
2πr
Force
acting
on
a
20 cm
=
1.4
2π(4
length
of
×
10
wire:
−5
F
=
BIl
×
10
)
−2
×
2.8
×
20
×
10
−6
=
7.84
×
10
N
Key points
ᔢ
When
two
other,
a force
straight
is
current-carrying
conductors
are
placed
parallel
to
each
experienced.
ᔢ
If
the
currents
are flowing
in
the
same
ᔢ
If
the
currents
are flowing
in
opposite
direction,
directions,
the force
the force
is
is
attractive.
repulsive.
57
6.6
The
electromagnet
Learning outcomes
The
An
On
completion
of
this
section,
be
able
electromagnet
current
describe
flows
the
principle
of
use
the
of
the
magnetic
solenoid
shown
effect
in
of
a
Figure
current.
6.6.1,
a
When
magnetic
is
produced.
electromagnet
and
When
a
ferromagnetic
material
such
as
iron
is
the
introduced
uses.
through
to:
field
ᔢ
makes
you
a
should
electromagnet
appreciate
its
lines.
The
principle
inside
result
by
the
is
solenoid,
in
which
an
an
there
increase
in
is
a
the
electromagnet
concentration
magnetic
flux
of
density.
The magnetic field produced
is
used
it
has
ᔢ
it
is
easily
ᔢ
it
is
able
B
a
high
that
=
material
The effect of introducing a
soft iron core
ᔢ
by
the
N
Figure 6.6.2
by a solenoid
Recall
is
I
Figure 6.6.1
iron
This
S
N
I
Soft
field
works.
ferrous
S
magnetic
construct
to
electromagnets
because:
permeability
magnetised
and
concentrate
the
µnI.
to
magnetic
The
magnetic
flux
relative
demagnetised
field
density
at
permeability
lines.
the
of
centre
iron
µ
is
of
a
solenoid
is
approximately
given
2000.
r
Therefore,
inside
an
when
air-filled
significantly.
dielectric
relative
a
solenoid
This
(e.g.
ferromagnetic
can
be
ceramic)
permittivity
ε
the
magnetic
compared
between
of
material
with
the
ceramic
such
field
the
plates
can
be
iron
effect
of
as
as
density
an
high
of
is
introduced
increases
inserting
air-filled
as
a
capacitor
.
3000.
Recall
The
that
the
r
εA
capacitance
of
a
capacitor
is
given
by
C
=
.
d
The
permeability
magnetic
ability
to
field,
of
transmit
Electromagnets
ᔢ
Magnetic
ᔢ
CA
T
ᔢ
Used
ᔢ
Circuit
ᔢ
Scrapyards
ᔢ
Magnetic
ᔢ
Relays
ᔢ
Electric
the
electric
many
is
a
measure
permittivity
of
of
a
its
ability
material
is
to
a
transmit
measure
field.
different
imaging
practical
in
the
transportation
applications:
(MRI)
industry
to
levitate
trains
breakers
to
separate
door
and
move
metallic
objects
locks
bells
locks
are
are
attached
to
used
used
the
in
as
modern
a
door
locking
locking
and
the
systems.
mechanism
for
electromagnet
Powerful
doors.
is
A
metal
attached
to
a
of
scanners
electromagnets
58
material
resonance
Electromagnets
is
an
have
Magnetic door
plate
a
whereas
the
its
Chapter
door
frame.
contact
in
with
many
the
When
one
stores.
store.
The
electromagnet
the
pole
A
electromagnet
of
the
and
presses
inside
release
energised,
electromagnet.
customer
person
is
the
the
a
This
button
store
a
metal
type
that
presses
the
of
alerts
button
plate
to
Magnetic fields
makes
operation
someone
6
is
seen
inside
de-energise
the
door
.
Relays
A
magnetic
that
uses
relay
an
is
shown
in
electromagnet
Figure
in
one
6.6.3.
circuit
A
to
magnetic
switch
relay
on
a
is
a
device
iron
armature
secondar y
to
circuit.
is
The
closed
in
relay
the
fi rst
the
soft
the
electromagnet.
closes
a
iron
consists
the
on
Electric
is
contacts
magnetic
switch
core
relay
a
is
large
of
circuit
a
two
cur rent
magnetised.
The
in
upper
the
that
end
allows
of
soft
the
circuit.
a
circuits.
fl ows
The
second
it
cur rent
separate
small
When
through
iron
cur rent
is
moves
main
switch
solenoid
ar mature
ar mature
The
the
the
attracted
upwards
advantage
circuit
to
be
secondary
and
of
circuit
to
and
using
used
to
electromagnet
circuit.
bells
Figure 6.6.3
An
a
electric
current
The
by
bell
starts
shown
flowing
electromagnet
the
hammer
electromagnet
armature
the
is
to
process
in
is
in
and
attracts
this
Figure
the
the
process
repeats
to
its
When
electromagnet
soft
and
demagnetised
return
6.6.3.
and
original
iron
the
armature.
metal
The
bell
is
A magnetic relay
pressed,
is
struck
broken.
strip
circuit
is
magnetised.
The
circuit
springy
position.
switch
become
electric
the
the
is
The
causes
closed
the
again
and
itself.
bell
push
springy
metal
soft
strip
iron
armature
contact
screw
C
electromagnet
hammer
gong
Figure 6.6.4
An electric bell
Key points
ᔢ
An
electromagnet
ᔢ
When
ᔢ
ᔢ
a ferrous
makes
core
is
Electromagnets
can
be
Electromagnets
have
electric
locks,
door
use
place
of
inside
turned
many
bells
the
on
uses,
and
magnetic
a
solenoid,
and
off,
it
unlike
including
magnetic
effect
MRI
of
a
greatly
current.
increases
permanent
the field.
magnets.
machines, CAT
scanners,
relays.
59
7
Electromagnetic
7
.
1
Faraday’s
Learning outcomes
and
induction
Lenz’s
Electromagnetic
Experiment
On
completion
should
be
able
of
this
section,
state
to:
Consider
Faraday’s
law
electromagnetic
1
a
solenoid
attached
to
a
sensitive
galvanometer
as
shown
in
7.1.1.
of
induction
magnet
into
ᔢ
induction
you
Figure
ᔢ
laws
the
moved
coil
use Faraday’s law to determine the
magnitude of an induced e.m.f.
S
ᔢ
state
the
ᔢ
Lenz’s
law
direction
discuss
of
Lenz’s
and
an
law
induced
as
N
determine
an
e.m.f.
induced
example
causes
of
conservation
of
e.m.f.
current
0
energy.
to flow
sensitive
galvanometer
S
Figure 7.1.1
As
the
bar
on
the
galvanometer
.
magnet
is
slowly
moved
towards
the
coil,
a
deflection
is
seen
N
goes
to
to
zero.
As
the
If
the
motion
magnet
passes
of
the
magnet
through
the
stops,
coil
and
the
deflection
exits
on
the
right,
coil
datalogger
a
Figure 7.1.2
deflection
in
the
Experiment
Consider
induced
resistor
direction
is
seen
on
the
galvanometer
.
2
another
of
opposite
known
experiment
resistance
in
which
(Figure
a
solenoid
is
connected
to
a
7.1.2).
current
A
datalogger
samples
of
is
the
connected
voltage
across
across
the
the
resistor
.
resistor
The
over
a
datalogger
period
of
takes
time.
Since
the
I
1
resistance
bar
is
magnet
known,
is
the
dropped
passes
through
shows
the
passes
through
the
current
from
a
solenoid
variation
of
the
in
the
height
and
above
exits
current
circuit
in
the
through
the
can
be
determined.
solenoid.
the
resistor
The
bottom.
with
magnet
Figure
time
The
as
the
7.1.3
magnet
0
time
Experiment
the
solenoid.
3
I
2
In
of
this
two
Figure 7.1.3
the
A
experiment
copper
magnets
on
of
In
AB.
as
the
the
of
to
In
wire
in
deflects
the
galvanometer
.
magnets,
order
them.
galvanometer
no
in
explain
each
wire,
If
is
the
one
the
there
is
on
and
As
in
the
is
is
wire
When
the
above
ends
moves
wire
you
need
a
is
to
of
a
stiff
the
piece
poles
of
downwards
is
direction.
horizontally
between
of
between
magnitude
galvanometer
motion
the
the
opposite
moved
the
to
down
the
direction.
experiments
relative
up
greater
wire
deflection
connected
7.1.4.
deflects
the
is
moved
Figure
galvanometer
movement
the
a
The
shown
galvanometer
upwards
the
wire
moved
The
the
faster
deflection
between
the
poles
observed.
see
the
conductor
link
and
a
between
magnetic
B
field.
A
magnetic
magnetic
Figure 7.1.4
60
fields
field
are
exists
around
represented
by
a
bar
lines
magnet.
called
In
6.1
magnetic
it
was
flux.
seen
that
Chapter
7
Electromagnetic
induction
Magnetic flux
Definition
The
total
amount
magnetic field
φ
=
where
The
is
of
magnetic flux
given
φ
through
an
area
A
at
right
angles
to
a
by
BA
B
is
unit
the
of
magnetic
magnetic flux
magnetic
flux
when
flux
a
density
is
flux
the
of
the
weber
density
of
magnetic field.
(Wb).
1 T
1
weber
passes
is
defined
perpendicularly
as
the
through
2
an
If
area
the
of
1 m
magnetic
magnetic
flux
field
is
acts
given
at
an
φ
by
=
angle
of
θ
to
the
area
A,
then
the
total
BA cos θ
B
area
A
area
A
θ
normal
B
total flux
φ
Figure 7.1.5
In
=
B A
Experiment
a
that
the
wire
them
an
part
on
electric
In
and
on
is
an
the
is
=
wire
end
the
A
of
other
end
circuit,
predict
used.
=
The
a
B A cos θ
tiny
as
when
thumb
index
of
of
the
current
is
finger
it
wire
it
speed
of
with
of
the
a
the
which
If
induced
is
is
were
that
to
connected,
current,
the
pushes
negatively
result
wire
induced
direction
in
The
the
the
This
Fleming’s
that
becomes
charged.
an
the
Using
force
wire
wire.
When
downwards.
upwards.
the
points
the
wire.
density
charge.
move
galvanometer
in
in
the
field,
flux
the
negative
called
the
points
is
of
field.
positively
ends
v
experience
end
electrons
end
magnetic
and
moves
This
one
magnetic
inside
electrons
direction
The
body
flux,
to
a
the
magnetic
current
the
in
is
the
becomes
This
the
the
wire.
between
flows.
conductor
.
the
on
B
carries
the
move
moves
electrons
conventional
induced
is
AB
magnetic
to
where
in
the
that
cuts
them
charge
moving
see
AB
body
BQv,
the
the
electric
to
r ule
charged
F
is
we
current
order
hand
the
rule,
towards
e.m.f.
wire
causes
downwards,
that
charged
Q
in
the
that
a
body
electron
hand
as
force
field,
moves
means
left
a
charged
Each
3,
force
when
experiences
magnetic
φ
Defining magnetic flux
experience
Recall,
total flux
form
an
cur rent
Fleming’s
of
the
direction
of
force
the
right
field
acting
field.
current
The
second
When
This
wire
finger
AB
means
Fleming’s
them
charged
When
force
left
and
wire
moves
that
towards
points
the
hand
end
the
AB
upwards,
of
other
is
experienced
the
the
end
moved
by
the
direction
the
conventional
rule,
B
in
the
This
becomes
in
induced
inside
moves
experience
end
of
the
positively
horizontally
electrons
the
electrons
current
electrons
wire.
of
between
the
wire
it
current.
move
a
force
wire
force
upwards.
downwards.
that
Using
pushes
becomes
negatively
charged.
poles
is
of
zero.
the
magnet,
Therefore
no
the
e.m.f.
Figure 7.1.6
is
induced
The
effect
in
of
the
producing
electromagnetic
Fleming’s right hand rule
wire.
an
electric
current
using
magnetism
is
called
induction
61
Chapter
7
Electromagnetic
induction
Magnetic flux
coil
area
having
linkage
Definition
N turns
A
Magnetic flux
linkage
refers
to
the flux
linking
or
passing
through
a
coil
and
is
B
numerically
Flux
total flux
flux
φ
=
linkage
Figure 7.1.7
=
B A
N
N φ
–
φ
equal
linkage
number
– flux
of
to
=
Nφ
turns
passing
Nφ
of
the
through
coil
the
coil
Defining magnetic flux
linkage
Faraday’s
and
Lenz’s
laws
Δφ
Faraday ’s
law
is
expressed
as
E
=
–N
where
E
is
the
induced
e.m.f.,
Δt
Definition
N
a
Faraday’s
law
of
is
the
time
number
of
turns
of
the
coil,
Δφ
is
the
change
in
magnetic
flux
in
Δt
electromagnetic
ΔB
induction
states
that
the
Faraday ’s
magnitude
law
is
also
expressed
as
E
=
–NA
,
where
N
is
the
number
Δt
of
the
induced
to
the
rate
flux
of
e.m.f.
is
change
proportional
of
of
magnetic
turns
change
of
in
the
coil,
A
magnetic
is
the
flux
cross-sectional
density
in
a
time
area
of
the
coil,
ΔB
is
the
Δt
linkage.
The
magnitude
magnetic
field
the
be
induced
increasing
the
length
ᔢ
increasing
the
speed
ᔢ
the
magnitude
magnetic
of
at
in
a
conductor
moving
through
a
by:
conductor
which
the
situated
in
conductor
the
is
magnetic
moving
field
through
the
field
increasing
The
e.m.f.
increased
ᔢ
magnetic
Definition
of
can
field
strength
of
the
can
be
of
the
induced
e.m.f.
increased
increasing
the
cross-sectional
ᔢ
increasing
the
number
ᔢ
increasing
the
rate
ᔢ
ensuring
that
the
at
area
turns
which
magnetic
in
a
field.
coil
of
wire
situated
in
a
by:
ᔢ
of
magnetic
of
the
field
of
the
wire
coil
in
the
magnetic
is
coil
field
changes
perpendicular
to
the
plane
of
the
coil.
Lenz’s
law
states
that
the
induced
The
e.m.f.
(or
current)
acts
in
such
negative
induced
direction
to
produce
effects
the
change
causing
e.m.f.
in
the
the
equation
rate
of
for
change
Faraday ’s
of
law
magnetic
implies
flux
is
a
direct
consequence
law
is
used
N
of
the
principle
to
predict
the
direction
S
N
S
X
of
conservation
of
of
the
induced
the
Lenz’s
energy.
current.
S
N
X
Y
N
Y
induced
induced
current
current
(a)
(b)
Figure 7.1.8
When
Figure
the
62
that
linkage.
it.
Lenz’s
S
(−)
opposes
to
law
oppose
sign
a
the
Applying Lenz’s law
north
7.1.8(a),
solenoid.
pole
lines
The
of
of
a
magnet
magnetic
induced
approaches
flux
current
in
are
the
cut.
a
solenoid,
An
solenoid
e.m.f.
flows
as
is
in
in
induced
a
in
direction
to
Chapter
produce
a
north
pole
to
oppose
the
motion
of
the
magnet
coming
7
Electromagnetic
towards
suspension
the
point
solenoid.
When
lines
a
of
north
pole
magnetic
solenoid.
This
direction
to
the
induction
moves
flux
time
are
the
produce
a
away
from
again
cut,
induced
south
the
and
current
pole
to
solenoid,
an
e.m.f.
as
is
in
Figure
induced
in
the
solenoid
attract
the
magnet
7.1.8(b),
in
flows
the
in
moving
a
away
from
solenoid.
Figure
7.1.9
angles
to
shows
a
metal
sheet
oscillating
freely
in
a
plane
at
S
right
N
a
magnetic
oscillations.
to
the
the
in
rate
poles
it.
According
of
of
The
field.
change
the
to
of
metal
Faraday ’s
flux
magnet,
magnetic
The
law,
linkage.
magnetic
field
between
sheet
As
flux
the
comes
an
induced
the
is
to
metal
cut
north
and
pole
rest
after
e.m.f.
sheet
an
is
is
south
few
proportional
moves
e.m.f.
and
a
between
induced
pole
is
not
Figure 7.1.9
uniform
parts
of
at
the
currents
The
of
the
fields
of
sheet.
Since
result,
Since
in
it.
the
energy
is
metal
oppose
of
The
the
energy
being
is
of
dissipated,
are
a
induced
conductor
,
of
of
the
the
in
eddy
induced
metal
different
currents
sheet.
decreases.
can
currents
derived
is
flow
motion
oscillations
eddy
e.m.f.s
sheet
direction
oscillations
amplitude
This
different
the
The
that
considerations.
metal
sheet.
in
a
sheet.
amplitude
reduction
As
flowing
magnetic
energy
the
edges.
metal
begin
produces
Hence
the
from
the
also
be
dissipate
the
explained
thermal
oscillation
amplitude
of
of
the
in
terms
energy
the
in
metal
oscillations
decreases.
Induced
Consider
angles
In
a
to
time
Change
e.m.f.
a
straight
a
the
a
straight
conductor
magnetic
t,
in
in
field
distance
magnetic
flux
of
of
length
field
travelled
linked
conductor
l
moving
strength
by
the
with
B
with
(Figure
speed
v
at
v
l
right
7.1.10).
conductor
the
a
conductor
=
vt
=
φ
=
B(vt
=
BA
magnetic field
×
l)
=
Bvtl
density
B
into
of flux
the
paper
Bvtl
Induced
e.m.f.
E
=
rate
of
change
of
magnetic
flux
=
=
Blv
Figure 7.1.10
t
Key points
ᔢ
The
the
φ
ᔢ
amount
=
numerically
Faraday’s
the
of
is
magnetic flux
given
φ
through
an
area
A
at
right
angles
to
by
BA
Magnetic flux
is
ᔢ
total
magnetic field
law
induced
linkage
equal
of
refers
to
to
the flux
is
or
passing
through
a
coil
and
Nφ
electromagnetic
e.m.f.
linking
induction
proportional
to
the
states
rate
of
that
change
the
of
magnitude
of
magnetic flux
linkage.
ᔢ
Lenz’s
to
law
states
produce
that
effects
to
the
induced
oppose
the
e.m.f.
(or
change
current)
causing
acts
in
such
a
direction
it.
63
7
.2
Motors
Learning outcomes
On
completion
of
this
and
generators
A
section,
simple d.c.
motor
F
you
axis
should
be
able
to:
C
ᔢ
describe
and
applications
explain
of
simple
B
electromagnetic
S
induction:
–
a
simple
d.c.
D
motor
split-ring
N
–
a
simple
a.c.
A
motor
commutator
carbon
brush
coil
–
a
d.c.
–
an
generator
F
a.c.
generator.
carbon
brush
current
Figure 7.2.1
An
electric
7.2.1
coil
of
wire
end
This
is
called
The
with
the
of
the
result,
A
As
is
coil
coil
a
pivot.
changes
a
d.c
power
on
of
a
the
end
is
The
supply
a
to
left
and
in
acting
force.
an
vertical,
coil
an
the
by
copper
.
hand
the
the
so
that
upward
a
The
can
When
downward
two
forces
direction
line
brush
the
up
through
vertical
position,
to
the
section
force.
The
downward
anticlockwise
rule
coil.
flowing
carbon
blocks,
springs.
brushes
stops
experiences
an
a
pivot.
of
anticlockwise
one
direction,
on
carbon
left
on
Figure
rectangular
ring
These
overshoots
from
experiences
rotate
Fleming’s
current
change
reverses
and
the
the
split
rectangular
in
a
mounted
a
experiences
rotate
is
of
commutator
upward
coil
the
it
is
of
forces
AB,
to
and
halves
of
of
situated
the
are
coil
to
an
from
of
a.c.
with
A
to
simple
in
pressed
regularly
coil
force
As
the
the
the
an
coil
inertia,
coil
of
coil
T
wo
induction.
consists
half
supply.
section
the
diagram
wire
connected
force.
motor
.
the
right
to
It
of
the
section
force.
As
a
direction.
motor
direction
direction.
64
now
anticlockwise
a.c.
its
The
against
direction
When
the
the
brushes
downward
an
of
field.
power
commutator
in
shows
in
d.c.
cause
motor
.
commutator .
tightly
a
electromagnetic
connected
experiences
continues
Each
are
flowing
upward
the
on
coil
T
wo
brushes
and
a.c.
7.2.2
the
vertical.
is
is
of
d.c.
magnetic
to
CD
result
CD
the
copper
.
wire
commutator
now
rectangular
on
a
of
through
current
simple
Figure
is
a
the
The
AB
flows
couple
simple
in
pressed
section
in
a
split-ring
are
is
application
of
connected
gap
then
other
.
coil
the
coil
the
coil.
and
a
the
coil
determine
current
until
the
are
produce
an
situated
of
to
is
diagram
brushes
used
force.
a
called
brushes
the
motor
shows
Each
be
A simple d.c. motor
two
forces
function
result,
the
wire
motor
.
The
supply.
to
section
similarly
to
the
continues
An
to
a.c.
and
of
the
rotate
supply
a
left
of
The
current
hand
experience
the
coil
slip
in
a
mounted
rings
power
Fleming’s
will
of
is
rings.
initially
split-ring
to
coil
slip
slip
that
CD
couple
combination
the
consists
The
connected
Suppose
a
It
field.
against
According
produce
coil
is
power
time.
B.
The
a.c.
magnetic
tightly
experienced.
direction.
a
rotates
rings
and
commutator
an
rule
an
in
the
in
the
anticlockwise
Chapter
7
Electromagnetic
induction
F
axis
C
B
S
D
N
A
coil
slip
ring
F
to
Figure 7.2.2
A
A
generator
rotated
the
The
in
power
As
the
of
side
A.
depends
on
the
a
the
coil
of
circuit.
rotates
the
in
is
of
induction.
induction,
induced
coil
hand
coil
rule,
is
7.2.4
main
is
in
it.
rotation
can
magnetic
an
be
when
The
and
a
coil
magnitude
the
causes
the
the
how
a
an
an
by
or
that
number
by
some
e.m.f.
a
CD
is
is
of
of
using
will
to
flow
output
in
of
D
coil
same
a
d.c.
d.c.
is
If
produced.
right
direction.
from
the
voltage
is
Fleming’s
the
is
no
mechanical
upwards.
flow
in
is
generated.
current
moves
flowing
dynamo
there
anticlockwise
current
current
is
resistor
,
side
current
the
7.2.3)
rotate
field,
to
in
the
induced
vertical,
to
determined
rotated
and
(Figure
difference
made
connected
is
shows
generator
coil
the
is
downwards
Figure
axis
electromagnetic
speed
The
The
current
the
commutator
with
d.c.
motor
.
generator
moves
the
simple
d.c.
the
varies
of
electromagnetic
e.m.f.
split-ring
time.
of
an
of
a
the
right
When
law
field,
Suppose
AB
Fleming’s
of
in
direction
rule.
application
coil.
that
output
hand
to
e.m.f.
supply
means.
The
another
magnetic
the
to
supply
generator
Faraday ’s
construction
similar
the
a
induced
turns
the
is
to
in
power
A simple a.c. motor
simple d.c.
According
a.c.
Then
By
to
C
to
zero.
B
The
direction
all
generator
time.
of
rotation
(horizontal)
induced
F
e.m.f.
in
coil
C
B
S
D
split-ring
N
A
commutator
carbon
time
brush
coil
F
carbon
brush
Y
output
orientation
voltage
the
of
coil
magnetic field
X
Figure 7.2.3
A simple d.c. generator
Figure 7.2.4
The output voltage from a
d.c. generator
65
Chapter
7
Electromagnetic
induction
A
simple
The
of
an
the
is
a.c.
motor
.
difference
is
made
to
magnetic
connected
current
can
the
is
coil
rule,
an
and
varies
to
a
by
an
and
e.m.f.
a
in
an
a.c.
generated.
current
by
CD
will
shows
a
using
is
moves
flow
the
how
or
dynamo
simple
power
mechanical
is
positions,
7.2.6
no
means.
If
the
direction.
upwards.
D
to
direction
the
The
C
similar
the
the
the
to
B
flow
to
of
voltage
A.
side
of
in
a.c.
of
rotation
(horizontal)
F
C
B
S
D
N
A
coil
slip
F
Y
X
output
Figure 7.2.5
voltage
A simple a.c. generator
induced
e.m.f.
in
coil
time
orientation
the
Figure 7.2.6
66
of
coil
magnetic field
The output voltage from an a.c. generator
moves
hand
the
the
sides
coil
generator
time.
axis
in
the
AB
When
the
The
Suppose
right
current
of
The
generator
rule.
Fleming’s
that
rotates
direction
the
to
circuit.
coil
of
hand
Then
By
of
output
As
right
is
generator
.
in
output
produced.
from
a.c.
supply
Fleming’s
anticlockwise
side
current
change
generator
illustrates
is
some
resistor
,
the
a.c.
7.2.5
there
determined
Figure
with
simple
that
rotate
induced
changes.
a
Figure
is
rotated
CD
of
field,
be
downwards
AB
generator
construction
main
coil
a.c.
ring
Chapter
Factors
that
affect
the
speed
of
rotation
of
a
d.c.
or
a.c.
7
Electromagnetic
induction
motor:
Exam tip
1
The
magnitude
of
the
field
2
The
magnitude
of
the
current
between
the
flowing
magnetic
in
the
poles.
If
coil.
you
are
asked
operation
3
The
Factors
or
a.c.
number
that
of
affect
turns
the
in
the
of
magnitude
of
the
e.m.f.
induced
in
the
coil
in
a
d.c.
1
Draw
the
a
simple
north
magnet,
The
speed
2
The
number
3
The
magnitude
of
of
turns
of
in
the
the
field
between
the
magnetic
poles.
Key points
a
d.c.
Label
3
Show
4
motor,
ᔢ
In
an
ᔢ
When
using
a.c.
a
a
the
current
split-ring
motor,
coil
is
the
is
made
to flow
in
the
same
direction
all
there
rotated
sketch
south
coil
and
and
label
poles
the
of
the
split-
edges
the
of
the
direction
of
coil.
the
current
coil.
Mention
Fleming’s
predict
the
left
hand
direction
of
rule
the
the
acting
on
the
coil.
commutator.
are
in
the
the
force
by
and
commutator.
2
to
time
the
coil.
in
In
explain
motor:
rotation.
ring
ᔢ
to
d.c.
coil.
motor:
1
a
a
slip
rings.
magnetic field
an
e.m.f.
is
induced
in
it.
67
7
.3
Examples
Learning outcomes
on
electromagnetic
Faraday’s disc
Consider
On
completion
of
this
section,
be
able
provide
explanations for
the
disc
the
situated
disc
between
rotates,
an
the
e.m.f.
poles
is
of
two
induced
magnets
between
of
of
the
disc
and
its
circumference.
As
the
disc
the
centre
rotates,
the
various
electromagnetic
induction.
rule.
and
In
are
set
up
inside
the
it
experience
example
pushed
electrons
copper
As
rotation
electrons
examples
copper
7.3.1).
to:
of
ᔢ
a
you
(Figure
should
induction
at
shown,
towards
the
between
centre
the
the
of
force
the
disc
and
according
electrons
centre
the
centre
a
of
the
results
the
to
Fleming’s
experience
disc.
in
a
The
an
hand
upward
force
accumulation
potential
circumference
left
of
of
difference
the
disc.
being
The
flow
of
disc
electrons
towards
electrons
equals
the
the
centre
stops
magnetic
when
force
the
acting
electric
on
force
acting
on
the
them.
mA
Magnet
Consider
carbon
Figure 7.3.1
moving
Figure
in
a
coil
7.3.2.
brushes
Faraday’s disc
S
A
coil
B
Figure 7.3.2
When
the
frequency.
the
magnet
When
oscillation
variation
of
and
the
after
of
the
a
is
displaced
resistor
the
is
magnet
amplitude
resistor
was
of
gently,
it
begins
connected
becomes
the
oscillating
terminals
damped.
oscillation
connected
to
to
of
the
the
Figure
magnet
terminals
with
AB
7.3.3
with
A
and
a
at
constant
time
shows
time,
t,
the
before
B.
amplitude
t
time
Figure 7.3.3
As
the
magnet
resistor
The
current
dissipated
magnet
the
68
moves
completes
in
the
flowing
it.
amplitude
of
the
circuit
through
This
through
towards
the
thermal
coil.
the
coil,
resulting
resistor
energy
Therefore,
oscillation
of
an
in
the
is
if
e.m.f.
a
causes
derived
thermal
magnet
is
current
will
induced
flowing
thermal
from
the
in
it.
The
through
energy
to
movement
energy
is
being
reduce
over
a
it.
be
of
the
dissipated,
period
of
Chapter
time.
and
A
a
If
a
the
of
oscillation
smaller
of
the
search
search
value
magnet
current-carrying
Consider
a
resistor
is
used,
reduces
solenoid
more
more
energy
will
be
7
Electromagnetic
induction
dissipated
rapidly.
placed
near to
coil
the
coil
situation
as
shown
where
in
a
current-carrying
Figure
solenoid
is
placed
next
to
7.3.4.
search
coil
solenoid
Figure 7.3.4
Figure
7.3.5(a)
shows
the
variation
of
current
flowing
through
the
solenoid
I
with
time.
directly
The
magnetic
proportional
to
field
the
strength
current
B
produced
flowing
by
through
the
it.
solenoid
The
is
variation
of
t
the
magnetic
The
An
search
e.m.f.
field
coil
is
is
strength
situated
induced
electromagnetic
B
in
with
in
the
the
The
will
magnetic
search
induction.
time
coil
be
similar
field
of
the
graph
produced
according
direction
to
to
by
Faraday ’s
induced
(b).
the
(a)
solenoid.
law
current
of
is
B
determined
using
Lenz’s
law.
dφ
Induced
e.m.f.
E
=
t
–
dt
Graph
(c)
shows
the
variation
of
the
induced
e.m.f.
in
the
search
coil
with
(b)
time.
A
A
washer
light
Figure
aluminium
7.3.6.
washer
resting on
A
jumps
washer
large
and
a
rests
direct
E
solenoid
on
current
immediately
one
end
of
suddenly
falls
a
solenoid
flows
in
the
as
shown
solenoid.
t
in
The
back.
(c)
aluminium
Figure 7.3.5
washer
I
solenoid
I
Exam tip
Figure 7.3.6
In
the
given
A
it.
magnetic
Since
the
field
is
produced
current
by
increases
a
to
solenoid
a
large
when
value
a
current
quickly,
the
flows
through
changes
rapidly.
The
aluminium
washer
is
situated
an
field.
An
e.m.f.
is
induced
in
the
washer
.
Since
and
a
conductor
,
a
current
flows
in
it
because
of
the
in
the
current
produces
a
magnetic
field
that
in
e.m.f.
mind
in
a
that
produced
by
the
solenoid.
This
is
in
agreement
an
e.m.f.
conductor
is
when
the
with
is
cut. Always
The
that
an
alternating
magnetic
current
field
of
laws.
aluminium
induced
opposes
be
changing
remember
induced
may
application
Lenz’s
magnetic flux
is
you
magnetic
induced
magnetic
unfamiliar
Faraday’s
Keep
field
examination
Lenz’s
law.
produces
a
changing
The
magnetic field.
repulsive
flowing
e.m.f.
is
force
in
the
causes
the
solenoid
induced
in
the
washer
becomes
washer
to
move
steady,
and
it
upwards.
there
is
no
When
the
changing
current
flux.
No
falls.
69
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
3
Calculate
accompanying CD.
that
rod
1
a
Explain
and
b
2
meant
the
by
terms
current-carrying
solenoid
produced
by
a
produced
by
a flat
diagram
shows
1.5 cm.
density
It
is
the
in
a
9.5 mT formed
poles. The
view
uniform
in
conductor
the
coil.
acceleration
rolling,
of
when
the
the
rod,
assuming
current
in
the
5.0 A.
[5]
4
A
strip
of
aluminium foil
shown
is
hung
over
a
wooden
peg
below.
[3]
of
a
[3]
square
of
magnetic field
between
current
initial
without
[3]
circular
top
is
the
slides
magnetic field
as
a
ii
south
the
magnetic field:
produced
side
by
[2]
iii
The
of
is
magnetic flux
Sketch
i
what
it
magnetic
wire
is
wire
of flux
north
and
2.0 A.
A
d.c.
that
power
a
supply
current flows
is
connected
through
the
between
P
and Q
so
aluminium foil.
1.5 cm
M
N
5
a
State
b
Explain
Two
what
long
is
the
observed
with
the
aluminium foil. [1]
observation.
straight
wires
P
[4]
and Q
are
parallel
to
each
2.0 A
other. The
6.0 A. The
the force
L
Determine
the
in
wire
separation
per
unit
of
P
is
the
length
4.0 A
wires
on
and
is
wire Q
that
in Q
is
2.2 cm. Calculate
due
the
current
in
O
wire
a
current
sizes
and
electromagnetic forces
directions
that
act
on
of
6
the
the
sides
a
P.
[4]
Calculate
wire
LM
of
the
resistance
diameter
per
0.06 mm
metre
and
of
a
copper
resistivity
−8
and
b
NO
Why
do
sides
of
the
no
MN
square
of
wire.
electromagnetic forces
and OL
of
the
1.7
[6]
act
on
the
b
square?
[2]
×
10
copper
It
made
is
turns
3
An
aluminium
rod
of
mass
60 g
is
placed
across
Ω m.
The
on
horizontal
copper
tubes
that
are
a
low voltage
supply. The
aluminium
the
centre
of
and
perpendicular
winding
cardboard
to
one
construct
layer
tube
25 mm. The
rod
with
of
of
a
solenoid.
close-packed
length
solenoid
is
18 cm
and
connected
to
a
battery
of
e.m.f.
9.0 V
and
in
negligible
lies
internal
across
used
connected
series
to
is
two
diameter
parallel
by
a
[2]
wire
the
resistance.
uniform
Calculate:
magnetic field
of
a
permanent
magnet
as
shown
in
i
the
the
resistance
of
the
wire
used
to
make
the
diagram.
solenoid
The
magnetic field
acts
over
a
region
ii
8.0 cm
×
6.0 cm. The
magnetic flux
[4]
measuring
density
of
the
magnetic flux
density
at
the
centre
of
the
the
solenoid.
field
between
the
poles
is
[3]
0.25 T.
7
A
particle
has
a
mass
of
m
and
a
charge
of
+q.
8.0 cm
a
State
on
the
this
i
a
ii
an
iii
a
magnitude
particle
when
and
it
is
direction
at
rest
of
the forces
in:
gravitational field
[2]
copper
6.0 cm
electric field
of field
strength
E
[2]
tubes
b
State
this
magnetic field
magnetic field
the
of flux
magnitude
particle
when
it
and
is
density
direction
moving
B.
of
with
[1]
the force
a velocity
on
v
in
aluminium
of
the
a
permanent
direction
normal
to:
rod
magnet
70
i
a
ii
an
gravitational field
iii
a
electric field
of field
magnetic field
of flux
[2]
strength
density
E
B.
[2]
[2]
Revision questions
3
7
8
a
Define
the
b
Positive
narrow
tesla.
ions
are
a
travelling
beam. The
magnetic field
in
[3]
ions
of flux
semicircular
arc
through
enter
a
density
as
B
shown
a vacuum
region
and
of
are
in
11
An
electron
enters
a
a
density
uniform
travelling
with
magnetic field
7
.5 mT,
in
a
of
a
speed
uniform
direction
at
of
3.0
×
10
−1
m s
magnetic flux
right
angles
to
the
field.
deflected
below.
a
Sketch
field
the
and
path
show
of
the
the
electron
direction
of
in
the
magnetic
the field.
[2]
uniform
b
Explain
why
the
path
travelled
by
the
electron
is
magnetic field
circular.
[2]
detector
c
Explain
why
the
speed
of
the
electron
is
unchanged.
[2]
d
Calculate
the force
e
Calculate
the
acting
on
the
electron.
[2]
10.2 cm
radius
of
the
path
of
the
electron
the field.
beam
of
f
positive
Calculate
the
electric field
strength
required
to
ions
provide
an
equal force
to
that
provided
by
the
magnetic field.
g
The
ions,
travelling
5
1.62
×
when
in
[3]
10
a
speed
Explain
of
of
a
how
it
is
particular
possible
speed
by
to
the
select
use
of
electrons
electric
and
−1
m s
the
with
[2]
,
are
detected
diameter
of
the
at
arc
a fixed
in
the
magnetic fields.
detector
[3]
magnetic
−31
12
field
is
A
particle
has
mass
of
9.
1
×
10
kg
and
a
charge
10.2 cm.
−19
of
i
State
the
direction
of
the
magnetic field.
−1.60
[1]
The
positive
ions
have
a
mass
×
10
Calculate
If
the
the
and
a
charge
of
magnetic flux
+3.2
density
×
10
show
a
diagram
the
An
electron
is
6
6.5
×
10
of
path
strength
similar
taken
travelling
magnetic flux
the force
b
the
right
density
acting
c
the
radius
of
a
uniform
path.
magnetic field
of
of flux
the
radius
of
curvature
of
its
by
to
the
is
the
increased,
one
with
above
A
a
to
ions.
a velocity
angles
to
a
proton
speed
density
[1]
angles
the
magnetic field
v
the
circular
of
path
mass
enters
1.67
a
circular
the
10
uniform
2.2 mT. The
to
×
proton
kg
travelling
magnetic field
of flux
is
right
travelling
magnetic field. The
path
while
in
the
with
at
proton
travels
magnetic field. The
along
radius
curvature
of
the
path
of
the
proton
is
4.2 cm.
of
a
electron
acceleration
of
of
5.0 mT. Calculate:
on
[4]
−27
13
Explain
why
when
the
proton
travelling
travels
inside
along
the
a
circular
magnetic field.
[2]
[3]
b
centripetal
in
0.30 T. Calculate
path
a
a velocity
−1
m s
circular
of
at
with
the
−1
m s
travelling
density
a
9
is
C.
[3]
magnetic field
sketch
10
It
−19
kg
magnetic field.
iii
×
C.
of
−26
2.99
10
7
4.5
ii
×
the
electron
described
by
Explain
why
change
while
the
speed
of
the
proton
does
not
[2]
travelling
inside
the
magnetic field.
the
[2]
electron.
10
In
[2]
the velocity
selector
of
a
mass
14
positive
ions
5
1.2
×
field
10
of
travelling
with
a
speed
pass
120 mT
undeflected
which
is
Calculate
a
Define
b
State:
the
speed
v.
[5]
magnetic flux density
and
its SI
unit.
[3]
of
−1
m s
c
spectrometer,
through
perpendicular
a
to
magnetic
an
i
Faraday’s
ii
Lenz’s
law
[2]
electric
law.
[2]
field.
c
a
Draw
a
labelled
diagram
to
show
the
A
straight
wire AB
magnetic field.
between
the
two fields.
Write
down
an
expression for
moved
State
at
right
three factors
angles
that
to
affect
a
the
[2]
magnitude
b
is
relationship
the force
of
the
induced
e.m.f.
between A
and
B.
exerted
[3]
by
c
by
d
the
Write
magnetic field.
down
the
Hence
an
expression for
[1]
the force
electric field.
calculate
electric field.
the
15
exerted
a
strength
of
the
[2]
what
is
meant
by
electromagnetic
induction.
[1]
electric field
Explain
b
Describe
an
[2]
experiment
electromagnetic
to
demonstrate
induction.
[4]
71
16
A
solenoid
is
designed
of flux
density
length
of
the
25 mT
to
at
solenoid
respectively. The
produce
its
are
a
magnetic field
centre. The
1.50 cm
current flowing
radius
and
22
and
large
solenoid
Calculate
50.0 cm
through
A
solenoid
the
is
50 cm
long
magnetic flux
when
a
current
of
and
has
density
80
turns.
inside
2.0 A flows
in
the
it.
[3]
the
−3
23
solenoid
is
A
thin
copper
ring
encloses
an
area
of
1.8
×
10
−2
m
.
10.0 A.
The
plane
of
the
ring
is
normal
to
a
uniform
magnetic
Calculate:
field. The
a
the
minimum
number
of
turns
per
unit
magnetic field
strength
−2
that
must
be
used
total
length
of
wire
required for
this
A
small
copper
disc
of
diameter
50 mm
rotates
24
axis
at
12.0
revolutions
per
second
in
a
of flux
density
1.4
×
10
the
4.0
metal framed
pivots
about
at
right
angles
Explain
why
an
axle
and
the
to
the
e.m.f.
rim
Calculate
window.
T. The
plane
of
e.m.f.
is
of
the
generated
disc
[3]
window
is
1.2 m
edge
high
and
and faces
0.8 m
due
wide.
south.
the
magnetic flux
(Horizontal
through
component
of
the
the
closed
Earth’s
=
18 μT. Vertical
component
of
the
rotation.
between
when
it
magnetic field
=
45 μT.)
[2]
the
rotates.
The
window
is
opened
through
an
angle
of
90°
[3]
a
time
of
0.90 s. Calculate
the
average
e.m.f.
Calculate:
induced
i
the
magnetic flux
ii
the
potential
the
rim
and
cut
every
difference
the
axle
of
revolution
maintained
the
between
25
[3]
A
circular
It
is
coil
placed
horizontal
coil
of
cross-sectional
area
3.5
×
10
in
the
window frame.
[3]
[3]
disc.
−4
A
the
magnetic field
in
18
. Calculate
on
b
b
a
−1
T s
magnetic
Earth’s
a
10
a vertical
magnetic field
acts
×
ring.
It
−2
field
in
A
a
an
of
design.
[3]
17
rate
[3]
induced
the
at
length
constant
b
increases
of
so
diameter
that
its
140 mm
plane
magnetic field
of
is
has
600
turns.
perpendicular
uniform flux
to
a
density
2
m
and
50 mT.
90
turns
a
The
is
placed
plane
the
of
in
the
a
uniform
coil
is
at
magnetic field.
right
magnetic field. Calculate
flux
density
if
the flux
angles
the
axis
to
magnetic
linkage for
the
coil
is
−4
2.0
b
×
The
10
coil
density
coil
Wb.
is
[3]
now
placed
0.38 T. The
and
the
in
magnetic field
angle
between
magnetic field
is
the
of flux
axis
of
30°. Calculate
the
uniform
magnetic
the
coil
field
flux
19
A
linkage for
circular
coil
is
coil
placed
of
at
the
coil.
radius
right
[3]
1.5 cm
angles
has
to
a
1800
turns. The
magnetic field
of
a
flux
density
field
is
75 mT. The
reversed
in
a
direction
time
of
of
the
25 ms. Calculate
ends
of
magnitude
the
of
the
induced
the
coil
in
e.m.f.
when
magnetic flux
this
passing
through
the
position.
[2]
the
b
average
Calculate
magnetic
across
The
coil
is
rotated
axis
in
time
through
90°
about
a vertical
the
coil.
a
of
90 ms.
[5]
Calculate:
20
A
straight
wire
of
length
15 cm
is
travelling
at
a
i
the
change
in
magnetic flux
linkage
produced
−1
constant
field
the
21
An
speed
of flux
ends
of
aircraft
of
2.5 m s
density
the
has
in
a
uniform
45 mT. Calculate
magnetic
the
e.m.f.
across
wire.
a
by
ii
[3]
wing
span
of
50 m.
It
the
is
26
is flying
this
With
the
what
is
rotation
average
[2]
e.m.f.
induced
in
the
coil
when
rotated.
aid
of
a
it
[2]
clearly
labelled
diagram,
explain
−1
horizontally
vertical
at
600 km h
component
of
the
in
a
region
Earth’s
where
the
magnetic field
−5
is
4.5
×
10
induced
72
T. Calculate
between
one
the
wing
potential
tip
and
difference
the
other.
[3]
meant
by
the
Hall effect.
[8]
Revision questions
27
A
rectangular
thick
field
and
slice
carries
of flux
a
of
a
semiconductor
current
density
0.6 T
of
is
120 mA. A
acts
on
the
28
2.2 mm
magnetic
shown
in
the
diagram. A
potential
Draw
b
Explain
a
a
Use
difference
is
produced
across
the
the
diagram
operation
of
of
a
a
simple
simple
d.c.
d.c.
motor.
motor.
[4]
[5]
the
semiconductor
laws
of
electromagnetic
induction
to
of
explain
7
.65 mV
simple
semiconductor
29
as
a
3
the
operation
of
a
simple
a.c.
generator.
due
[5]
to
the
Hall
effect. The
main
charge
carriers
in
the
−19
semiconductor
Calculate
volume.
the
are
electrons
number
Explain
your
of
=
=
−1.6
×
carriers
10
per
calculation.
B
I
(e
charge
=
b
C).
Sketch
a.c.
unit
the
variation
generator
with
of
the
time.
induced
e.m.f.
in
the
[3]
[8]
0.6 T
120 mA
d
2.2 mm
73
Module
Answers
to
selected
structured
the
1
Practice
multiple-choice
questions
questions
can
and
be found
exam
to
on
questions
4
the
A
cell
is
connected
of
e.m.f.
4.0 V
to
a
and
negligible
network
of
internal
resistors
resistance
shown
below.
accompanying CD.
Multiple-choice questions
4 kΩ
10 kΩ
1
A
generator
produces
120 W
of
power
at
a
potential
4.0 V
of
12 kV. The
cables
in
a
the
of
power
total
is
transmitted
resistance
through
6 Ω. What
is
the
A
B
overhead
power
loss
10 kΩ
6 kΩ
cables?
0.6 mW
b
600 W
c
6 W
d
0.2 mW
C
2
The
diagram
below
shows
the
relationship
between
The
a
direct
current
I
in
a
conductor
and
the
V
across
it. When
V
<
1.5 V,
the
through
the
conductor
is
potential
difference
between C
between C
and A
and
B
is
is
V
V
current
and
. What
2
is
flowing
difference
1
the
difference
potential
potential
the
value
V
negligible.
−
V
1
a
?
2
−1.0 V
b
+1.0 V
c
+0.40 V
d
−0.40 V
I /mA
5
A
wire
of
length
0.6 m
is
travelling
at
a
speed
of
−1
500
12 ms
perpendicular
to
a
magnetic field
of flux
−5
density
400
e.m.f.
3.0
×
10
T. What
generated
is
between
the
the
magnitude
ends
of
the
−4
a
2.
16
×
10
of
the
wire?
−4
V
b
1.20
d
0V
×
10
V
300
−4
c
200
6
4.20
A flat
×
10
circular
V
coil
turns. A
current
What
the
has
of
a
radius
2.0 A
of
passes
2.5 cm
through
and
the
120
coil.
100
the
is
magnetic field
strength
at
the
centre
of
coil?
0
a
0
1
2
3
4
2.0 mT
b
3.0 mT
c
1.5 mT
d
6.0 mT
5
V/ V
7
Two
identical
connected
Which
of
the following
statements
about
in
capacitors
series
with
of
a
capacitance
potential
C
are
difference
of
V.
the
The
total
energy
stored
in
the
capacitors
is
E
1
conductor
is
correct?
Two
a
It
obeys Ohm’s
law
when
V >
1.5 V
and
identical
=
4 V,
its
resistance
is
of
capacitance
C
are
when
connected
V
capacitors
in
parallel
with
a
potential
difference
of
V.
10 Ω
The
total
energy
stored
in
the
capacitors
is
E
2
b
It
obeys Ohm’s
law
when
V
>
1.5 V,
but
its
E
1
resistance
is
constant.
What
is
the
value
of
?
E
2
c
It
does
its
not
obey Ohm’s
resistance
is
law,
but
when
V
>
law,
but
when
V =
1.5 V
a
It
does
not
resistance
obey Ohm’s
is
b
¼
c
4V
Two
horizontal
2.0 mm. The
Which
i
10 Ω
What
metal
lower
potential
d
2
of
the following
plates
plate
should
is
be
are
at
a
separated
potential
applied
to
the
by
of
Kirchhoff ’s first
ii
−8 V.
upper
is
correct?
its
conservation
3
4
10 Ω
8
d
½
Kirchhoff ’s
of
iii
Kirchhoff ’s
of
is
a
consequence
of
the
energy.
second
conservation
plate
law
law
is
a
consequence
of
the
charge.
second
law
states
that
the
−1
to
create
upwards
a
74
an
in
−14 V
electric field
the
space
b
of
strength
between
−2 V
c
the
3000 V m
algebraic
plates?
+2 V
in
d
+14 V
a
circuit
p.d.s
a
i
only
c
iii
only
sum
is
around
b
i
d
i,
of
the
equal
the
and
ii
e.m.f.s
to
the
loop.
ii
only
and
iii
around
algebraic
any
sum
loop
of
the
Module
9
The
diagram
B
the
in
below
plane
of
shows
the
two
paper. A
parallel
is fixed
wires A
and
B
and
is free
13
a
Explain
b
A
what
is
1
Practice
meant
by
the
exam
term
questions
drift velocity
to
[2]
move.
sample
area
of
A. The
a
conductor
main
charge
has
a
cross-sectional
carriers
are
electrons
and
I
each
A
I
per
has
unit
a
charge
volume
of
e. There
and
the
are
drift
n
charge
velocity
of
carriers
the
B
charge
When
the
opposite
same
current
directions,
I
passes
which
way
through
does
B
each
wire
current
in
Upwards
out
b
Away from A
c
Towards A
d
Downwards
in
of
the
paper
in
the
plane
the
plane
of
of
the
the
is
I, flowing
v.
Derive
through
an
the
expression for
the
conductor.
[4]
move?
14
a
carriers
The
I–V
characteristic of
a thermistor
is
shown
below.
I / mA
paper
paper
40
10
When
a
an
electron,
uniform
angles
to
out
of
the
travelling
magnetic field
its
path,
paper
which
in
a vacuum,
of flux
way
is
density
it
B
enters
at
30
right
deflected?
20
a
In
the
direction
of
B
into
b
In
the
direction
opposite
c
In
a
a
B
parabolic
into
a
path
circular
path
10
direction
perpendicular
to
B
into
a
circular
path
d
In
the
direction
of
B
into
a
parabolic
0
path
0
2
4
6
8
V/ V
Structured questions
The
11
a
Define
electric
b
Define
the
c
A
charge.
thermistor
is
connected
as
shown
below.
[2]
9V
current
for
a
SI
of
period
unit
of
charge.
2.
1 A flows
of
[2]
through
a
75 W
light
bulb
480 s.
200 Ω
Determine:
i
the
charge
that flowed
ii
the
energy
dissipated
iii
the
potential
iv
the
number
the
lamp.
through
the
bulb
[2]
R
in
the
difference
of
lamp
across
electrons
[2]
the
lamp
that flowed
[2]
through
The
12
A
battery
of
has
0.6 Ω. The
resistance
an
ammeter
e.m.f.
battery
is
70 mA. Calculate:
is
of
9V
and
internal
connected
across
resistance
a
resistor
a
the
current flowing
through
the
200 Ω
through
the
thermistor
b
the
current flowing
c
the
potential
d
the
value
a
Explain
resistor
[2]
[2]
of
difference
across
the
thermistor
[1]
12 Ω. Calculate:
a
the
current
b
the
potential
in
the
circuit
difference
c
the
power
dissipated
d
the
power
supplied
in
by
across
the
12 Ω
resistor
15
force
the
the
12 Ω
resistor
[2]
battery
[2]
battery.
of
the
power
is
dissipated
in
the
and
b
Define
c
State
the
the fraction
of
R.
[3]
[2]
[2]
e
reading
[2]
difference
terminal
between
potential
electromotive
difference.
[4]
the volt.
[1]
Kirchhoff ’s first
physical
law
on
and
which
second
each
is
laws
and
state
based.
[6]
the
[1]
d
In
the
circuit
negligible
internal
below,
internal
resistance
batteries
resistance.
of
2 Ω.
P
and Q
Battery
have
R
has
an
19
6 Ω
I
A
I
1
a
State:
4 Ω
2
i
Faraday’s
ii
Lenz’s
law
of
electromagnetic
induction
[1]
I
3
law.
[1]
R
b
Using
3 Ω
a
bar
explain
12 V
magnet
how
Lenz’s
and
law
coil
can
as
be
an
example,
considered
an
3V
example
of
the
law
of
conservation
of
energy.
[3]
internal
P
2 Ω
resistance
c
A flat
circular
coil
of
radius
1.8 cm
consists
of
Q
1.5 V
450
turns. The
uniform
B
i
is
placed
magnetic field
Calculate
the
coil
the
at
of flux
right
angles
density
magnetic flux
to
a
35 mT.
passing
through
coil.
[2]
Determine:
ii
i
the
currents
I
,
I
1
ii
the
potential
iii
the
terminal
and
I
2
The field
[8]
to
3
difference
potential
across A
and
difference
of
B
[2]
battery
do
is
reduced
so. Calculate
to
zero
the
and
e.m.f.
it
takes
induced
0.
15 s
in
the
coil.
[3]
R.
20
A
student
wants
to
construct
a
solenoid
to
produce
a
[2]
magnetic field
16
a
State Coulomb’s
b
Define
the
law for
terms
electrostatic
charges.
electric field strength
[2]
The
and
solenoid
has
38.0 cm. The
electric potential.
of flux
a
density
radius
solenoid
of
wire
of
25 mT
1.50 cm
carries
a
at
and
its
centre.
length
current
of
of
11 A.
[2]
Calculate:
c
State
the
relationship
between
electric field
a
strength
and
electric
potential.
the
minimum
that
d
Two
point
charges A
and
B
have
charges
−5 μC. A
and
B
are
10 cm
is
required for
of
this
turns
per
unit
length
solenoid
[3]
30 μC
b
and
number
[1]
the
total
length
of
wire
required
to
construct
the
apart. Calculate:
solenoid.
i
the
electric field
strength
at
the
21
between
the
two
[3]
midpoint
charges
Using
a
diagram,
explain
the
origin
of
the
Hall
effect.
[3]
[7]
ii
the
electric
between
potential
the
two
at
the
midpoint
charges.
[3]
On
your
diagram
indicate
the
direction
of
the
Hall
voltage.
17
A
12.0 μF
and
is
capacitor
then
is
charged
discharged
by
through
a
a
12.0 V
1.2 MΩ
d.c.
resistor.
22
a
Calculate
being
the
charge
on
the
capacitor
just
Calculate
c
After
the
5.0 s,
constant for
the
circuit.
the
charge
on
ii
the
p.d.
iii
the
current
across
in
capacitor
the
the
capacitor
b
[3]
a
Define
the
terms
magnetic flux density
and
negative
ions
travelling
at
−1
pass
0.
18 T
a
the
Write
undeflected
which
is
through
perpendicular
to
a
magnetic
an
electric
an
to
Write
labelled
magnetic
and
to
show
the
orientation
electric fields.
equation for
the
an
diagram
the force
[2]
acting
on
the
electric field.
equation for
ions
[1]
the force
acting
on
the
ions
the
due
tesla.
to
the
magnetic field.
[1]
[4]
d
b
spectrometer,
m s
Draw
due
[3]
c
18
of
of
[3]
circuit.
10
field.
a
the
mass
×
field
[1]
calculate:
i
a
2.2
[2]
time
In
5
before
discharged.
b
[1]
supply
Sketch
the
magnetic flux
pattern
due
to
a
Calculate
the field
strength
E
of
the
electric field.
long
[2]
straight
c
Write
an
density
d
Two
wire
carrying
equation
at
a
straight
separated
a
to
distance
wires X
distance
a
current
I.
determine
of
r from
and Y,
of
r. X
[3]
the
the
each
magnetic flux
wire.
of
carries
length
a
l
carries
a
current
of
I
.
2
the
same
I
and
1
I
a
Define
the
b
Derive
an
are
current
of
in
are flowing
the farad.
expression for
two
Explain
ii
State
why
wire X
whether
exerts
the force
is
a force
[3]
c
Three
initially
capacitance
on
attractive
down
wire Y.
wire Y.
[4]
or
[1]
an
expression for force
[2]
capacitors
uncharged
capacitors
of
in
2
repulsive.
76
and
1.5 μF,
0.25 μF
and
2.2 μF
are
direction.
i
Write
capacitance
series.
connected
iii
term
I
1
and Y
23
[2]
acting
on
[2]
to
a
12 V
battery
as
shown.
Module
1
Practice
exam
questions
12 V
2.2 μF
0.80 cm
1.5 μF
0.25 μF
8.5 cm
Calculate:
i
the
capacitance
of
the
1.5 μF
and
0.25 μF
ii
the
iii
[2]
combined
the
total
capacitance
charge
supplied
iv
the
charge
on
the
2.2 μF
v
the
charge
on
the
1.5 μF
in
by
the
the
circuit
[1]
battery
capacitor
and
the
[2]
[1]
the
coil
is
end
of
i
stored
in
the
1.5 μF
The
diagram
below
shows
a
circuit
capacitor.
of four
the
coil
a
Explain
[2]
ii
Each
connected
resistance
to
of
a
battery
of
e.m.f.
into
is free
a
the
a
of
why
the
this
of
the
straight
diameter
loops
circuit
to
is
of
8.5 cm
0.80 cm. The
such
that
the
lower
move.
current flows
loop
long
has
of
throught
loops
of
the
the
coil
coil,
the
decreases.
occurs.
coil
[4]
may
be
wire. When
considered
the
current
as
is
similar
9.0 V
through
the
coil,
a
mass
of
0.35 g
and
is
internal
wire
connected
flowing
resistors
the
separation
a
24
of
separation
When
[2]
energy
loop
and
0.25 μF
capacitors
vi
Each
in
series
hung from
the free
end
of
the
coil
in
order
0.
1 Ω
to
return
original
4 Ω
the
A
the
loops
separation
of
the
of
current flowing
coil
to
their
0.80 cm. Calculate
through
the
coil.
[4]
D
B
26
9V
uniform
electric field
metal
exists
plates. The
between
two
separation
of
the
plates
4 Ω
4 Ω
0.
1 Ω
A
horizontal
4 Ω
is
12 mm
and
a
potential
difference
of
1000 V
is
C
applied
across
the
plates. A
particle O
−19
4.8
at
×
10
the
plate
and
lower
by
of
charge
−27
C
mass
plate
the
and
5.0
is
×
10
kg
starts from
moved vertically
to
rest
the
top
electric field.
Calculate:
a
the
equivalent
resistance
of
the
external
through
the
battery
circuit
[3]
b
the
current flowing
c
the
potential
[2]
1000 V
A,
d
B, C
the
and
a
the
12 mm
resistors
[4]
current flowing
Define
across
D
resistors A,
25
difference
B, C
through
and
term
each
of
the
D.
O
[4]
magnetic flux
density
and
Calculate:
give
b
its SI
Sketch
field
a
unit.
[3]
diagram
around
a
to
long
show
the
straight
a
the
electric field
b
the
work
c
the
gain
strength
between
the
plates
done
on O
by
the
electric field
[2]
current-carrying
conductor.
in
gravitational
in
kinetic
potential
energy
[2]
of O
c
A
coil
of
wire
[2]
magnetic
consisting
suspended from
a fixed
of
two
point
loops
as
[2]
is
d
the
gain
e
the
speed
energy
of O
[1]
shown.
of O
when
it
reaches
the
top
plate.
[2]
77
8
Alternating
8.
1
Alternating
Learning outcomes
currents
currents
Alternating
Electricity
On
completion
of
this
section,
be
able
use
the
terms
frequency,
of
when
and
root mean square value
applied
or
to
an
electrons
in
one
graphically
use
the
is
transmitted
differs
from
a
using
direct
an
alternating
current
(d.c.)
in
current
that
reverses
direction
regularly
with
time.
A
direct
the
current
direction
as
shown
only.
in
An
alternating
Figure
current
can
be
represented
8.1.1.
alternating
current
shown
can
be
represented
mathematically
by
the
voltage
following
ᔢ
home
current
alternating
The
current
your
alternating
peak
flows
value
An
to:
flow
ᔢ
reaching
you
(a.c.).
should
currents
equation
x
=
x
sin ωt
equation.
to
0
represent
or
an
alternating
current
Equation
voltage
I
=
I
sin ωt
ω
=
2πf
0
ᔢ
discuss
the
advantages
of
using
−1
I
alternating
currents
and
I
voltages for
the
–
the
value
peak
of
the
current
(at
time
t)/A
ω
–
angular frequency/rad s
f
–
frequency/Hz
high
transmission
–
the
ω
–
angular frequency/rad s
t
–
time/s
of
value
of
the
current/A
0
−1
electrical
energy.
The
peak
value
of
the
current
I
is
the
maximum
value
of
the
current.
0
current
The
period
T
of
the
alternating
current
is
defined
as
the
time
taken
for
I
0
one
complete
cycle.
I
r.m.s.
The
frequency
The
frequency
f
is
the
number
of
complete
cycles
generated
per
second.
0
T
and
period
are
related
by
the
following
equation:
time
1
f
=
T
Figure 8.1.1
Power
An alternating current
The
current
with
time.
time.
and
considerations
This
times
supplied
and
As
a
the
means
when
by
the
voltage
result
the
that
it
is
a.c.
at
there
a
supplied
power
are
times
maximum.
mains
varies
by
a
supplied
power
also
when
Figure
with
station
varies
the
vary
supplied
8.1.2
sinusoidally
sinusoidally
shows
power
how
with
is
the
zero
power
time.
Definition
The
root
mean
square
value
of
an
alternating
current,
I
, is
the
value
of
the
r.m.s.
steady
direct
resistive
current
which
delivers
the
load.
power
Equation
I
0
I
=
r.m.s.
2
√
I
–
root
mean
square
current/A
r.m.s.
I
–
0
time
Figure 8.1.2
78
peak
value
of
current/A
same
average
power
as
the
a.c.
to
a
Chapter
8
Alternating
currents
Example
An
alternating
power
supply
is
represented
by
the
following
equation
sin ωt
can
in
volts:
V
=
230 sin 314 t
State:
a
the
peak
b
the
r
.m.s.
c
the
frequency
a
V
=
value
of
value
the
of
of
power
the
the
supply
power
power
supply
supply.
230 V
0
V
230
0
b
V
=
=
=
163 V
r
.m.s.
2
2
√
c
√
Comparing
the
voltage
expression
with
V
=
V
we
see
that
0
ω
=
f
=
2πf
=
314
314
∴
=
50 Hz
2π
Transmission of
Electrical
energy
distributed
supplied
by
by
is
electrical
generated
transmission
the
power
in
power
cables
station
energy
is
to
stations.
homes
given
by
P
This
and
=
energy
factories.
IV.
The
is
then
The
power
power
loss
in
the
2
transmission
cables
transmit
the
and
cur rent.
high
power
by
power,
The
By
using
is
down
by
stepping
ver y
is
this
to
cur rent
current
using
up
=
I
voltage
high
R.
is
the
and
the
square
at
That
the
power
or
a
designed
is
to
say,
voltages
loss
in
cur rent
losses
station
cur rent
is
high
power
of
power
low
voltages.
that
smaller,
The
network
transmitted
and
field
operate
magnetic
is
used
because
transformers
stepping
electromagnetic
magnetic
not
at
high
P
transmission
energy
doing
the
a
by
in
low
to
voltage
a
given
low
cur rents.
transmission
flowing
the
either
transmit
for
and
the
can
through
transmission
them.
lines
minimised.
Alternating
of
at
proportional
making
are
for
given
The
a.c.
electrical
reason
cables
power
is
is
with
(see
down
induction,
produced
direct
by
it
which
easily
be
stepped
T
ransformers
voltages.
the
current,
can
8.2).
is
They
operate
possible
alternating
however
,
are
using
because
current.
because
up
very
an
or
stepped
efficient
the
alternating
T
ransformers
there
is
at
principle
no
do
changing
field.
Key points
ᔢ
In
an
with
ᔢ
The
alternating
current
the flow
of
electrons
reverses
direction
regularly
time.
peak
value
of
an
alternating
current
is
the
maximum
value
of
the
current.
ᔢ
The
root
mean
square
I
value
of
an
alternating
current
is
the
value
of
r.m.s.
the
to
ᔢ
steady
a
direct
resistive
Electrical
current
which
delivers
the
same
average
power
as
the
a.c.
load.
energy
is
transmitted
using
a.c.
at
high
voltages.
79
8.2
The
transformer
Learning outcomes
The
simple
T
ransfor mers
On
completion
of
this
section,
be
able
are
devices
used
to
change
the
voltage
supply.
Figure
8.2.1
shows
a
simple
iron-cored
explain
of
a
the
simple
principle
of
operation
V
s
use
the
relationship
ᔢ
ideal
discuss
the
input
output
V
V
p
s
I
p
s
transformer
energy
transformer
are
s
p
=
V
p
an
I
p
I
s
=
N
for
alternating
transformer
N
ᔢ
an
transformer
.
to:
I
ᔢ
of
you
power
should
iron-cored transformer
and
losses
state
in
how
a
they
minimised.
Figure 8.2.1
A
a
A simple iron-cored transformer
transformer
soft
coil,
the
iron
the
soft
the
in
field,
the
because
in
the
The
core.
two
transmitted
field
When
alternating
iron
magnetic
that
consists
core.
an
soft
they
primary
the
and
an
secondary
e.m.f.
electrically
primary
is
to
a
is
in
it.
each
secondary
not
wound
in
the
It
this
coil
to
field
be
in
noted
Energy
via
with
field
around
primary
alternating
should
other
.
work
magnetic
to
magnetic
situated
from
do
coil
connected
induced
the
varying
is
alternating
isolated
coil
secondary
coil
T
ransformers
produce
a
supply
produces
the
core.
not
secondary
primary
alternating
alternating
are
iron
do
a
current
Since
coils
from
of
an
the
is
magnetic
direct
induce
currents
an
e.m.f.
coil.
coil
has
N
turns
of
copper
wire
and
the
secondary
coil
p
has
N
turns
of
copper
wire.
If
we
apply
an
alternating
voltage
V
s
on
the
p
primary
coil
we
get
an
alternating
voltage
V
on
the
secondary
coil.
The
s
currents
in
the
primary
and
secondary
coils
are
I
and
I
p
an
ideal
transformer
the
input
power
is
equal
to
respectively.
For
s
the
output
power
.
Equation
For
an
ideal
Power
input
I
V
p
I
transformer
=
power
=
I
p
current
–
voltage
–
current
–
voltage
in
power
losses)
output
V
s
–
(no
s
the
primary
coil/A
p
V
across
primary
coil/V
p
I
in
the
secondary
coil/A
s
V
across
the
secondary
coil/V
s
Equation
There
N
is
a
relationship
that
relates
the
voltages
across
the
coils
to
the
V
p
p
number
=
N
of
turns
in
each
coil.
The
ratio
of
the
voltages
is
equal
to
the
V
s
N
s
–
ratio
number
of
turns
of
wire
on
p
primary
V
coil
–
voltage
–
number
we
up
across
primary
of
can
or
the
number
adjust
step
the
down
a
of
turns.
output
This
voltage.
voltage
means
that
Therefore
depending
on
the
a
if
we
adjust
transformer
turns
the
can
turns
either
ratio,
step
ratio.
coil/V
p
For
a
step-up
transformer
N
>
N
s
N
of
turns
of
wire
and
the
transformer
will
step
up
the
p
on
s
voltage.
secondary
V
–
coil
voltage
across
secondary
coil/V
For
a
the
voltage.
step-down
transformer
,
N
<
s
s
80
N
and
p
the
transformer
will
step
down
Chapter
8
Alternating
currents
Exam tip
ᔢ
Either
use
peak
throughout
ᔢ
Whenever
are
ᔢ
more
ᔢ
A
the
Energy
In
ideal
a
real
about
ᔢ
up
efficient.
energy
step-up
coil
of
the
a
than
the
step-down
voltage
a
by
use
transformer,
on
coil
or
values
ensure
primary
on
the
the
power
ensure
primary
certain factor,
that
there
coil.
transformer,
than
Remember
r.m.s.
two.
but
it
will
output
that
coil.
reduce
cannot
be
input.
a transformer
has
transformer
,
a
a
calculations
mix
secondary
same factor.
in
transformer
the
the
not
of
diagram
on
power
losses
Thermal
diagram
step
that
the
Do
secondary
a
turns
will
by
than
98%
a
the
sketching
current
greater
An
on
are fewer
transformer
throughout
calculations.
sketching
turns
Whenever
there
values
the
100%
there
are
efficiency
power
T
ransformers
is
lost
losses.
lose
because
and
of
A
some
the
therefore
real
has
no
power
transformer
power
because
resistance
of
of
the
can
the
losses.
be
following:
windings
of
the
2
coils
(P
=
I
R).
In
order
to
reduce
this,
the
windings
are
made
of
thick
copper
.
ᔢ
Thermal
up
as
energy
the
made
of
soft
iron
demagnetises
magnetic
ᔢ
Thermal
in
a
in
flux
the
core
of
continually
instead
more
energy
angles
currents
lost
of
easily.
steel
Soft
the
transformer
.
reverses
because
iron
is
The
direction.
soft
iron
therefore
core
(The
magnetises
said
to
have
warms
core
a
is
and
high
permeability.)
changing
right
is
magnetic
is
lost
in
magnetic
to
begin
the
eddy
field,
magnetic
flowing.
currents.
e.m.f.s
field.
These
As
are
Since
currents
the
soft
induced
the
are
soft
in
iron
the
iron
known
core
soft
core
as
is
eddy
is
situated
iron
a
core
at
conductor
,
currents.
2
These
order
is
of
currents
to
reduce
made
iron.
up
of
the
energy
energy
numerous
e.m.f.s
currents
cause
are
produced
still
to
be
losses
thin
due
sheets
induced
cannot
dissipated
flow
in
to
of
the
easily
(P
eddy
I
R)
currents,
metal
soft
=
rather
iron
between
the
the
than
core,
the
in
but
sheets
iron
soft
a
single
the
of
core.
iron
In
core
block
induced
metal.
Example
An
ideal
transformer
secondary
is
turns
connected
secondary
to
coil.
to
is
the
the
constructed
number
primary
Calculate
V
coil
the
N
p
such
that
primary
and
a
current
4 Ω
in
the
turns
ratio
is
resistor
the
of
1 : 15.
is
primary
the
A
number
240 V
connected
to
of
supply
the
coil.
15
p
=
V
of
=
N
s
1
s
240
15
Key points
=
V
1
s
240
V
oltage
across
secondary
coil
V
=
=
ᔢ
16 V
A
transformer
can
step
up
or
step
s
15
down
V
flowing
through
4 Ω
resistor
=
=
=
R
I
V
p
=
p
I
voltage.
4 A
ᔢ
Transformers
ᔢ
Energy
only
work
on
a.c.
4
is
lost
in
a
transformer
V
s
s
as
I
a
result
of
resistance
of
eddy
currents,
V
s
I
alternating
16
s
Current
an
s
=
the
copper
windings
p
V
p
and
16
Therefore,
current
flowing
in
primary
coil
×
=
the
changing
magnetic field
4
=
0.27 A
in
the
iron
core.
240
81
8.3
Semiconductors
Learning outcomes
p-type
and
n-type
Semiconductor
On
completion
of
this
section,
be
able
describe
the
electrical
semiconductors
distinguish
n-type
and
between
as
silicon
silicon
In
p-type
and
materials
explain
a
its
pure
process
silicon
the formation
of
discuss
layer
at
a
p-n
the flow
p-n
biased
of
junction
or
current
discuss
diode
is forward-
reverse-biased
the
I–V
p-n
three
recall
that
a
name
implies,
nor
a
good
have
conductivities
conductor
.
Pure
that
elements
the
four
are
called
its
electrons
electrons
an
conductive
valence
are
impurity
shell
(e.g.
in
intrinsic
held
its
semiconductors .
outermost
tightly
(another
properties.
in
If
an
is
valence
covalent
element)
phosphorus)
or
In
is
bonds.
added
element
added
to
shell.
to
having
In
the
five
silicon,
an
is
formed.
Four
of
the
valence
shell
electrons
are
used
for
n-type
bonding
is
one
left
over
.
There
is
in
which
is
its
formed.
means
that
allows
therefore
valence
The
one
shell
three
bond
for
an
excess
is
conduction.
(e.g.
valence
short
of
boron)
shell
one
is
If
of
an
negative
element
charge
added
electrons
electron.
to
are
This
having
silicon,
used
a
for
p-type
bonding.
to
as
a
hole.
There
is
therefore
an
excess
of
deficiency
positive
is
charge
carriers
diode
junction
two
its
are
of
transistor
p-n
which
allows
for
conduction.
is
The
basically
all
enhance
electrons
(holes)
ᔢ
germanium
doping,
(electrons)
material
characteristic
junction
state,
in
there
referred
the
the
when
This
ᔢ
as
insulator
junction
carriers
the
good
the
and
ᔢ
a
there
called
to
material
depletion
and
atom,
electrons
ᔢ
neither
properties
a
of
them
to:
such
ᔢ
materials,
you
make
should
materials
amount
of
impurity
added
to
the
silicon
is
approximately
1
part
junctions.
6
impurity
atoms
affects
Definitions
silicon
A
p-type
material
is
one
in
which
of
charge
carriers
are
10
produce
the
is
6
parts
pure
semiconductor
.
electron–hole
resistivity
of
the
approximately
pairs
doped
for
This
means
conduction.
semiconductor
.
that
The
The
only
amount
1
in
10
added
resistivity
of
0.1–60 Ω m.
the
In
majority
to
a
pure
metal
such
as
copper
free
electrons
are
the
main
charge
carriers.
holes.
Each
copper
atom
produces
at
least
one
conduction
electron.
Copper
−8
An
n-type
the
material
majority
of
is
one
charge
in
which
carriers
therefore
has
a
low
resistivity
(ρ
=
1.68
×
10
Ω m).
are
electrons.
The
p-n
junction
movement
of
‘holes’
Definition
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
electrons
The
depletion
either
there
side
are
of
no
region
the
net
p-n
is
the
region
junction
charge
‘holes’
on
where
n-type
carriers.
p-type
depletion
layer
+
movement
–
of
electrons
direction
diffusion
drift
Figure 8.3.1
The p -n junction
n
Forward-biased junction
Consider
what
material.
The
across
–
the
p-type
material.
drift
p-n
current
positive
The
into
n
gained
Reverse-biased junction
After
of
a
p-type
in
the
the
material
n-type
p-type
This
and
the
has
respect
across
acquires
prevents
time,
diffuse
material
with
electrons
electrons
some
(holes)
n-type
potential
material.
82
when
carriers
boundary
material
movement
Figure 8.3.3
happens
charge
is
placed
material
material.
against
(electrons)
The
charge
an
n-type
diffuse
carriers
in
+
the
p
current
current
p
Figure 8.3.2
of
to
a
lost
the
the
further
across
the
some
p-type
movement
electrons
of
The
charge
movement
boundary
material.
boundary.
negative
p-n
of
electrons
with
n-type
a
prevents
material
respect
across
holes
the
acquires
This
p-type
holes
and
and
into
to
the
stops
further
has
the
n-type
boundary.
because
of
the
Chapter
potential
difference
across
the
junction.
This
potential
difference
is
8
Alternating
currents
called
I /A
the
contact
become
or
bar rier
depleted
of
potential.
electrons
The
and
region
holes
is
around
called
the
the
boundary
depletion
that
region
has
or
forward
depletion
layer.
The
depletion
region
is
about
1 µm
thick.
current
breakdown
voltage
A
diffusion
cur rent
concentration
of
is
a
holes
current
and
that
occurs
electrons
in
a
because
p-n
of
junction
a
difference
(Figure
in
the
8.3.1).
leakage
Forward-biased
p-n
V/ V
current
0.7 V
avalanche
junction
current
If
a
cell
is
connected
across
a
p-n
junction
so
that
the
positive
terminal
is
reverse
connected
to
the
the
cell.
in
the
Electrons
of
potential,
p-type
material
‘Holes’
terminal
same
the
n-type
biased.
of
to
the
cell.
electrons
time,
depletion
‘holes’
region
material
(Figure
p-type
in
the
When
are
are
then
and
the
8.3.2),
the
material
n-type
the
pushed
from
decreases
is
said
repelled
by
the
material
from
in
are
voltage
the
the
terminal
junction
are
applied
pushed
negative
repelled
is
n-type
p-type
width.
A
to
to
be
forward-
the
than
p-type
the
drift
to
cur rent
Figure 8.3.4
terminal
The I–V characteristic of a
semiconductor diode
negative
the
barrier
material.
n-type
voltage
connected
positive
by
greater
is
At
material.
flows
the
The
across
collector
collector
the
n
base
p-n
junction.
applied
This
across
a
is
p-n
a
current
that
occurs
when
a
potential
difference
p
base
is
n
junction.
emitter
emitter
Reverse-biased
If
a
cell
is
connected
connected
to
the
to
the
p-type
biased.
p-n
across
n-type
material
‘Holes’
in
the
Figure 8.3.5
junction
a
p-n
junction
material
(Figure
p-type
and
8.3.3),
the
the
material
so
that
the
negative
junction
are
positive
terminal
is
attracted
said
to
is
to
the
terminal
An n-p -n transistor
is
connected
be
collector
reverse-
collector
p
negative
base
base
terminal
of
the
cell.
Electrons
in
the
n-type
material
are
attracted
n
to
p
the
positive
terminal
of
the
cell.
Electrons
and
‘holes’
move
away
from
emitter
emitter
the
junction
potential
the
A
drift
p-n
and
also
the
depletion
increases.
current
junction
ceases
is
region
Eventually,
to
referred
increases
electrons
in
and
width.
The
‘holes’
stop
barrier
moving
and
Figure 8.3.6
A p -n-p transistor
flow.
to
as
a
junction
diode.
The
junction
diode
is
Key points
used
to
rectify
alternating
currents.
ᔢ
The
I–V
characteristic of
a
A
semiconductor
neither
semiconductor diode
good
Figure
8.3.4
difference
shows
across
configuration.
When
the
the
the
Between
potential
conducting.
A
drift
in
reverse-biased
a
potential
minimal
particular
characteristic
is
0 V
current
the
the
V
positive,
and
difference
When
The
I
diode
now
flows
across
configuration.
that
called
no
increases
difference
current
voltage,
0.7 V
,
does
the
a
diode.
diode
the
is
is
current
beyond
through
The
flow
of
the
diode
breakdown
the
flows
0.7 V
,
the
through
the
the
diode
ᔢ
negative,
flow
the
is
the
diode
current.
diode
is
to
ᔢ
a
very
large
a
A
p-type
are
amount
of
adding
an
to
a
impurity
enhance
properties.
material
majority
of
of
is
one
charge
in
which
carriers
suddenly
An
n-type
material
is
one
in
the
majority
are
of
charge
electrons.
junction transistor
junction
be
n-type
made
transistor
such
materials
transistor
.
It
can
that
is
a
essentially
p-type
(Figure
8.3.5).
also
made
be
made
material
This
such
is
type
that
up
of
two
p-n
sandwiched
of
an
transistor
n-type
junctions.
between
is
called
material
is
It
two
p-type
materials
(Figure
8.3.6).
This
type
of
depletion
either
where
two
an
The
on
side
there
p-n-p
are
the
no
is
the
p-n
net
region
junction
charge
sandwiched
transistor
The
junction
transistor
is
is
essentially
a
region
of
carriers.
n-p-n
ᔢ
between
called
a
current.
ᔢ
can
is
nor
holes.
carriers
The
process
semiconductor
which
The
the
conductive
the
zero.
At
ᔢ
conducts
is
controlled
diode.
begins
practically
leakage
voltage,
the
material
conductor
insulator.
Doping
its
is
good
potential
forward-biased
it.
current
called
When
in
a
two
p-n
junctions.
transistor
.
83
8.4
Rectification
Learning outcomes
Rectification
Rectification
On
completion
of
this
section,
is
a
process
whereby
an
a.c.
voltage
is
converted
into
a
d.c.
you
voltage.
should
ᔢ
use
be
a
able
to:
diode for
half-wave
Half-wave
rectification
ᔢ
use
the
wave
bridge
rectifier for full-
A
single
how
rectification
diode
this
voltage.
ᔢ
represent
half-wave
is
rectification
discuss
the
smoothing
use
a
of
be
used
The
graphs
The
in
to
rectify
input
Figure
an
supply
8.4.2
a.c.
to
show
voltage.
the
the
Figure
circuit
input
is
an
8.4.1
shows
alternating
difference
across
the
load
(output
voltage
voltage).
and
When
the
the
input
graphically
voltage
ᔢ
can
achieved.
and fullpotential
wave
rectification
a
capacitor for
rectified
a.c.
wave.
is
conducts
diode
is
positive
an
current
always
either
an
voltage
is
zero
positive
into
half-cycle),
current.
reverse-biased
output
a.c.
(positive
electric
a
and
in
or
When
does
this
not
The
diode
input
conduct
half-cycle.
zero.
d.c.
the
the
an
Notice
input
is
forward-biased
voltage
is
negative,
electric
that
voltage
has
current.
the
output
been
and
the
The
voltage
converted
is
from
voltage.
diode
input
voltage
a.c.
load
time
Figure 8.4.1
Half-wave rectification
output
voltage
time
Figure 8.4.2
Graphical representation of half-wave rectification
Full-wave
Four
diodes
manner
,
how
and
the
is
be
A
used
process
is
achieved.
potential
Consider
cycle).
can
the
this
rectification
the
rectify
called
The
flows
an
in
across
when
a.c.
full-wave
graphs
difference
situation
current
to
the
through
the
,
When
rectification.
Figure
input
D
voltage.
8.4.4
load
voltage
because
it
Figure
show
(output
is
is
used
the
in
this
8.4.3
shows
input
voltage
voltage).
positive
(positive-half
forward-biased.
The
2
current
supply,
then
the
flows
current
through
flows
the
load.
through
D
On
.
its
return
Consider
back
the
to
the
situation
a.c.
when
power
the
4
input
D
,
voltage
because
is
it
negative
is
(negative-half
forward-biased.
The
cycle).
current
A
current
then
flows
flows
through
through
the
load.
3
On
its
return
back
to
the
a.c.
power
supply,
the
current
flows
through
D
.
1
Notice
a.c.
voltage
output
84
that
full-wave
into
voltage
a
d.c.
graphs
rectification
voltage
for
is
than
both
much
more
half-wave
methods.)
efficient
at
rectification.
converting
(Compare
an
the
Chapter
8
Alternating
currents
P
D
D
1
2
input
S
voltage
Q
D
D
4
load
3
time
R
Figure 8.4.3
Full-wave rectification
output
voltage
Function of the
In
Figure
8.4.3
a
capacitor
capacitor
is
inserted
in
parallel
with
the
load.
time
The
output
function
of
this
capacitor
is
to
help
smooth
the
output
voltage
so
that
voltage
it
has
fewer
ripples.
When
the
output
voltage
increases,
the
capacitor
with
charges.
to
τ
As
soon
discharge
=
CR.
as
the
through
When
a
output
the
larger
load.
voltage
The
begins
time
capacitance
is
to
fall,
constant
placed
in
for
the
the
parallel
capacitor
circuit
with
the
begins
is
smoothing
capacitor
load,
the
time
time
constant
time
to
even
smaller
increases.
discharge.
by
The
This
size
increasing
of
means
the
the
that
ripples
size
of
the
in
the
capacitor
the
output
takes
a
voltage
longer
is
made
Figure 8.4.4
Graphical representation of
full-wave rectification
capacitor
.
Key points
ᔢ
Rectification
is
a
process
whereby
an
a.c.
voltage
is
converted
into
a
d.c.
voltage.
ᔢ
Half-wave
ᔢ
Full-wave
ᔢ
A
rectification
capacitor
output
ᔢ
rectification
inserted
in
is
is
achieved
achieved
parallel
by
by
with
using
a
single
using four
the
load
diode.
diodes.
helps
reduce
the
ripples
in
the
voltage.
Increasing
the
circuit. This
size
of
reduces
the
the
capacitor
ripples
in
increases
the
output
the
time
voltage
constant
in
the
even further.
85
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
5
4
a
Explain:
accompanying CD.
i
why
the
supply
alternating
1
a
Explain
what
is
meant
by
the
root
mean
of
an
alternating voltage.
why for
An
alternating voltage
V is
primary
and
not
coil
direct
must
be
current
a
constant
power
input,
the
[2]
output
[2]
current
b
the
square
ii
(r.m.s.) value
to
current
represented
by
must
decrease
if
the
input voltage
the
increases.
equation
V =
120sin (120πt),
where
V
is
b
in volts
and
t
is
in
[2]
measured
The
graph
below
shows
the variation
with
seconds.
time t
of
the
current
I
in
the
primary
coil
of
a
p
Determine:
c
transformer.
i
the
peak voltage
ii
the
r.m.s. voltage
iii
the frequency.
The
[1]
alternating voltage
such
that
the
I
[1]
P
[1]
mean
is
applied
power
across
output from
a
resistor
0
the
t
resistor
is
1.2 kW. Calculate
the
resistance
of
the
resistor.
2
An
[2]
alternating voltage
20 V
and frequency
of
supply
has
a
peak value
of
60 Hz.
i
Sketch
Determine:
a
the
time
period
b
the
root
c
the
mean
square voltage
ii
[2]
Sketch
time
power
dissipated
in
a
resistor
when
graph
the
to
show
the variation
magnetic flux
in
the
core
with
of
the
transformer.
[2]
mean
a
of
a
of
[2]
graph
the
to
show
e.m.f.
the variation
induced
in
the
with
secondary
the
coil.
alternating
resistor
of
supply
is
connected
resistance
in
series
with
[3]
a
2.0 Ω
[2]
6
An
ideal
coil.
It
is
transformer
used
to
has
4000
convert
a
turns
mains
on
its
supply
of
primary
220 V
r.m.s.
3
Explain
what
is
meant
by
the following
terms:
to
a
A
semiconductor
b
Doping
c
A
d
An
e
Depletion
a
Describe
an
The
semiconductor
4
[1]
region
[1]
structure
an
iron-cored
and
principle
of
coil
The
peak
power
The
transformer
to
a
has
a
transformer
turns
8V
Assuming
ratio
input voltage
that
the
the
the
6.0 V.
is
connected
rectified
to
a
using
resistor
a
R.
number
of
turns
on
the
secondary
transformer.
a
diagram
c
Sketch
a
graph
the
is
to
of
[3]
a full-wave
show
rectifier.
the variation
[4]
voltage
across
the
with
time
resistor.
t
[2]
Add
of
N
has
/N
is
circuit
component X
to
your
diagram
in
b
=
show
how
the voltage
across
the
resistor
can
20
p
a value
transformer
a
60 W.
be
sinusoidal
of
Sketch
to
the
of
[4]
s
and
peak value
operation
transformer.
input
transformer
and
b
d
b
the
rectifier
Calculate
of
of
a
[1]
semiconductor
the
output from
full-wave
a
n-type
having
[1]
[1]
p-type
alternating voltage
smoothed.
[1]
of
e
Explain
f
On
how
component X
works
in
the
circuit.
[2]
ideal,
r.m.s.
your
graph
in
c
show
what
happens
to
the
calculate:
voltage
i
the
r.m.s.
value for
the
output
ii
the
mean
iii
the
r.m.s. value
of
the
input
iv
the
r.m.s.
of
the
output
voltage
is
power
input
current
capacitor
when
component X
added.
[2]
[2]
current.
For
an
ideal
transformer:
[1]
a
explain
of
b
why
the
coils
are
wound
on
a
core
made
iron
explain
core.
86
the
[1]
7
value
across
[2]
[1]
why
thermal
energy
is
generated
in
the
[2]
Revision questions
8
a
Explain
for
the
why
it
is
efficient
necessary
to
use
transmission
of
high voltages
electrical
Using
the
4
graph:
energy.
i
determine
the
peak
ii
determine
the
root
iii
calculate
voltage
[1]
[3]
b
Explain
why
it
current
when
is
advantageous
transmitting
to
use
electrical
alternating
energy.
[2]
In
a
real
transformer,
the
core
on
which
the
secondary
coils
are
wound
is
energy
losses
due
to
currents
induced
in
[1]
alternating
[2]
capacitor C
has
a
capacitance
of
10.0 μF.
a
single
discharge
of
the
capacitor
through
R,
use
the
determine
resistor
the
change
diagram
in
potential
to:
the
change
in
charge
the
iv
core.
the
laminated. This
the
reduces
of
primary
For
and
the frequency
square voltage
voltage.
The
9
mean
Explain:
difference
a
how
these
currents
arise
in
the
core
v
b
why
laminating
the
core
reduces
energy
determine
to
the
currents.
of
charger for
a
car
the
calculate
resistor
A
each
capacitor
[2]
[2]
vi
10
on
losses
plate
due
[1]
[3]
battery
operates
using
a
the
average
current
R
in
the
[2]
120 V
r.m.s.
vii
a.c.
supply. The
The
battery
current
11
An
charger
drawn from
alternating
resistor
the
is
provides
90%
the
supply
I
is
is
given
in
at
and
in
is
of
R.
[2]
the
series with
t of the
t
value
[4]
expression
amps
the
14.0 V.
supply.
connected
by the
measured
d.c.
efficient. Calculate
a.c.
is
10 A
R. The variation with time
resistor
where
charger
estimate
I
=
a
current
8.2 sin
measured
I
in
(377t),
in
seconds.
a
Calculate
the frequency
b
Calculate
the
c
Given
that
resistor
A
sinusoidal
to
a
bridge
output
and
a
a
of
the
of
the
power
exceed
value
of
the
dissipated
350 W,
in
calculate
the
the
R.
[2]
consisting
rectifier
[2]
[2]
alternating voltage
rectifier
supply.
current.
mean
cannot
minimum
12
r.m.s.
is
supply
of four
connected
is
connected
diodes. The
to
a
resistor
R
capacitor C.
Sketch
a
diagram
of
the
circuit
described
above.
b
State
the
c
The
[3]
the function
of
the
capacitor
in
circuit.
[1]
variation
difference
V
with
time
across
the
t
of
the
resistor
potential
R
is
shown
below.
V/V
15
10
5
0
10
20
30
40
50
time/ms
87
9
Analogue
9.
1
Transducers
Learning outcomes
electronics
Input devices
The
On
completion
of
this
section,
LDR,
electronic
should
ᔢ
be
able
explain
the
use
of
the
light-
resistor
(LDR),
thermistor
and
microphone
The
the
light-dependent
light-dependent
when
exposed
describe
the
operation
light-emitting
buzzer
are
used
as
input
devices
in
resistor
(LDR)
to
resistor
(LDR)
sunlight.
is
When
a
device
placed
in
whose
an
resistance
enclosure
decreases
(complete
devices
darkness),
ᔢ
microphone
the
A
input
and
circuits.
to:
dependent
as
thermistor
you
and
diode
the
relay
of
the
(LED),
as
the
sunlight,
an
LDR
its
its
is
resistance
resistance
shown
in
is
approximately
decreases
Figure
to
10 M Ω.
When
approximately
100
exposed
Ω.
The
to
bright
symbol
for
9.1.1.
output
devices.
Figure 9.1.1
The symbol for an LDR
The thermistor
A
thermistor
is
A
thermistor
having
resistance
a
device
decreases
whose
negative
with
increasing
Figure
varies
temperature
is
Figure 9.1.2
The symbol for a thermistor
A
in
resistance
thermistor
The
shown
a
with
temperature.
coefficient
temperature.
The
is
one
whose
symbol
for
a
9.1.2.
microphone
microphone
signals.
The
Figure 9.1.3
is
a
device
circuit
that
symbol
converts
for
a
sound
microphone
waves
is
into
shown
in
electrical
Figure
9.1.3.
The symbol for a microphone
Output devices
The
LED,
buzzer
and
relay
are
used
as
output
devices
in
electronic
circuits.
The
A
light-emitting diode
light-emitting
flows
through
through
anode
the
88
is
it
in
diode
it.
(LED)
Since
one
begins
is
direction.
approximately
diode
it
to
(LED)
is
a
a
The
1.8 V
conduct,
device
diode,
LED
more
the
an
that
emits
electric
only
begins
positive
potential
light
current
than
to
the
difference
when
can
conduct
it
current
flow
when
cathode.
across
a
only
the
When
remains
Chapter
at
1.8 V
,
whatever
an
the
symbol
for
usually
connected
LED
is
magnitude
shown
in
of
the
Figure
current
9.1.4.
A
flowing
through
protective
it.
resistor
9
Analogue
electronics
The
is
protective
in
series
with
an
LED.
The
resistor
limits
the
current
resistor
flowing
through
the
LED
and
prevents
damage
to
it.
+
Figure 9.1.4
LED
–
anode
(forward
cathode
biased)
The symbol for an LED
Figure 9.1.5
LED with a protective
resistor
The
The
buzzer
buzzer
is
a
device
connected
to
it.
connected
to
a
in
Figure
A
solid-state
an
sound
electric
sounder
.
The
when
a
oscillator
symbol
power
circuit
for
the
supply
which
sounder
is
is
is
shown
The symbol for a buzzer
is
a
device
controlled
by
electromagnet,
the
produces
of
relay
relay
are
that
consists
9.1.6.
Figure 9.1.6
The
It
switch
closed’
that
an
the
relay
contacts.
(NC).
For
is
energised.
For
is
energised.
The
a
a
consists
of
one
electromagnet.
becomes
Relays
can
normally
normally
symbol
be
open
a
more
a
energised
closed
for
or
When
and
‘normally
relay,
relay,
relay
is
switch
current
the
the
can
open’
in
or
close
contacts
open
Figure
NO
Figure 9.1.7
(NO)
which
through
either
contacts
shown
contacts
flows
close
or
the
open
‘normally
when
the
when
the
relay
relay
9.1.7.
NC
The symbol for a relay
Key points
ᔢ
The
LDR,
thermistor
and
microphone
buzzer
relay
are
used
as
input
devices
in
electronic
circuits.
ᔢ
The
LED,
and
are
used
as
output
devices
in
electronic
circuits.
89
9.2
Operational
Learning outcomes
Properties of
An
On
completion
of
this
amplifiers
section,
operational
be
able
and
describe
the
properties
of
ᔢ
operational
compare
one
output.
(op-amp)
is
Sometimes
an
integrated
there
are
circuit
more
than
(IC).
just
It
has
two
operational
the
amplifier
properties
of
in
a
single
integrated
circuit.
There
will
also
be
connections
the
for
ideal
amplifier
amplifier
to:
amplifiers
ᔢ
ideal operational
you
inputs
should
an
supplying
operational
a
power
to
amplifier
the
and
operational
the
symbol
amplifier
.
used
to
Figure
represent
9.2.1
shows
an
it.
real
+V
s
op-amp
an
with
ᔢ
use
ᔢ
understand
an
op-amp
ideal
as
a
op-amp
inverting
input
comparator
output
+
the
term
saturation
non-inverting
input
V
s
Figure 9.2.1
The
An operational amplifier
properties
ᔢ
It
has
an
ᔢ
Essentially
of
an
ideal
infinite
we
operational
input
are
amplifier
are
as
follows.
impedance.
saying
that
no
current
enters
it
through
either
of
its
inputs.
ᔢ
It
ᔢ
This
has
an
is
there
infinite
the
were
output
ᔢ
It
has
ᔢ
There
gain
only
would
zero
will
It
of
ᔢ
It
ᔢ
It
has
be
ᔢ
If
an
a
real
or
difference
(limited
by
negative
between
the
feedback
the
power
input
supply
(see
9.3).
signals,
If
the
voltage).
impedance.
voltage
drop
bandwidth.
for
all
the
which
across
slew
output
change
instantly
operational
(The
the
frequencies
infinite
in
sudden
change
A
no
infinite
amplifies
has
slight
gain
positive
the
output
circuit
of
the
amplifier
.
frequencies
change
a
saturate
output
operational
ᔢ
open-loop
without
rate.
were
an
is
any
slew
to
of
an
amplifier
is
the
range
constant.)
input
caused
made
without
amplifier
in
(The
voltage
bandwidth
gain
signal
rate
by
the
a
of
by
an
step
input
the
same
amplifier
change
in
voltage,
is
the
the
amount.
the
rate
input
output
of
voltage.)
would
delay.
behaves
as
follows.
12
ᔢ
The
input
impedance
ᔢ
The
open-loop
ᔢ
The
output
ᔢ
The
bandwidth
ᔢ
The
slew
is
approximately
10
Ω
+V
s
5
gain
is
approximately
10
for
d.c.
signals.
+
+
impedance
rate
is
is
is
approximately
100 Ω
limited.
limited.
–
V
output
+
Figure
9.2.2
shows
how
power
is
connected
to
an
operational
amplifier
.
Note
V
that
all
voltages
are
measured
with
respect
to
a
common
potential
(earth).
0V
Using
An
an operational
operational
amplifier
can
amplifier
be
used
to
as
a
comparator
compare
two
voltages.
In
V
s
Figure
This
Figure 9.2.2
9.2.2,
is
because
90
operational
there
is
no
amplifier
feedback
is
said
amplifier
to
either
of
the
to
connection
Wiring an operational
operational
amplifier
the
inputs.
be
in
from
open-loop
the
output
mode.
of
the
Chapter
In
order
to
determine
the
output
voltage,
you
simply
multiply
9
Analogue
electronics
the
Equation
difference
by
the
in
voltage
open-loop
between
the
inverting
and
non-inverting
terminals
gain.
The
+
ᔢ
If
V
>
V
the
output
voltage
will
be
positive.
(Positive
saturation
+ V
)
output
amplifier
is
voltage
given
of
an
operational
by
s
+
+
ᔢ
If
V
<
V
the
output
voltage
will
be
negative.
(Negative
saturation
− V
)
V
=
A
out
s
(V
−
V
)
0
+
ᔢ
If
V
=
V
the
output
voltage
will
be
zero.
V
–
output
voltage/V
out
When
the
operational
amplifier
is
set
up
as
shown
in
Figure
9.2.3,
it
is
being
A
–
open-loop
V
–
non-inverting
gain
V
–
inverting
0
+
5
used
as
a
comparator
.
The
gain
+
Suppose
V
=
1.5 V
and
amplifier
is
V
=
=
A
out
The
be
amplifier
is
typically
10
and
and
the
supply
voltage
to
or
−30 000 V
.
the
−15 V
.
5
(V
−
V
)
=
10
(1.5
−
1.8)
=
−30 000 V
output
−15 V
.
The
voltage
of
Therefore,
output
is
said
the
the
to
operational
output
be
amplifier
voltage
will
be
can
either
−15 V
and
Exam tip
not
saturated
Always
Normally,
when
an
operational
amplifier
is
being
used
as
a
the
divider
inputs
is
circuits
fixed
at
are
a
used
to
particular
provide
voltage
the
(the
voltages
at
reference
its
inputs.
voltage).
at
the
other
terminal
is
then
compared
with
the
can
use
physical
you
electrical
property
were
to
working
components
design
for
temperature
of
the
value
icing
an
some
aircraft
wings
of
an
whose
resistance
reference
practical
circuits.
manufacturer
.
aircraft
should
varies
For
Y
ou
never
with
told
drop
of
this
be
exceeded. The
saturate
to
this
output
value
will
when
operational
loop
amplifier
is
in
open-
mode.
suppose
that
below
as
voltage.
some
example,
were
voltage
One
the
W
e
supply
amplifier,
The
always
voltage
the
operational
cannot
of
note
comparator
,
the
potential
input/V
input/V
0
maximum
+15 V
operational
1.8 V
+15 V
+
V
the
−
operational
Then
of
the
a
particular
+15 V
or
simple
below
In
else
circuit
this
this
to
A
the
light
value.
circuit,
divider
.
of
A
the
wings
an
LED
possible
voltage
thermistor
is
would
or
sound
circuit
at
used
the
as
occur
.
is
a
buzzer
illustrated
inverting
the
Y
ou
are
if
the
in
device.
is
to
design
temperature
Figure
terminal
sensing
asked
drops
9.2.3.
fixed
The
a
by
the
resistance
potential
of
V
the
+
V
thermistor
varies
with
temperature.
The
thermistor
is
attached
to
the
+
output
wing
LED.
of
the
aircraft.
When
the
non-inverting
terminal
will
is
light
to
indicating
output
temperature
terminal
greater
saturate
The
than
+15 V
.
to
the
of
varies
the
This
pilot
of
the
the
as
that
is
thermistor
well.
voltage
will
circuit
If
at
the
cause
there
the
a
varies,
voltage
buzzer
to
a
buzzer
voltage
the
at
or
the
non-inverting
terminal,
to
potential
the
at
inverting
the
is
connected
sound
or
the
op-amp
the
LED
–15 V
to
problem.
Figure 9.2.3
Another
from
an
use
of
a
comparator
alternating
input
is
in
voltage.
the
production
Figure
9.2.4
of
a
shows
square-wave
how
a
sine
voltage
Using an op -amp as a
comparator.
wave
−
can
be
converted
into
a
square
wave.
As
soon
as
V
>
0,
the
output
Key points
−
saturates
to
−9 V
.
When
V
<
0,
the
output
saturates
to
+9 V
.
ᔢ
V
An
ideal
operational
amplifier
/ V
has
an
infinite
input
impedance,
+
an
output
time
0
T
infinite
open-loop
impedance,
gain,
zero
infinite
+9 V
T
bandwidth
and
infinite
slew
rate.
2
ᔢ
It
has
two
inputs:
inverting
and
non-inverting.
+
V
/ V
o
V
ᔢ
An
operational
amplifier
can
be
V
o
used
–9 V
+9
loop
as
a
comparator
in
open-
mode.
time
0
T
T
ᔢ
When
used
as
a
comparator,
the
2
–9
output
the
voltage
voltage
of
cannot
the
exceed
power
supply.
Figure 9.2.4
91
9.3
Inverting
Learning outcomes
and
non-inverting
Feedback
Feedback
On
completion
of
this
section,
be
able
represent
an
inverting
and
of
to
the
the
process
input
of
taking
signal
a
being
fraction
fed
into
of
an
the
output
amplifier
.
signal
There
and
are
op-amp
with
a
feedback.
There
is
negative
feedback
and
positive
two
feedback
non-
When
inverting
it
to:
kinds
ᔢ
is
you
adding
should
amplifiers
negative
feedback
is
used,
there
is
a
reduction
in
gain
of
the
single
amplifier
.
There
are
numerous
advantages
to
using
negative
feedback.
input
These
ᔢ
use
the
earth
ᔢ
in
derive
of
an
concept
the
an
of
the
inverting
and
ᔢ
increased
ᔢ
less
ᔢ
greater
the
distortion
gain
When
positive
reduction
explain
the
bandwidth
meaning
of
an
explain
for
a
the
of
gain
gain–frequency
typical
stability.
in
feedback
is
used,
there
is
increased
instability
and
a
bandwidth.
and
amplifier
The
ᔢ
operating
non-inverting
amplifier
ᔢ
bandwidth
amplifier
expression for
inverting
include:
virtual
gain of
an
inverting
amplifier
curve
op-amp
The
gain
of
an
amplifier
is
the
ratio
of
the
output
voltage
to
the
input
voltage.
ᔢ
determine
bandwidth from
gain–frequency
curve.
a
Figure
9.3.1
shows
an
inverting
amplifier
with
negative
feedback.
R
f
I
R
X
V
in
V
out
+
Figure 9.3.1
An inverting amplifier
Assumptions:
ᔢ
Since
the
slightest
open-loop
difference
terminals
will
gain
in
cause
of
the
operational
potential
the
between
output
voltage
amplifier
the
to
is
inverting
be
high,
and
saturated.
the
non-inverting
In
order
for
the
+
output
ᔢ
No
has
Since
(0 V),
be
the
current
a
high
the
enters
input
potential
in
saturated.
Potential
operational
potential
the
true
of
order
The
the
at
the
the
the
point
difference
terminals
not
of
the
to
be
saturated,
operational
X
non-inverting
inverting
output
of
the
(inverting
across
R
=
terminal
terminal
operational
terminal)
V
has
−
is
is
to
difference
across
R
=
0
−
no
current
enters
Current
in
R
the
V
=
out
operational
current
in
R
f
V
–
0
0
–
V
in
out
=
R
R
f
92
at
be
called
0
f
Since
=
amplifier
earth
0 V
.
amplifier
in
Potential
V
V
since
it
impedance.
at
for
amplifier
amplifier:
a
potential
This
not
virtual
has
to
be
earth
to
Chapter
V
9
Analogue
electronics
V
in
out
=
V
+
in
R
R
f
V
out
−V
R
=
V
out
R
in
V
f
R
R
out
V
oltage
f
f
gain:
–
=
V
R
in
X
The
gain of
a
non-inverting
amplifier
R
Figure
9.3.2
shows
a
non-inverting
amplifier
with
negative
feedback.
Assumptions:
ᔢ
Since
the
slightest
open-loop
difference
terminals
will
gain
in
cause
of
the
operational
potential
the
between
output
voltage
amplifier
the
to
is
inverting
be
high,
and
saturated.
the
non-inverting
In
order
for
the
Figure 9.3.2
Non-inverting amplifier
with negative feedback
+
output
ᔢ
No
of
the
current
operational
enters
the
amplifier
terminals
not
of
to
the
be
saturated,
operational
V
=
V
amplifier
since
it
gain
has
a
high
input
impedance.
bandwidth
Since
the
potential
at
the
non-inverting
terminal
is
V
,
the
potential
at
in
the
inverting
the
output
terminal
has
to
be
V
also.
This
has
to
be
true
in
order
for
in
The
of
the
potential
at
operational
X
is
V
.
We
amplifier
can
write
not
the
to
be
saturated.
potential
at
X
in
terms
of
in
the
output
divider
voltage
by
thinking
of
the
resistors
as
making
up
a
potential
frequency
circuit.
R
V
(
=
in
R
Figure 9.3.3
+
)
R
The frequency response
V
out
curve of an amplifier
f
V
R
+
R
out
f
(
=
V
)
R
in
gain
V
R
out
V
oltage
f
gain:
=
1
+
V
R
in
bandwidth
The frequency
Figure
9.3.3
obtained
by
shows
response of
the
varying
frequency
the
an op-amp
response
frequency
of
the
of
an
input
op-amp.
signal
and
The
graph
is
measuring
frequency
the
gain
of
amplifier
the
is
op-amp
the
range
at
of
each
input
frequencies
frequency.
for
which
The
the
bandwidth
gain
of
an
of
an
Figure 9.3.4
amplifier
The effect of negative
feedback on gain and bandwidth
remains
and
constant.
constant.
which
the
At
gain
At
low
higher
starts
to
frequencies,
frequencies,
decrease
the
the
(i.e.
gain
gain
the
of
the
amplifier
decreases.
point
at
The
which
is
high
frequency
there
is
a
at
sudden
Key points
change
A
in
the
logarithmic
logarithmic
by
is
slope)
scale
scale
compressing
plotted
graph
on
shows
a
is
is
called
typically
allows
the
for
larger
a
large
scale
bandwidth
cut- off
used
values
logarithmic
how
the
is
on
it
break
the
range
and
if
or
of
gain
to
be
determined
and
values
expanding
is
frequency.
frequency
to
the
be
expressed
from
a
shown
smaller
as
axes.
on
a
values.
decibels.
frequency
A
ᔢ
graph
Figure
9.3.4
shows
the
effect
on
gain
and
bandwidth
is
a fraction
of
and
Gain
adding
being fed
The
the
process
the
it
to
into
output
the
an
of
taking
signal
input
signal
amplifier.
response
ᔢ
curve.
Feedback
when
The
voltage
gain
of
an
of
a
inverting
negative
R
amplifier
feedback
is
used.
As
the
gain
is
reduced,
the
bandwidth
is
f
–
increases.
R
ᔢ
The
voltage
gain
non-
R
f
inverting
amplifier
is
1
+
.
R
ᔢ
When
the
negative feedback
gain of the
and the
is
used,
amplifier decreases
bandwidth
increases.
93
9.4
The
summing
amplifier
and
the
voltage follower
Learning outcomes
The
A
On
completion
of
this
section,
summing
be
able
music
describe
the
amplifier
ᔢ
solve
as
use
a
of
the
relating
amplifier
amplifier
as
a
combines
is
often
several
required
inputs
that
into
several
one
output
signals
signal.
(guitar
,
In
piano
the
use
of
etc.)
can
be
combined
by
using
a
summing
amplifier
.
gain of
a
summing
amplifier
to
order
for
the
output
not
to
be
saturated,
the
potential
at
the
inverting
circuits
and
describe
voice
The voltage
In
ᔢ
it
inverting
summing
problems
summing
amplifier
industry,
to:
and
ᔢ
amplifier
you
the
should
summing
the
voltage follower.
op-amp
non-inverting
potential
must
at
also
Assume
shown
the
be
at
that
in
terminals
of
the
op-amp
non-inverting
terminal
zero
(virtual
each
Figure
potential
input
source
is
is
must
0 V
,
the
be
the
same.
potential
at
Since
the
the
point
X
earth).
positive
and
currents
are
flowing
as
9.4.1.
R
f
I
f
R
I
1
1
X
V
1
V
R
I
o
2
2
+
V
2
R
I
3
3
V
3
Figure 9.4.1
Using
A summing amplifier
Kirchhoff ’s
first
I
+
law,
we
I
I
1
V
1
–
V
0
2
–
=
3
V
0
+
R
+
2
3
–
1
write,
I
f
0
0
+
R
can
2
–
V
0
=
R
R
3
f
R
R
f
V
=
0
–
(
+
all
the
resistors
have
the
same
V
=
0
f
V
1
+
R
R
–(V
)
3
R
2
value,
V
2
1
If
R
f
V
3
then
+
1
V
+
2
V
)
3
Example
An
ideal
circuit
operational
shown
Calculate
94
the
in
amplifier
Figure
output
9.4.2.
voltage.
is
used
in
its
inverting
mode
in
the
mixer
Chapter
9
Analogue
electronics
12.0 kΩ
+9 V
3.0 kΩ
X
+0.60 V
V
8.0 kΩ
o
+
+0.81 V
4.2 kΩ
–9 V
+0.30 V
Figure 9.4.2
R
R
f
V
=
–
0
(
R
f
V
f
+
V
1
–
=
Using
a
summing
analogue
In
the
world
digital
2
3
A
A
a
of
simple
inputs.
of
the
bit
(0.81)
+
(0.30)
8
amplifier to
can
device
also
values
different
to
of
bits.
The
be
as
an
4.2
)
perform digital to
convert
analogue
The
signals
a
into
digital
Figure
resistances
resistor
4R
an
R
are
to
(DAC)
A,
scaled
to
the
is
B
converted
converter
an
analogue
constructed
and
C
are
represent
to
the
least
into
(ADC).
signal.
into
9.4.3.
to
be
analogue
corresponds
corresponds
to
digital
signal
converter
in
have
analogue
converted
shown
the
resistor
analogue
called
can
digital
amplifier
The
(MSB).
a
amplifier
3-bit
summing
12
+
−4.47 V
electronics,
by
signals
summing
12
(0.60)
3
conversion
signals
Digital
(
3
R
12
=
0
)
V
R
1
V
+
2
R
most
using
the
the
signal.
digital
weights
significant
significant
bit
(LSB).
3
Since
Each
there
is
three
combination
Suppose
at
are
an
V
a
logic
input
is
1
inputs,
will
at
there
correspond
an
input
represented
by
is
are
to
2
a
=
8
unique
represented
0 V
.
possible
Suppose
output
by
the
logic
a
+5 V
output
combinations.
voltage.
and
that
voltage
of
a
logic
the
0
DAC
.
out
1
V
is
determined
by
V
out
=
–
out
(
V
+
1
V
1
+
V
2
)
3
2
4
1
Therefore,
if
the
input
digital
signal
is
010,
V
=
–
out
(
0
+
Table 9.4.1
1
(5)
+
(0)
2
4
)
A
B
C
V
/V
out
=
T
able
9.4.1
illustrates
V
all
the
possible
−2.5 V
combinations
of
input
and
0
0
0
0.00
0
0
1
−1.25
0
1
0
−2.50
0
1
1
−3.75
1
0
0
1
0
1
1
1
0
1
1
1
output.
R
1
R
A
–
Bit0
B
–
Bit1
(MSB)
V
2R
2
V
out
V
3
C
–
Bit3
(LSB)
Figure 9.4.3
analogue
−5.00
output
4R
+
−6.25
−7
.50
−8.75
A simple digital to analogue converter
95
Chapter
9
Analogue
electronics
The voltage follower
The
of
circuit
this
in
circuit
Figure
is
1.
9.4.4
The
represents
output
a
voltage
voltage
is
equal
follower
.
to
the
The
input
voltage
gain
voltage.
+
V
in
V
out
Figure 9.4.4
The
low
A voltage follower (buffer)
voltage
Cascade
Several
the
be
follower
impedance
It
is
is
used
often
to
connect
referred
to
as
high
a
impedance
to
of
one
circuits
acts
can
as
the
be
connected
input
of
together
another
.
The
in
such
a
way
amplifiers
are
that
said
to
cascaded
Suppose
three
amplifier
circuits
have
voltage
gains
A
,
A
1
respectively.
A
sources
buffer
.
amplifiers
amplifier
output
circuit
loads.
=
A
×
When
A
1
×
2
cascaded,
the
overall
voltage
gain
is
and
A
2
3
given
by
A
3
amplifier
1
amplifier
2
amplifier
3
V
V
in
out
voltage
gain
= A
voltage
gain
= A
1
Figure 9.4.5
voltage
gain
= A
2
3
Cascaded amplifiers
Example
Figure
9.4.6
voltage
gain
shows
of
the
a
cascade
amplifier
circuit.
Calculate
the
overall
circuit.
450 kΩ
600 kΩ
1200 kΩ
100 kΩ
60 kΩ
120 kΩ
+
V
+
V
o
+
Figure 9.4.6
The
first
amplifier
The
voltage
is
non-inverting.
R
450
f
gain
A
=
1
+
=
1
+
=
5.5
1
R
The
second
The
voltage
amplifier
is
100
inverting.
R
600
f
gain
A
=
–
=
–
=
–10
2
R
The
third
amplifier
The
voltage
is
60
inverting.
R
1200
f
gain
A
=
–
=
–
=
−10
2
R
Overall
voltage
gain
=
A
×
1
96
120
A
×
2
A
=
3
5.5
×
−10
×
−10
=
550
Chapter
The transfer
Figure
gain
9.4.7
of
an
characteristic of
shows
a
circuit
operational
operational
amplifier
that
amplifier
.
is
+V
can
be
The
and
−V
s
an operational
used
power
.
A
to
measure
supply
signal
the
is
Analogue
electronics
amplifier
open-loop
connected
generator
9
to
the
connected
to
the
+V
s
s
input
of
the
operational
input
voltage
V .
An
amplifier
.
oscilloscope
An
is
oscilloscope
used
to
is
measure
used
the
to
measure
output
the
voltage
V
i
.
o
signal
+
The
frequency
of
the
signal
generator
is
set
to
a
desired
value.
The
input
V
V
generator
o
V
voltage
V
is
then
adjusted
to
a
very
low
value
( μV)
and
the
corresponding
s
i
output
voltage
V
is
measured.
Since
the
open-loop
gain
of
the
operational
o
5
amplifier
is
A
the
plot
of
very
large
output
(10
+
),
the
voltage
V
output
saturates
against
input
when
voltage
V
V
o
transfer
characteristic
of
the
−
>
is
V
+
or
known
V
<
V
as
the
i
operational
amplifier
.
Figure
9.4.8
shows
Figure 9.4.7
the
transfer
characteristic
of
an
operational
amplifier
.
The
open-loop
Circuit used to measure the
gain
open-loop gain of an operational amplifier
of
the
linear
The
an
operational
region
circuit
of
in
amplifier
the
Figure
operational
is
found
by
determining
the
gradient
of
the
graph.
9.4.7
amplifier
.
can
The
be
used
to
frequency
plot
of
the
the
frequency
input
signal
response
is
varied
of
V
and
0
the
corresponding
gain
(V
/V )
o
of
the
operational
amplifier
is
measured.
i
supply
Figure
9.4.9
shows
the
variation
of
the
open-loop
gain
with
frequency
+
+V
of
s
saturation
the
input
signal.
gain
(V
–
) / μV
V
+
6
10
saturation
5
V
10
s
supply
–
4
10
3
Figure 9.4.8
10
The transfer characteristic
of an operational amplifier
2
10
10
1
2
1
10
3
10
4
10
5
10
6
10
10
frequency/Hz
Figure 9.4.9
Example
A
student
is
asked
to
design
a
circuit
to
combine
two
signals
V
and
1
V
to
form
signal
V
2
according
to
the
relationship
V
0
minimum
100 kΩ.
This
The
input
Design
design
circuit
requires
output
signal
of
R
=
–
0
(
to
the
a
for
both
achieve
use
of
a
summing
signals
this
should
–4V
1
be
no
.
The
2
less
than
goal.
summing
amplifier
amplifier
with
two
with
two
inputs
is
inputs.
given
by
Key points
R
f
V
–3V
0
resistance
a
=
f
V
+
V
1
)
,
ᔢ
2
R
R
1
A
summing
several
where
R
is
amplifier
combines
2
the
feedback
resistance
and
R
f
and
1
R
represent
the
inputs
into
one
output
input
2
signal.
resistances
for
the
two
inputs.
ᔢ
Comparing
this
equation
with
V
=
–3V
0
–4V
1
we
The voltage follower
or
R
3
and
used
to
connect
=
high
4
impedance
sources
to
low
R
1
also
2
know
that
R
and
R
1
must
be
greater
than
next
step
is
to
impedance
loads.
The
a
100 k Ω
2
ᔢ
choose
values
of
resistors
that
satisfy
the
conditions
above.
possible
solution
is
R
=
f
1200 kΩ,
R
=
1
400 kΩ
and
R
=
gain
equal
of
One
is
f
=
The
buffer
R
f
R
We
circuit
get
2
to
each
of
the
cascade
product
amplifier
in
amplifier
of
the
the
is
gains
circuit.
300 kΩ
2
97
9.5
Analysing
op-amp
Example
Learning outcomes
Figure
On
completion
of
this
section,
9.5.1
be
able
to:
ᔢ
analyse
simple
ᔢ
analyse
the
is
amplifier
shows
used
feedback,
in
the
of
of
variation
an
ideal
with
frequency
amplifier
operational
f
of
circuit
shown
in
amplifier
the
voltage
Figure
(op-amp).
gain
The
G,
op-amp
9.5.2.
circuits
gain,
response
the
you
without
should
circuits
G
amplifier
6
10
circuits
to
input
signals,
using
5
10
timing
diagrams
4
10
ᔢ
understand
the
term
clipping
3
10
when
applied
to
amplifiers.
2
10
10
1
2
1
10
3
10
4
10
5
10
6
10
10
frequency,
f /Hz
Figure 9.5.1
1 MΩ
+12 V
10 kΩ
+
input
output
–12 V
Figure 9.5.2
Calculate:
a
the
gain
b
the
bandwidth
c
the
peak
A,
of
the
of
output
amplifier
the
amplifier
voltage
for
an
input
signal
of
peak
value
0.1 V
and
V
in
frequency:
+0.2 V
i
100 Hz
ii
1
5
time
d
A
×
10
signal
Hz.
generator
waveform
is
produced
attached
by
the
to
the
signal
input
of
generator
–0.2 V
is
the
amplifier
.
shown
in
The
Figure
9.5.3.
4
The
frequency
output
of
the
signal
is
1
×
10
Hz.
Sketch
the
waveform
of
the
signal.
Figure 9.5.3
6
R
1
f
a
Gain
A
=
×
10
2
–
=
–
=
−10
=
−100
3
R
10
×
10
i
2
b
A
gain
of
10
frequency
c
i
From
4
corresponds
response
the
feedback
Peak
98
is
a
bandwidth
of
10
Hz
(according
to
the
curve).
graph,
(Remember
to
G
with
=
100
when
feedback,
the
f
=
gain
100 Hz.
is
reduced.
The
gain
with
100.)
output
voltage
=
G
×
peak
input
voltage
=
100
×
0.1
=
10 V
Chapter
9
Analogue
electronics
5
ii
From
Peak
the
graph,
output
G
=
voltage
10
=
when
G
×
f
=
peak
1
×
10
input
Hz
voltage
=
10
×
0.1
=
1.0 V
2
d
The
peak
saturate
output
to
voltage
+12 V
.
The
will
be
output
peak
of
10
×
signal
the
0.2
is
=
20 V
.
The
output
will
‘clipped’.
output
output
voltage
saturates
output
at
+12 V
+12 V
time
–12 V
Figure 9.5.4
Example
A
particular
18 °C
can
and
be
grows
The
plant
21 °C
used
to
(Figure
LEDs
positive
L
At
a
the
that
plant
monitor
the
the
to
ambient
survive.
A
temperature
temperature
student
of
the
be
designs
room
between
a
where
circuit
the
that
plant
9.5.5).
and
L
1
is
requires
for
emit
light
when
the
output
from
the
appropriate
op-amp
2
and
high.
temperature
of
The
thermistor
18 °C
the
has
potential
a
negative
difference
temperature
across
R
is
coefficient.
4.5 V
.
+12 V
+12 V
L
2
+5 V
T
+
–12 V
X
+12 V
L
1
+
R
+4 V
–12 V
0V
Figure 9.5.5
The
potential
at
the
non-inverting
inputs
of
both
saturates
to
−12 V
,
op-amps
is
4.5 V
.
+
output
of
the
op-amp
at
the
top
because
V
The
−
<
V
.
Since
L
2
is
reverse-biased,
L
will
not
light.
The
output
of
the
op-amp
at
the
bottom
2
+
saturates
to
+12 V
,
because
V
−
>
V
.
Since
L
is
forward-biased,
L
1
L
lights
up
at
the
lower
temperature
limit
will
light.
1
of
18 °C.
1
Consider
When
what
the
temperature
thermistor
across
the
difference
potential
top
happens
increases
at
X
saturates
the
of
and
thermistor
across
when
+12
V
.
temperature
room
increases,
resistance
decreases.
resistor
increases
to
the
its
the
R
the
+5.0 V
,
L
is
the
decreases.
same
increases
beyond
Since
At
of
now
can
therefore
be
used
as
an
increases.
temperature
the
output
of
of
potential
the
difference
potential
+4.5 V
.
When
the
forward-biased,
2
L
room
The
time,
beyond
the
the
the
op-amp
L
will
at
the
light.
2
indicator
to
determine
when
the
2
temperature
of
the
room
exceeds
21 °C.
99
10
Digital
10.
1
Logic
electronics
gates
Digital
Learning outcomes
systems
variables.
On
completion
should
ᔢ
be
able
describe
of
this
section,
used
to:
the function
following
you
logic
NAND, OR,
circuits,
gates:
NOR,
of
the
NOT, AND,
EXOR,
in
low
ᔢ
A
high
use
truth
tables
to
of
no
represent
more
than
logic.
can
variable
is
Logic
have
a
involves
only
voltage.
two
the
use
possible
T
ypically,
a
of
binary
values.
5 V
power
In
digital
supply
is
circuits.
voltage
of
voltage
0 V
of
represents
+5 V
logic
represents
0.
logic
1.
EXNOR
circuits
can
be:
the
ᔢ
function
using
variable
binary
digital
A
operate
binary
the
ᔢ
Digital
ᔢ
A
combinational
–
the
output
of
the
circuit
is
determined
by
the
two
current
inputs
inputs.
ᔢ
sequential
–
inputs
previous
and
Sequential
the
digital
ᔢ
synchronous
ᔢ
asynchronous
input
are
circuits
the
performed
NOT
circuits
–
a
–
of
the
circuit
A
B
1
0
0
1
are
clock
the
classified
input
circuit
is
NOT
used
to
drive
are
(ICs)
building
by
a
can
be
blocks
particular
built
of
logic
to
digital
gate
is
perform
the
logic
The
represented
the
current
by
circuit
operations
changes
in
the
functions.
logic
by
a
Logic
function
tr uth
table
Truth table for a NOT gate
A
B
gate
inverts
the
logic
NOT gate
state
of
the
gate
Table 10.1.2
Truth table for an AND gate
A
B
C
0
0
0
A
C
100
all
driven
circuits.
inverter
.
AND
by
as:
operations
Figure 10.1.1
an
determined
gate
Table 10.1.1
The
is
outputs.
signals.
Integrated
gates
output
0
1
0
1
0
0
1
1
1
B
Figure 10.1.2
AND gate
input.
It
is
sometimes
called
Chapter
NAND
Digital
electronics
gate
Table 10.1.3
Truth table for a NAND gate
A
B
C
0
0
1
0
1
1
A
A
C
C
B
B
1
0
1
1
1
0
Figure 10.1.3
OR
10
NAND gate
gate
Table 10.1.4
Truth table for an OR gate
A
B
C
0
0
0
0
1
1
1
0
1
A
C
B
Figure 10.1.4
1
1
NOR
OR gate
1
gate
Table 10.1.5
Truth table for a NOR gate
A
B
C
0
0
1
0
1
0
A
A
C
C
B
B
1
0
0
1
1
0
Figure 10.1.5
EXOR
NOR gate
gate
Table 10.1.6
Truth table for an EXOR gate
A
B
C
0
0
0
0
1
1
1
0
1
1
1
0
A
C
B
Key points
Figure 10.1.6
EXOR gate
ᔢ
Digital
use
ᔢ
EXNOR
gate
Table 10.1.7
Truth table for an EXNOR gate
B
C
0
0
1
ᔢ
A
C
1
0
1
0
0
Binary
Digital
Digital
the
0
only
and
circuits
can
or
possible
1.
be
sequential.
sequential
or
two
logic
circuits
can
be
asynchronous.
C
B
B
Figure 10.1.7
Logic
gates
are
the
building
EXNOR gate
blocks
1
has
logic
synchronous
ᔢ
1
logic
involves
logic.
combinational
A
0
electronics
binary
values:
ᔢ
A
of
of
digital
circuits.
1
ᔢ
The
is
logic function
represented
of
using
a
a
logic
truth
gate
table.
101
10.2
Equivalent
Learning outcomes
logic
Equivalent
NAND
On
completion
of
this
section,
be
able
NOR
gates
be
used
to
gates
form
all
are
called
possible
‘universal
logic
gates.
gates’.
Figure
This
is
10.2.1
because
each
illustrates
how
to:
all
ᔢ
and
logic
you
can
should
gates
construct AND
gates from
other
logic
gates
can
be
made
by
using
only
NAND
gates
or
only
NOR
NOR
gates.
gates
ᔢ
construct OR
gates from
NAND
NOT
gates
ᔢ
redesign
NOR
a
circuit
gates
or
to
contain
NAND
only
gates
AND
ᔢ
reduce
gate
circuits
to
the
minimum
count.
NAND
OR
NOR
EXOR
EXNOR
Figure 10.2.1
Equivalent logic gates
Redesigning
a
circuit to
use only
NOR
gates
Example
Consider
using
the
only
Figure 10.2.2
102
logic
NOR
circuit
gates
in
and
Figure
10.2.2.
minimise
the
Draw
number
the
of
equivalent
gates
in
the
circuit
circuit.
Chapter
10
Digital
electronics
cancel
Figure 10.2.4
Minimise the number of
cancel
gates
Figure 10.2.3
Replace all logic gates with NOR gates only
Redesigning
a
circuit to
use only
NAND
gates
Example
Y
ou
are
provided
a
Replace
b
Minimise
by
a
all
using
The
AND
equivalent
Figure 10.2.6
b
See
the
the
a
with
a
logic
quad-NAND
gates
number
single
gates,
of
in
Figure
gates
quad-NAND
NOT
NAND
gate
gates
chip.
so
10.2.5
that
the
with
logic
NAND
circuit
gates.
could
be
made
chip.
and
(Figure
the
OR
gate
are
replaced
with
their
Figure 10.2.5
10.2.6).
Replace all logic gates with NAND gates only
Figure
10.2.7.
cancel
cancel
cancel
Key points
ᔢ
All
logic
using
NOR
ᔢ
Figure 10.2.7
gates
only
can
NAND
be
made
gates
or
by
only
gates.
In
order
of
logic
to
minimise
the
number
Minimise the number of gates
any
gates
two
perform
used
sequential
in
a
circuit,
gates
that
inverting functions
are
removed.
103
10.3
Logic
gates
Learning outcomes
and
completion
of
this
section,
diagrams
be
able
are
very
useful
in
analysing
digital
circuits.
They
you
illustrate
should
diagrams
Timing diagrams
Timing
On
timing
the
logic
states
of
the
logic
circuit
various
inputs
and
outputs
over
a
period
of
to:
time.
ᔢ
understand
diagrams.
how
to
use
timing
Example
Consider
in
Figure
10.3.1.
The
inputs
are
I
and
I
1
the
output
is
and
2
X.
I
P
1
Q
I
2
X
R
Figure 10.3.1
The
truth
I
table
I
is
illustrated
P
Q
R
X
0
0
1
0
1
0
1
1
0
0
0
1
0
1
0
0
0
1
1
1
0
1
1
1
2
0
The
inputs
I
and
1
In
order
to
I
are
below:
shown
in
the
timing
diagram
in
Figure
10.3.2.
2
draw
the
timing
diagram
for
the
output
X,
the
inputs
I
and
1
must
be
considered.
logic
1
logic
0
I
1
time
I
2
time
X
time
Figure 10.3.2
104
Timing diagram
I
2
Chapter
10
Digital
electronics
Example
Consider
the
logic
circuit
in
Figure
10.3.3.
The
inputs
are
I
and
I
1
the
output
is
and
2
X.
I
P
1
I
2
X
Q
S
R
Figure 10.3.3
I
I
P
Q
R
S
X
0
1
1
1
1
0
0
1
1
0
1
1
0
1
0
1
1
0
1
0
1
1
0
0
0
0
1
1
2
0
The
inputs
I
and
I
1
In
order
to
draw
are
shown
in
the
timing
diagram
in
Figure
10.3.4.
2
the
timing
diagram
for
the
output
X,
the
inputs
I
and
1
must
be
I
2
considered.
I
1
time
I
2
time
X
time
Figure 10.3.4
Timing diagram
Key point
ᔢ
Timing
diagrams
are
useful
in
analysing
logic
circuits.
105
10.4
Applications
Learning outcomes
of
completion
of
this
section,
be
able
discuss
the
systems
ᔢ
describe
gates
to
in
are
found
in
all
aspects
in
of
homes
modern
and
society.
industry
age.
Engineers
have
used
digital
systems
to
solve
We
live
various
in
the
problems
to:
and
ᔢ
systems
systems
you
digital
should
gates
Application of digital
Digital
On
logic
application
homes
and
applications
real
world
of
industry
of
enhance
our
quality
of
life.
digital
Applications of digital
systems
in the
home
logic
ᔢ
T
elevision
–
flat
ᔢ
Music
ᔢ
Video
ᔢ
Computers
ᔢ
Refrigerators
ᔢ
Microwave
ᔢ
W
ashing
ᔢ
Home
screens,
LEDs,
LCDs,
plasma
problems.
–
–
CD
HD
players,
TV
,
–
MP3
DVD
laptops,
players
desktops,
ovens
security
systems
T
elecommunications
cellular
tablets
machines
Applications of digital
ᔢ
players
systems
–
PBXs,
in
digital
industry
phones,
cellular
technology,
phones
ᔢ
Data
communications
ᔢ
Control
ᔢ
Cable
ᔢ
The
ᔢ
Car
systems
–
–
switches,
industrial
routers,
wireless
access
points
plants
TV
internet
manufacturing
Using
logic
gates to
solve
real
world
problems
Example
The
circuit
opens
a
energised.
This
when
ᔢ
A
logic
ᔢ
B
ᔢ
C
ᔢ
D
is
is
is
is
logic
logic
logic
other
which
will
Figure
In
happen
Any
106
in
door
.
a
10.4.1
order
for
can
only
and
b
are
represents
the
lock
happen
also
to
when
logic
a
combination
open,
1.
c
is
This
the
a
logic
can
lock
solenoid
1,
only
system
coil
which
has
can
happen
to
that
be
only
when:
1
0
1
0.
combination
sound
a
will
buzzer
.
cause
c
to
be
logic
0
and
d
to
be
logic
1,
Chapter
10
Digital
electronics
+5 V
A
B
a
c
d
b
C
D
solenoid
lock
0V
Figure 10.4.1
Example
The
circuit
in
Figure
10.4.2
represents
a
simple
alarm
system
in
a
car
.
A
D
1
push-button
switch
is
attached
to
each
of
the
four
doors
of
a
car
.
all
at
logic
inputs
of
the
When
D
2
the
doors
of
the
car
are
closed,
D
,
D
1
the
alarm
is
armed,
a
logic
1
is
,
D
2
sent
and
D
3
to
are
0.
When
4
one
of
the
AND
gate.
D
3
When
any
door
is
opened,
a
logic
1
is
sent
to
the
other
input
of
the
AND
D
4
gate.
The
output
of
the
AND
gate
is
a
logic
1
and
the
buzzer
sounds.
armed
signal
Example
The
circuit
in
temperature
inserted
in
Figure
of
10.4.3
water
the
tank
in
to
a
represents
tank
and
measure
the
the
a
system
water
that
level.
temperature
monitors
A
and
the
thermistor
the
water
0V
is
level
is
Figure 10.4.2
measured
below
the
a
be
float
open
water
switched
element
on
the
the
and
is
was
there
overheat
the
and
not
was
and
the
water
of
closed
tank
when
When
the
output
relay
in
would
switch..
value
switch,
normally
of
a
pre-set
float
level
by
level
AND
the
is
higher
gate
water
taken
no
burn
temperature
water
in
is
tank
a
water
the
the
and
falls
level
logic
1.
switched
account,
the
the
than
becomes
heater
into
of
of
The
on.
heater
the
If
the
could
heating
out.
+5 V
240 V
0 V
Figure 10.4.3
Key points
ᔢ
Applications
ᔢ
Logic
gates
of
can
digital
be
systems
used
as
can
control
be found
circuits
in
in
homes
real
and
world
industry.
problems.
107
10.5
Sequential
Learning outcomes
circuits
Sequential
In
On
completion
of
this
section,
a
combinational
be
able
explain
the
operation
of
consisting
gates
or
two
of
two
NOR
NAND
the
present
describe
the
A
ᔢ
combine
the
inputs.
In
operation
inputs
of
store
the
circuits
flip-flop
or
bistable
output
is
at
sequential
all
times
circuits,
dependent
the
output
is
on
the
dependent
and
the
previous
previous
are
latch
is
element.
output.
output.
flip-flops
a
circuit
and
that
Flip-flops
are
The
A
memory
are
has
the
memory
element
able
two
basic
to
elements
store
stable
one
states.
building
bit
It
blocks
used
is
of
is
information.
said
of
in
to
be
sequential
a
circuits.
triggered
to
sequential
gates
a
ᔢ
the
a fliprequired
flop
of
to:
on
ᔢ
circuit,
you
combination
should
circuits
There
are
different
types
of
flip-flops
(SR,
JK,
D
and
T
flip-flops).
bistable
triggered
flip-flops)
to
make
bistables
a
3-bit
(T
binary
The SR flip-flop
(set–reset flip-flop)
counter
An
ᔢ
draw
a
circuit
construction
to
of
show
a
half-adder
SR
flip-flop
(Figure
the
is
constructed
from
a
pair
of
cross-coupled
NOR
gates
10.5.1).
–
and
The
inputs
are
called
S
(set)
and
R
(reset).
The
outputs
are
Q
and
its
use
two
half-adders
and
an OR
complement
of
Q.
(i.e.
When
Q
=
1,
Q
=
0
and
when
Q
=
0,
Q
=
is
1).
to
The
construct
Q
–
operation
the
ᔢ
–
Q .
–
explain
circuit
operates
as
follows:
a full-adder.
ᔢ
When
both
latched
ᔢ
When
in
the
inputs
its
R
last
are
low,
the
output
Q
does
not
change
and
remains
state.
input
is
set
to
low
and
the
S
input
is
set
to
high,
the
Q
R
Q
output
ᔢ
When
ᔢ
When
set
the
output
Q
is
of
R
to
input
the
the
logic
R
is
latch
and
1.
S
set
is
to
high
reset
inputs
to
and
logic
are
set
the
S
input
is
set
to
low,
the
Q
0.
to
high,
the
output
is
unpredictable.
S
This
Figure 10.5.1
state
is
an
invalid
state.
SR flip -flop using NOR
T
able
10.5.1
shows
that
there
are
only
two
stable
output
states
for
the
SR
gates
flip-flop.
Table 10.5.1
Truth table for an SR flip-flop
Example
–
S
R
Q
Q
0
0
0
1
0
1
1
0
1
0
1
1
Consider
flip-flop.
latch
ᔢ
In
ᔢ
In
ᔢ
In
ᔢ
In
ᔢ
In
a
sequence
Suppose
sequence
sequence
sequence
2,
S
S
3,
4,
=
S
S
of
=
0
=
=
changes
0
and
and
1
0
R
and
and
R
=
R
R
that
=
0,
=
=
1
Q
0,
0,
take
place
initially.
will
Q
Q
hold
will
will
at
The
be
its
set
hold
its
the
inputs
output
last
to
Q
of
will
output,
an
be
SR
set
to
which
was
0.
which
was
1.
which
was
0.
1.
last
output,
invalid
An
Table 10.5.2
–
Sequence
S
R
Q
Q
1
0
1
0
1
2
0
0
0
1
as
sequence
sequence
SR
flip-flop
shown
Figure
in
5,
6,
S
S
can
Figure
10.5.3
=
=
0
0
and
and
also
be
R
R
=
=
1,
0,
Q
Q
will
will
constructed
be
reset
hold
using
its
to
last
0.
output,
cross-couple
NAND
gates
10.5.2.
shows
the
circuit
symbol
for
an
SR
flip-flop.
S
Q
Q
S
3
1
0
1
0
4
0
0
1
0
5
0
1
0
1
6
0
0
0
1
R
Q
Q
R
Figure 10.5.2
gates
108
0.
SR flip -flop using NAND
Figure 10.5.3
flip -flop
Circuit symbol for an SR
Chapter
10
Digital
electronics
Table 10.5.3
The T flip-flop
(toggle flip-flop)
T
The
circuit
10.5.4.
the
If
If
the
clock
the
T
symbol
T
input
input
for
input
a
T
is
flip-flop
high,
the
(toggle
T
flip-flop
is
low,
the
T
flip-flop
holds
is
The
data
able
to
is
high)
in
can
or
Q
Q
next
Figure
(toggles)
(current
(next
state)
state)
0
0
0
hold
state
0
1
1
hold
state
1
0
1
toggle
1
1
0
toggle
when
its
previous
value.
Q
counter
fl ip-fl ops
used
count
counter
to
binary
several
register
state
in
Circuit symbol for a T flip -flop
3-bit
When
changes
shown
Q
V
clock
The
is
changes.
T
Figure 10.5.4
flip-flop)
for
the
the
be
are
storing
for m
of
number
positive
negative
connected
and
1s
of
edge
and
0s.
clock
edge
together,
shifting
data
A
counter
pulses
triggered
triggered
the
the
for m
an
is
ar riving
(i.e.
(i.e.
they
from
a
r egister .
register
at
its
input
input
a
exter nal
that
clock
changes
changes
A
source.
is
input.
from
from
A
low
high
to
n
low).
An
states.
n-bit
binar y
counter
consists
of
n
3-bit
binar y
counter
consists
of
three
A
fl ip-fl ops
and
fl ip-fl ops
has
and
2
distinct
has
eight
3
states
(i.e.
2
=
8).
I
Q
T
Q
Q
0
Figure
10.5.5
shows
of
a
edged
The
outputs
T
of
the
circuit
flip-flops.
signal
the
The
(i.e.
counter
diagram
output
the
are
input
Q
,
all
operation
at
of
2
logic
the
0,
Figure
for
3-bit
binary
when
changes
from
and
1
10.5.6
a
changes
Q
0
initially
Q
1
A 3-bit binary counter
consisting
negative
Q
Q
Q
Figure 10.5.5
Q
T
V
V
V
clock
Q
T
Q
.
the
counter
input
logic
Assuming
1
to
the
detects
logic
0).
outputs
are
2
shows
the
timing
diagram
for
the
counter
.
Table 10.5.4
T
able
10.5.4
shows
the
sequential
truth
table
for
the
3-bit
binary
Sequential truth table for a
counter
.
3-bit binary counter
clock
Clock
pulse
Q
Q
0
Q
1
2
pulse
0
Q
1
0
1
0
1
0
1
0
0
0
Q
1
1
0
0
1
0
0
0
0
1
1
0
0
2
0
1
0
3
1
1
0
4
0
0
1
5
1
0
1
6
0
1
1
7
1
1
1
1
1
0
0
0
0
Q
1
1
1
1
2
count
stage
Figure 10.5.6
0
1
2
3
4
5
6
7
The timing diagram for a 3-bit binary counter
109
Chapter
10
Digital
electronics
Adding
binary
When
adding
Figure
10.5.7
1-bit
numbers
binary
shows
how
numbers,
1-bit
0
Figure 10.5.7
The
0
sum
and
numbers
0
0
1
+0
+1
+0
0
1
1
carry
bit
are
produced.
added.
carry
1
0
a
are
1
+1
sum
0
Adding 1-bit binary numbers
method
Figure
a
binary
can
10.5.8
be
gives
extended
an
to
example
any
of
number
adding
of
two
n-bit
3-bit
binary
numbers.
numbers.
101+
011
1000
Figure 10.5.8
The
A
Adding two 3-bit numbers
half-adder
half-adder
(Figure
is
an
arithmetic
circuit
used
to
add
two
1-bit
numbers
10.5.9).
A
S
B
A
S
half-
adder
B
C
C
(a)
(b)
Figure 10.5.9
The
circuit
added.
the
The
other
A half-adder: (a) typical diagram, (b) logic diagram
has
two
circuit
produces
inputs
has
the
two
A
and
B,
outputs.
carry
(C).
110
Truth table for a half-adder
A
B
S
(sum)
C
(carry)
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
One
T
able
half-adder
.
Table 10.5.5
which
represents
output
10.5.5
the
produces
shows
the
two
the
bits
sum
truth
to
be
(S)
table
and
for
a
Chapter
10
Digital
electronics
The full-adder
A
full-adder
full-adder
10.5.10).
The
circuit
A
carry
above
it
circuit
made
full-adder
input
in
is
adds
the
for
is
the
cascade.
three
by
one-bit
using
typically
a
full-adder
T
able
two
numbers
half-adders
component
is
10.5.6
(A,
from
the
shows
in
an
Carry
OR
cascade
output
truth
A
and
and
a
carry
the
B
table
of
gate
A
(Figure
adders.
from
for
in).
a
the
adder
full-adder
.
S
fullB
(a)
adder
A
C
C
in
B
out
carry
carry
(b)
in
out
sum
C
S
in
(sum)
half-
adder
sum
A
half-
C
out
adder
B
carry
out
Key points
Figure 10.5.10
A full-adder: (a) typical diagram, (b) logic diagram
ᔢ
A flip-flop
meaning
Table 10.5.6
is
a
that
bistable
is
has
device,
two
stable
Truth table for a full-adder
states.
A
B
C
S
(sum)
in
C
out
ᔢ
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
A flip-flop
of
ᔢ
A T flip-flop
the
ᔢ
is
A
clock
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
to
store
3-bit
changes
input
binary
constructed
1
able
one
bit
information.
state
changes.
counter
by
when
using
can
be
three T flip-
flops.
ᔢ
A half-adder is an arithmetic circuit
used to add two 1-bit numbers.
ᔢ
A full-adder
1-bit
circuit
adds
three
numbers.
111
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
5
be
ii
Derive
accompanying CD.
an
expression for
the
gain
of
b
In
a
[4]
particular
circuit
R
=
10 kΩ
and
R
1
1
a
State
three
characteristics
of
an
ideal
An
=
100 kΩ.
2
operational
amplifier.
b
the
amplifier.
i
Calculate
the
ii
Determine
gain
of
the
amplifier.
[1]
[3]
amplifier
circuit for
a
microphone
is
the
output
voltage
when
shown
V
=
+1.
1V
[1]
in
below.
iii
Determine the output voltage when
V
= + 2.0 V
[2]
in
c
A
sinusoidal voltage
V
=
2.0 sin ωt
is
applied
to
in
the
input
of
the
amplifier.
+
i
Sketch
a
graph
to
show
the
variation
of
the
variation
of
the
150 kΩ
input
ii
V
voltage
Sketch
a
with
graph
to
time.
show
[2]
the
out
R
output voltage
V
with
time.
[3]
out
3
X
Three voltage
V
=
to
−2V
o
State
the
type
of feedback
used
in
State
[1]
two
advantages
of
using
this
type
of
feedback.
iii
State
iv
The
the
•
type
output
of
amplifier
potential
being
difference
used.
V
is
The
a
•
−
3V
,
V
All
−
and
V
2
an
are
to
be
3
output:
V
2
circuit
input
than
[2]
produce
1
Design
this
amplifier.
ii
V
1
combined
i
signals
3
using
the following
resistance
of
the
information:
inputs
must
be
[5]
greater
10 kΩ
the
resistor values
must
be
less
than
500 kΩ
[1]
4
3.8 V
Two
circuits
P
and Q
are
cascaded
to
produce
an
out
for
of
1
a
potential
difference
across
the
resistor
output
R
V
.
o
50 mV. Calculate:
the
gain
of
the
amplifier
Q
P
[2]
120 kΩ
2
2
a
The
the
resistance
diagram
below
of
the
shows
resistor X.
an
[2]
inverting
amplifier.
+9 V
8 kΩ
–0.5 V
R
V
o
2
+
+
+12 V
–9 V
R
1
P
+
a
Calculate
V
[3]
o
V
in
–12 V
V
b
out
State
the
name
of
the
circuit Q
and
state
its
gain.
[2]
c
Suggest
d
Determine
a
practical
the
use for
circuit Q.
minimum voltage
which
[1]
causes
V
o
to
i
Explain
virtual
why
the
point
P
is
referred
to
as
saturate
to
−9 V.
[2]
a
earth.
[3]
5
200 kΩ
The figure
below
shows
a
cascade
amplifier
circuit.
1200 kΩ
100 kΩ
1200 kΩ
50 kΩ
120 kΩ
V
+
i
+
V
o
+
112
Revision questions
a
Calculate
the
overall
gain
b
Calculate
the
output
voltage
of
the
V
,
circuit.
when
[4]
V
o
=
7
a
The
diagram
below
shows
an
diagrams
can
be
to
illustrate
constructed
how
the following
using
NAND
gates
logic
only.
i
[2]
6
Use
gates
20 mV.
5
amplifier
i
NOT
gate
ii
AND
gate
[1]
iii
OR
i
Draw
[1]
circuit.
b
gate
a
[2]
truth
table for
the
logic
circuit
shown
16.0 kΩ
below.
4.0 kΩ
+12 V
[4]
A
E
8.0 kΩ
C
16.0 kΩ
D
G
+
V
A
V
–12 V
B
F
V
B
out
V
C
ii
Replace
NAND
all
the
gates.
circuit
Hence
components
minimise
the
with
number
of
gates.
a
State
b
For
the
type
of
amplifier
circuit
shown.
[1]
8
each
may
be
of
the
inputs A,
considered
as
a
B
and C,
single
the
input
[5]
a
Explain
b
State
what
is
meant
by
a
bistable
device.
[2]
amplifier
the function
of
bistable
devices
in
digital
amplifier.
circuits.
When
the
potential
amplifier
is
V
is
not
. Write
saturated,
down
an
the
expression for
c
Draw
d
Using
=
−(G
out
are
V
1
+
G
A
V
2
+
G
B
V
3
),
where
G
C
,
G
1
and
This
a
amplifier
what
[3]
circuit
is
uses
meant
by
State
9
term
truth
effects
of
negative feedback
on
table,
a
explain
how
an SR flip-flop
latch.
[5]
a
diagram
of
a
3-bit
binary
counter
using
Use
a
truth
[3]
table
binary
to
explain
the
operation
of
a
counter.
[5]
a
Add
the following
b
Draw
c
Construct
d
Draw
binary
numbers:
101
and
011 [3]
[2]
potentials
V
,
V
A
+1.0 V. Copy
and
and
B
V
are
either
complete
the following
/V
V
/V
V
B
/V
V
C
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
logic
a
diagram
truth
of
a
half-adder.
table for
a
[2]
half-adder.
[3]
table.
be
A
a
0V
C
[3]
V
as
an
10
input
Draw
3-bit
[2]
two
used
T flip-flops.
b
negative
amplifier.
The
a
negative feedback.
the
be
3
feedback
or
[1]
G
2
constants.
Explain
e
an SR flip-flop.
the form:
V
d
circuit for
out
can
c
the
V
out
in
[1]
output
a
diagram
to
show
constructed from
how
a full-adder
half-adders
and
give
can
an
/V
example
out
11
a
to
Construct
a
explain
truth
its
operation.
table for
[4]
the following
circuit.
[4]
A
E
D
b
f
Explain
how
analogue
the
circuit
converter.
can
C
B
be
used
as
a
digital
to
[3]
You
an
are
LED
doors.
the
Each
doors
switch
door
are
asked
inside
is
closes
and
the
a
circuit
has
a
logic
car. The
a
switch
closed. When
opened,
Construct
design
sports
door
are
closed,
logic
to
a
LED
truth
to
the
the
LED
LED
door
to
is
is
not
control
car
has
open
two
when
opened,
lights. When
table for
the
which
lights. When
does
meet
a
circuit
sports
the
either
both
doors
light.
the
logic
and
requirements.
draw
a
[5]
113
12
The
A
diagram
and
below
shows
a
digital
circuit
with
inputs
15
The
diagram
below
shows
an
amplifier
circuit.
B.
420 kΩ
C
A
+15 V
12 kΩ
E
+
V
B
in
–15 V
V
out
D
a
Construct
b
The timing diagram
of
the
a
truth
inputs A
diagram
to
table for
and
show
below
the
circuit.
[4]
0 V
illustrates the variation
B. Copy
the
digital
and
variation
complete
of
the
a
State
the
b
Derive
type
of
circuit.
[1]
the
output
E.
[3]
of
c
the
an
expression for
the voltage
gain
circuit.
Explain
the
explain
why
[5]
concept
of
a virtual
earth
and
A
the
inverting
terminal
is
at
this
potential.
d
Calculate
e
Explain
[3]
the
what
voltage
is
gain
meant
by
of
the
amplifier.
saturation
of
[2]
an
B
amplifier.
f
[2]
Calculate
amplify
the
maximum
before
input voltage
saturation
of
the
it
output
can
occurs.
[2]
E
16
The
diagram
below
shows
an
inverting
amplifier.
120 kΩ
13
For
the
NOR
circuit
gates.
below,
Minimise
replace
the
all
the
number
of
logic
logic
gates
gates
with
in
+15 V
1.5 kΩ
the
circuit.
V
[5]
in
X
V
out
+
A
D
–15 V
F
C
0V
E
B
The
input voltage
V
is
35 mV.
in
14
Construct
a
truth
table for
the following
logic
[5]
A
a
State
b
Calculate
the
c
State
voltage
at
the
point X.
[1]
circuit.
and
the
current
explain
in
the
the
1.5 kΩ
current
in
resistor.
the
[2]
120 kΩ
C
resistor.
[2]
B
d
Calculate
the
output
voltage
e
D
V
.
[2]
out
F
Calculate
the
maximum
input voltage
before
E
saturation
114
occurs.
[2]
Revision questions
17
a
b
Explain
what
applied
to
State
two
an
is
meant
by
negative feedback
when
amplifier.
advantages
[2]
of
using
21
The following
amplifier
diagram
being
used
shows
as
a
and
5
operational
summing
amplifier.
negative feedback.
50 kΩ
[2]
c
Explain
what
is
meant
by
the
bandwidth
of
an
+12 V
20 kΩ
amplifier.
18
a
[2]
State four
properties
of
an
ideal
10 kΩ
operational
+
V
amplifier.
1
[4]
–12 V
V
out
V
2
b
Sketch
a
graph
curve for
an
of
a
typical frequency
operational
response
amplifier.
[3]
0V
c
Label
the
d
Illustrate
bandwidth
what
bandwidth
feedback
of
is
and
happens
the
gain
to
of
the
amplifier
the
gain
when
amplifier.
[1]
and
a
negative
used.
State
the
type
of feedback
being
used
in
the
circuit.
[2]
b
[1]
Calculate
the
output
voltage
V
when
the
input
out,
19
The
diagram
below
illustrates
a
summing
amplifier.
voltages
are
V
=
–2.0 V
and
1
V
=
+
3.0 V.
[3]
2
R
c
f
Calculate
voltages
the
are
new
set
to
output voltage
–4.0 V.
if
both
input
[2]
I
f
I
R
1
1
V
1
X
R
I
V
2
2
o
+
V
2
R
I
3
3
V
3
a
Derive
an
expression for
the voltage
gain
of
amplifier.
b
State
c
Explain
used
20
a
Draw
[6]
one
as
a
use
how
a
Explain
this
type
summing
of
analogue
diagram
to
show
can
the
be
a
used
sine
to
of
can
how
a
single
a
circuit.
[6]
operational
square
input voltage.
the
[1]
be
converter.
generate
wave
operation
amplifier.
amplifier
to
voltage from
b
of
a
digital
amplifier
the
wave
[3]
[3]
115
Module
2
Practice
exam
questions
d
Multiple-choice questions
Answers
to
selected
structured
the
multiple-choice
questions
questions
can
and
be found
to
on
the
accompanying CD.
1
An
alternating
steady
current
current
dissipate
heat
of
at
of
r.m.s.
I flowing
equal
magnitude
through
rates. What
4 A
identical
is
and
a
resistors
the value
of I?
(+)
4
A
a
√
2
A
b
transformer
secondary
a.c.
supply
5.0 Ω
What
a
3
4 A
c
2 A
d
√
(–)
2 A
2
is
the
is
of
supply
can
efficient. The
primary
in
the
2.2 A
the following
be
to
connected
b
Which
the
current
0.
11 A
correct
100%
to
connected
resistor
is
is
turns
the
the
primary
into
a
the
coil
and
4
220 V
Which
of
the following
operational
is
true
about
an
ideal
amplifier?
a
secondary
i
It
has
an
ii
It
has
a
iii
It
has
an
infinite
open
loop
gain.
coil.
high
output
impedance.
coil?
d
illustrates
d.c.
of
1 : 20. A
0.22 A
circuits
converted
is
primary
across
c
ratio
turns
supply
2.0 A
how
an
with
a
i
only
c
i
and
ii
infinite
input
only
impedance.
b
i
and
d
i,
ii
iii
and
only
iii
a.c.
the
5
polarity?
An
alternating voltage
connected
to
the
of
input
of
0.
15 V
the
peak value
amplifier
is
circuit
below.
a
1 MΩ
+12 V
10 kΩ
+
input
–12 V
(+)
output
(–)
b
What
a
6
is
the
peak
+15 V
Which
of
b
output voltage?
+12 V
the following
c
is
not
−15 V
an
negative feedback?
(+)
116
Increased
b
Less
c
Reduced
d
Greater
bandwidth
distortion
(–)
c
(+)
a
(–)
bandwidth
operating
stability
d
advantage
−12 V
of
using
Module
7
An voltage
inverting
of
2.0 V
is
amplifier
applied
shown
to
the
input
of
a
non-
ii
State
below.
and
explain
transformer
b
A
step-up
2
that
Practice
three
is
questions
design features
makes
transformer
exam
it very
of
the
efficient.
connected
to
a
[6]
240 V
r.m.s.
+9 V
a.c.
supply. The
output
of
the
transformer
is
+
12 kV
and
is
connected
to
a
lighted
street
sign.
r.m.s
i
Explain
–9 V
and
the
distinction
peak value
of
an
between
r.m.s.
value
alternating voltage.
[2]
10 kΩ
ii
Calculate
the
peak
value
of
the
primary
2.0 V
voltage.
V
[2]
o
N
iii
Calculate
the
p
ratio
for
N
the
transformer.
[2]
s
5 kΩ
iv
v
Given
that
the
street
sign
is
to
transformer.
the
Calculate
coil
What
is
the
output
voltage
V
?
c
o
a
8
4.0 V
b
The following
equivalent
to
6.0 V
c
combination
what
type
of
of
9.0 V
NOR
logic
d
gates
2.0 V
is
Explain
a.c.
12
a
gate?
at
of
the
the
why
current flowing
12 mA,
calculate
through
the
the
power
input
[2]
current flowing
in
the
primary
transformer.
electrical
[2]
power
is
transmitted
using
high voltages.
Explain
what
is
semiconductor
b
Distinguish
c
A
[5]
meant
and
by
the
between
an
intrinsic
term
n-type
doping
and
[2]
p-type
materials.
[2]
p-type
placed
i
material
against
Using
a
AND
b
OR
c
NOT
d
State
the
The following
diagram
shows
a
half-adder.
If A
=
B
=
1,
what
is
the value
of
the
sum
and
material
layer
explain
and
what
explain
is
how
are
layer forms.
meant
by
the
the
an
approximate value for
it forms.
depletion
thickness
[3]
of
layer.
[1]
1
d
and
n-type
XOR
ii
9
an
each other. A depletion
diagram,
depletion
a
and
State
and
explain
what
happens
to
the
depletion
carry?
region
A
when
a
p-n
junction
is:
i
forward-biased
[2]
ii
reverse-biased.
[2]
sum
B
e
Distinguish
between
diffusion
current
and
drift
current.
[2]
carry
a
Sum
=
1, Carry
=
f
State
one
g
Sketch
an
use
of
a
p-n
junction
I–V
characteristic
of
diode.
a
p-n
[1]
junction
0
diode.
b
Sum
=
0, Carry
c
Sum
=
1, Carry
d
Sum
=
0, Carry
=
13
=
=
[2]
0
The figure
below
shows
a
non-inverting
amplifier.
1
1
+9 V
10
Which
of
the following
is
true
about
a flip-flop?
+
i
It
is
a
ii
It
can
monostable
device.
–9 V
be
made
using
cross-coupled
R
NAND
f
gates.
iii
It
can
store
one
‘bit’
of
V
memory.
out
V
in
a
i
only
c
ii
and
iii
only
b
i
d
i,
and
ii
ii
only
and
iii
R
1
Structured questions
11
a
i
Draw
used
a.c.
a
to
diagram
convert
supply.
of
a
a
transformer
120 V
a.c.
that
supply
can
into
a
be
12 V
[2]
a
Derive
the
an
expression for
non-inverting
the
amplifier.
closed-loop
gain
A
of
[5]
117
b
The
resistors
R
and
R
f
respectively
in
are
500 kΩ
practical
Determine
the
gain
ii
Determine
the
output
=
50 kΩ
i
Complete
circuit.
i
V
and
a
truth
table
with
two
inputs
and
1
a
A
of
the
one
amplifier.
voltage
ii
[2]
Draw
by
when
+100 mV.
iii
[2]
output for
a
the
Draw
the
diagram
timing
an
logic
of
the
gate.
logic
[3]
gate
represented
diagrams.
equivalent
[1]
logic
circuit
to
represent
in
iii
Determine
the
maximum
value
of
V
the
such
logic
gate
in
ii
using
only
NAND
gates.
[2]
in
that
the
V
is
not
saturated.
[2]
out
iv
Suppose
V
15
=
0.9 sin (2π(20)t). Sketch
a
a
i
Draw
to
the
symbol
and
truth
tables for
an OR
graph
in
gate
and
an
EXOR
gate.
[6]
show:
ii
1
the
variation
of
V
with
time
Draw
a
circuit
diagram
to
show
how
an OR
[2]
in
gate
2
the
variation
of
V
with
time.
can
be
constructed from
a
number
of
[3]
out
NAND
c
You
are
asked
to
design
to
a
circuit
to
signals
v
and
v
1
.
v
2
is
the
output
Redesign
the
microphone
and
circuit
below
so
that
it
may
be
signal from
1
constructed
a
[3]
combine
iii
two
gates.
v
is
a
signal from
an
using ONLY
NAND
gates.
Reduce
electric
2
the
guitar. The
two
signals
v
and
1
follows:
V
=
3v
o
the
−
2v
1
circuit. The
inputs
must
circuit
to
be
,
are
added
where
V
the
minimum
gate
count.
[6]
as
is
the
output
of
A
o
resistance
greater
to
2
2
input
v
circuit
than
of
both
100 kΩ.
signal
Design
a
B
satisfy
these
conditions.
[4]
C
14
a
An SR flip-flop
i
Draw
a
is
called
diagram
of
a
bistable
an
device.
SR flip-flop
using
only
b
NAND
ii
Explain
iii
What
gates.
what
is
i
Draw
a
circuit
is
meant
the function
by
the
term
of flip-flops
bistable
in
with
the
how
it
aid
truth
performs
Explain,
with
of
a
half-adder.
table,
Explain
its function
and
that function.
the
aid
of
a
[4]
diagram,
how
two
[2]
half-adders
inputs
a
digital
circuits?
The
of
[1]
ii
b
diagram
[2]
to
an SR flip-flop
are
and
I
1
I
can
be
used
to
build
a full-adder.
. The
2
[2]
—
outputs
are Q
and Q . Using
table
below,
as
electronic
an
explain
how
the
sequential
the SR flip-flop
truth
operates
latch.
[5]
—
Sequence
I
I
1
Q
16
The figure
below
together. The
a
negative
shows
output
edged
three T flip-flops
changes
signal
(i.e.
when
the
input
1
to
logic
0). The
outputs
are Q
2
0
0
3
1
0
4
0
0
5
0
1
1
I
T
clock
Q
T
Q
Q
is
of
a
logic
Q
Q
Q
1
2
0
Copy
and
complete
, Q
0
and Q
Q
T
Q
Q
outputs Q
diagram
2
0
a
timing
and Q
1
V
1
V
0
V
1
0
The
, Q
0
1
detects
changes from
Q
logic
6
connected
input
2
number
c
the
below
gate. The
shows
output
two
of
inputs
the
logic
the
truth
table
to
show
and Q
1
the
[4]
2
P
Clock
gate
pulse
Q
Q
0
R.
Q
1
2
0
0
0
0
1
1
0
0
P
2
3
4
Q
5
b
The
outputs from
a
are
connected
to
the
inputs
R
of
118
the
circuit
below. When
an
input
receives
Module
a
logic
an
1,
input
a
potential
receives
applied. Q
, Q
0
and Q
+5 V
0,
is
a
the
apply voltages
of
V
,
0V
V
0
circuit
Practice
exam
questions
applied. When
potential
2
to
Q
0
of
logic
1
respectively
Q
a
2
is
and
V
1
2
below.
Q
1
2
20 kΩ
10 kΩ
40
kΩ
10 kΩ
+12 V
V
+
out
–12 V
0V
0V
i
State
the
ii
State
an
type
of
circuit
equation
shown
relating
V
above.
to
the
[1]
three
out
input voltages
V
,
V
0
iii
Copy
and
Q
Q
to
Q
1
V
0
1
0
0
1
1
1
0
0
1
0
1
how
analogue
the
[2]
2.
the following
0
Explain
V
1
complete
2
iv
and
[2]
/V
out
−2.50
−6.25
circuit functions
converter
table.
(DAC).
as
a
digital
[3]
119
11
The
particulate
nature
of
electromagnetic
radiation
11.
1
The
photoelectric
Learning outcomes
The
photoelectric
Consider
On
completion
of
this
section,
be
able
and
describe
ᔢ
state
experiment
connected
the
photoelectric
to
weak
with
ᔢ
the
explain
observations
photoelectric
the
function,
a
ultraviolet
gold
in
Figure
leaf
11.1.1.
A
electroscope.
zinc
plate
When
the
is
negatively
plate
terms
radiation
the
plate
slowly
loses
its
is
charge.
effect
source
the
shown
to
to:
exposed
ᔢ
the
effect
you
charged
should
effect
associated
of
monochromatic
electromagnetic
radiation
effect
photon,
work
threshold frequency,
negatively
cut-off wavelength
charged
ᔢ
state
Einstein’s
equation
to
zinc
explain
ᔢ
define
the
the
photoelectric
plate
effect
–
–
–
–
–
–
–
–
electronvolt.
metal
rod
insulator
thin
gold
foil
gold
the
electroscope
Figure 11.1.1
The
zinc
plate
has
a
negative
it
the
electroscope.
gold
leaf
is
repels
touching
incident
moves
inwards
charge
as
leaks
away.
Demonstrating the photoelectric effect
electroscope
object
leaf
negative
on
the
the
the
zinc
charge.
electrons
The
gold
horizontal
plate,
the
When
away
leaf
plate.
gold
it
rises,
When
leaf
is
from
placed
the
on
indicating
to
fall.
a
(UV)
This
gold
plate
that
ultraviolet
begins
the
horizontal
leaf
of
charged
radiation
is
phenomenon
Definition
is
Photoelectric
effect
known
a
metal
surface
electromagnetic
are
is
exposed
radiation,
emitted from
the
the
Experiments
resulted
When
as
in
photoelectric
performed
the
to
following
effect
investigate
the
photoelectric
effect
have
observations:
to
electrons
ᔢ
surface.
Electrons
are
emitted
electromagnetic
ᔢ
If
the
intensity
emitted
ᔢ
per
Electrons
radiation
of
the
second
are
is
immediately
radiation,
only
above
even
radiation
also
upon
the
is
being
exposed
intensity
increased,
of
the
the
to
radiation
number
of
is
weak.
electrons
increases.
emitted
a
if
when
minimum
the
frequency
frequency
of
the
(threshold
incident
frequency
f
).
0
If
the
frequency
emitted,
ᔢ
If
the
even
frequency
frequency
of
ᔢ
the
Electrons
are
below
the
of
required
emitted
energies
120
if
is
intensity
the
for
from
minimum
of
incident
electron
electrons
emitted
range
this
the
frequency
radiation
radiation
emission,
is
is
electrons
are
high.
greater
the
no
than
maximum
the
minimum
kinetic
energy
increases.
with
zero
a
up
range
to
a
of
kinetic
maximum
energies.
value.
The
kinetic
Chapter
Classical
explain
wave
why
theor y
there
electromagnetic
with
high
intense
of
kinetic
electrons
to
the
a
until
particle
Einstein’s
1905
to
Low
of
a
how
wave
to
of
electromagnetic
radiation
cannot
only
ver y
emission
electromagnetic
of
kinetic
the
Also,
the
energy
questions
explain
wave
It
nature
intensity
emission.
intensity
the
These
particulate
electrons
theor y,
intensity
the
effect.
low
of
The
instantaneous
electron
increase
attempted
instead
Einstein
the
increasing
should
or
emission
allow
produce
photoelectric
Albert
to
will
frequency.
Einstein
photoelectric
cause
sur face.
able
not
model
to
According
theor y,
but
the
frequency
able
radiation
radiation
electrons
unanswered
is
metal
be
wave
electromagnetic
emitted
the
not
explain
threshold
energies.
from
should
according
In
a
radiation
electromagnetic
radiation
using
is
cannot
11
of
the
remained
photoelectric
effect
model.
equation
provided
an
explanation
for
the
photoelectric
Definition
effect.
of
as
He
proposed
being
made
that
up
of
the
a
electromagnetic
stream
of
tiny
radiation
bundles
or
can
be
packets
of
thought
energy.
A
Each
packet
is
discrete
and
is
therefore
referred
to
as
a
quantum.
photon
is
a
quantum
electromagnetic
packets
are
called
emitted
electron
1
=
radiation.
photons
photon
E
of
These
h f
E
2
=
k
m
v
e
2
Equation
Surface
metal
electrons
require
The
surface
the
least
amount
of
the
to
be
of
a
photon
E
is
given
by
equation
removed.
E
These
more
energy
energy
electrons
energy
(more
work
to
=
hf
require
be
E
–
energy
h
–
the
of
a
photon/J
removed
done).
Planck
constant
−34
(6.63
Figure 11.1.2
f
–
×
10
frequency
J s)
of
the
incident
radiation/Hz
When
a
a
single
the
single
amount
conserved
means
it
from
energy
The
photon
electron
in
that
the
of
of
on
energy
the
all
surface
minimum
surface
(Figure
equivalent
energy
emitted
the
surface
interaction
the
metal
the
reaches
the
to
hf
between
supplied
and
any
to
of
the
11.1.2).
to
the
the
the
metal,
The
surface
photon
remaining
interacts
the
will
amount
Energy
electron.
be
used
will
with
delivers
electron.
and
electron
it
photon
be
to
the
is
This
remove
kinetic
Definition
electron.
energy
needed
to
free
the
electron
from
the
surface
is
The
called
the
work
function
(φ).
If
the
energy
of
the
photon
is
greater
work function
energy
the
work
function,
the
electron
will
escape
from
the
surface
and
1
E
=
is
then
transferred
to
the
electron
as
kinetic
is
to free
the
an
minimum
electron
the
surface
of
the
metal.
energy,
2
mv
2
k
energy
needed
the
from
remaining
φ
than
e
Einstein’s
equation
to
explain
the
photoelectric
effect
is
given
below.
Equation
hf
=
φ
+
E
hf
–
energy
of
φ
–
energy
needed
k max
E
–
incident
kinetic
photon/J
to free
energy
of
electron from
emitted
surface
(work function)/J
electron/J
k max
121
Chapter
11
The
particulate
nature
of
electromagnetic
An
electron
radiation
will
only
be
emitted
if
the
energy
of
the
photon
is
greater
than
Definition
the
its
The
threshold frequency
(f
)
is
work
function.
frequency
(E
=
The
hf).
of
the
photon
Therefore,
energy
the
higher
is
directly
the
proportional
frequency
of
the
to
incident
the
0
radiation,
minimum frequency
of
the
surface.
electromagnetic
radiation
electrons
to
be
greater
Thus
a
the
energy
of
the
frequency
is
photons
required
arriving
in
order
at
for
the
metal
electrons
This
minimum
frequency
is
called
the
threshold
frequency
to
(f
be
).
0
emitted.
The
intensity
power
the
of
incident
energy
per
on
the
incident
unit
unit
surface
radiation
P
the
per
electromagnetic
Equation
I
is
minimum
required
emitted.
for
the
incident
area
time
(P
radiation
per
unit
increases
Therefore,
more
photons
more
photon–electron
=
the
E/t).
surface.
number
increasing
the
Increasing
of
the
radiation
(I
Therefore,
represents
time.
the
time.
electromagnetic
on
=
P/A).
the
photons
intensity
of
the
of
the
is
of
photons
intensity
striking
of
defined
intensity
number
the
is
Power
the
the
the
as
incident
incident
incident
surface
incident
as
defined
per
radiation
unit
causes
=
A
I
–
intensity
of
incident
emitted
electromagnetic
from
to
strike
the
the
metal
surface.
interactions.
metal
As
a
This
means
result,
more
that
there
electrons
will
will
be
be
surface.
−2
radiation/W m
If
the
frequency
of
the
incident
radiation
is
below
the
threshold
−1
P
–
incident
power/W
or J s
(gives
frequency,
a
measure
of
the
photons
incident
on
have
sufficient
per
unit
will
be
emitted.
This
is
because
energy
to
allow
the
electrons
to
the
photons
overcome
the
will
work
the
function.
surface
electrons
number
not
of
no
Increasing
the
intensity
of
the
incident
radiation
will
have
no
time)
effect
because
it
only
means
that
more
photons
are
striking
the
metal
2
A
–
surface
area/m
surface
the
per
unit
electrons
The
time
to
electrons
the
the
metal,
further
Definition
them
emitted
For
The
cut-off
wavelength
(λ
)
a
because
from
from
the
metal,
a
work
of
a
have
surface
range
the
of
are
energy
lower
is
a
not
function
energy
surface
there
will
metal
much
have
still
have
kinetic
not
therefore
given
the
radiation)
maximum
away
and
they
overcome
electromagnetic
having
and
is
and
leave
(when
kinetic
the
ones
needed
metal
kinetic
enough
to
energy
the
exposed
to
require
more
The
to
remove
allow
surface.
energies.
closest
to
the
them.
energy
electrons
surface
of
Electrons
to
remove
energies.
threshold
frequency
( f
).
The
metal
has
0
is
0
a
corresponding
cut- off
wavelength
(λ
).
In
order
for
electrons
to
be
0
the
maximum
incident
wavelength
electromagnetic
required for
electrons
to
of
the
emitted,
the
wavelength
be
than
of
the
incident
electromagnetic
radiation
must
radiation
be
lower
the
cut- off
wavelength.
emitted.
The
and
work
the
function
cut- off
of
a
metal
wavelength
as
is
related
shown
to
in
both
the
the
threshold
frequency
equation.
Equation
c
φ
=
hf
h
=
0
λ
0
φ
–
work function
h
–
the
f
–
threshold frequency/Hz
–
speed
–
cut-off
Planck
/J
constant/J s
0
−1
c
λ
of
light/m s
wavelength/m
0
The
Definition
energy
extremely
1 eV
an
is
the
energy
electron
potential
as
it
transformed
moves
difference
of
by
through
1 V.
a
energies
unit
is
associated
small.
of
defined.
electron
The
photons
moves
with
SI
are
This
photons
unit
for
much
unit
through
is
a
in
electromagnetic
energy
smaller
called
the
potential
is
the
than
joule
the
electronvolt
difference,
(J),
joule,
radiation
but
(eV).
energy
since
another
is
is
the
suitable
When
an
transferred.
Since
−19
an
electron
has
a
charge
of
1.6
×
10
C,
when
an
electron
−19
1 eV
=
1.6
×
10
J
through
a
potential
difference
of
1 V
,
the
work
−19
W
122
=
QV
=
1.6
×
10
done
−19
×
1
=
1.6
×10
J
is:
is
moved
Chapter
11
The
particulate
nature
of
electromagnetic
radiation
Example
15
Ultraviolet
radiation
of
frequency
4.83
×
10
Hz
is
incident
on
a
metal
−18
surface.
The
work
function
of
the
metal
surface
is
1.92
×
10
J.
Calculate:
a
the
energy
b
the
maximum
metal
of
a
photon
kinetic
from
this
energy
of
radiation
the
photoelectrons
emitted
−34
a
from
the
surface.
Energy
of
a
photon
=
hf
=
6.63
×
10
15
×
4.83
×
10
−18
=
b
hf
=
f
+
3.20
×
10
J
E
kmax
E
=
hf
−φ
=
3.20
kmax
−18
E
×
10
−18
–
1.92
×
−18
10
=
1.28
×
10
J
kmax
Example
−19
The
work
function
illuminated
Electrons
with
are
of
a
metal
is
2.30
electromagnetic
emitted
from
the
×
10
radiation
J.
of
The
metal
wavelength
surface
is
200 nm.
surface.
Calculate:
a
the
energy
b
the
threshold
c
the
maximum
of
a
photon
having
frequency
kinetic
of
a
the
energy
wavelength
of
200 nm
metal
of
the
electrons
emitted
from
the
metal
surface.
–34
hc
6.63
×
10
8
×
3.0
×
10
−19
a
Energy
of
a
photon
=
=
9.95
×
10
J
–9
λ
200
×
10
–19
φ
2.30
×
10
6.63
×
10
14
b
Threshold
frequency
=
=
3.47
×
10
Hz
–34
h
c
hf
=
φ
+
E
kmax
E
=
hf
−φ
=
9.95
kmax
−19
E
×
10
−19
–
2.30
×
10
−19
=
7.65
×
10
J
kmax
Key points
ᔢ
When
are
a
metal
ᔢ
Classical
ᔢ
Einstein
the
ᔢ
1
surface
emitted from
wave
the
theory
proposed
photoelectric
electronvolt
through
a
p.d.
a
exposed
cannot
particle
to
electromagnetic
(photoelectric
explain
model
of
the
radiation,
electrons
effect).
photoelectric
electromagnetic
effect.
radiation
to
explain
effect.
(eV)
of
is
surface
is
the
energy
transformed
by
an
electron
as
it
moves
1 V.
123
11.2
Investigating
Learning outcomes
the
Investigating the
Figure
On
completion
of
this
section,
be
able
describe
shows
how
the
effect
plate
is
exposed
to
photoelectric
ultraviolet
(UV)
effect
light.
can
be
Electrons
investigated.
are
emitted
A
from
to:
the
ᔢ
11.2.1
photoelectric
effect
you
zinc
should
photoelectric
an
experiment
of
investigate
the
zinc
plate.
The
emitted
electrons
are
attracted
to
the
gauze
because
to
photoelectric
its
the
positive
positive
electric
potential.
terminal
of
the
This
variable
is
because
d.c.
the
gauze
is
connected
to
supply.
effect
ultraviolet
ᔢ
provide
an
explanation
of
light
the
experiment
gauze
ᔢ
explain
the
terms
current
and
explain
how
photoelectric
+
stopping potential
zinc
plate
ᔢ
used
to
the
measure
results
the
can
be
Planck
constant.
Figure 11.2.1
The
Apparatus to investigate the photoelectric effect
ammeter
measures
the
When
the
a
small
very
gauze
difference
is
between
is
As
the
photoelectric
Reversing
to
the
zero.
reaching
current
on
this
will
the
gauze.
to
the
the
potential
zero
is
gauze
This
called
voltmeter
and
the
the
the
plate
potential
the
leaving
gauze.
zinc
increased,
because
radiation
causes
has
repels
gauze
zero
a
the
photoelectric
the
surface
remains
the
of
unchanged,
photoelectric
negative
the
is
potential
electrons
made
even
current
implies
difference
the
is
is
to
much
electrons
UV
The
plate
respect
how
photoelectric
of
zinc
constant.
gauze
the
current
The
of
supply
gauze
of
with
the
the
remain
d.c.
The
potential
potentiometer
,
reduce
and
constant.
of
cur rent .
the
matter
number
instance,
plate.
the
No
intensity
of
between
potential
plate
the
the
photoelectric
the
to
zinc
current
zinc
If
the
A
as
positive
remains
polarity
In
surface.
adjusting
to
the
decrease.
zinc
long
photoelectric
detected.
the
dependent
zinc.
respect
is
the
difference
a
current
the
to
at
current
photoelectric
current
measures
potential
that
that
stopping
current
with
emitted
more
from
negative
eventually
no
electrons
causes
the
potential
(V
the
by
reduces
are
photoelectric
).
s
photoelectric
current / μA
B
increasing
intensity
of
illumination
A
V
S
potential
difference / V
Figure 11.2.2
The
graph
Increasing
B.
This
labelled
the
graph
photoelectric
124
A
in
intensity
has
the
current
Figure
of
the
same
is
11.2.2
UV
shows
radiation
characteristic
greater
.
The
the
result
produces
shape
stopping
as
A.
of
the
the
experiment.
graph
However
,
potential
is
the
labelled
the
same
Chapter
because
the
energy
of
the
photon
and
the
work
function
11
The
remains
particulate
fixed,
nature
of
electromagnetic
radiation
so
V
/ V
S
the
In
maximum
the
experiment
known.
of
the
The
zinc
radiation
A
kinetic
graph
same
metal
is
of
above,
the
and
stopping
will
the
apparatus
and
varied
energy
remain
frequency
can
be
Planck
the
also
the
to
constant.
stopping
potential
of
used
incident
determine
The
potential
against
constant.
UV
the
frequency
in
frequency
each
is
of
case
plotted
radiation
work
is
as
is
function
the
incident
measured.
shown
in
f
Figure
0
11.2.3.
frequency/Hz
According
to
Einstein’s
equation:
Figure 11.2.3
1
hf
=
φ
+
Graph of stopping
potential against freq uency
2
mv
(1)
2
The
work
function
of
the
φ
zinc
=
metal
is
given
by:
hf
(2)
0
The
maximum
kinetic
1
energy
the
=
electrons
is
given
by:
eV
(3)
s
Equations
(2)
eV
and
=
hf
(3)
–
V
(1)
and
rearranging
it,
=
h
f
–
f
s
(4)
0
e
Equation
Equation
0
h
∴
into
hf
s
Comparing
emitted
2
mv
2
Substituting
of
(4)
e
with
y
=
mx+
c,
h
the
gradient
of
the
straight
line
=
m
=
e
hf
0
the
y-intercept
=
c
=
−
e
So
according
to
Equation
(4),
if
a
graph
of
V
is
plotted
against
f,
the
s
gradient
will
be
h/e,
the
x-intercept
will
be
f
and
the
y-intercept
will
be
0
−hf
/e
0
Applications of the
One
of
(Figure
the
best
photoelectric
applications
11.2.4).
As
of
the
effect
photoelectric
electromagnetic
radiation
effect
(e.g.
is
light)
in
the
strikes
photocell
glass
bulb
radiation
the
collector
metal
metal
are
cathode,
attracted
This
device
in
the
energy
wrist
towards
has
photocopiers
is
electrons
the
even
manufacture
and
produces
a
of
emitted
collector
numerous
and
watches,
are
from
and
digital
camera.
photovoltaic
electrical
calculators
a
applications.
and
energy.
the
surface.
small
It
can
current
be
Another
cells.
electrons
produced.
in
light
important
These
orbiting
is
found
devices
Photovoltaic
satellites
The
cells
the
meters,
vacuum
application
harness
can
be
solar
found
–
in
+
Earth.
Figure 11.2.4
A photocell
Key points
ᔢ
The
photoelectric
photoelectric
radiation
ᔢ
The
is
effect
current
can
when
be
investigated
the frequency
of
by
measuring
the
incident
the
electromagnetic
varied.
stopping
potential
can
be
used
to
determine
the
work function
of
the
metal.
ᔢ
The
Planck
constant
photoelectric
effect
can
be
determined from
the
data
obtained from
experiments.
125
11.3
Examples
on
A
completion
of
this
section,
parallel
ᔢ
be
solve
able
1.2
relating
effect
to
of
ultraviolet
radiation
of
wavelength
240 nm
and
−2
of
800 W m
−4
to:
problems
beam
you
intensity
should
photoelectric
Example
Learning outcomes
On
the
×
10
is
incident
normally
on
a
zinc
surface
of
area
−2
m
.
The
work
function
of
zinc
is
4.3 eV
.
the
Calculate:
photoelectric
effect.
a
the
energy
b
the
power
c
the
number
d
the
threshold
e
the
maximum
of
a
of
photon
the
of
of
ultraviolet
radiation
photons
incident
frequency
kinetic
incident
of
radiation
on
per
the
surface
second
on
the
surface
zinc
energy
of
the
emitted
electrons.
–34
hc
a
Energy
of
a
photon
E
6.63
=
×
8
10
×
3.0
×
10
=
–9
λ
240
×
10
−19
=
8.29
×
10
J
−4
b
Power
c
Energy
P
=
IA
=
incident
800
per
×
1.2
second
×
on
10
=
surface
0.096 W
=
0.096 J
−19
One
photon
has
an
energy
of
=
8.29
×
10
J
0.96
Number
of
photons
incident
per
second
on
surface
=
–19
8.29
×
10
17
=
d
Work
function
of
zinc
Work
function
in
φ
=
1.16
10
4.3 eV
−19
joules
×
=
4.3
×1.6
×
−19
10
=
6.88
×
10
J
–19
φ
6.88
×
10
6.63
×
10
15
Threshold
frequency
of
zinc
f
=
=
=
h
1
e
Maximum
kinetic
energy
of
1.04
×
10
Hz
–34
0
the
photoelectrons
2
mv
2
=
hf
–
φ
=
8.29
×
10
=
1.41
×
10
−19
−19
−
6.88
×
10
−19
J
Example
In
the
when
photoelectric
it
is
illuminated
frequency.
E
of
effect,
The
the
with
variation
emitted
electrons
are
emitted
electromagnetic
with
electrons
frequency
is
shown
f
in
of
from
radiation
the
Figure
a
metal
of
a
particular
maximum
kinetic
11.3.1.
kmax
8
–19
E
/10
×
J
kmax
×
6
×
4
×
×
2
0
0
4
8
12
16
20
14
/10
Figure 11.3.1
126
surface
Hz
energy
Chapter
Use
the
graph
11
The
particulate
nature
of
electromagnetic
radiation
to
find:
a
the
threshold
b
the
Planck
c
the
work
The
frequency
constant
function
photoelectric
of
the
equation
metal.
is
hf
=
φ
+
E
kmax
Rearranging
the
equation,
E
=
hf
−
φ
kmax
The
graph
in
Figure
11.3.1
shows
a
plot
of
E
against
f.
Therefore
the
kmax
gradient
the
a
of
work
The
the
line
function
threshold
is
of
equal
the
to
kinetic
Therefore,
the
Planck
constant
and
the
y-intercept
is
φ,
metal.
frequency
maximum
the
corresponds
energy
of
the
to
the
point
photoelectrons
is
on
the
graph
where
the
zero.
14
threshold
frequency
=
6.8
×
10
Hz
–19
0
–
6.0
×
10
−34
b
Planck
constant
=
=
14
6.8
×
–
16
×
14
c
Using
the
point
(16
×
10
6.52
=
hf
−
E
−19
,
=
J s
10
10
6.0
×
10
),
the
work
function
−34
φ
×
14
10
(6.52
×
is
given
14
10
×
16
×
10
by
−19
)
−
6.0
×
10
kmax
−19
=
4.43
×
10
J
Example
1
6
/10
In
an
experiment
photoelectric
to
effect,
investigate
the
the
–1
4
m
λ
wavelength
λ
of
the
3
electromagnetic
surface
E
is
and
the
radiation
incident
maximum
measured.
The
on
kinetic
variation
a
energy
with
E
kmax
of
metal
2
kmax
1/λ
is
shown
in
Figure
11.3.2.
1
Determine:
a
the
work
b
the
value
function
of
the
metal
surface
–4
of
the
Planck
–3
–2
–1
0
1
2
3
4
constant.
–19
E
Figure 11.3.2
hc
Starting
with
Einstein’s
φ
+
(
E
kmax
1
the
equation
we
E
=
hf
=
λ
)
E
φ
get
J
hc
=
equation
λ
Rewriting
/10
kmax
=
+
λ
hc
hc
1
Therefore,
plotting
a
graph
of
against
E
will
obtain
a
straight
line
kmax
λ
1
with
φ
gradient
and
y-intercept
hc
.
hc
1
−19
a
When
=
0,
φ
=
–E
.
Therefore,
φ
=
4.0
×
10
J
kmax
λ
−19
b
Using
the
points
(2
×
10
,
6
3
×
10
−19
)
and
(–4
×
10
,
0),
6
3
the
gradient
of
the
straight
line
×
10
–
0
=
–19
2
×
10
–19
–(–4
×
10
)
24
=
5
×
10
1
24
Therefore,
=
5
×
10
hc
1
−34
h
=
=
24
5
×
10
6.67
×
10
J s
8
×
3
×
10
127
Revision
Answers
found
1
A
to
on
questions
the
that
require
questions
calculation
can
6
c
be
the
number
the
surface
d
the
threshold frequency
e
the
maximum
accompanying CD.
1.2 mW
laser
produces
blue
light
of
of
photons
incident
per
second
on
[2]
of
zinc
[2]
wavelength
kinetic
energy
of
the
emitted
−7
4.75
×
m.
10
electrons.
[2]
Calculate:
5
a
the frequency
of
blue
light
a
State
a
b
the
energy
of
c
the
number
the
laser.
a
photon
of
the
effect
that
blue
light
clean
metal
surface
photons
emitted
is
electrons
illuminated
when
with
[2]
electromagnetic
of
produces
[2]
per
radiation.
[1]
second from
b
The variation
with frequency
f
of
the
maximum
[2]
kinetic
energy
E
of
the
emitted
electrons from
kmax
2
A
photocell
is
shown
in
the
diagram
a
below.
metal
surface
maximum
glass
bulb
is
shown
below.
kinetic
radiation
energy
of
emitted
–19
collector
metal
electrons
E
/ 10
J
kmax
2.65
vacuum
–
As
the
metal
inside
electromagnetic
+
the
glass
radiation,
bulb
is
exposed
electrons
are
to
emitted
and
9
then
attracted
to
the
collector
plate. The
13
14
ammeter
frequency/10
detects
a
the
6
b
a
small
current
quantity
of
of
charge
reaching
the
collector
in
seconds
the
[2]
number
collector
of
photoelectrons
during
this
Hz
12 μA. Calculate:
reaching
i
State
ii
Use
the
the
threshold frequency.
graph
to
estimate
the
[1]
Planck
constant.
the
time.
iii
[2]
Calculate
[3]
the
work function
energy for
the
metal.
3
a
Explain
what
is
meant
by
the
photoelectric
effect.
c
[3]
b
Explain
what
is
meant
by
the
Copy
the
showing
higher
terms:
[2]
diagram
the
above. Sketch
effect
of
work function
using
a
a
new
metal
graph
that
has
a
energy.
[2]
14
i
threshold frequency
ii
c
work function
Write
down
an
d
[2]
energy.
equation for
why for
electrons
[2]
the
Explain
photoelectric
with
a
2.65
×
are
range
a frequency
emitted from
of
kinetic
of
13
the
×
10
metal
energies from
Hz
surface
zero
to
−19
effect.
4
A
parallel
[2]
beam
of
ultraviolet
radiation
6
of
10
a
Explain
b
Calculate
J.
what
[2]
is
meant
by
a
photon.
[1]
−2
wavelength
is
incident
−4
8.4
×
10
255 nm
normally
and
on
intensity
a
zinc
of
600 W m
surface
of
having
area
a
the
energy
wavelength
of
of
a
photon
of
red
light
650 nm.
[2]
−2
m
. The
work function
of
zinc
is
4.3 eV.
7
Describe
an
experiment
to
show:
Calculate:
a
a
the
energy
b
the
power
of
a
photon
of
ultraviolet
radiation
the
particulate
nature
of
electromagnetic
[2]
radiation
of
the
radiation
incident
on
the
b
[2]
128
[4]
surface
the
wave
nature
of
electromagnetic
radiation.
[6]
Revision questions
8
a
What
evidence
provide
about
does
the
the
photoelectric
nature
of
effect
State
three
observations
State
what
is
b
State
three
meant
by
the
photoelectric
effect.
[2]
[2]
photoelectric
a
electromagnetic
radiation?
b
10
6
concerning
with
the
effect.
[3]
c
The
the
experimental
photoelectric
radius
of
an
atom
observations
associated
effect.
is
[3]
approximately
−10
c
Explain
the
photoelectric
photoelectric
effect
using
Einstein’s
equation.
[4]
2.0
×
10
surface
m. A
and
lamp
provides
is
placed
energy
above
at
a
a
rate
metal
of
−2
d
Explain
0.80 W m
the following:
. An
electron
requires
a
minimum
−19
energy
i
Doubling
the
intensity
of
of
radiation
electrons
doubles
emitted from
a
Increasing
energy from
equal
the frequency
of
the
wave
cannot
b
a
be
but
which
has
a
For
increases
of
the
the
emitted
laboratory
explained
requires
particular
light
to
that
theory,
maximum
emitted from
that
the
electron
the
can
a
circular
of
the
area
which
atom. On
the
has
a
basis
having
electrons.
the
wave
in
of
which
taken for
of
an
to
be
emitted from
the
metal
surface.
on
your
answer.
[5]
light
light
nature.
a
time
which
theory
explanation
the
[3]
demonstration
by
an
estimate
kinetic
Comment
Describe
be
incident
electron
radiation
a
to
[3]
the
9
J
metal
surface.
energy
10
the
radius
ii
×
surface. Assume
collect
number
5.76
incident
metal
electromagnetic
of
[4]
wavelength
of
640 nm,
calculate:
i
its frequency
ii
the
energy
of
iii
the
rate
emission
power
of
of
[2]
a
15 W.
photon
of
[2]
photons for
a
light
[2]
129
11.4
Millikan’s
Learning outcomes
oil
completion
of
this
section,
1909
ᔢ
be
able
describe
Robert
Millikan
experiment
performed
experiments
to
measure
the
elementary
you
electric
should
experiment
Millikan’s oil drop
In
On
drop
charge
e.
Figure
11.4.1
shows
a
diagram
of
Millikan’s
apparatus.
to:
Millikan’s
oil
temperature-controlled
drop
oil
bath
experiment
ᔢ
discuss
the
evidence
in
Millikan’s
calibrated
oil
drop
experiment for
quantisation
of
atomiser
the
scale
charge.
X-rays
low-power
microscope
brass
oil
parallel
droplets
chamber
metal
plates
Figure 11.4.1
T
wo
up
Millikan’s apparatus
horizontal
so
that
centre.
An
hole.
the
atomiser
.
oil
the
the
drop.
terminal
was
the
were
plates
as
scale
The
drops
some
viewed
specks.
v
The
create
got
of
the
using
was
was
The
used
was
a
oil
in
plate
fine
mist
by
diameter
had
of
oil
friction
droplets
a
they
through
between
The
due
+
D
forces
to
r
–
ρ
U
a
the
air
at
its
above
came
through
a
low-power
and
the
of
measure
focused
determined
on
by
microscope.
droplets
the
the
one
were
out
the
microscope
distance
of
hole
the
region
in
included
travelled
particular
measuring
The
seen
oil
by
drop
time
t
1
fall
set
hole
droplets
as
passed
were
small
plates.
eyepiece
to
20 cm
upper
charged
the
illuminated
microscope
velocity
to
between
was
that
approximately
apart.
used
oil
region
microscope
calibrated
oil
of
plates
1.5 cm
Eventually
droplets
between
the
Some
entered
The
were
atomiser
the
and
metal
they
a
an
and
for
it
its
to
1
known
distance
x,
when
there
was
no
electric
field
applied
plates.
acting
and
radius
on
drag
of
oil
a
falling
force
D
oil
due
drop
to
air
are
its
weight
(Figure
W,
the
upthrust
U
11.4.2).
drop
–
density
of
the
–
density
of
air
oil
0
ρ
a
W
η
–
Figure 11.4.2
g
–
–
of
air
acceleration
due
to
gravity
Forces acting on a falling
v
oil drop
viscosity
–
terminal
velocity
1
4
Weight
of
oil
drop
=
volume
of
drop
×
density
of
oil
×
g
=
3
π r
ρ
3
Upthrust
=
weight
of
=
volume
air
displaced
4
of
drop
×
density
of
air
×
g
=
force
=
6 π r η v
(Stokes’
3
π r
3
Drag
g
0
ρ
g
a
law)
1
Weight
of
oil
drop
4
130
upthrust
4
3
π r
3
=
ρ
g
0
=
drag
force
3
π r
3
+
ρ
g
a
+
6 π r η v
(1)
1
Chapter
An
electric
moved
field
upward
E
was
with
a
then
applied
terminal
between
velocity
v
.
the
The
plates
so
terminal
11
The
that
particulate
the
by
measuring
the
time
t
for
velocity
of
electromagnetic
radiation
drop
v
2
determined
nature
was
2
the
oil
drop
to
move
through
a
2
known
The
distance
electric
y
force
F
acting
on
an
oil
drop
is
Eq,
where
q
is
the
charge
on
+
E
the
oil
plates
drop
and
acting
on
Weight
d
is
an
of
4
and
E
the
oil
oil
=
V/d,
separation
drop
drop
where
+
when
drag
of
it
is
force
4
3
π r
ρ
3
g
+6 π r η v
0
Equation
the
the
=
potential
plates.
moving
π r
ρ
3
(1)
is
difference
Figure
upwards
upthrust
+
11.4.3
with
electric
a
across
shows
the
the
terminal
U
F
forces
E
velocity.
force
D
W
3
=
2
Subtracting
V
g
+
Eq
(2)
–
a
from
Equation
(2)
Figure 11.4.3
gives:
Forces acting on an oil drop
moving upward with terminal velocity
6 π r η v
=
Eq
−
6 π r η v
2
1
Eq
=
6 π r η(v
–
v
–
v
2
6 π r η(v
2
q
)
1
)
1
=
E
The
electric
field
The
terminal
strength
velocity
v
E
was
was
measured
measured
using
using
v
1
The
terminal
velocity
radius
r
of
an
oil
=
=
V/d
x/t
1
v
was
measured
using
1
v
2
The
E
=
y/t
2
drop
was
determined
by
2
calculation
using
Equation
(1).
9ηv
1
r
=
√
How
ᔢ
ᔢ
Millikan
X-rays
were
charge
on
A
ᔢ
A
used
the
eliminate
low
–
oil
to
ionise
)g
a
accuracy of
the
air
inside
the
his
experiments
chamber
to
increase
the
drops.
convection
vapour
ρ
0
improved the
constant-temperature
to
2(ρ
pressure
enclosure
surrounded
the
apparatus
in
order
currents.
oil
was
used
in
the
experiment
to
reduce
evaporation.
Conclusions
e
Figure
11.4.4
Millikan’s
illustrates
the
type
of
results
obtained
when
–
elementary
charge
performing
experiment.
electric
4e
charge
After
the
ᔢ
measuring
the
charges
on
hundreds
of
oil
drops
Millikan
concluded
3e
following:
Electric
of
a
charge
unique
is
quantised.
elementary
All
charge
electric
charges
are
integral
multiples
e
2e
e
−19
ᔢ
The
magnitude
Therefore,
since
of
the
charge
is
fundamental
quantised,
charge
charge
e
can
=
1.6
only
×
10
exist
as
C.
e,
2e,
number
3e,
oil
4e,
of
droplets
etc.
Figure 11.4.4
Results of the experiment
Key points
ᔢ
Millikan’s
ᔢ
All
experiments
electric
charges
are
showed
integral
that
electric
multiples
of
charge
a
is
unique
quantised.
elementary
charge
e
−19
ᔢ
The
magnitude
of
the fundamental
charge
e
=
1.6
×
10
C.
131
11.5
Emission
Learning outcomes
and
Line
If
On
completion
of
this
section,
absorption
spectra
white
be
able
distinguish
between
emission
discuss
how
evidence for
through
light
is
a
diffraction
made
up
of
grating,
different
a
spectrum
colours,
of
each
wavelength
(red,
orange,
yellow,
blue,
green,
having
indigo,
range
from
400 nm
(violet)
to
750 nm
violet).
(red).
spectra
line
spectra
the
light
discrete
isolated
from
a
lamp
that
contains
sodium
vapour
is
allowed
to
pass
provide
a
diffraction
grating,
a
spectrum
is
observed.
However
,
unlike
the
energy
spectrum
in
White
wavelengths
through
levels
passed
absorption
If
ᔢ
is
seen.
respective
The
and
is
to:
its
ᔢ
light
you
colours
should
spectra
atoms.
colours.
obtained
This
specifically,
using
spectrum
it
is
called
white
is
a
light,
called
line
a
this
line
spectrum
spectr um
emission
contains
(Figure
only
11.5.1).
narrow
diffraction
slit
grating
lamp
More
spectr um
line
sodium
specific
spectrum
screen
(a)
Figure 11.5.1
Suppose
produced
E
A line emission spectrum
white
is
light
is
observed.
passed
A
through
series
of
dark
a
cool
lines
gas
and
against
the
a
spectrum
coloured
background
1
E
is
2
of
observed.
white
(Figure
allowable
The
light.
dark
This
lines
type
of
correspond
spectrum
to
is
missing
called
a
colours
line
of
absorption
spectrum
Beyond
occurs.
this
The
point
the
seen)
missing
colours
ionisation
electron
is
removed
white
from
are
atom)
–
(b)
(dark
levels
lines
the
spectr um
orbits
energy
within
spectrum
11.5.2).
line
(fixed
the
cool
gas
diffraction
screen
atom.
light
e.g.
sodium
Figure 11.5.2
electron
grating
vapour
A line absorption spectrum
Explaining
line
spectra
E
2
In
photon
of
order
to
explain
line
spectra,
light
needs
to
be
thought
of
as
being
light
made
up
photons,
as
described
in
the
photoelectric
effect.
It
is
also
released
known
that
orbital
electrons
of
the
atoms
can
absorb
light
energy
that
E
is
1
incident
particular
(c)
increasing
wavelength
So,
for
unique
Line
from
a
From
can
are
spectra
absorb
vapour
produced
spectrum.
elements
only
sodium
that
experiments,
are
provide
a
Scientists
present.
present
by
in
can
evidence
is
observed
light
produces
lamp.
observe
the
bodies
for
was
emit
lamp
neon
This
celestial
and
it
a
Each
a
discrete
certain
element
as
used
the
Since
only
fixed
wavelengths
are
being
energy
produced
a
that
is
a
identify
determine
what
Sun.
levels
in
of
produces
and
to
atoms
wavelengths.
spectrum
spectrum
principle
such
of
that
in
isolated
atoms.
Discrete energy levels and
the line spectrum of an isolated atom
132
element
line
elements
Figure 11.5.3
them.
example,
different
which
on
a
line
spectrum,
Chapter
this
suggests
that
the
only
discrete
or
fixed
energy
levels
11
are
The
particulate
possible
nature
of
electromagnetic
radiation
in
Definition
isolated
energy
atoms
is
(Figure
11.5.3(a)
and
(b)).
It
can
therefore
be
said
that
the
quantised
A
When
an
electron
falls
from
a
higher
energy
level
E
to
a
lower
line
of
energy
emission
discrete
spectrum
bright
lines
is
on
a
a
series
dark
2
level
E
,
a
photon
of
light
is
emitted
(Figure
11.5.3(b)).
The
background.
energy
1
of
this
photon
is
exactly
equal
to
E
−
E
2
between
E
and
E
2
the
greater
is
.
The
greater
the
difference
1
the
energy
of
the
emitted
photon
(shorter
1
wavelength).
The
movement
of
an
electron
from
one
energy
level
to
Definition
another
levels,
is
called
different
a
transition.
transitions
Since
are
there
possible.
are
several
This
discrete
explains
why
energy
only
certain
A
wavelengths
are
present
in
the
spectrum
of
a
hot
gas
like
line
absorption
continuous
Observing
are
fixed
values.
a
If
orbits
Figure
series
an
Figure
of
E
that
to
is
is
level
E
.
be
can
with
with
1
can
called
lines
supplied
energy
it
electrons
11.5.3(b)
horizontal
atom
level
11.5.3(a),
occupy.
an
energy
this
that
level
values
an
inside
These
energy
energy,
If
seen
orbits
falls
atom,
have
diagram .
associated
electron
electron
the
may
to
is
with
jump
back
fixed
It
from
of
light
is
dark
bright
is
spectrum
a
crossed
lines.
energy
made
each
energy
by
there
up
of
one.
energy
level
E
2
photon
spectrum
sodium.
,
a
1
emitted.
The
wavelength
of
this
photon
is
given
Equation
by
hc
λ
=
hc
E
–
E
2
E
1
–
E
2
An
atom
needs
to
be
supplied
with
the
exact
amount
of
energy
E
–
E
2
order
for
an
electron
to
undergo
a
transition
from
E
to
E
1
will
not
take
place
if
too
little
or
too
much
energy
is
.
The
=
hf
or
E
1
–
E
2
=
1
λ
in
1
transition
E
–
lower
energy
level/J
1
2
supplied.
If
too
little
E
–
higher
–
the
energy
level/J
2
energy
is
supplied,
energy
level.
the
electron
may
not
be
able
to
make
it
to
a
higher
h
Planck
constant
−34
one
are
of
the
used
If
too
much
allowable
to
energy
energy
determine
the
is
levels
energy,
supplied,
within
the
the
frequency
electron
atom.
or
The
will
not
occupy
equations
wavelength
of
a
shown
(6.63
Figure
Each
or
emitted
11.5.3(c)
line
in
a
line
shows
a
possible
indicates
a
10
J s)
8
c
photon
λ
absorbed
×
–
speed
of
light
(3
–
wavelength/m
×
10
−1
m s
)
spectrum.
transition
line
on
spectrum
the
energy
for
level
this
particular
atom.
diagram.
Key points
The
concept
used
to
of
explain
discrete
how
energy
line
levels
emission
inside
and
line
an
atom
can
absorption
therefore
spectra
be
occur
ᔢ
(Figure
A
line
emission
spectrum
is
a
11.5.4).
series
discrete
energy
dark
levels
ᔢ
A
a
of
line
absorption
continuous
Line
bright
lines
on
a
background.
crossed
ᔢ
discrete
by
bright
dark
spectra
spectrum
is
spectrum
lines.
provide
evidence for
photon
E
=
photon
discrete
E
atoms.
energy
levels
in
isolated
h f
=
h f
ᔢ
Each
of
is
a
line
emitted
result
Absorption
spectrum:
Emission
of
Electron
moves
to
a

higher
energy
level.
moves
to
Photon
is
absorbed.

Photon
Explaining line spectra
as
a
change
is
Specific
energies
of
an
(lines)
observed,
therefore
there
are
level.
energy
levels.
emitted.
ᔢ
Figure 11.5.4
absorbed
energy
photon
a
are
energy
discrete

photon
spectrum:
Electron
lower
a
energy. The
or
an
electron.

represents
specific
–
E
E
2
=
hf
is
used
photon
when
undergoes
from
to
calculate
1
the frequency
E
to
2
an
of
an
an
emitted
electron
energy
change
E
1
133
11.6
Examples
Learning outcomes
of
line
spectra
Example
0
On
completion
of
this
section,
you
–0.378 eV
n = 6
should
ᔢ
use
be
able
the
to:
relationship
hf
=
E
−
2
solve
E
n = 5
–0.544 eV
n = 4
–0.850 eV
to
1
problems.
–1.51 eV
n = 3
energy
–3.40 eV
n = 2
n = 1
Figure 11.6.1
Figure
a
the
of
b
when
part
an
of
of
the
the
the
pattern
electrons
visible
Energy
of
emitted
level
makes
a
diagram
transition
photon
does
not
of
a
hydrogen
from
lie
n
in
=
the
2
to
atom.
n
visible
=
1,
region
spectrum.
wavelength
of
a
photon
that
could
be
emitted
atom.
of
fall
spectrum
energy
emitted
minimum
hydrogen
the
the
electron
electromagnetic
Sketch
when
a
that
Calculate
The
shows
wavelength
the
from
c
Energy level diagram of a hydrogen atom
11.6.1
Show
–13.6 eV
the
to
lies
n
visible
=
emission
spectrum.
(This
occurs
2.)
between
photon
line
400 nm
when
an
(violet)
electron
to
750 nm
makes
an
n
(red).
=
2
to
n
=
transition:
E
=
E
−
E
2
Energy
is
difference
a
scalar
is
1
=
−13.6
=
−10.2 eV
quantity,
so
−
(−3.40)
only
the
magnitude
of
the
energy
needed:
−19
E
W
avelength
n
=
1
of
=
photon
10.2
×
1.6
emitted
×
when
10
an
−18
J
=
1.632
electron
×
makes
10
an
n
J
=
transition:
hc
λ
=
E
–
E
2
1
–34
6.63
×
10
8
×
3.0
×
10
=
–18
1.632
×
10
−7
This
It
134
wavelength
lies
in
the
=
1.22
=
122 nm
does
not
ultraviolet
×
lie
10
in
region
the
of
m
range
the
400 nm
to
electromagnetic
750 nm.
spectrum.
2
to
1
Chapter
b
Minimum
hydrogen
energy
E
=
wavelength
atom
level
of
a
photon
corresponds
diagram.
The
to
that
the
largest
can
largest
be
emitted
transition
possible
11
The
from
is
from
nature
of
electromagnetic
radiation
the
possible
transition
particulate
on
E
the
=
0
to
−13.6 eV
.
−19
E
−
E
2
W
avelength
of
=
13.6
×
1.6
×
10
−18
J
=
2.176
×
10
J
1
photon
emitted:
hc
λ
=
E
–
E
2
1
–34
6.63
×
8
10
×
3.0
×
10
=
–18
2.176
×
10
−8
=
c
The
wavelength
of
a
9.14
photon
×
10
m
emitted
from
transition:
–34
6.63
n
=
3
to
n
=
2,
λ
×
8
10
×
3.0
×
10
=
=
658 nm
–19
(3.40
–
1.51)
×1.6
×
10
–34
6.63
n
=
4
to
n
=
2,
λ
×
10
8
×
3.0
×
10
=
=
488 nm
=
435 nm
=
411 nm
–19
(3.40
–
0.850)
×1.6
×
10
–34
6.63
n
=
5
to
n
=
2,
λ
×
10
8
×
3.0
×
10
=
–19
(3.40
–
0.544)
×1.6
×
10
–34
6.63
n
=
6
to
n
=
2,
λ
×
10
8
×
3.0
×
10
=
–19
(3.40
Now
that
spectrum
all
the
can
be
wavelengths
sketched
as
have
–
0.378)
been
shown
in
×1.6
×
calculated,
Figure
10
the
line
emission
11.6.2.
violet
411 nm
red
435 nm
488 nm
658 nm
increasing
Figure 11.6.2
wavelength
The line spectrum of an isolated hydrogen atom
Example
The
diagram
below
shows
some
of
the
outer
energy
0
levels
of
a
mercury
atom.
ionisation
–1.6
–3.7
energy/eV
–5.5
–10.4
Figure 11.6.3
Calculate
–10.4 eV
The
the
ionisation
energy
=
ionisation
energy
completely
E
Outer energy levels of a mercury atom
required
–10.4
energy
from
–
0
an
to
=
energy
in
joules
for
an
electron
situated
for
an
atom.
electron
For
completely
the
is
the
electron
remove
it
energy
at
required
energy
from
the
level
atom
to
the
remove
–10.4 eV
,
it
the
is
–10.4 eV
.
−19
Therefore,
in
level.
energy
in
joules
=
10.4
×
1.6
×
10
−18
=
1.66
×
10
J
135
11.7
Wave–particle
Learning outcomes
The
The
On
completion
should
ᔢ
be
able
explain
of
this
section,
wave–particle duality
photoelectric
It
can
only
is
made
be
cannot
explained
by
be
explained
assuming
that
using
the
classical
wave
electromagnetic
theory.
radiation
to:
the
wave–particle
up
of
particles
(photons).
Similarly
line
spectra
can
be
explained
duality
a
photon
model.
matter
The
ᔢ
effect
you
using
of
duality
describe
and
interpret
photoelectric
electromagnetic
evidence
provided
by
the
provides
radiation.
This
evidence
means
for
that
the
particulate
electromagnetic
nature
of
radiation
has
electron
characteristics
diffraction for
effect
the
wave
nature
that
make
it
seem
like
particles.
of
However
,
it
is
also
known
that
electromagnetic
radiation
(or
particles
electromagnetic
ᔢ
discuss
interference
diffraction
as
and
evidence
of
patterns.
nature
of
These
are
can
be
properties
diffracted
that
are
and
produce
associated
interference
with
waves.
the
Light
wave
waves)
is
a
form
of
electromagnetic
radiation.
Experimental
evidence
electromagnetic
shows
that
light
can
be
diffracted
and
can
produce
interference
patterns
radiation
ᔢ
use
the
relation for
the
de
Broglie
(Y
oung’s
double
wave.
the
In
slit
case
experiment).
of
the
These
photoelectric
experiments
effect,
the
show
idea
of
that
light
light
being
a
is
a
wave
h
cannot
wavelength
λ
be
used
to
explain
the
effect.
In
order
to
explain
the
experimental
=
p
observations
of
made
particles
wave
dual
of
in
tiny
some
nature,
concept
can
the
photoelectric
called
instances
which
be
is
photons.
and
as
referred
applied
to
effect,
all
a
to
light
Light
particle
as
the
in
has
can
to
be
therefore
others.
Light
wave–particle
electromagnetic
thought
of
behave
is
said
duality .
as
as
to
being
a
have
The
a
same
radiation.
Exam tip
1
The
2
photoelectric
Interference
diffraction
3
Which
4
Under
and
de
a
h
λ
one
and
is
the
conditions,
other
evidence for
provide
the
evidence for
particulate
the
wave
nature
nature
of
of
radiation.
correct? They
certain
under
called
In
provides
radiation.
electromagnetic
Equation
effect
electromagnetic
are.
electromagnetic
conditions
wave–particle
both
it
behaves
as
a
if
a
radiation
behaves
wave. This
as
a
particle
phenomenon
is
duality.
Broglie
1922
Louis
particle
under
de
under
certain
Broglie
certain
suggested
conditions,
that
wave
particles
is
might
able
to
behave
behave
like
a
as
wave
conditions.
=
p
He
λ
–
wavelength
of
proposed
The
h
–
the
Planck
the
equation
shown
that
relates
wavelength
and
momentum.
matter/m
equation
suggested
that
a
particle
is
able
to
have
a
wavelength.
constant
−34
(6.63
×
10
J s)
−1
p
–
momentum/kg m s
Electron diffraction
Experimental
have
mass
Newtonian
All
136
these
evidence
and
charge
mechanics
features
shows
and
can
show
that
can
be
also
that
an
electrons
deflected
be
applied
electron
behave
by
to
like
electric
the
behaves
particles.
and
movement
as
if
it
They
magnetic
were
of
a
fields.
electrons.
particle.
Chapter
However
,
Figure
electrons
are
11.7.1
capable
shows
of
an
experiment
producing
a
that
diffraction
11
demonstrates
particulate
nature
of
electromagnetic
radiation
that
pattern.
evacuated
electron
The
tube
gun
heater
rings
on
electrons
fluorescent
screen
graphite
positive
1 kV
electrode
Figure 11.7.1
An
electron
towards
on
Demonstrating electron diffraction
a
with
they
a
gun
thin
waves,
must
Graphite
number
and
be
de
Electrons
therefore
is
of
a
carbon
travel
through
of
the
the
to
space
electrons
would
towards
be
the
exhibit
atoms
a
able
This
is
to
which
are
pattern
a
of
then
rings
phenomenon
produce
a
experiment
wave–particle
like
material.
arranged
behaves
a
wave.
spacing
were
seen
are
waves.
diffraction
the
electrons,
diffraction
projected
is
produced
associated
diffraction
provides
pattern
evidence
to
theory.
graphite
graphite,
comparable
like
of
A
Diffraction
electrons
polycrystalline
layers,
layers
beam
screen.
since
Broglie’s
a
graphite.
behaving
many
dots
of
phosphor-coated
support
If
produces
layer
graphite
uniform
a
like
screen.
target
the
The
the
waves
particles,
their
of
a
Increasing
increases
pass
that
of
Since
grating.
wavelength
layers
to
consists
planes.
diffraction
When
similar
crystal
in
between
the
Each
as
occurs.
behaving
on
duality
a
the
light.
large
there
The
are
electrons
through
of
of
the
electrons
is
graphite.
uniform
the
speed
momenta.
distribution
of
the
This
of
electrons
causes
Exam tip
their
h
wavelength
to
decrease
since
λ
=
.
As
a
result
the
diameter
of
The
the
electron
diffraction
experiment
p
concentric
rings
provides
decreases.
nature
6
electron
small
the
particles
wave
such
as
electrons.
Example
An
evidence for
of
is
travelling
at
a
speed
of
5.5
×
10
−1
m s
.
Calculate
the
−31
wavelength
of
this
electron.
(Mass
of
an
electron
=
9.11
×
10
kg,
the
−34
Planck
constant
=
6.63
×
10
–34
h
h
J s)
6.63
×
10
−10
λ
=
=
=
=
–31
mv
p
9.11
×10
1.32
×
10
m
6
×
5.5
×
10
Key points
ᔢ
Wave–particle
wave
and
duality
particle
refers
to
the
idea
that
light
and
matter
have
both
properties.
h
ᔢ
de
Broglie
stated
that
the
wavelength
of
a
particle
is
given
by
λ
=
p
ᔢ
Electron
diffraction
provides
evidence
that
particles
are
able
to
behave
as
waves.
ᔢ
Interference
and
electromagnetic
ᔢ
The
diffraction
evidence for
the
wave
nature
of
particle
nature
of
radiation.
photoelectric
electromagnetic
provide
effect
provides
evidence for
the
radiation.
137
11.8
X-rays
Learning outcomes
X-ray
When
On
completion
of
this
section,
ᔢ
be
able
explain
the
someone
has
a
broken
arm
or
leg
the
most
common
way
to
view
you
the
should
production
injury,
without
doing
an
internal
examination,
is
to
get
an
X-ray.
to:
process
X-rays
are
X-rays
are
simply
shadow
pictures
of
internal
structures
inside
the
body.
of X-ray
produced
by
bombarding
a
metal
surface
with
electrons
that
production
have
ᔢ
explain
the
origins
of
line
been
−μx
use
the
a
large
potential
difference.
When
a
particle
is
accelerated,
electromagnetic
radiation
is
produced
spectra
(called
ᔢ
through
and
charged
continuous X-ray
accelerated
relationship
I =
I
Bremsstrahlung
charged
e
particle,
the
radiation).
shorter
the
The
greater
wavelength
the
of
acceleration
the
of
the
electromagnetic
0
radiation
for
the
attenuation
of X-rays
produced.
in
matter
–
ᔢ
discuss
the
use
radiotherapy
of X-rays
and
high
in
imaging
+
voltage
supply
in
medicine.
heated filament
(cathode)
vacuum
cooled
anode
X-ray
window
X-rays
Figure 11.8.1
Figure
11.8.1
produced
In
this
result
The
using
kinetic
the
Figure
X-ray
The
the
of
11.8.2
a
of
an
X-ray
process
from
the
tube.
called
surface
Electrons
ther mionic
of
the
are
emission .
cathode
as
a
in
in
this
the
process.
very
large
converted
the
into
process
potential
are
accelerated
difference
When
the
electrons
decelerations.
into
thermal
radiation.
The
electromagnetic
towards
(20–100 kV).
Most
energy.
radiation
spectrum.
of
The
strike
the
Some
of
X-rays
the
the
metal
electrons
metal
kinetic
produced
The
the
energy
kinetic
lies
then
in
the
exit
window.
shows
a
typical
spectrum
of
the
X-rays
produced
in
an
tube.
produced
distribution
lines.
There
is
These
also
wavelength
electrons
138
is
X-ray
spectrum
the
escape
by
spectrum
intensity
is
diagram
cathode
large
undergo
converted
continuous
used.
very
energy
region
X-ray
simple
produced
a
they
is
through
a
heated
electrons
electrons
energy
X-ray
a
heating.
electrons
surface,
of
shows
from
process,
of
anode,
gain
A simple diagram showing how X-rays are produced
is
a
shows
of
lines
are
cut- off
emitted
converted
into
two
distinct
wavelengths
and
components.
a
characteristic
wavelength.
when
all
X-rays.
the
series
of
This
the
of
metal
minimum
kinetic
There
sharp
energy
of
is
a
high-
target
being
wavelength
the
incident
Chapter
11
The
particulate
nature
of
electromagnetic
radiation
intensity
High-intensity
characteristic
spikes
of
the
are
metal
Electrons
from
anode.
continuous
E
2
the
cathode
spectrum
supply
for
energy
the
electrons
in
wavelength
the
atom
metal
to
be
excited.
E
1
Sharp
cut-off:
orbital

determined

all
the
by
energy
the
of
operating
the
electron
voltage
electron
is
converted
into
thermal
energy.
(a) Excitation
Figure 11.8.2
Typical spectrum of X-rays during production
orbital
A
a
continuous
frequency
of
spectrum
range.
decelerations
The
and
is
one
in
electrons
this
is
what
which
all
hitting
gives
frequencies
the
rise
metal
to
the
are
target
possible
have
continuous
a
electron
within
wide
range
spectrum.
E
2
When
the
electrons
orbits
level
of
strike
the
the
metal
(excitation).
metal
atoms
When
surface,
become
these
the
electrons
excited
electrons
fall
and
from
that
jump
a
high
to
are
a
found
higher
energy
in
energy
level
to
a
photon
lower
energy
level
(de-excitation)
photons
of
energy
are
emitted
released
(Figure
hc
11.8.3).
This
gives
rise
to
the
line
spectrum
(spikes).
The
atoms
E
that
=
λ
E
1
make
up
orbital
the
electrons
correspond
metal
metal
is
to
used
positions
target
can
only
photons
for
the
because
have
of
distinct
occupy
fixed
different
anode,
different
the
energy
energy
energy
levels.
levels
being
high-intensity
metal
atoms
have
This
so
different
released.
spikes
means
If
occur
different
a
that
spikes
different
at
energy
different
(b)
De-excitation
levels.
Figure 11.8.3
Excitation of orbital electrons
Attenuation of X-rays
The
intensity
through
of
matter
.
X-rays
Figure
decreases
11.8.4
exponentially
shows
what
as
the
happens
to
radiation
the
passes
intensity
of
a
absorbing
parallel
beam
of
X-rays
having
an
initial
intensity
I
as
it
passes
through
medium
x
0
µx
an
absorbing
medium.
The
intensity
at
the
point
P
is
given
by
I
=
I
e
,
0
where
µ
is
called
the
linear
absorption
coefficient
of
the
material.
X-rays
P
initial
Example
intensity
I
0
An
X-ray
tube
operates
at
50 kV
and
the
current
through
it
is
1.1 mA.
μx
I
=
I
e
0
Calculate:
Figure 11.8.4
a
the
electrical
power
Attenuation of X-rays
input
through matter
b
the
speed
c
the
cut- off
a
Power
b
Gain
of
the
electrons
wavelength
of
when
the
they
X-rays
hit
the
=
IV
=
1.1
×
10
inside
the
tube
emitted.
−3
input
target
3
×
50
×
10
=
55 W
Equation
in
kinetic
energy
1
of
electrons
=
loss
in
electrical
potential
energy
2
mv
=
–μx
QV
I
2
=
I
e
0
2QV
2
v
I
=
–
intensity
m
of X-rays
at
a
distance
−2
x/W m
−2
I
2QV
–
initial
intensity
of X-rays/W m
–
linear
absorption
0
v
=
√
m
μ
coefficient/
−1
m
–19
2
v
×
1.6
×
10
3
×
50
=
–31
√
9.11
×
×
10
x
–
distance
travelled
through
10
medium/m
8
v
=
1.33
×
10
−1
m s
139
Chapter
11
The
particulate
nature
of
electromagnetic
c
The
radiation
minimum
kinetic
energy
wavelength
of
the
is
incident
the
wavelength
electrons
is
emitted
converted
when
into
all
the
X-rays.
hc
E
=
λ
hc
QV
=
λ
–34
hc
Cut- off
wavelength
λ
6.63
=
×
8
10
×
3.0
×
10
=
–19
1.6
QV
3
×10
×
50
×
10
−11
=
2.49
×
10
m
Example
−1
The
linear
absorption
coefficient
µ
for
bone
and
muscle
are
2.9 cm
and
−1
0.95 cm
respectively
.
A
parallel
X-ray
beam
of
intensity
I
is
incident
on
0
some
bone
muscle
of
tissue
thickness
passing
through
of
thickness
1.3 cm.
the
4.0 cm.
Calculate
muscle
tissue
the
and
Below
the
intensity
bone.
muscle
of
Give
the
the
tissue
X-ray
is
some
beam
answer
in
after
terms
of
I
0
µx
Intensity
after
passing
through
the
muscle
tissue
=
I
=
e
I
0
−(0.95
=
I
×
4.0)
e
=
0.0224I
0
0
µx
Intensity
after
passing
through
the
bone,
I
=
I
e
0
−(2.9
=
0.0224 I
×
1.3)
e
−4
=
5.16
×
10
I
0
0
Uses of X-rays
X-ray
X-rays
of
the
and
are
used
body.
are
inside
body
This
tissue.
X-ray
to
They
therefore
the
tissue.
soft
photography
produce
are
able
have
means
As
a
very
to
shadow
pass
easily
different
that
result
photographs
penetrating
bones
there
is
when
through
densities.
are
a
more
of
the
Bone
body.
tissue
effective
contrast
internal
compared
at
between
structures
with
visible
Different
is
denser
absorbing
bone
and
X-ray
computed tomography
(CT)
rotating X-ray
detectors
fan-shaped
X-ray
beam
data
motorised
platform
rotating X-ray
Figure 11.8.5
140
A CT scanner
source
than
soft
X-rays
soft
photographs.
light
tissues
than
tissue
on
Chapter
When
using
It
X-rays
no
(CT)
scanners
the
body.
patient
to
are
The
be
obtained.
X-ray
and
the
Powerful
picture
ᔢ
a
able
the
the
the
lies
on
inside
by
of,
is
for
a
example,
a
shows
a
particulate
broken
two -dimensional
11.8.5
toroidal
which
of
scanner
fast,
a
the
detailed
movable
an
there
source
images
of
very
a
scanner
nature
of
electromagnetic
radiation
leg
picture.
computed
is
and
the
three-dimensional
table.
an
of
X-ray
detectors
are
a
slice
source
are
taken.
images
of
multiple
images
to
produce
provides
being
detailed
of
CT
pictures
the
scanner
.
body
may
multiple
about
patient
slices
a
for
the
and
rotated
The
the
body
allows
(doughnut-shaped)
image
slice
This
and
combine
region
Advantages:
Figure
produce
X-ray
computers
of
to
technique
The
of
image
photograph
depth.
patient
multiple
axis
the
an
The
scanner
.
Inside
detectors.
patient
along
is
make
of
positioned
T
omography
be
to
film,
perception
tomography
of
used
photographic
gives
CT
are
11
are
is
the
shifted
taken.
three-dimensional
investigated.
three-dimensional
pictures
of
the
body
ᔢ
Disadvantage:
greater
exposure
to
X-rays
than
standard
imaging
techniques
ᔢ
Uses:
to
detect
solid
tumours
and
other
problems
in
the
abdomen
and
chest
Radiotherapy
X-rays
at
a
are
also
person’s
used
body
in
to
radiotherapy.
kill
cancer
High-energy
cells
and
keep
X-rays
them
are
from
directed
growing
and
multiplying.
Key points
ᔢ
X-rays
have
ᔢ
The
are
been
produced
by
accelerated
typical
continuous
spectrum
bombarding
through
of X-rays
distribution
of
a
a
large
metal
surface
potential
produced
wavelengths
in
and
electrons
that
difference.
an X-ray
a
with
series
tube
of
consists
sharp
of
a
high-intensity
lines.
ᔢ
The
intensity
ᔢ
X-rays
body,
the
are
of X-rays
used
to
decreases
obtain
three-dimensional
treatment
of
exponentially
shadow
images
pictures
using CT
of
as
it
passes
internal
scanners
through
structures
and
in
of
matter.
the
radiotherapy
in
cancers.
141
12
Atomic
12.
1
Atomic
structure
Early theories of the
The
completion
of
this
section,
be
able
give
a
of
brief
account
of
the
of
Thomson,
the
atom
due
Rutherford
to
cut
describe
and
J.
the
to
atoms.
suggest
They
that
matter
thought
that
if
was
made
you
were
in
up
half
with
and
a
then
particle
continue
that
was
cutting
so
the
small
resulting
that
it
could
halves,
not
be
further
.
Thomson
He
was
knew
investigating
hydrogen
that
he
atom.
This
was
the
dealing
nature
Experiments
particle
became
with
a
of
the
particles
particle
showed
known
as
that
the
much
in
cathode
the
smaller
particle
was
than
negatively
electron
the Geiger–Marsden
experiment
describe
end
any
charged.
ᔢ
earliest
the Geiger–Marsden
experiment
interpret
the
called
Bohr
rays.
ᔢ
were
particles
something
would
divided
J.
ᔢ
Greeks
small
early
you
theories
very
to:
to
ᔢ
Ancient
atom
you
up
should
radioactivity
structure
Learning outcomes
On
and
the
In
nuclear
model
of
the
1903
there
the
Thomson
must
‘plum
be
suggested
positive
pudding’
that
charges
model
of
since
inside
the
the
the
atom.
atom
atom
In
this
was
as
electrically
well.
the
He
atom
neutral,
proposed
was
visualised
atom.
as
being
a
positively
throughout
charged
sphere
with
negative
electrons
distributed
it.
electrons
–
–
–
positive
sphere
–
–
Figure 12.1.1
The ‘plum pudding’ model of the atom
Geiger–Marsden
In
1909
Hans
Rutherford,
(α-particles)
helium
of
The
to
stopped
was
the
very
a
of
a
tiny
used
and
Ernest
now
lost
at
the
the
to
electrons.)
thin
to
in
reach
α-particles
of
foil.)
light
of
area,
was
α-particle
very
A
Each
was
thin
ensure
is
In
sheet
to
that
an
the
a
the
the
of
in
direction
which
helium
of
alpha
nucleus,
experiment
gold
evacuated
gold
would
glass
foil.
The
a
Ernest
particles
i.e.
fine
a
beam
experimental
time
foil.
have
(If
lost
screen,
an
chamber
.
the
all
their
coated
α-particle
This
was
experiment
energy
with
struck
zinc
the
done
had
been
and
sulphide
glass
seen.
α-particles
reducing
very
under
experiment
12.1.2.
microscope.
beam
(An
performed
gold
Marsden,
famous
two
a
Figure
was
flash
collision
foil
absorbed,
142
in
air
,
narrow
small
The
in
to
has
fired
α-particles
short
fixed
screen,
a
was
experiment
allow
a
scattered.
that
shown
performed
A
were
atom
is
and
performed
α-particles
set-up
Geiger
experiment
was
the
This
uncertainty
prevent
the
used.
too
in
many
α-particles
were
ensured
the
that
there
scattering
α-particles
scattered
angle.
from
only
was
being
once.
Chapter
12
Atomic
structure
and
radioactivity
+90°
moveable
table
microscope
gold foil
alpha
source
180°
microscope
–90°
zinc
sulphide
gold foil
screen
(b)
(a)
Top view
Side view
Figure 12.1.2
The
The Geiger–Marsden experiment
experimental
ᔢ
Most
ᔢ
Some
ᔢ
A
of
of
small
the
the
results
were
α-particles
went
α-particles
number
of
as
straight
were
particles
follows
(Figure
through
deflected
were
12.1.3):
the
gold
foil.
slightly.
deflected
by
angles
up
to
almost
180°.
The
ᔢ
following
Based
on
model),
the
gold
is
the
it
what
model
was
foil
was
of
actually
the
expected
with
little
atom
that
or
no
expected
at
most
the
of
to
time
the
happen:
(the
‘plum
particles
pudding’
would
pass
through
deflection.
number
alpha
of
particles
0
–180°
+180°
angular
Figure 12.1.3
deviation
The results of the experiment
143
Chapter
12
Atomic
structure
and
radioactivity
Conclusions of the
The
experiment
positive
region:
charge
the
Consider
shown
suggests
and
that
most
of
most
its
of
mass
the
are
atom
is
empty
concentrated
in
space
a
very
but
its
small
nucleus
the
in
experiment
path
Figure
alpha
of
four
α-particles
heading
towards
a
gold
nucleus
as
12.1.4.
particles
gold
nucleus
α
1
α
2
α
3
α
4
Figure 12.1.4
The
α
The path of α-particles approaching the nucleus
-particle
is
furthest
away
from
the
gold
nucleus.
It
therefore
4
experiences
undergoes
The
α
-
the
very
and
α
2
The
little
electrostatic
(repulsive)
force
and
consequently
deflection.
-particles
approach
closer
to
the
gold
nucleus.
They
3
therefore
larger
smallest
experience
larger
electrostatic
forces
and
consequently
undergo
deflections.
α
-particle
approaches
the
gold
nucleus
head- on.
The
electrostatic
1
force
acts
in
travelling,
a
and
direction
the
opposite
α-particle
is
to
that
in
therefore
which
repelled
the
in
α-particle
the
is
opposite
direction.
Most
of
the
that
the
very
small
The
fact
close
the
to
α-particles
atom
compared
that
180°
nucleus.
only
in
of
the
positively
of
atom
charged
straight
the
small
of
size
most
through
empty
number
that
nucleus
the
conclusions
model
a
went
mainly
with
suggests
The
concentrated
The
consists
of
the
of
of
a
gold
The
foil.
size
of
This
the
suggests
nucleus
is
atom.
α-particles
the
contains
the
space.
mass
of
positive
were
the
deflected
atom
charge
and
is
by
angles
contained
the
charge
in
is
nucleus.
this
experiment
(Figure
12.1.5).
nucleus
with
led
The
Rutherford
atom
negatively
to
consists
charged
develop
of
a
a
planetary
central
electrons
orbiting
−9
the
nucleus.
He
calculated
the
diameter
of
the
atom
(~10
m)
and
the
−14
diameter
There
of
the
were
nucleus
two
electromagnetic
electrons
in
According
accelerating
be
that
does
144
the
not
this
to
m).
problems
with
this
radiation
with
only
model
Maxwell’s
will
(~10
emit
electrons
happen.
of
the
atom
theory
of
model.
undergo
an
eventually
centripetal
radiation.
fall
atom
wavelengths.
electromagnetism,
electromagnetic
would
Firstly,
specific
acceleration.
charged
The
towards
emits
Secondly,
effect
the
particles
of
this
nucleus,
would
but
this
Chapter
12
Atomic
structure
and
radioactivity
electron
Figure 12.1.5
Later
are
The nuclear model of the atom (Rutherford’s model)
Niels
Bohr
observed
suggested
within
and
that
the
was
orbital
why
the
atom
able
the
to
electrons
electrons
(Figure
provide
occupy
a
model
do
not
to
fall
discrete
explain
into
the
energy
why
line
nucleus.
levels
or
spectra
He
‘shells’
12.1.6).
electrons
negatively
nucleus
charged
containing
positively
charged
protons
discrete
energy
levels
Figure 12.1.6
Table 12.1.1
(shells)
Bohr’s model of the atom
The relative masses and charges of subatomic particles
Relative
mass
Relative
1
Electron
charge
–1
1840
Proton
1
+1
Neutron
1
0
Key points
ᔢ
In J. J. Thomson’s
electrons
ᔢ
The
α-scattering
pudding’
ᔢ
In
In
Niels
charged
the
atom
throughout
experiments
is
a
positive
sphere
with
negative
it.
provided
evidence
against
this
‘plum
model.
Rutherford’s
electrons
ᔢ
model
distributed
model
orbiting
Bohr’s
model
nucleus
the
around
but
of
nucleus
is
very
small
and
positively
charged
with
it.
the
occupy
atom
electrons
discrete
energy
orbit
the
small
positively
levels.
145
12.2
Nuclear
Learning outcomes
reactions
Notation
An
On
completion
of
this
section,
atom
be
able
nucleus
represent
of
a
central
atoms
consists
of
nucleus
protons
and
surrounded
neutrons.
by
The
orbital
protons
electrons.
inside
the
to:
nucleus
ᔢ
consists
represent
you
The
should
used to
nuclear
reactions
are
positively
charged.
The
neutrons
inside
the
nucleus
are
using
uncharged.
nuclear
equations
Most
ᔢ
appreciate
proton
number
conserved
ᔢ
describe
fission
that
in
the
and
nucleon
and
energy
nuclear
the
mass
of
the
of
the
positive
charge
is
atom
located
is
concentrated
the
nucleus
contribute
in
the
nucleus.
in
the
The
nucleus.
electrons
All
of
orbiting
are
reactions
processes
of
number,
negatively
is
equal
A
nuclide
charged.
to
the
In
very
a
positive
little
neutral
charge
mass
atom,
in
the
to
the
the
atom.
negative
The
electrons
charge
of
the
are
electrons
nucleus.
nuclear
nuclear fusion.
protons
is
a
and
different
type
of
atom
neutrons.
whose
The
nuclei
following
have
notation
specific
is
used
numbers
to
of
represent
nuclides:
A
X
Z
X
represents
the
symbol
A
represents
the
nucleon
Z
represents
the
proton
A
=
Z
+
N,
where
N
is
for
the
nuclide.
number
number
the
or
or
mass
number
atomic
number
of
of
number
the
of
atom.
the
atom.
neutrons.
Example
23
A
sodium
atom
is
represented
by
Na.
11
Nucleon
Proton
number
number
(see
or
12.3)
atomic
Number
of
electrons
Number
of
protons
Number
of
neutrons
=
=
or
mass
number
number
=
=
23
11
11
11
=
23
−
11
=
12
Nuclear fission
Heavy
nuclei
such
unstable.
They
projected
towards
nuclear
makes
more
An
the
fission .
it
even
stable
example
two
as
can
a
uranium
decay
uranium
When
more
the
of
a
1
are
a
nucleus
+
The
reaction
it
→
92
by
can
induce
with
the
shown
nucleus
plutonium
However
,
collides
92
U
and
uranium
is
krypton
235
n
0
92)
accompanied
fission
fragments
=
neutron
unstable.
fragments,
(Z
spontaneously.
+
a
the
nucleus
a
in
the
and
a
(Z
=
94)
neutron
type
of
then
of
called
nucleus
splits
several
equation
barium
are
is
decay
uranium
emission
141
Kr
36
if
into
it
two
or
neutrons.
below.
In
this
nucleus.
1
Ba
+
3
56
n
+
energy
0
Definition
Energy
Nuclear fission
whereby
an
bombarded
is
an
induced
unstable
by
a
process
nucleus
of
is
released
kinetic
radiation.
is
uranium
neutron. The
energy
The
nucleus
splits
into two or
more
as well
as
several
chain
process.
emitted
produce
reaction.
in
even
In
The
particles
a
the
more
as
energy
well
reaction
fission
nuclear
as
is
in
the
form
electromagnetic
can
collide
reactions.
reactor
,
but
it
is
carefully
controlled.
A
material
neutrons.
the
146
and
fission
emitted
the
with
The
same
more
result
reaction
is
an
takes
stable
place,
fragments
the
the
neutrons
nuclei
uncontrolled
in
of
neutrons
being
released
in
the
reaction.
is
used
to
absorb
some
of
Chapter
In
nuclear
reactions,
mass
and
charge
are
always
conserved.
This
12
Atomic
structure
and
radioactivity
means
Definition
that
the
total
total
mass
mass
number
number
on
on
other
one
side
side
of
the
of
the
equation
must
equal
the
equation.
Isotopes
Consider
the
mass
(nucleon)
numbers
on
either
side
of
the
fission
same
equation.
mass
1
The
to
sum
the
of
sum
Consider
+
the
of
the
235
mass
the
92
+
numbers
mass
atomic
=
141
on
numbers
(proton)
+
(3
×
1)
left-hand
on
the
numbers
A
side
of
the
right-hand
on
either
=
side
of
an
atomic
element
number
+
92
=
36
+
56
+
(3
×
0)
of
the
Z
the
is
sum
of
the
atomic
numbers
on
left-hand
side
of
=
the
Definition
equation.
fission
equation.
whereby
92
equation
is
the
sum
of
the
atomic
numbers
on
the
right-hand
side
of
the
light
by
is
the
nuclei
combining
process
become
with
more
other
equal
light
to
different
equal
stable
The
but
the
numbers.
Nuclear fusion
0
have
236
equation
side
of
nuclei
to form
a
heavier
stable
equation.
nucleus,
of
accompanied
by
the
release
energy.
Nuclear fusion
Light
nuclei
can
become
more
stable
by
combining
with
other
Key points
light
2
nuclei
in
a
process
called
nuclear
fusion .
An
isotope
of
hydrogen
is
H
1
(deuterium).
T
wo
of
these
nuclei
can
combine
to
form
a
more
ᔢ
stable
Nuclear fission
is
an
induced
4
nucleus
(
He).
In
this
process
energy
is
process
released.
whereby
an
unstable
2
nucleus
Fusion
is
the
process
by
which
stars
such
as
the
Sun
produce
energy.
is
bombarded
neutron. The
example
of
a
fusion
reaction
is
shown
in
the
equation
2
H
+
H
→
He
1
the
mass
more
nucleus
splits
into
stable fragments
+
as
several
as
neutrons.
energy
2
ᔢ
Consider
or
4
well
1
a
below.
two
2
by
An
(nucleon)
numbers
on
either
side
of
the
fusion
Nuclear fusion is the process
equation.
whereby light nuclei become
2
+
2
=
4
A
=
4
more stable by combining with
The
to
sum
the
of
sum
Consider
the
of
the
mass
the
numbers
mass
atomic
1
numbers
(proton)
+
1
on
=
left-hand
on
the
numbers
side
of
the
right-hand
on
either
equation
side
side
of
2
of
the
Z
the
equal
other light nuclei to form a heavier
equation.
fusion
=
is
stable nucleus, accompanied by
the release of energy.
equation.
ᔢ
2
Nucleon
and
The
sum
of
the
atomic
numbers
on
left-hand
side
of
the
equation
is
the
sum
of
Table 12.2.1
the
atomic
numbers
on
the
right-hand
side
of
the
are
proton
number
conserved
in
equal
nuclear
to
number,
energy
reactions.
equation.
Comparison of nuclear fission and nuclear fusion
Fission
Similarities
Both
Fusion
processes
Charge
The
and
total
release
mass
rest
are
mass
energy.
conserved
of
the
in
the
resulting
process.
nuclide(s)
is
less
than
the
total
rest
mass
than
that
of
the
original
nuclide(s).
The
binding
energy
(see
12.3)
of
the
resulting fragments
is
greater
of
the
original
nuclide(s).
Differences
Most
of
the
energy
is
carried
away
by
massive
fragments.
Fission
can
Most
of
the
energy
is
carried
away
by
light
fragments.
be
easily
initiated
by
neutron
Fusion
is
difficult
to
achieve.
bombardment.
There
are
energy
Conditions
required
for
to
a
process
sustained
occur
manner
numerous
and
The fission
in
down
the
reactor.
masses
rate
can
release
possible
that
of
be
can
combinations
be
controlled
neutrons
of
produced.
in
by
the
slowing
nuclear
The
energy
reaction
Extremely
density
and
are
of
masses
always
high
produced
are
in
the
a
high
same.
temperatures
plasma
random fusion
the
and
required
collisions
to
to
allow for
occur.
147
12.3
Binding
energy
Learning outcomes
Mass defect
In
On
completion
of
this
section,
order
and
should
be
able
to
understand
why
energy
is
released
in
the
processes
of
fission
you
fusion
we
need
to
look
at
the
nucleus
of
an
atom
more
closely.
to:
12
Consider
a
stable
carbon
atom,
C.
There
are
six
protons
and
six
neutrons
6
ᔢ
understand
the
concept
of
mass
inside
defect
the
referred
to
nucleons.
ᔢ
appreciate
between
the
nucleus.
as
The
nucleons.
The
mass
protons
and
Therefore,
number
is
neutrons
the
often
inside
carbon
referred
the
nucleus
to
as
nucleus
contains
the
nucleon
are
twelve
number
.
association
energy
and
mass
Consider
the
following
hypothetical
experiment.
The
mass
of
a
carbon-12
2
represented
by
E = mc
nucleus
there
ᔢ
illustrate
graphically
the
binding
energy
per
are
six
nucleon
very
describe
the
strong
per
relevance
of
and
nucleon
to
use
the
nuclear fission
of
atomic
mass
degree
protons
However
,
unit
(u)
as
neutrons.
of
accuracy
.
These
Inside
nucleons
the
are
nucleus,
held
together
short-range
that
they
are
nuclear
an
forces.
infinite
Suppose
distance
you
apart.
separate
The
total
the
mass
nucleons
(protons
accuracy
.
Y
ou
and
neutrons)
is
now
measured
of
all
with
of
would
expect
that
the
mass
of
six
a
neutrons
would
this
nucleons
is
is
be
not
exactly
the
greater
equal
case.
than
to
the
mass
Measurements
the
mass
of
the
of
the
show
carbon-12
and
that
carbon-12
the
nucleus.
total
nucleus.
mass
This
of
seems
a
to
unit
six
degree
nuclear
the
ᔢ
and
high
binding
six
fusion
so
individual
high
energy
protons
a
number
the
ᔢ
with
nucleon
nucleons
with
measured
variation
by
of
is
contradict
the
law
of
conservation
of
mass.
The
missing
mass
is
called
energy.
the
In
to
mass
order
be
defect
to
separate
supplied.
The
the
nucleons
energy
in
supplied
the
nucleus
accounts
completely,
for
an
energy
apparent
has
increase
in
2
mass.
Definition
In
energy
The
mass
defect
difference
of
between
a
nucleus
the
mass
is
of
Einstein’s
are
and
constituent
the
total
mass
of
,
he
states
that
mass
and
energy
energy
the
required
binding
binding
nucleus
to
completely
energy
energy
divided
by
per
the
of
the
nucleon
total
separate
the
nucleons
of
a
nucleus
nucleus.
is
number
equal
of
to
the
binding
energy
of
the
nucleons.
10
2
=
mc
the
The
E
=
its
called
Equation
E
the
The
nucleons.
equation
interchangeable.
Binding
nucleus
famous
mc
E
–
energy/J
m
–
mass
9
56
−1
c
–
speed
of
light/m s
VeM/noelcun
defect/kg
Fe
8
26
238
7
U
92
4
He
2
6
rep
5
ygrene
4
gnidnib
3
2
2
H
1
1
0
0
20
40
60
80
100
120
140
160
180
200
nucleon
Figure 12.3.1
148
Binding energy per nucleon against nucleon n umber
220
240
number
is
Chapter
Figure
12.3.1
nucleon
energy
shows
number
per
the
for
variation
common
nucleon,
the
of
binding
nuclei.
greater
is
The
the
energy
larger
stability
per
the
of
nucleon
value
the
of
12
Atomic
structure
and
radioactivity
against
binding
nucleus.
An
isotope
56
of
iron
Fe
has
the
highest
binding
energy
per
nucleon.
This
means
that
26
it
is
the
most
separate
The
the
stable
nucleus
nucleons
binding
energy
on
the
graph.
It
requires
the
most
energy
curve
can
be
used
to
region
determine
whether
to
completely.
fusion
or
fission
is
likely
to
of
maximum
stability
occur
.
10
Consider
the
following
fission
fission
reaction.
9
1
235
n
+
uranium
fragments
(A
Kr
nucleus
=
141
curve,
energy
of
of
the
2
is
following
H
+
splits
92).
be
seen
uranium
in
fusion
into
From
is
two
the
that
fragments
released
2
the
greater
nucleus.
the
process.
reaction.
4
H
1
235)
=
can
two
original
energy
=
A
n
0
→
He
1
8
7
6
5
4
gnidnib
Consider
(A
it
3
ygrene
Therefore
the
the
+
rep
energy
1
Ba
56
and
binding
that
+
36
binding
than
141
→
92
VeM/noelcun
The
92
U
0
2
3
fusion
2
2
T
wo
light
nuclei
of
H
(A
=
2)
fuse
together
1
1
4
to
form
He
(A
=
4).
can
be
seen
From
the
binding
energy
2
0
curve,
it
that
the
binding
energy
0
of
the
resulting
product
is
greater
than
20
40
60
80
100
120
140
160
180
nucleon
2
binding
energy
of
the
two
200
H
nuclei.
220
240
the
number
Therefore
1
Figure 12.3.2
energy
is
released
Unified
Mass
used
The
and
as
in
the
atomic
energy
another
relative
mass
are
unit
atomic
unit
(u)
interchangeable.
of
Binding energy, fission and fusion
process.
Therefore,
the
unit
of
mass
can
be
energy.
mass
A
of
an
atom
is
defined
by
r
mass
A
of
the
atom
=
12
r
one-twelfth
of
the
mass
of
a
C
atom
6
12
The
relative
atomic
mass
of
C
is
exactly
12.
6
The
unified
atomic
mass
unit
(u)
is
defined
as
1/12
the
mass
of
the
12
carbon
atom
Key points
C.
6
12
Mass
of
one
carbon
atom
12
=
g
=
kg
23
6.02
×
=
12 u
ᔢ
The
energy
required
6.02
×
10
completely
12
of
−27
Therefore,
1 u
to
26
10
=
=
1.66
×
10
a
separate the
nucleus
is
called
nucleons
the
binding
kg
26
12
×
6.02
×
10
energy
of
the
nucleus.
−19
1 eV
=
1.6
×
10
J
ᔢ
−19
1 MeV
=
1.6
×
10
6
×
10
=
1.6
×
10
The
is
−13
change
in
mass
of
1.0 kg
has
an
energy
equivalence
nucleus
number
=
mc
8
=
1.0
×
(3.0
×
10
)
2
1.6
Therefore,
1 u
=
9.0
×
10
×
10
×
9
×
nucleon
energy
divided
by
the
of
total
of
nucleons.
J
Energy
nuclei
16
=
per
binding
16
ᔢ
–27
energy
the
of:
2
E
to
J
the
A
binding
equal
is
released
when
heavy
undergo fission.
10
MeV
–13
1.6
×
10
ᔢ
Energy
nuclei
1 u
=
is
released
when
light
undergo fusion.
931 MeV
149
12.4
Calculating
energy
changes
Example
Learning outcomes
238
Calculate
the
binding
energy
per
nucleon
of
the
nucleus
U
using
the
92
On
completion
of
this
section,
you
following
should
ᔢ
be
able
appreciate
between
data.
to:
the
Mass
of
a
proton
energy
and
Mass
of
a
neutron
by
1.00728 u
=
1.00867 u
mass
2
represented
=
association
E = mc
238
and
use
Mass
of
U
nucleus
=
238.05076 u
92
this
equation
to
solve
problems.
1 u
=
931 MeV
T
otal
Mass
mass
defect
of
of
nucleons
the
nucleus
=
(92
=
239.93558 u
=
×
1.00728)
239.93558
−
+
(146
×
238.05076
1.00867)
=
1.88482 u
238
Binding
energy
of
the
nucleus
U
in
MeV
=
1.88482
×
=
1755 MeV
931
92
1755
Binding
energy
per
nucleon
=
=
7.373 MeV
238
Example
A
possible
fission
238
1
U
is:
144
+
n
92
The
reaction
→
90
Ba
0
+
56
binding
energy
1
Kr
+
2
36
per
n
0
nucleon
of
the
released
in
particles
is
as
follows:
235
U
7.62 MeV
Ba
8.38 MeV
92
144
56
90
Kr
8.67 MeV
36
a
Estimate
b
Calculate
the
c
State
forms
a
Firstly,
two
the
the
energy
mass
of
energy
binding
multiplying
the
equivalent
the
energy
binding
of
this
of
fission
this
answer
each
energy
in
joules.
energy.
in
a
is
nucleus
per
reaction
transformed
is
nucleon
determined
by
the
into.
by
nucleon
number
.
6
Binding
energy
of
U-235
=
7.62
×
10
9
×
235
=
1.7907
×
10
6
Binding
energy
of
Ba-144
=
8.38
×
10
×
144
=
1.2067
×
10
6
Binding
energy
of
Kr-90
=
8.67
×
10
released
=
(7.803
×
10
eV
8
×
90
=
8
Energy
eV
9
7.803
×
10
eV
9
+
1.2067
×
10
)
9
−
(1.7907
×
10
)
8
=
1.963
×
10
eV
−19
1 eV
=
1.6
×
10
J
8
∴
Energy
released
=
1.963
×
−19
10
×
1.6
2
b
Using
E
=
mc
,
E
m
=
2
c
–11
3.141
m
×
10
=
8
(3.0
×
10
2
)
–28
=
3.49
×
10
kg
–28
Mass
150
equivalence
=
3.49
×
10
kg
×
10
−11
=
3.141
×
10
J
Chapter
c
The
Ba)
energy
and
is
in
the
form
electromagnetic
of
kinetic
energy
of
the
fragments
(Kr
12
Atomic
structure
and
radioactivity
and
radiation.
Example
Consider
the
4
following
9
He
+
2
1
Be
→
4
The
mass
reaction:
12
n
+
C
0
of
each
6
particle
in
the
reaction
is
as
follows:
4
He
4.00260 u
Be
9.01212 u
n
1.00867 u
2
9
4
1
0
12
C
12.00000 u
6
Calculate:
a
the
mass
b
the
energy
a
Mass
defect
equivalence
defect
of
this
=
(4.00260
+
=
0.00605 u
=
0.00605
=
1.0043
×
10
=
1.0043
×
10
=
9.04
10
mass.
9.01212)
–
(1.00867
+
12.00000)
–27
×
1.66
×
10
kg
–29
kg
2
b
E
=
mc
–29
8
×
(3.0
×
10
2
)
–13
×
J
151
Revision
Answers
found
1
to
on
a
questions
the
how
gases
discrete
Several
are
require
calculation
can
be
5
accompanying CD.
Explain
of
that
questions
at
of
in
the
the
energy
provide
levels
emission
The
diagram
X-rays
emission
pressure
electron
lines
shown
lines
low
7
spectrum
below
produced
in
shows
a
typical
an X-ray
spectrum
of
the
tube.
intensity
evidence for
in
atoms.
spectrum
of
[3]
hydrogen
below.
0
wavelength
410
434
486
656
Provide
wavelength/nm
a
b
Calculate
the
energy
of
the
photons
A
explanations for
continuous
the following:
spectrum
of
wavelengths
is
associated
produced.
with
c
The
each
line
energy
atom
are
illustrate
spectral
in
the
levels
shown
the
lines
of
spectrum.
an
electron
below. Copy
transitions
in
that
[3]
[4]
in
the
a
b
The
c
There
spectrum
has
a
sharp
cut-off.
[1]
hydrogen
diagram
produce
are
spikes
in
the
spectrum.
[2]
and
the four
b
6
[4]
An
electron
potential
de
is
accelerated from
difference
Broglie
of
wavelength
rest
through
5.0 kV. Calculate
of
the
a
the
electron.
[4]
–0.60
–0.87
7
a
State
what
is
meant
by
the
de
Broglie
–1.36
wavelength.
–2.42
b
An
electron
potential
[1]
is
accelerated from
difference
of
rest
through
a
920 V.
–5.45
–19
energy/ 10
i
Calculate
the final
momentum
of
the
J
electron.
ii
[2]
Calculate
the
de
Broglie
wavelength
of
the
electron.
c
Describe
nature
an
of
[2]
experiment
to
demonstrate
the
wave
electrons.
[5]
–21.80
d
Explain
what
is
meant
by
the
wave–particle
duality.
2
Distinguish
between
a
line
emission
spectrum
and
a
8
line
absorption
spectrum.
a
Explain
Explain
how X-rays
are
what
is
meant
by
the
nucleon
number
[4]
and
3
[3]
produced
in
an X-ray
b
machine.
[4]
the
Sketch
with
nuclear
a
binding
labelled
nucleon
graph
number
energy
to
of
of
show
the
the
nucleus.
the variation
binding
energy
nucleon.
4
In
an X-ray
tube,
the
potential
difference
cathode
and
the
metal
target
is
the
speed
of
the
electrons
when
[3]
Hence
explain
they
hit
fusion
of
target
the
cut-off
wavelength
of
the X-rays
emitted.
fission
of
small
nucleon
numbers
nuclei
[3]
having
large
nucleon
numbers
[3]
releases
152
having
energy
[3]
ii
b
nuclei
the
releases
metal
why:
42 kV. Calculate:
i
a
per
between
c
the
[3]
energy.
[3]
Revision questions
7
238
9
a
Outline
the
model
of
the
atom
proposed
by
12
Calculate
the
binding
energy
of
the
nucleus
U
92
J. J. Thompson,
b
In
the
the
Rutherford
α-scattering
supervision
projected
Sketch
of
α-particle
in
Bohr.
experiments
Rutherford,
towards
diagrams
and
a
to
thin
the following
performed
α-particles
sheet
illustrate
[6]
of
the
in
under
MeV.
Use
The
α-particle
nucleus
of
a
is
The
Mass
of
a
proton
path
of
Mass
of
a
neutron
Mass
of
uranium
1.00728 u
=
1.00867 u
an
directly
towards
nucleus
=
238.05076 u
=
931.3 MeV
the
atom.
passes
=
instances:
moving
gold
α-particle
data:
gold foil.
[1]
13
ii
the following
were
1 u
i
[6]
close
to
the
nucleus
of
Calculate
the
energy
released
in
the following
a
reaction.
gold
atom.
[6]
[1]
241
237
Am
→
Np
95
iii
The
α-particle
from
the
passes
nucleus
of
a
some
gold
distance
4
+
93
He
2
away
atom.
Use
[1]
the following
mass
data:
241
c
Explain
how
provides
of
the
the
α-particle
evidence for
the
scattering
existence
experiment
and
small
nucleus.
size
Am
241.06687 u
Np
237
.05843 u
237
[4]
4
He
d
Give
an
an
estimate for
atom
and
the
the
radius
radius
of
an
of
the
nucleus
atom.
4.00241 u
of
[2]
14
Calculate
the
energy
released
reaction.
10
State
what
is
meant
by
each
of
[6]
216
Rn
86
Proton
b
Nucleon
the following
the following:
220
a
in
number
→
4
Po
+
84
He
2
[1]
Use
number
the following
mass
data:
[1]
220
c
An
isotope
[1]
Rn
219.9642 u
Po
215.9558 u
216
d
Binding
energy
of
e
Binding
energy
per
a
nucleus
[2]
4
11
Consider
2
H
1
The
the fusion
2
+
nucleon
H
masses
→
each
4.00241 u
1
He
+
2
of
He
reaction:
3
1
[1]
n
0
particle
are
as follows:
2
H
2.01355 u
He
3.01493 u
1
3
2
1
n
1.00867 u
0
Calculate:
a
the
mass
defect
b
the
energy
released
[2]
in
the
reaction
in
joules.
[3]
153
12.5
Radioactivity
Learning outcomes
Radioactive decay
In
On
completion
of
this
section,
1896
by
should
be
able
Henri
a
uranium
relate
radioactivity
to
instability
more
There
beta
discuss
the
a
nucleus
spontaneous
stable
ᔢ
identify
nature
the
of
nuclear
origins
environmental
a
this
photographic
phenomenon
plate
was
affected
‘radioactivity ’.
of
an
are
nucleus.
three
particles
types
( β)
In
atom
of
and
this
is
unstable,
process
it
will
ionising
gamma
ionising
disintegrate
radiation.
ray
photons
radiation
They
( γ).
are
is
to
produce
alpha
Radioactive
produced.
particles
decay
is
( α),
the
and
random
process
whereby
an
unstable
nucleus
attempts
decay
to
become
of
the
stable
by
disintegrating
into
another
nucleus
and
one
or
more
and
hazards
three
ionising
radiations.
of
If
background
you
were
to
observe
the
atoms
in
a
sample
of
radium
you
would
make
radiation
some
ᔢ
that
called
and
spontaneous
random
He
nuclear
a
ᔢ
observed
compound.
to:
Whenever
ᔢ
Becquerel
you
describe the operation of simple
radiation detectors (G–M tube,
cloud chamber and spark counter).
interesting
would
still
we
not
say
the
not
able
predict
that
same
you
be
It
to
predict
which
of
speed
is
up
the
unaffected
is
a
and
decay
by
you
it
would
decay
radium
If
when
atom
radioactive
sample
cannot
process.
observations.
were
to
would
decay
it
If
process.
to
single
did
is
if
decay
is
you
would
the
atom,
decay
what
Also,
you
Radioactive
external
a
it
This
strongly,
process.
conditions
decay.
next.
random
heat
observe
a
could
meant
were
realise
is
you
you
when
to
take
that
spontaneous
nucleus.
Definition
Detectors of
Radioactive decay
and
an
random
unstable
process
nucleus
become
stable
another
nucleus
one
or
more
particles,
is the
by
of
beta
The Geiger–Muller tube
whereby
attempts
to
Figure
disintegrating
and
emitting
the following:
particles,
into
any
alpha
gamma
rays.
tube
12.5.1
thin
cylindrical
wire
unpredictable.
predict
or
which
It
is
decay
is
impossible
nucleus
will
decay
When
ionising
atoms
become
and
charged
when.
positive
large
ions
nature:
the
around
the
are
is
external
not
to
affected
the
temperature
nucleus
and
radiation
the
ions
are
to
ions
This
of
There
result
larger
reduce
would
to
to
vapour
is
is
a
zero
and
acts
as
they
a
the
with
anode.
vapour
at
and
low
tube.
A
cathode.
argon
towards
cathode.
and
of
a
cylindrical
window,
the
and
The
produce
electrons
travel
anode
more
causes
further
even
moving
In
agent.
electric
If
would
order
The
bromine
slowly.
the
ionisation.
electrons
pulses.
quenching
collide
the
G–M
cathode.
pulse.
cathode,
current
the
accelerate
atoms
current
as
anode
mica
towards
the
as
the
the
thin
prevents
multiple
of
produced
electrons
with
acts
‘avalanche’
around
particular
acts
bromine
end
argon
an
large
the
ions
collide
when
then
than
create
the
accelerate
more
tube
across
electrons
This
which
and
one
applied
with
positive
were
bromine
ions
the
gas
at
tube.
tube
of
through
free
collide
axis
argon
is
enters
positive
(G–M)
metal
window
400 V
The
The
the
with
mica
about
particles.
it
along
filled
thin
neutralised
to
positive
molecules.
be
prevent
argon
This
decay
prevents
process
a
cylindrical
lies
is
ionised.
number
positive
released.
is
anode.
The
this,
Spontaneous
the
Geiger–Muller
electrons
The
the
next
tube
difference
field
to
which
potential
anode
a
sealed
There
towards
the
a
pressure.
more
nature:
of
The
accelerating
Exam tip
shows
consists
The
the
Random
radiation
spontaneous
by
them
from
reaching
the
cathode.
conditions
(e.g.
pressure).
The
output
measures
at
which
of
the
it
a
G–M
number
receives
tube
of
is
connected
pulses
it
to
receives.
a
scaler
A
or
ratemeter
.
ratemeter
A
measures
scaler
the
rate
pulses.
anode
cathode
mica
to
window
argon
gas
bromine
Figure 12.5.1
154
A G–M tube
+
and
vapour
pulse
counter
Chapter
The
cloud
Figure
soaked
is
placing
the
in
the
cloud
The
into
its
the
The
and
alcohol
ionising
In
ring
alcohol
ions
it.
are
of
of
the
as
on
tiny
of
top
the
these
the
and
chamber
diffuses
chamber
,
The
detector
,
of
readily
ionising
produced.
up
type
the
vapour
When
condenses
shows
this
at
evaporates
regions
with
vapour
radiation
felt
temperature
The
cooler
air
,
chamber
.
The
alcohol
base.
supersaturated
supersaturated
sites,
a
tracks.
vapour
.
ice
downwards
as
alcohol.
with
dry
becomes
shows
seen
with
saturates
Atomic
structure
and
radioactivity
chamber
12.5.2
radiation
12
the
is
and
act
sites.
droplets
of
is
chamber
reduced
by
continuously
the
radiation
ions
ionising
chamber
as
The
air
there
travels
through
nucleation
path
travelled
by
condensation.
lid
felt
ring
saturated
light
with
vapour
radioactive
black
source
screen
solid
sponge
Figure 12.5.2
carbon
dioxide
A cloud chamber
high
voltage
+
gauze
The
A
spark
spark
counter
counter
(Figure
12.5.3)
is
a
visible
and
audible
way
of
wire
demonstrating
the
ionisation
effect
of
different
types
of
radiation.
It
Figure 12.5.3
consists
above
a
applied
of
a
metal
thin
at
wire
the
gauze
(cathode)
(anode).
anode.
The
When
a
which
gauze
is
is
situated
earthed
radioactive
and
substance
a
a
few
high
is
A spark counter
millimetres
voltage
brought
is
close
to
Key points
the
gauze,
jump
the
air
between
in
the
that
wire
region
and
the
becomes
ionised.
This
causes
a
spark
to
gauze.
ᔢ
Radioactive
decay
spontaneous
Background
radiation
process
whereby
nucleus
If
a
G–M
tube
is
placed
in
front
of
a
radioactive
source,
clicks
are
source
The
the
is
that
the
removed
G–M
time.
tube
It
is
is
source
from
is
in
front
detecting
called
emitting
a
of
type
background
ionising
the
of
G–M
radiation.
tube,
radiation
However
,
clicks
that
are
exists
still
if
radiation
surroundings
(cosmic
is
random
radiation
from
space,
minerals
from
in
the
by
sources
of
ᔢ
cosmic
ᔢ
radioactive
ᔢ
radon
background
radiation
gas
from
materials
that
radiation
outer
that
us
all
any
or
one
more
particles,
nuclear
power
ᔢ
medical
stations
and
of
earth,
etc.).
ᔢ
Random
nature:
the
It
still
found
in
rocks
ᔢ
predict
which
next
when.
or
Spontaneous
process
air
weapons
the
radiation
formation
can
cause
cell
damage.
Prolonged
exposure
can
to
immediate
real
count
The
of
detect
cancers,
a
mutations
reading
radioactive
rate
actual
is
is
will
to
decay
nature: the decay
not
affected
by
external to the
nucleus
and
pressure).
A G–M
of
count
a
source
source,
rate
on
of
a
a
G–M
or
even
present.
the
tube
death.
This
background
radioactive
even
Background
though
means
that
radiation
source
is
there
in
must
is
order
first
determined
by
cloud
counter
can
chamber
be
used
or
to
radiation
ionising
radiation.
no
to
be
tube,
lead
detect
us
decay
impossible
facilities.
spark
causes
is
testing
ᔢ
to
following:
particles,
nucleus
(e.g. temperature
Background
the
beta
rays.
conditions
ᔢ
into
emitting
are:
naturally
in
and
the
space
are
accumulates
become
disintegrating
nucleus
unpredictable.
The
unstable
to
heard.
around
radiation .
detected
an
attempts
another
alpha
radioactivity
the
the
gamma
Background
is
random
heard,
stable
indicating
and
get
the
measured.
ᔢ
Background
radiation
radioactivity
is
random
detected from
the
subtracting
surroundings.
the
background
reading
from
the
reading
measured
on
the
G–M
tube.
155
12.6
Types
of
Learning outcomes
radiation
Types of
There
On
completion
should
ᔢ
be
able
describe
the
as
this
section,
nature
of
β-particles
different
and
types
are
three
ᔢ
Alpha
particles
ᔢ
Beta
ᔢ
Gamma
particles
ᔢ
of
ionising
radiation:
ray
( α)
( β)
photons
( γ)
of
Whenever
ionising
types
radiation
you
to:
α-particles,
γ-rays
of
ionising
radiation
passes
through
matter
,
it
causes
electrons
to
be
radiation
distinguish
between
β-particles
and
reference
to
effect
of
fields,
and
α-particles,
γ-rays
charge,
electric
with
mass,
and
speed,
‘knocked
out’
example,
a
‘knock’
As
it
of
beta
atoms.
particle
electrons
ionises
This
the
out
air
moving
of
it
results
air
in
through
molecules
loses
the
some
of
air
and
its
formation
will
leave
energy
of
have
a
enough
trail
and
ions.
of
For
energy
ions
slows
to
behind
it.
down.
magnetic
penetrating
properties.
Alpha
particles
(α)
4
2+
An
alpha
particle
is
a
helium
nucleus
(
He
).
It
consists
of
two
protons
2
and
two
three
neutrons,
forms
of
and
ionising
centimetres
in
ions.
12.6.1
of
Figure
α-particles
length
are
in
a
have
of
a
positive
radiation.
because
most
illustrates
cloud
observed.
concentration
emitted
air
has
ions.
of
It
means
also
a
the
what
chamber
.
This
similar
As
charge.
is
means
is
the
most
result,
α-particles
energy
is
seen
Short
that
It
when
thick
the
that
used
of
α-particles
of
travel
in
observing
tracks
most
up
intense
a
the
few
producing
the
almost
produce
the
of
only
movement
the
a
same
high
α-particles
that
are
energies.
241
Americium-241
(
Pa)
produces
α-particles.
α-particles:
Figure 12.6.1
Beta
A
short

all
thick
tracks
approximately
the
same
length
Alpha particles moving in a cloud chamber
particles
beta

particle
is
(β)
a
fast-moving
electron.
It
has
a
mass
of
m
and
charge
e
−19
of
e
(1.6
Figure
×
β-particles
than
10
12.6.2
the
C).
β-particles
illustrates
in
a
ones
cloud
what
chamber
.
produced
by
are
is
the
less
seen
The
ionising
when
tracks
than
observing
are
much
α-particles.
the
movement
longer
and
of
thinner
α-particles.
90
Strontium
(
Sr)
produces
β-particles.
β-particles:
Figure 12.6.2
Gamma
Gamma
has
156
no
longer,

tracks
thinner
of
tracks
varying
lengths
Beta particles moving in a cloud chamber
ray
photons
radiation
mass

and
(γ)
consists
no
of
charge.
high-energy
Figure
12.6.3
electromagnetic
illustrates
what
waves.
is
seen
It
when
Chapter
observing
very
thin
weakly
the
movement
and
not
ionising
well
of
γ-radiation
defined.
compared
The
with
in
a
reason
alpha
or
cloud
for
beta
chamber
.
this
is
that
The
tracks
γ-radiation
12
Atomic
structure
and
radioactivity
are
is
particles.
60
Cobalt
(
Co)
produces
γ-rays.
γ-ray
Figure 12.6.3
α-particles,
the
different
G–M
effect
It
can
ᔢ
α-particles
ᔢ
β-particles
ᔢ
γ-rays
a
One
are
sheet
of
are
the
the
are
be
are
and
way
on
materials
tube.
thin

tracks
of
tracks
varying
lengths
properties
β-particles
12.6.4).
observe

Gamma radiation in a cloud chamber
Penetrating
(Figure
photons:
very
γ-rays
to
count-rate
placed
shown
stopped
stopped
most
penetrate
investigate
the
matter
measured
between
the
differently
penetrating
on
source
a
properties
G–M
of
the
tube
is
to
when
radiation
and
the
that:
by
by
a
thin
a
thin
penetrating
sheet
sheet
but
of
paper
of
(2 mm)
aluminium
most
of
the
(1–10 mm)
radiation
is
stopped
by
lead.
γ-ray
photons
+
β-particles
β
α-particles
γ
paper
Figure 12.6.4
aluminium
lead
α
The penetrating properties of ionising radiation
–
Effect of
electric fields
Figure 12.6.5
Behaviour of α-particles,
β-particles and γ-rays in an electric field
Figure
as
12.6.5
they
pass
shows
the
through
an
paths
taken
electric
and
moves
toward
the
negative
and
moves
toward
the
positive
than
the
have
no
α-particle
charge
and
and
is
field.
plate.
pass
α-particles,
An
plate.
deflected
therefore
by
A
α-particle
β-particle
The
more
β-particles
has
has
β-particle
in
through
the
the
a
a
electric
positive
negative
has
electric
and
a
charge
charge
smaller
field.
field
γ-rays
The
mass
γ-rays
undeviated.
α
Effect of
Figure
12.6.6
as
they
to
the
is
magnetic fields
pass
paths
shows
the
through
and
experienced
a
the
taken
magnetic
points
by
paths
into
the
α-particles
by
field.
plane
and
α-particles,
The
of
magnetic
the
the
β-particles
paper
.
field
A
β-particles.
is
γ
γ-rays
perpendicular
magnetic
The
and
force
direction
of
β
the
force
charge
is
and
determined
therefore
by
pass
Fleming’s
through
left-hand
the
rule.
magnetic
The
field
γ-rays
have
no
undeviated.
The
magnetic field
and
Figure
12.6.6
separately
deflection
is
with
a
composite
each
compared
type
to
of
diagram.
radiation.
β-particles
in
It
illustrates
The
the
the
α-particles
same
results
show
magnetic
is
into
perpendicular
the
paper
obtained
very
field.
points
little
Figure 12.6.6
Behaviour of α-particles,
β-particles and γ-rays in a magnetic field
157
Chapter
12
Atomic
structure
and
radioactivity
The
A
inverse
point
the
source
inverse
square
of
γ-rays
square
of
law for
emits
the
in
γ-rays
all
distance
directions.
d
from
the
The
intensity
I
varies
as
source.
1
I
∝
2
d
The
intensity
T
able
12.6.1
Table 12.6.1
1
of
the
γ-rays
compares
the
thus
decreases
properties
of
with
the
distance
various
from
ionising
the
source.
radiations.
Comparison of the properties of α-particles, β-particles and γ-rays
Nature
α-particles
β-particles
γ-rays
Helium
Fast-moving
Photon
electron
electromagnetic
nuclei
of
radiation
2
3
Charge
+2e
Mass
4m
−e
None
1
p
m
(
or
e
4
Energy
None
m
)
p
1836
3–7 MeV
1–2 MeV
1–2 MeV
Monoenergetic
Range
Monoenergetic
from
emission
given
nuclide
of
of
from
energies from
given
from
given
nuclide
nuclide
0
to
a
maximum
5
Speed
6
Range
Up
in
air
to
0.01c
3–10 cm
(0.01–0.9)c
c
1–2 m
Order
of
kilometres
7
Ionising
Strongly
ionising
Weakly
ionising
Very
property
8
Penetrating
Stopped
ability
sheet
of
by
a
thin
Stopped
paper
thin
Affected
an
by
by
sheet
a
Most
of
of
it
is
stopped
by
aluminium
sheet
lead
(1–10 mm)
(1–10 cm)
Yes
Yes
No
Yes
Yes
No
(2 mm)
9
little
ionisation
of
a
electric
field
10
Affected
by
a
magnetic field
Dangers of
Ionising
DNA
the
of
radiation
can
cell
to
even
Although
cause
158
affect
can
the
divide.
developing
and
ionising
radiation
damage
DNA
functioning
Ionising
cancer
,
of
radiation
radiation
in
the
the
cell.
can
sickness
nucleus
It
lead
can
to
of
also
cells.
mutations,
(nausea,
Damaged
affect
fatigue
the
ability
increased
and
loss
of
of
risk
hair)
sterility.
α-
severe
and
β-particles
damage
to
the
cannot
skin.
travel
As
far
into
α-radiation
living
is
tissue,
strongly
they
ionising
can
it
can
Chapter
cause
can
severe
be
surface
caused
by
burns.
γ-rays,
Direct
because,
damage
to
although
internal
less
organs
ionising,
of
they
the
are
12
Atomic
structure
and
radioactivity
body
more
penetrating.
In
2011
an
devastated
to
radiation
radiation
of
leaks.
leaks
developed
and
not
An
Nuclear
may
had
take
on
the
have
protested
accident
of
or
the
for
radioactive
act
Caribbean
facilities
to
deal
precautions
to
their
against
could
with
any
when
of
use
from
the
the
seriously
type
the
impact
of
nuclear
handling
the
energy.
Caribbean
United
Caribbean
leading
impact
Consequently,
nuclear
of
that
damaged,
countryside.
use
The
tsunami
were
determine
the
waste
region.
a
plants
years
rethink
terrorist
caused
surrounding
leaders
route
to
9.0
power
some
had
the
Safety
magnitude
have
livelihood
have
It
have
transportation
Japan.
of
Japan.
nations
Caribbean
a
earthquake
parts
on
Sea
as
Kingdom
the
to
welfare
community
does
disaster
.
radioactive
material
ᔢ
Forceps
ᔢ
Always
it
ᔢ
at
should
turn
a
be
used
when
radioactive
handling
source
away
radioactive
from
your
material.
body
and
never
point
anybody.
Radioactive
material
should
be
stored
in
lead
containers
when
not
in
use.
ᔢ
Avoid
eating
or
drinking
where
radioactivity
experiments
are
being
performed.
ᔢ
A
radiation
detector
should
be
present
in
order
to
monitor
radiation
levels.
ᔢ
Radioactive
waste
materials
should
never
be
disposed
of
in
an
ordinary
bin.
Key points
ᔢ
Three
types
gamma
of
ionising
radiation
ᔢ
An
α-particle
ᔢ
A
ᔢ
γ-rays
ᔢ
α-particles
are
the
ᔢ
α-particles
are
stopped
ᔢ
β-particles
ᔢ
γ-rays
β-particle
sheet
is
is
consist
are
of
alpha
particles,
beta
particles
and
helium
nucleus.
a fast-moving
of
are
the
a
electron.
high-energy
strongest
stopped
most
by
by
a
a
electromagnetic
ioniser
thin
thin
penetrating
and
sheet
sheet
but
waves.
γ-rays
of
of
most
are
paper
the
the
ioniser.
(2 mm).
aluminium
of
weakest
(1–10 mm).
radiation
is
stopped
by
a
lead.
ᔢ
α-particles
ᔢ
γ-rays
ᔢ
The
are
and
not
intensity
square
are
rays.
of
the
β-particles
deflected
of
by
are
deflected
electric
γ-rays from
distance from
a
or
point
the
by
electric
and
magnetic fields.
magnetic fields.
source
is
inversely
proportional
to
the
source.
159
12.7
Radioactive
Learning outcomes
decay
Decay
equations
Alpha-decay
On
completion
should
ᔢ
be
able
represent
of
this
section,
you
to:
α-,
The
β-
and
following
equation
γ-decay
illustrates
A
A−4
X
→
ᔢ
simple
define
the
nuclear
terms
A
=
activity
and
The
and
mass
number
decreases
of
the
by
parent
two.
For
234
graphically
and
He
2
nuclide
decreases
by
four
and
the
atomic
example:
230
Th
use
α-decay.
recall
−λN
recognise,
+
Z−2
→
4
Ra
90
ᔢ
during
equations
number
decay constant,
happens
4
Y
Z
using
what
+
He
88
2
represent
solutions
of
the
Beta-decay
λt
decay
law
based
on
x
=
x
e
for
0
The
activity,
number
particles
and
of
following
equation
illustrates
received
count
A
rate
A
X
→
define
ᔢ
use
happens
during
β-decay.
0
Y
Z
ᔢ
what
undecayed
+
e
Z+1
−1
half-life
The
0.693
the
relation
λ
mass
number
of
the
parent
nuclide
remains
the
same
and
the
atomic
=
number
T
increases
by
one.
For
example:
1/2
14
14
C
→
During
β-decay
electron
a
is
not
+
e
7
neutron
(β-particle).
β-particle
0
N
6
The
−1
spontaneously
β-particle
produced
from
1
is
→
emitted
electrons
1
n
into
from
orbiting
a
the
the
proton
and
nucleus.
nucleus
of
an
The
an
atom.
0
p
0
changes
+
e
1
−1
Gamma-decay
The
following
equation
illustrates
A
A
X
→
mass
same,
number
since
photon
is
the
γ-ray
emitted,
The
per
activity
second.
sample
is
A
of
The
a
SI
atomic
photon
the
Activity, decay
+
Z
and
has
proportional
is
to
the
the
γ-decay.
γ
no
of
mass
becomes
constant
radioactive
during
0
number
nucleus
unit
happens
0
X
Z
The
what
and
sample
parent
and
more
no
nuclide
charge.
stable
and
remain
Since
less
a
the
γ-ray
excited.
half-life
is
becquerel
number
the
of
the
(Bq).
number
The
nuclei
N
of
nuclei
activity
present
of
in
a
decaying
radioactive
the
sample.
Equation
A
A
=
λ
–λN
is
the
∝
proportionality
N
therefore
constant
in
the
equation
A
and
=
is
−λN
called
the
decay
constant
−1
A
–
activity/Bq
λ
–
decay
or
s
−1
constant/s
The
−1
or
min
h
,
decay
λ
is
the
probability
of
decay
of
a
nucleus
per
unit
time.
etc.
Suppose
N
constant
or
−1
–
number
of
undecayed
over
160
the
number
of
radioactive
nuclei
present
nuclei
a
period
of
time
as
shown
in
Figure
12.7.1.
in
a
sample
is
recorded
Chapter
number
Atomic
structure
and
radioactivity
of
N
radioactive
nuclei
12
0
in
λt
N
sample
=
N
e
0
1
T
=
1
N
half-life
of
radioactive
nuclide
0
2
2
1
N
0
4
1
N
0
8
1
N
0
16
time/t
0
2T
T
4T
3T
5T
1
1
1
1
1
2
2
2
2
2
Figure 12.7.1
It
can
be
nuclei
seen
that
present
in
the
the
graph
is
sample
exponential
at
time
t
=
0
in
is
nature.
N
.
As
The
nuclei
number
decay,
of
the
0
number
remaining
decreases
over
time.
The
half-life
T
of
a
radioactive
1/2
substance
to
is
decrease
Half-life
the
to
can
average
half
also
of
be
its
substance
to
to
ᔢ
After
one
half
of
is
terms
the
initial
(T
),
for
the
number
of
undecayed
nuclei
value.
in
then
its
half-life
taken
initial
defined
radioactive
decrease
time
of
time
activity.
taken
for
The
the
half-life
activity
of
of
a
the
sample
value.
the
number
of
undecayed
nuclei
present
is
1/2
N
/2.
0
ᔢ
After
two
half-lives
(2 T
),
the
number
of
undecayed
nuclei
present
is
1/2
N
/4.
0
ᔢ
After
three
half-lives
(3 T
),
the
number
of
undecayed
nuclei
present
1/2
is
N
/8.
0
N
T
1/2
N
T
1/2
0
T
N
1/2
0
0
N
0
2
Deriving the decay
Let
the
time
The
t
number
be
8
equation
undecayed
nuclei
present
in
a
radioactive
sample
at
N
activity
nuclei
of
4
of
the
sample
is
proportional
to
the
number
of
undecayed
present.
dN
Therefore,
–
∝
N
dt
dN
–
=
λN
dt
Separating
the
variables,
dN
=
−λN dt
dN
=
−λ dt
1
N
Integrating
both
sides
of
the
equation,
1
∫
dN
=
ln N
=
N
∫
−λ dt
−λt
−λt
N
=
e
N
=
e
+
+
c
c
−λt
c
×
e
161
Chapter
12
Atomic
structure
and
radioactivity
c
Let
e
=
A
Then
N
=
Ae
N
=
N
N
=
N
λt
At
time
t
=
0,
let
0
λt
Therefore
e
0
This
equation
determine
decay
is
the
known
number
constant
is
as
of
known.
the
decay
undecayed
The
equation.
nuclei
equation
It
can
present
can
also
be
at
be
used
any
to
time,
written
in
once
terms
the
of
λt
activity:
A
=
A
e
0
Equation
λt
N
=
N
e
0
N
–
number
λ
–
decay
–
initial
–
time/s
of
undecayed
nuclei
at
time
t
−1
N
constant/s
number
of
undecayed
nuclei
0
t
The
relationship
decay
between
half-life
and the
constant
λt
The
decay
equation
is
given
by
N
=
N
e
.
0
At
time
t
=
0,
N
=
N
0
N
0
At
time
t
=
T
,
N
=
1/2
2
N
0
λT
Therefore
=
e
=
e
1/2
2
1
λT
1/2
2
T
aking
natural
logarithms
on
both
sides
of
the
equation:
1
λT
ln
1/2
=
ln (e
)
=
− λT
2
−ln 2
ln e
1/2
ln 2
=
λT
1/2
ln 2
T
=
0.693
=
1/2
λ
λ
Equation
ln 2
=
T
λ
–
T
0.693
=
1/2
λ
half-life
1/2
λ
–
decay
constant
Example
24
Sodium-24
pure
(
sample
Na)
of
is
a
radioactive
sodium-24
has
a
isotope
mass
of
with
a
half-life
of
5.0 g.
Calculate:
a
the
number
b
the
decay
c
the
initial
of
sodium-24
constant
activity
present
λ
A
of
0
162
atoms
the
sample
in
the
sample
15.0
hours.
A
Chapter
d
the
activity
of
the
sample
after
45
hours
e
the
activity
of
the
sample
after
30
hours.
12
Atomic
structure
and
radioactivity
23
(Avogadro
constant
N
=
6.02
×
10
per
mole)
A
23
a
24 g
contains
6.02
×
10
sodium-24
atoms.
23
5.0
×
6.02
×
10
23
5.0 g
=
contains
1.25
×
10
sodium-24
atoms
200
b
Half-life
=
15
0.693
×
60
×
60
=
54 000 s
0.693
−5
λ
=
=
=
1.28
×
10
s
54 000
T
1/2
c
A
=
−λN
−5
A
=
−1.28
×
23
10
×
1.25
×
10
0
18
=
1.60
×
10
Bq
45
d
Number
of
half-lives
=
=
3
15
After
3
half-lives,
the
activity
of
the
sample
n
1
=
A
0
(
2
)
,
where
n
is
the
number
of
half-lives
3
1
18
=
1.60
×
10
(
×
2
)
17
=
e
Activity
of
the
sample
2.0
×
after
10
30
Bq
hours
is
given
by:
λt
A
=
A
e
0
–5
18
=
1.60
×
10
=
4.02
×
10
−(1.28
×
×
10
×
30
×
3600)
e
17
Bq
Key points
ᔢ
Mass
ᔢ
The
number
activity
A
and
of
a
atomic
number
radioactive
are
sample
conserved
is
the
in
nuclear
number
of
reactions.
nuclei
decaying
per
second.
ᔢ
The
activity
present
in
ᔢ
The
decay
ᔢ
The
half-life
number
of
of
the
a
radioactive
sample
is
proportional
to
the
number
of
nuclei
sample.
constant
of
a
λ
is
the
probability
radioactive
undecayed
nuclei
substance
to
of
is
decrease
decay
the
to
of
a
nucleus
average
half
of
its
time
per
unit
taken for
initial
time.
the
value.
λt
ᔢ
Radioactive
decay
can
be
expressed
mathematically
by
x
=
x
e
0
0.693
ᔢ
Half-life
and
decay
constant
are
related
by
the
equation
=
T
.
1/2
λ
163
12.8
Measuring
Learning outcomes
half-life
Experiment to
measure the
half-life of
radon-220
220
Radon-220
On
completion
of
this
section,
be
able
measured
describe
an
experiment
a
is
a
gas
with
laboratory.
a
half-life
When
it
of
55
decays,
it
seconds.
emits
Its
the
half-life
of
with
can
G–M
a
very
thin
mica
window
can
be
used
to
measure
substance
the
activity
of
a
232
Rn
with
a
is
one
of
the
decay
products
of
an
isotope
of
thorium
Th.
86
All
90
the
short
A
gas.
220
radioactive
half-life
α-particles.
to
the
measure
in
to:
tube
ᔢ
Rn)
86
be
should
(
you
other
nuclides
in
the
decay
series
have
half-lives
that
are
either
much
half-life.
longer
or
much
shorter
than
radon-220,
so
they
do
not
contribute
to
the
232
activity
of
the
sample
form
thorium
of
the
gas.
The
Th
is
present
in
a
solid,
in
the
90
220
of
hydroxide
and
Rn
is
a
gas,
which
means
that
the
two
86
can
be
Figure
easily
separated.
12.8.1
half-life
of
illustrates
radon-220.
the
The
apparatus
that
experiment
is
can
be
used
performed
in
to
a
measure
fume
the
cupboard.
clip
valves
ratemeter
G –M
tube
squeeze
bottle
thorium
radon
hydroxide
gas
and
air
powder
Figure 12.8.1
G–M
Experiment to measure the half-life of radon-220
1
A
tube
is
2
Some
3
The
screw
clips
4
The
bottle
is
thorium
used
to
measure
hydroxide
are
(which
the
background
is
solid)
a
is
count
placed
in
rate.
a
squeeze
bottle.
opened.
squeezed
so
that
it
forces
radon-220
into
the
collecting
vessel.
5
The
screw
radon-220
collecting
6
7
The
8
The
A
are
then
produced
closed.
by
the
This
prevents
thorium
any
hydroxide
freshly
from
produced
entering
the
vessel.
after
closing
the
screw
clips,
the
scalar
and
stop
watch
started.
rate
164
gas
Immediately
are
9
clips
count
rate
R
background
to
find
graph
of
the
A
is
recorded
count
actual
against
rate
is
count
time
every
is
10
seconds
subtracted
rate
A
plotted.
for
from
a
the
period
of
5
measured
minutes.
count
Chapter
12
Atomic
structure
and
radioactivity
Sample data
A
student
and
performed
obtained
Background
Table 12.8.1
Time/s
the
an
experiment
following
count
rate
=
to
measure
the
half-life
of
radon-220
results.
2
counts
per
second
Sample data
−1
Measured
count
rate
Corrected
−1
count
rate
ln
A/s
−1
R/s
A/s
10
53
51
3.93
20
47
45
3.81
30
35
33
3.50
40
36
34
3.53
50
32
30
3.40
60
29
27
3.30
70
21
19
2.94
80
23
21
3.04
90
19
17
2.83
100
18
16
2.77
110
15
13
2.56
120
13
11
2.40
130
13
11
2.40
140
10
8
2.08
150
9
7
1.95
160
9
7
1.95
170
8
6
1.79
180
9
7
1.95
190
8
6
1.79
200
6
4
1.39
210
6
4
1.39
220
5
3
1.
10
230
4
2
0.69
240
5
3
1.
10
250
5
3
1.
10
260
5
3
1.
10
270
4
2
0.69
280
4
2
0.69
290
3
1
0
300
3
1
0
165
Chapter
12
Atomic
structure
and
radioactivity
Measuring
The
half-life directly from the
half-life
plotting
(Figure
all
can
the
be
measured
data
points
directly
on
the
from
graph,
a
curve
a
decay
curve
curve.
of
After
best-fit
is
drawn
12.8.2).
–1
activity/s
50
40
30
20
10
0
50
100
150
200
250
300
time/s
Figure 12.8.2
Several
ᔢ
ᔢ
points
The
of
The
The
activity
curve
the
is
decrease
on
by
not
is
worth
indicate
from
a
activity
curve
that
at
is
value
time.
through
decay
the
its
with
noting:
This
indicates
that
the
number
decaying.
pass
the
half
are
decreasing
nuclei
curve
with
point
interest
does
half-life
start
Any
of
radon-220
about
ᔢ
Plotting a decay curve to measure half-life
the
points.
radioactive
curve
time
chosen
is
all
t
is
=
decay
constant.
0
and
in
the
measured.
It
order
time
This
to
The
is
is
a
random
not
be
for
process.
necessary
measure
taken
will
fluctuations
its
half-life.
value
the
half-life
to
of
radon-220.
−1
Initial
activity
=
50 s
:
−1
time
taken
for
activity
to
reduce
to
25 s
to
reduce
to
20 s
to
reduce
to
10 s
=
60
−
10
=
50 s
=
75
−
25
=
50 s
=
130
−1
Initial
activity
=
40 s
:
−1
time
taken
for
activity
−1
Initial
activity
=
20 s
:
−1
time
taken
for
activity
50
Average
half-life
+
50
+
−
75
55
=
=
51.7 s
3
Measuring
Consider
the
half-life
decay
law
by
plotting
a
straight
line
equation:
−λt
A
=
A
e
0
T
aking
log
(ln
e)
on
both
sides
(natural
logarithms),
e
−λt
ln A
=
ln A
e
0
−λt
ln A
=
ln A
–
ln e
–
λt ln e
0
ln A
=
ln A
0
ln A
=
ln A
−
0
166
λt
(but
ln e
=
1)
to
=
55 s
Chapter
If
a
graph
gradient
of
ln A
will
be
against
t
obtained,
gradient
is
drawn,
a
straight
line
with
a
12
Atomic
structure
and
radioactivity
negative
where:
=
λ
and
y-intercept
=
ln A
0
Figure
12.8.3
shows
a
plot
of
ln A
against
t
–1
ln A/s
4.0
3.0
2.0
1.0
0
50
100
150
200
250
300
time/s
Figure 12.8.3
The
data
found
half
by
the
From
Plotting a straight line to measure half-life
points
length
the
are
drawing
of
a
plotted
large
the
and
a
triangle
line
of
best
line
3.60
of
line
Therefore,
drawn.
The
is
gradient
greater
is
than
–1.0
=
λ
=
Half-life
=
that
the
half-life
measured
to
plot
a
by
0.693
=
line
of
straight
best
fit
=
54.6
seconds
0.0127
measured
the
−0.0127
0.0127
λ
easier
is
hypotenuse
–240
0.693
is
fit
its
fit.
=
35
half-life
best
that
graph:
Gradient
Notice
of
such
by
both
line
methods
method
rather
than
a
is
is
more
curve
of
not
the
same.
accurate
best
The
because
it
fit.
Key points
ᔢ
The
rate
ᔢ
The
half-life
is
The
radon-220
measured
half-life
activity
ᔢ
of
to
be
decrease
half-life
straight
can
over
can
a
can
period
measured
by
also
half
be
its
be
of
by
determined
experimentally. The
count
time.
measuring
initial
the
average
time
taken for
the
value.
calculated from
the
gradient
of
a
suitably
plotted
line.
167
12.9
Uses
of
radioisotopes
Learning outcomes
Uses of
Isotopes
On
completion
of
this
section,
radioisotopes
are
atoms
of
the
same
element
that
have
the
same
12
number
should
be
able
but
discuss
the
in
different
mass
number
.
For
example,
uses
tracers, for
13
C,
of
of
carbon.
Isotopes
have
the
same
number
14
C
6
and
C
6
of
are
6
protons
and
so
radioisotopes
the
as
a
to:
isotopes
ᔢ
atomic
you
carbon
dating
and
same
Because
number
of
this,
of
a
electrons.
physical
This
method
means
is
they
usually
are
used
chemically
to
separate
similar
.
them.
radiotherapy.
Some
isotopes
of
elements
are
radioactive
and
have
many
useful
applications.
Radioactive dating
Radioisotopes
can
archaeological
years.
lead.
of
A
When
The
the
very
of
age
small
of
rock
but
When
photosynthesis.
animals
absorb
carbon-14
is
determined
to
of
is
be
on
in
the
of
ratio
atoms
a
forms
is
during
4500
stable
million
isotope
comparing
age
in
and
in
the
has
the
of
ratio
all
In
way,
of
a
When
of
through
plants
they
discovered
number
half-
living
dioxide
this
atmosphere
a
carbon
carbon-14.
the
a
of
dioxide
found
plants.
The
of
by
β-emitter
absorb
these
found
half-life
sample.
and
they
decays.
the
a
carbon
is
radioactive
but
artefacts
has
determined
stable
alive,
feed
carbon-12
can
Carbon-14
the
replaced
and
eventually
proportion
are
Animals
it
present
Carbon-12
plants
rocks
decays
lead
comparing
of
date
Uranium-238
atoms.
some
not
by
number
to
sample
constant
years.
organisms.
the
a
carbon-14
5730
used
uranium-238
uranium-238
contains
life
be
expeditions.
and
die,
the
artefact
carbon-14
is
atoms
to
present.
Tracers
Radioisotopes
added
can
to
be
the
detected.
Iodine-131
is
responsible
gland
the
requires
a
T
racers
Iron-59
is
amount
engine
in
a
worn
material
worn
Radioactive
radioactive
water
leaks
radiation
be
with
is
be
amount
radioisotope
thyroid
the
gland
body ’s
functioning.
uptake
by
a
from
half-life
are
the
using
distribution
is
of
(e.g.
is
the
of
a
radioisotope
emits
radiation
the
located
in
the
metabolism.
The
patient
thyroid
used
short
45
from
days.
of
is
gland
neck
The
is
and
is
thyroid
injected
is
30
before
days).
oil
in
is
detect
half-life
the
tumours
It
is
the
iron-59.
with
measured
used
pipe,
is
iron
the
by
the
then
in
added
the
body.
measure
mass
test
oil.
The
be
the
emits
and
is
period,
run
any
Immediately
amount
of
calculated.
water
.
ground
the
process,
engine
underground
to
water
of
the
(iron-56),
The
engine
can
to
initial
test.
During
leaks
the
surface
The
the
in
manufacturing
measured.
component
to
and
During
deposited
underground
detected
clots
non-radioactive
engine
engine
also
a
of
determined
time
of
the
with
blood
components.
component
tracers
from
small
located.
and
locate
made
activity
isotope
can
isolated.
the
off
the
to
engine
period
the
The
A
The
and
growth
proper
component
off
test,
traced
and
uniform
extended
the
for
component
the
tracers.
studied.
tracer
.
used
β-emitter
wear
with
be
a
as
be
detector
.
be
over
after
as
iodine
of
metal
to
can
iodine
activity
an
used
regulating
also
a
of
together
168
It
radiation
can
be
used
for
radioactive
using
the
can
material
pipes.
When
radiation.
and
the
A
the
The
leak
can
Chapter
Thickness
12
Atomic
structure
and
radioactivity
control
source
rollers
β-particles
sheets
of
with
a
long
materials
half-life
(e.g.
the
are
used
thickness
to
of
monitor
paper
in
the
thickness
paper
mills
or
of
β -particles
of
the
sheet
thickness
Figure
on
of
sheets
12.9.1
one
side
of
shows
of
a
of
paper
aluminium).
how
sheet
of
this
is
paper
achieved.
and
a
A
source
detector
is
of
β-particles
placed
on
the
is
placed
other
detector
side.
The
detector
thickness
of
the
controls
paper
the
sheet
spacing
increases,
between
the
the
detector
rollers.
When
measures
a
the
reduction
Figure 12.9.1
in
the
to
reduce
sheet
the
number
β-particles.
in
because
sheet.
air
unable
to
between
detector
them.
sends
When
detector
measures
The
detector
sends
a
an
the
a
signal
thickness
increase
signal
to
to
the
in
the
rollers
the
of
rollers
the
paper
number
to
Monitoring the thickness of
paper sheets
of
increase
the
them.
cannot
and
γ-rays
The
The
the
between
α-source
range
β-particles.
spacing
decreases,
spacing
An
the
of
be
would
are
too
detector
detect
used
not
this
in
a
the
α-particles
detector
.
and
measure
changes
because
the
penetrating,
would
any
for
reach
would
A
not
constant
thickness
be
the
only
cannot
stopped
reading
of
have
γ-source
and
by
a
be
short
used
the
would
paper
be
paper
.
Smoke detectors
Americium-241
has
a
very
detector
,
long
enabling
difference
detector
,
the
is
is
is
the
applied
they
alarm
used
in
half-life.
smoke
The
air
to
across
absorb
conduct
an
some
detectors.
α-particles
of
air
the
a
gap.
This
ionise
small
is
the
an
air
current
When
smoke
α-particles.
The
α-emitter
inside
when
a
the
potential
particles
current
that
smoke
enter
the
decreases
and
triggered.
Imaging
γ-rays
are
very
techniques.
pipes.
For
penetrating
For
this
photographic
example,
a
γ-ray
film
detect
any
Other
applications
blocks
γ-rays
in
faults
is
the
are
in
car
used
and
they
source
placed
the
is
on
welded
include
are
are
the
placed
the
check
in
to
on
other
non-invasive
check
one
side.
the
side
of
This
imaging
quality
the
of
pipe
welded
and
technique
is
a
able
to
joint.
detection
manufacturing
to
used
used
baggage
of
hairline
fractures
without
having
in
engine
industry.
at
airports
to
search
it.
Sterilisation
γ-rays
are
used
Key points
to
sterilise
food.
An
intense
beam
of
γ-rays
can
be
used
ᔢ
to
kill
microorganisms
on
vegetables,
meats
and
grains.
It
prolongs
Isotopes
element
shelf
life
of
these
items
and
reduces
the
need
for
using
refrigeration.
A
similar
method
uses
γ-rays
to
atoms
that
have
of
the
the
same
same
preservatives
atomic
or
are
the
sterilise
number
but
different
medical
mass
number.
instruments.
ᔢ
Cancer treatment
Radioisotopes
such
as
tracers,
Cobalt-60
is
source
of
γ-rays.
A
large
dose
of
γ-rays
is
focused
on
thickness
of
cancerous
cells
in
the
body.
The
high
dose
of
energy
is
able
to
uses
smoke
control,
detectors
and
kill
cancer
cancerous
many
dating,
the
imaging,
site
have
radioactive
treatment.
cells.
169
Revision
Answers
to
questions
that
require
questions
calculation
can
8
be
60
7
The
half-life
of
cobalt-60,
Co,
is
5.26
years. Work
27
found
on
the
accompanying CD.
out:
1
a
Explain
why
radioactive
spontaneous
b
Explain
and
what
is
decay
is
a
considered
random.
meant
by
the
terms
activity
Show
that
the
and
[3]
number
of
undecayed
nuclei
number
cobalt-60
[3]
decay constant.
c
the
of
protons
and
neutrons
in
a
nucleus
b
the
decay
c
the
activity
[2]
constant
of
10.0
of
cobalt-60
grams
of
[2]
cobalt-60.
[3]
N
212
8
An
isotope
of
bismuth,
Bi,
has
a
decay
constant
of
83
present
in
a
radioactive
sample
at
time
t
is
given
212
−2
1.
15
−λt
by
N
=
N
e
10
min
where
N
is
the
initial
number
present
in
the
sample.
produce
Explain
what
radioactive
is
meant
by
polonium
decays
(Po).
by
emitting
Polonium
a
then
β-particle
emits
an
[5]
α-particle
a
Bi
of
0
to
2
.
83
,
0
nuclei
×
−1
the
half-life
of
a
a
sample.
[2]
Write
decay
to
produce
two
an
nuclear
isotope
equations
of
to
lead
(Pb).
illustrate
the
two
processes.
[4]
212
b
Derive
an
equation
relating
the
decay
constant
λ
b
Calculate
c
Given
the
half-life
of
Bi
.
[2]
83
and
half-life
of
a
radioactive
sample,
starting
212
with
that
a
sample
of
Bi
contains
2.5 μg
of
the
83
−λt
the
decay
equation
N
=
N
e
.
[3]
isotope,
0
c
A
radioactive
isotope
has
a
half-life
of
15.0
this
isotope,
the
decay
ii
the
activity
activity
of
the
sample.
[3]
a
Describe
an
experiment
to
show
that
radioactive
calculate:
decay
i
the
hours.
9
For
calculate
constant
a
a
random
process.
[5]
[2]
b
23
of
is
sample
containing
6.02
×
On
a
single
diagram,
illustrate
the following:
10
nuclei.
[3]
i
radioactive
decay
is
an
ii
radioactive
decay
is
a
iii
background
exponential
random
process
[1]
process
[1]
226
3
A
radium
nuclide
Ra
emits
an
α-particle
to
88
produce
an
isotope
of
radon. State
the
number
c
protons
and
neutrons
in
the
isotope
of
radiation.
[1]
of
radon.
Explain
what
is
meant
by
the
term
background
[2]
radiation.
State
two
sources
of
background
radiation.
4
a
Explain
what
is
meant
by
the
term
constant.
[2]
10
b
A
[3]
decay
radioactive
isotope
has
a
half-life
of
18
a
Draw
a
diagram
to
show
the
path
of
a
stream
days. A
of
α-,
β-
and
γ-radiation
as
they
pass
through
a
12
sample
of
the
isotope
contains
3.0
×
10
atoms.
uniform
Calculate:
b
i
the
decay
ii
the
initial
constant for
iii
the
activity
activity
of
the
of
the
the
isotope
sample
sample
after
72
days.
Draw
a
diagram
[3]
of
[3]
magnetic field.
[2]
c
α-,
electric field.
State
β-
an
and
to
[3]
show
γ-radiation
the
as
path
they
of
a
pass
stream
through
a
[3]
instrument
that
can
be
used
to
detect
β-particles.
5
Distinguish
γ-ray
between
photons,
α-particles,
making
reference
β-particles
to
charge,
mass,
11
speed
and
penetration.
Radioisotopes
40
a
K
is
an
isotope
of
potassium.
It
has
a
half-life
×
argon
of
10
years
which
is
age
were
170
of
no
and
stable.
atoms
the
argon
decays
rock
In
to
by
atoms
a
to form
sample
argon
of
atoms
assuming
present
in
an
rock,
is
that
the
isotope
the
ratio
originally
also
be
be
dangerous
to
living
organisms
dangers
useful.
of
the
ionising
radiation
emitted
radioisotopes.
[5]
of
b
i
there
[6]
Discuss
two
useful
applications
of
radioisotopes.
1 : 3. Calculate
sample.
can
Discuss
by
potassium
the
they
of
9
1.26
can
[6]
but
6
[1]
and
ii
State
that
the
[4]
properties
make
them
of
these
suited for
radioisotopes
their
use.
[3]
Revision questions
12
State
the
number
they
changes
of
to
neutrons
the
that
number
occur
of
protons
within
and
nuclei
16
when
emit:
a
α-particles
[2]
b
β-particles
[2]
The
half-life
of
manganese-56
of
manganese-56
to
decay
a
State
of
has
a
mass
is
of
2.6
hours. A
1.0 μg
and
is
8
sample
known
β-particles.
the
the
nucleon
daughter
number
nuclide
and
proton
produced
in
number
the
decay
process.
c
γ-radiation.
[2]
[1]
b
Sketch
a
graph
to
show
the
decay
of
the
sample
16
13
A
radioactive
source
contains
7
.0
×
10
radioactive
over
a
15-hour
period.
[3]
7
nuclei
and
has
an
activity
of
4.5
×
10
Bq.
For
this
c
source,
Calculate
present
a
the
the
number
of
manganese-56
atoms
calculate:
decay
constant
b
the
half-life
c
the
time
2.0
×
in
the
sample.
[2]
[2]
d
Calculate
the
e
Determine
initial
activity
of
the
sample.
[2]
[2]
taken for
the
activity
to fall
the
time
at
which
the
number
of
15
to
manganese-56
is
equal
to
1.35
×
10
.
[3]
5
10
Bq.
[4]
17
14
Radioactive
of
one
or
more
radiation:
State
a
the
has
a
decay
of
usually
the following
α-particles,
type
of
range
results
of
the
types
β-particles
emission
in
or
emission
of
life
the
is
a
radioactive
years. A
sample
of
isotope
with
a
strontium-90
half-
has
of
4.5
×
10
an
Bq. Calculate:
γ-radiation.
−1
a
the
decay
b
the
mass
constant
in
s
[3]
that:
energies,
greatest
28.0
activity
rather
than
density
of
of
strontium-90
in
the
sample.
[4]
discrete
[1]
produces
of
9
ionisation
values
b
Strontium-90
ionisation
in
18
Radium-224
radium-224
a
has
has
a
a
half-life
mass
of
of
3.6
days. A
sample
of
1.82 mg. Calculate:
−1
medium
c
[1]
produces
the
least
density
of
ionisation
in
a
medium
is
not
the
decay
b
the
activity
affected
by
electric
and
magnetic fields
does
of
the
of
radium-224
in
s
[2]
sample.
[4]
Americium-241
not
directly
result
in
a
change
in
is
an
artificially
produced
radioactive
[1]
element. A
e
constant
[1]
19
d
a
particular
sample
of
americium-241
of
proton
5
mass
number
of
the
nucleus
has
g
15
a
range
produces
of
thick
The
radioactive
It
known
is
to
a
diagram,
half-life
be
of
gas
its
in
short
solid
explain
and
in
a
air
cloud
radon-220
a
used
you
Explain
to
an
activity
of
4.2
×
10
Bq. Calculate:
chamber.
is
sample
of
americium-241
atoms
present
in
[2]
b
the
decay
[2]
c
the
half-life
the
determine
the
the
constant
of
americium-241.
[2]
easily
nuclide. With
how
number
the
[1]
α-particles.
and
would
determine
the
[1]
emits
half-life
parent
how
radon-220.
collected
cm
tracks
have
separated from
of
a few
has
[1]
a
f
3.6 μg
data
aid
the
would
half-life.
[7]
171
Module
Answers
to
selected
structured
the
3
Practice
multiple-choice
questions
questions
can
and
be found
exam
to
on
questions
6
the
In
an
on
laboratory
an
oil
drop,
experiment
to
the following
measure
results
the
were
charge
obtained:
accompanying CD.
16.24,
16.51,
8.
12,
24.36,
24.35,
8.
12,
32.53,
32.48,
32.45
16.24
Multiple-choice questions
The
1
Which
of
the following
is
unit
value
true?
used for
do
the
elementary
i
The
photoelectric
effect
provides
the
wave
nature
of
electromagnetic
electron
diffraction
evidence for
the
Interference
and
wave
experiment
nature
diffraction
of
units
the
wave
nature
of
on
an
unit. What
magnitude
electron
as
of
measured
the
in
used?
21.
14
b
7
.83
c
8.
12
d
4.06
particles.
provides
The
light
in
a
beam
has
a
continuous
spectrum
of
evidence
wavelengths from
for
e
the SI
the
provides
7
iii
not
radiation.
a
The
charge
was
suggest for
evidence for
the
ii
charge
results
410 nm
to
680 nm. The
light
is
electromagnetic
incident
on
some
cool
hydrogen
gas
and
is
then
radiation.
viewed
a
i
only
b
i
and
ii
through
following
c
2
i,
The
ii
and
iii
potential
maintained
of
d
difference
at
the X-rays
ii
across
80 kV. What
is
and
iii
tube
cut-off
1.55
×
is
wavelength
emitted?
−11
a
10
m
b
8.00
×
10
×
the
a
Dark
b
Coloured
lines
c
Dark
d
Coloured
on
a
lines
lines
on
coloured
on
a
Which
wave
of
white
lines
m
d
1.55
×
10
the following
of
on
background
a
white
An
electron
level
E
than
E
cannot
be
electromagnetic
explained
by
the
moves from
within
an
atom.
an
E
c
Diffraction
is
the
level
–
of
E
E
2
to
E
energy
energy
the
photon
2
–
E
–
E
level
that
is
hc
1
c
d
hc
is
incident
on
a
zinc
the following
E
1
E
2
1
–
E
2
+
is
reaction.
3
H
→
He
1
What
radiation
2
the
energy
released
in
the
reaction?
plate
Use
are
–
1
1
2
p
photoelectrons
E
E
Consider
1
ultraviolet
higher
effect
Polarisation
When
wavelength
b
2
1
4
a
1
. What
a
photoelectric
at
released?
radiation?
9
d
energy
is
2
E
The
background
1
Interference
b
the
seen?
m
hc
a
of
is
background
−10
10
theory
that
background
black
2
3
grating. Which
spectrum
m
8
2.49
diffraction
describes
−11
−30
c
best
only
an X-ray
the
a
only
the following
data
to
work
out
your
answer.
emitted.
2
Mass
of
H
nucleus
=
2.01355 u
1
How
would
the
number
of
photoelectrons
emitted
Mass
per
second
N
and
the
maximum
kinetic
energy
E
of
proton
=
1.00728 u
of
3
the
photoelectrons
source
same
is
replaced
be
affected
with
a
less
when
intense
the
Mass
ultraviolet
source
of
of
=
1 u
E
a
Decreased
Increased
b
Increased
Decreased
c
Decreased
Unchanged
10
=
931.3 MeV
a
5.49 MeV
b
3.02 MeV
c
6.04 MeV
d
5621 MeV
Two
radioactive
activity
and Q
Unchanged
of
has
A at
a
electron
days
has
a
mass
9.
11
×
10
kg
and
2.85
×
10
is
the
de
wavelength?
−10
×
10
−10
m
b
2.85
×
10
−10
c
172
2.91
×
10
0.
12
P
has
days.
a
P
each
have
half-life
and Q
of
are
an
18
days
mixed
m
−11
m
total
d
2.91
activity
of
the
mixture
after
36
3
1
5
Broglie
A
4
1.82
=
of
and Q
is:
1
J. What
a
a
t
P
kinetic
−18
energy
time
half-life
together. The
An
samples
Decreased
−31
5
3.01494 u
wavelength?
N
d
He
2
the
×
10
m
b
A
8
c
A
16
d
A
8
Module
14
Structured questions
The
diagram
hydrogen
11
a
Briefly
describe
the
models
of
the
atom
below
3
Practice
represents
exam
energy
questions
levels
within
a
atom.
proposed
energy
energy/ eV
level
by:
i
J. J. Thomson
ii
Rutherford
iii
b
Neils
State
the
under
c
conclusion
empty
performed
supervision
how
these
that
space,
–0.54
4
–0.85
3
–1.5
2
–3.4
[3]
observations from
the
Explain
–0.37
5
[3]
Bohr.
experiments
6
[2]
the
of
an
scattering
and
Marsden,
Rutherford.
observations
with
alpha
by Geiger
atom
was
a very
led
[3]
to
the
composed
small
positive
mainly
of
nucleus.
[3]
(ground
d
State
the
approximate
i
the
diameter
of
a
gold
atom
[1]
ii
the
diameter
of
a
gold
nucleus.
[1]
a
a
Outline
Millikan’s
oil
drop
experiment
State
An
for
the
the
experimental
quantisation
of
1
–13.6
amount
electron
of
the
energy
ground
makes
a
required
to
ionise
an
state.
[1]
transition from
the
n
=
3
to
and
n
summarise
the
electron from
b
12
state)
value for:
evidence
it
=
2
state.
provides
charge.
[6]
i
Calculate
ii
Calculate
the frequency
of
the
photon
emitted.
b
State
the
two
measures
accuracy
of
his
taken
oil
by
drop
Millikan
to
[2]
improve
experiment.
the
wavelength
of
the
photon
[2]
emitted.
c
Explain
how
he
was
able
to
change
the
charge
iii
an
oil
[1]
on
drop.
In
which
region
Two
parallel
metal
plates
are
1.5 cm
electromagnetic
difference
between
the
would
the
radiation
be found?
[1]
apart. The
c
potential
the
[1]
spectrum
d
of
plates
is
An
electron
in
the
ground
state
of
a
hydrogen
1400 V.
atom
is
struck
by
a
photon. State
and
explain
−15
A
small
oil
drop
of
mass
of
7
.61
×
10
kg
remains
what
stationary
between
the
plates. The
density
of
i
is
negligible
Draw
the
ii
a
oil
comparison
diagram
to
show
that
the forces
of
the
acting
the
electric field
strength
electron
and
what
happens
the
photon
when
the
energy
of
the
photon
is:
oil.
i
8 eV
[2]
ii
15 eV.
[2]
on
between
plates.
15
a
the
charge
on
iv
Calculate
the
number
the
of
oil
drop.
electrons
[3]
attached
drop.
Explain
what
duality
[2]
Calculate
oil
the
[2]
iii
the
with
drop.
Calculate
the
in
to
the
to
air
happens
b
to
[2]
c
of
meant
by
the
electromagnetic
Describe
an
particles
can
Calculate
is
experiment
radiation.
which
demonstrate
the
wave–particle
wavelength
a
of
wave
a
[2]
demonstrates
that
nature.
particle
of
[3]
mass
−31
9.
11
13
a
Outline
the
experimental
evidence
that
to
that
b
electromagnetic
radiation
a
wave
[4]
ii
a
particle.
[4]
why
line
emission
spectra
16
a
the
existence
of
kg
when
20%
the
speed
Explain
what
is
inside
atoms.
with
a
speed
equal
of
light.
[3]
meant
by
the
photoelectric
effect.
[2]
discrete
State
three
experimental
observations
of
the
energy
photoelectric
levels
travelling
provide
b
evidence for
10
is:
i
Explain
×
suggests
effect.
[3]
[3]
c
Explain
energy
the
what
and
is
meant
by
the
terms
threshold frequency
photoelectric
effect.
work function
when
applied
to
[2]
173
d
Electromagnetic
3
10
radiation
of
intensity
8.5
×
19
Calculate
the
energy
released
in
the following
−2
W m
and
wavelength
240 nm
is
incident
on
reaction.
[6]
2
an
iron
surface
of
area
2.0 cm
. The
iron
235
surface
1
U
+
n
92
reflects
80%
of
the
electromagnetic
threshold frequency
of
iron
is
1.
1
×
10
Use
Hz.
Calculate:
i
the
intensity
absorbed
ii
the
by
power
of
the
the
of
electromagnetic
metal
the
radiation
surface
radiation
→
90
Ba
0
1
+
Kr
56
+
2
36
n
0
radiation.
15
The
144
[2]
incident
on
the following
data:
Mass
of
a
neutron
=
1.00867 u
Mass
of
uranium-235
Mass
of
barium-144
Mass
of
krypton-90
nucleus
nucleus
=
=
235.
124 u
143.923 u
the
surface
nucleus
=
89.920 u
[2]
7
iii
the
energy
of
a
photon
incident
on
the
surface
20
Calculate
the
binding
energy
of
the
Be
nucleus
in
4
[2]
iv
the
number
of
photons
incident
on
MeV.
Use
surface
Given
that
hitting
per
second.
only
the
photoelectrons,
the following
data:
[2]
0.001%
surface
[6]
the
of
result
the
in
the
production
the
photoelectric
vi
the
work function for
vii
the
stopping
of
a
proton
=
Mass
of
a
neutron
Mass
of
beryllium
1.00728 u
=
1.00867 u
of
determine:
v
Mass
photons
current
1 u
[2]
iron
=
nucleus
=
7
.01693 u
931.3 MeV
[2]
214
21
a
An
isotope
of
lead
Pb.
is
82
17
a
Outline
X-ray
the
potential for
principle
of
this
radiation.
production
[4]
of X-rays
in
tube.
i
Explain
ii
State
what
is
meant
by
the
term
isotope.
[1]
an
the
number
of
protons
and
neutrons
214
[6]
present
in
a
nucleus
of
Pb.
[2]
82
b
Sketch
a
diagram
to
show
the
spectrum
of
the
214
b
Pb
has
a
half-life
of
27
minutes. A
nucleus
of
82
X-rays
emitted.
Label:
214
Pb
decays
by
emitting
a
β-particle
to form
82
214
i
the
continuous X-ray
spectrum
[1]
element
214
X
.
Element
ii
the
characteristic X-ray
iii
the
cut-off
iv
Explain
spectrum
[1]
wavelength.
the
origin
of
two
each feature
of
→
is
of
Y
+
α
X
→
Z
+
β
[4]
tube
it
one
the
operates
at
100 kV
and
the
a
nuclear
reaction for
each
of
the
three
current
decays
through
by
methods:
X
Write
An X-ray
decay
83
[1]
spectrum.
c
can
X
83
described
above.
[5]
1.0 mA. Calculate:
214
c
i
ii
the
the
power
speed
input
of
the
A
sample
t
=
of
Pb
as
they
hit
a
mass
of
1.8 μg
at
time
82
[2]
electrons
has
0. Calculate:
the
i
target
the
number
of
atoms
in
the
sample
[2]
[3]
214
iii
the
cut-off
wavelength
of
the X-rays
ii
the
decay
iii
the
activity
constant
iv
the
time
4.5
×
of
Pb
emitted.
[2]
82
of
the
sample
at
time
t =
0
[2]
[3]
18
a
Explain
and
what
the
is
meant
nuclear
by
binding
the
nucleon
energy
of
the
number
Sketch
with
a
labelled
nucleon
diagram
number
of
to
the
show
nucleus.
a
binding
energy
per
[4]
what
is
meant
by
activity
of
the
sample
10
Bq.
is
[3]
[3]
Explain
what
is
meant
i
The
activity
ii
The
decay
The
half-life
iii
Explain
the
by
the following:
the variation
nucleon.
c
which
5
22
b
at
nuclear fusion
of
a
radioactive
constant
of
a
of
a
source
[1]
nuclide
[2]
nuclide
[2]
and
14
nuclear fission.
[3]
b
A
radioactive
radioactive
d
Using
the
sketch
in
b,
explain
why
energy
source
nuclei
contains
and
has
an
7
.5
×
10
activity
of
is
5
released
during
nuclear fusion
and
nuclear fission.
4.8
×
10
Bq.
For
this
source,
calculate:
[4]
i
the
decay
constant
ii
the
half-life
iii
the
time
4.0
×
taken for
[2]
[2]
the
activity
to fall
to
4
174
10
Bq.
[3]
Module
3
Practice
exam
of
plutonium
questions
239
23
a
Explain
what
b
Describe
is
meant
by
the
term
half-life.
[2]
29
The
radioactive
nuclide
Pu,
decays
94
an
experiment
to
measure
the
by
half-life
alpha-particle
emission
with
a
half-life
of
2.4
×
8.0
×
4
of
radon-220,
which
is
a
gas
with
a
short
10
half-life.
years. The
energy
of
the
alpha-particle
is
–13
10
[7]
c
Explain
how
determine
you
the
would
half-life
use
of
the
data
collected
to
radon-220.
[3]
J.
a
Write
an
b
State
the
in
24
a
Explain
what
b
Discuss
is
meant
by
a
radioisotope.
uses
of
number
nucleus
of
of
the
decay.
protons
the
decay
[2]
and
neutrons
product.
[2]
[1]
c
two
the
equation for
radioisotopes.
Calculate
the
decay
constant
of
the
plutonium
[6]
nuclide.
c
Indicate
the
properties
of
each
radioisotope
d
make
them
suited for
their
use.
The
plutonium
a
particular
device
a
Draw
b
A
a
labelled
diagram
of G–M
tube.
of
argon
and
bromine
is
is
inside
the G–M
tube.
used
as
State
power
a
power
required
source
by
the
1.8 W.
Calculate
the
minimum
rate
of
decay
of
1.8 W
of
sometimes
the
used
is
device. The
[5]
i
mixture
isotope
[2]
in
25
[2]
that
the function
plutonium for
it
to
produce
of
power.
each
gas.
ii
c
Explain
[3]
[3]
the
principle
of
operation
of
a G–M
Calculate
the
number
of
plutonium
atoms
tube.
needed
to
produce
the
activity
in
d
i.
[3]
[4]
30
26
a
Explain
the
principle
of
operation
of
a
In
the
context
a
chamber.
State
what
Sketch
a
pattern
the following
cloud
of
what
ionising
is
observed
radiations
is
when
an
atomic
meant
by
nucleus:
binding
present
how
it
energy,
and
arises.
[3]
each
in
b
State
what
c
State
the
is
meant
by
mass
difference.
[1]
a
relationship
between
binding
chamber:
energy
i
α-radiation
[2]
ii
β-radiation
[2]
iii
is
[6]
explain
b
of
cloud
γ-radiation
and
mass
‘Binding
energy
stability
of
a
of
difference.
the
nucleus
[1]
is
related
to
the
nucleus.’
[2]
d
Explain
what
is
meant
by
this
statement.
[2]
226
27
A
radium
nuclide
Ra
decays
by
alpha-particle
88
emission
of
a
to form
radium-226
Write
a
is
an
isotope
1600
nuclear
of
radon
Rn. The
half-life
years.
31
Fusion
is
the
equation for
the
decay.
[2]
of
three
Calculate
the
decay
constant
of
the
which
the Sun
energy
protons
Calculate
the
energy
released
in
the
that
is
process
equivalent
into
process
one
releases
an
helium
amount
to
the
mass
of
1
data.
=
data
of
radon-222
Mass
of
an
=
and
one
He
atom. Using
2
below,
calculate
the
energy
released
in
the
226.0254 u
reaction
Mass
4
H atoms
1
[3]
the
radium-226
difference
reaction,
between four
the following
of
produces
sequence
[2]
energy
Mass
a
radium
nuclide.
using
by
reactions. This
combines four
nucleus. The fusion
c
by
produces
separate fusion
effectively
b
process
energy. The Sun
in
MeV.
222.0175 u
1
Mass
of
H
=
1.00783 u
1
alpha
particle
=
4.0026 u
4
Mass
of
He
=
4.00260 u
[4]
2
28
a
Explain
and
b
what
decay
Show
that
half-life
is
meant
by
the
terms
half-life
constant.
the
T
of
decay
an
[2]
constant
isotope
are
λ
and
the
related
by
the
1/2
expression
λ
T
=
0.693
[3]
1/2
c
On
separate
diagrams,
sketch
a
graph
to
show:
i
that
radioactive
decay
is
exponential
[1]
ii
that
radioactive
decay
is
random.
[1]
iii
What
happens
radioactive
to
the
isotope
is
graph
in
heated
i
when
a
strongly?
[1]
175
13
Analysis
13.
1
Analysis
Learning outcomes
and
and
completion
of
this
interpretation
Plotting
In
On
interpretation
section,
experimental
be
able
way
of
draw
straight
line
graphs
is
is
often
by
collected.
drawing
a
This
data
suitable
must
graph
to
be
analysed.
determine
constants.
following
guidelines
should
be
used
when
drawing
graphs.
data
ᔢ
determine
data
data
using
The
experimental
ᔢ
analysing
to:
unknown
ᔢ
work,
you
One
should
graphs
gradients
Choose
a
scale
that
is
suitable
(e.g.
1 cm
to
5
units,
1 cm
to
10
units).
and
Do
not
use
scales
such
as
1 cm
to
3
units.
intercepts
ᔢ
ᔢ
rearrange
equations
to
obtain
The
at
straight
scale
should
be
chosen
such
that
the
points
being
plotted
occupy
a
least
half
the
graph
paper
.
line form.
ᔢ
The
and
x
and
the
y
ᔢ
The
graph
ᔢ
A
×
or
ᔢ
If
the
ᔢ
If
the
be
side
of
data
should
be
between
be
to
so
the
used
an
to
follow
that
with
height/m,
the
appropriate
represent
a
linear
there
are
quantities
being
plotted
temperature/°C).
title.
the
points
relationship,
equal
numbers
being
a
plotted.
line
of
of
data
best
fit
points
on
line.
appears
drawn
labelled
T/s,
given
be
appears
drawn
be
d/cm,
should
data
either
should
(e.g.
should
a
should
axes
units
to
follow
through
adjacent
a
the
non-linear
data
relationship,
points.
Do
not
use
a
a
smooth
straight
curve
line
points.
y-axis
T
wo
y
=
quantities
are
required
when
plotting
a
graph
( x-axis
and
y-axis).
In
mx + c
an
experiment
to
investigate
the
relationship
between
two
quantities,
y
2
only
B (x
, y
2
one
quantity
manipulated
y
quantity
1
A (x
(0,
, y
1
c)
be
is
variable .
changed
called
the
It
is
at
usually
a
time.
responding
plotting
This
variable .
on
quantity
the
is
The
x-axis.
called
The
the
responding
second
variable
is
)
1
usually
the
plotted
on
experiment,
The
x
can
)
2
general
the
all
y-axis.
other
equation
of
Since
only
quantities
a
straight
are
line
one
quantity
known
is
y
=
as
mx
is
being
controlled
+
c.
The
changed
in
variables .
gradient
x
1
or
2
slope
of
the
line
is
m.
The
intercept
on
the
y-axis
is
c.
Consider
x-axis
Figure
Figure 13.1.1
A straight line graph
A
and
13.1.1.
B
lie
on
the
straight
line.
The
coordinates
of
A
and
B
are
( x
,
y
1
and
(x
,
y
2
is
given
)
respectively.
The
gradient
–
y
–
x
2
(slope)
Plotting
linear
In
work
It
practical
quantities.
is
the
A
)
1
and
If
it
against
often
form
is
two
the
often
required
quantities
a
straight
x
non-linear
to
and
line
establish
y
are
graph
relationships
related
would
relationships
be
such
that
obtained
between
they
when
have
y
is
x
required
of
1
graphs from
relationship,
examples.
176
through
1
2
plotted
passing
=
x
linear
line
2
y
a
the
by:
Gradient
two
of
that
an
equation
of
expression
a
straight
be
rewritten
line.
The
so
table
that
it
shows
resembles
some
B
Chapter
Expression
What to
plot?
2
y
=
ax
T
=
kl
Constants
Gradient
a
and
b
a
(0,
b)
k
and
l
n
(0,
lg k)
a
and
b
a
(0,
b)
−k
(0,
ln A)
y-intercept
13
Analysis
and
interpretation
Exam tip
2
+
b
y
against
x
Recall
the
rules for
logarithms:
n
lg T
against
lg l
n
2
y
2
=
ax
2
+
bx
y
against
1
log
(A)
=
n log
b
A
b
x
x
2
log
(A)
+
log
b
(B)
=
log
b
(AB)
b
kt
N
=
Ae
ln N
against
t
A
and
k
A
3
log
(A)
–
log
b
Suppose
T
and
l
are
related
by
the
following
(B)
=
log
b
b
(
B
)
equation:
n
T
T
aking
log
on
both
=
kl
sides
of
the
equation
gives:
10
n
lg T
=
lg (kl
)
lg T
=
lg k
+
lg (l
lg T
=
lg k
+
n lg l
lg T
=
lg k
+
n lg l
Exam tip
n
)
1
log
is
usually
written
as
lg.
10
This
is
the
linear
form.
2
A
straight
line
graph
gradient
of
the
line
Suppose
N
and
t
will
is
are
n
be
and
related
obtained
the
by
if
lg T
y-intercept
the
is
is
following
plotted
against
lg l.
The
log
is
where
lg k
usually
written
as
ln,
e
e
=
2.718
equation:
kt
N
T
aking
log
on
both
=
sides
Ae
of
the
equation
gives:
e
kt
ln N
=
ln (Ae
)
ln N
=
ln A
+
ln N
=
ln A
–
kt ln e
ln N
=
ln A
–
kt
kt
A
straight
gradient
line
of
graph
the
line
will
is
−k
be
ln (e
This
obtained
and
the
)
if
ln N
is
y-intercept
is
the
plotted
is
linear
form.
against
t.
The
ln A
Example
T
wo
of
capacitors
2.2 µF
and
capacitors
the
are
capacitors
difference
The
V
results
are
the
connected
other
charged
are
an
using
the
shown
both
in
in
series.
One
unknown
a
discharged
across
are
is
battery
through
of
a
capacitors
the
table
capacitor
capacitor
e.m.f.
E.
10 M Ω
is
of
has
When
resistor
.
measured
a
capacitance
capacitance
fully
The
over
a
C.
The
charged,
potential
period
time.
below.
t/s
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
V/V
6.45
4.62
3.31
2.37
1.70
1.22
0.87
0.63
the
capacitors
The
potential
difference
V
across
varies
according
to
the
equation:
t/CR
V
=
V
e
0
where
C
represents
a
Plot
b
Calculate
a
graph
of
the
ln V
combined
against
the
gradient
the
value
of
capacitance
of
the
two
capacitors.
t
the
graph
and
hence
determine
the
time
constant.
c
Calculate
d
Determine
the
of
value
the
of
unknown
the
e.m.f.
E
capacitance.
of
the
battery.
177
Chapter
13
Analysis
and
interpretation
t/CR
a
V
=
V
e
0
T
aking
natural
logarithms
of
both
sides
of
the
equation,
(log
or
ln)
e
t/CR
ln V
=
ln (V
e
)
0
t/CR
ln V
=
ln V
+
ln (e
)
0
ln V
=
ln V
ln V
=
−(1/CR)t
–
t/CR
0
+
ln V
0
When
a
graph
of
The
gradient
of
The
y-intercept
ln V
the
=
against
straight
t
is
line
plotted,
=
a
straight
line
is
obtained.
−(1/ CR)
ln V
0
t/s
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
V/V
6.45
4.62
3.31
2.37
1.70
1.22
0.87
0.63
ln (V/V)
1.86
1.53
1.20
0.86
0.53
0.20
−0.
14
−0.46
ln V/ V
2.0
1.5
1.0
0.5
0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
–0.5
Figure 13.1.2
b
Choosing
(3.0,
2.0)
Gradient
two
and
of
points
(29.0,
straight
on
the
straight
line,
0.25)
line
0.25
–
2.0
29.0
–
3.0
=
−(1/ CR)
=
CR
=
=
−0.0673
−0.0673
1
0.0673
The
c
CR
time
=
CR
=
14.9 s
τ
=
CR
constant
=
14.9 s
14.9 s
6
C
×
10
×
10
=
14.9
14.9
C
=
6
10
C
The
178
combined
=
×
10
1.49 μF
capacitance
of
both
capacitors
=
1.49
μF
time/s
Chapter
1
For
capacitors
in
1
series,
total
interpretation
C
1
2
1
1
=
+
–6
×
and
+
C
1
4.9
Analysis
1
=
C
13
–6
10
2.2
×
C
10
2
1
1
1
=
−
–6
C
4.9
×
–6
10
2.2
×
10
2
1
5
=
2.166
×
10
=
4.6
10
C
2
−6
C
×
2
Therefore,
d
When
them
At
the
was
time
t
the
capacitance
capacitors
equal
=
0,
to
V
were
the
=
of
the
fully
e.m.f.
of
unknown
charged,
the
capacitor
the
battery
=
potential
4.6
μF
difference
across
E.
E
0
From
the
graph
ln V
=
2.2
=
e
=
9.0 V
0
2.2
V
0
Therefore
E
Key points
ᔢ
Experimental
data
can
be
used
to
determine
relationships
between
two
variables.
ᔢ
In
an
the
ᔢ
The
experiment,
the
manipulated
variable
is
the
variable
that
is
altered
in
experiment.
responding
some
action
ᔢ
Controlled
ᔢ
Some
is
(dependent)
taken
variables
non-linear
in
the
are
variable
those
equations
is
the
variable
that
changes
when
experiment.
kept
can
be
constant
throughout
rearranged
to
obtain
an
experiment.
linear
equations.
179
Analysis and interpretation: Practice exam questions
1
A
student
how
to
is
the
the
performed
current
potential
maintained
He
used
the
following
at
an
experiment
through
to
silicon
diode
difference
across
it
a
constant
circuit
in
is
when
temperature
Figure
13.
1.3
to
a
investigate
a
State
the
of
diode
80 °C.
obtain
and
filament
related
a
explain
lamp
b
Plot
graph
c
Comment
d
Determine
of
on
how
the
changes
lg (I/A)
resistance
with
against
whether
the
of
increasing
a
current. [3]
lg (V/V).
relationship
[6]
is
valid.
[2]
the
the
gradient
and
y-intercept
of
the
results.
line
e
I/mA
5
10
V/V
0.56
20
0.59
30
0.63
40
0.65
of
best fit.
Determine
[4]
the
values
of
n
and
k
[2]
60
0.66
0.68
3
An
experiment
is
performed
to
investigate
how
t
,
1/2
the
its
10 Ω
time
taken for
initial
a
capacitor
charge, varies
with
to
discharge
capacitance
to
C
of
half
the
series
capacitor.
In
an
experiment
six
different
capacitors
protective
(C)
resistor
are
obtain
+
available. The
the following
500
C/μF
circuit
in
Figure
13.
1.5
is
used
to
data.
666
1000
1500
2000
3000
mA
15.3
/s
t
19.8
28.7
45.6
60.4
87
.
1
1/2
Figure 13.1.3
The
current
difference
I in
the
diode
by
the
expression
V
is
related
to
the
+
potential
–
kV/T
I
=
I
e
0
+
Where
I
and
k
are
constants
and
T
is
C
the
0
temperature
a
Plot
a
measured
graph
of
in
kelvin.
ln (I/mA)
against
V
and
draw
a
line
100 kΩ
of
best fit.
[6]
b
Determine
the
gradient
c
Determine
the
value
of
and
k
the
and
y-intercept.
I
[4]
[4]
0
2
In
an
experiment
current
with
to
investigate
potential
the variation
difference for
of
a filament
Figure 13.1.5
lamp,
a
the
circuit
in
Figure
13.
1.4
was
Explain
how
the
data
in
the
table
could
be
used.
obtained.
variable
power
d.c.
b
State
the
experimenter
could
reduce
the
supply
random
+
[3]
how
errors
in
the
t
/s
measurement
of
t
.
[1]
1/2
–
c
Plot
a
graph
d
Draw
e
Determine
of
against
C/μF.
[6]
1/2
a
line
The variation
capacitor
through
best fit
the
of
with
a
of
through
gradient
potential
time
resistor
t
is
as
of
the
the
difference
the
given
the
points.
[1]
[2]
across
capacitor
by
data
line.
a
discharges
expression
−t/CR
Figure 13.1.4
V =
V
e
,
where
0
across
The following
results
were
1.25
2.40
is
the
initial
potential
difference
0
the
capacitor
and
R
is
the
combined
resistance
obtained.
of
V/V
V
3.20
4.20
5.50
6.20
f
the
100 kΩ
Show
resistor
that
the
and
time
t
the voltmeter.
for
the
potential
1/2
difference
I/A
0.85
1.20
1.34
1.56
1.81
to fall
to
half
to
e
its
initial value
is
given
by
1.84
t
=
CR ln 2.
[3]
1/2
It
is
suggested
that
I
and
V
are
related
by
the
g
Use
your
answer
to
calculate
the
value
of
R.
n
equation
I = kV
[2]
h
Determine
voltmeter.
180
the
value
of
the
resistance
of
the
[3]
Analysis
4
a
Einstein’s
equation
used
to
explain
and
interpretation
Practice
exam
questions
the
V
/V
−0.55
−0.90
−1.20
−1.50
−1.90
−2.50
−3.00
−7
.20
−9.60
−12.00
−14.80
−14.80
−14.80
in
photoelectric
effect
can
be
written
as
V
hf
=
φ
+
E
.
Explain
what
is
meant
by
/V
−4.40
out
each
max
of
the
symbols
used
in
the
equation.
[3]
b
Use
c
Determine
the
data
to
plot
a
graph
of
V
against
V
out
b
[5]
in
In an experiment to investigate the photoelectric
the
gain
of
the
amplifier from
your
effect, the wavelength λ of the electromagnetic
graph.
[2]
radiation incident on a metal surface is varied and
d
the stopping potential V
What
is
the
range
of
possible
input voltages for
is measured. The following
s
the
output
of
the
amplifier
not
to
be
saturated?
data was obtained from one such experiment.
[2]
Wavelength/nm
250
280
310
365
405
0.71
0.40
6
V
/V
2.41
1.72
1.20
In
an
experiment
radioactive
to
measure
sample,
the
half-life
the following
data
of
was
a
obtained.
s
Frequency/Hz
Time/min
1.0
Count
i
Draw
a
diagram
to
show
how
the
data
2843
2.0
3.0
4.0
5.0
6.0
2619
2420
2220
2040
2070
could
−1
rate/min
be
ii
obtained.
Copy
and
[4]
complete
the
table
to
include
The
frequency.
iii
Plot
a
background
count
rate
was
100
counts/min.
[4]
graph
of
stopping
potential
V
a
against
Use
the
data
to
plot
a
suitable
straight
line
graph.
s
frequency
iv
Use
your
f.
(Remember
[4]
graph
of
to
1
the
value
2
the
threshold frequency
the
Planck
3
the
work function
of
constant
the
to
allow for
the
background
count
rate.)
determine:
metal
[8]
[3]
b
Write
[2]
c
Determine
under
down
radioactive
investigation.
[2]
d
Determine
an
the
equation for
decay
the
constant
straight
λ
of
line.
[2]
the
sample.
the
[2]
half-life
of
the
radioactive
sample.
[2]
5
Figure 13.
1.6 shows how an operational amplifier is used
e
as a non-inverting amplifier. A student wishes to plot
Calculate
the
activity
of
a
sample
containing
15
the transfer characteristic (a plot of V
against V
out
2.0
).
×
10
atoms
of
the
radioactive
substance.
[3]
in
7
The
intensity
I
of
γ-rays
through
a
medium varies
V
in
according
V
to
the
equation:
out
+
μx
I
=
I
e
0
R
f
where
μ
is
medium
the
and
x
linear
is
the
absorption
distance
coefficient
travelled
of
inside
the
the
X
medium. An
linear
R
data
is
is
set
coefficient
up
of
to
measure
the
lead. The following
obtained.
Thickness
lead
experiment
absorption
of
0.
15
0.25
0.35
0.45
0.55
0.65
2338
1980
1690
1428
1220
1011
x/cm
Intensity
I
Figure 13.1.6
counts
a
Explain
any
how
other
order
to
you
circuit
plot
the
would
use
Figure
components
transfer
to
13.
1.6
collect
characteristic
and
data
of
second
in
a
the
amplifier.
data
was
obtained
in
one
particular
investigation.
V
/V
3.00
2.50
2.20
1.80
1.20
1.00
14.80
14.80
14.80
14.40
9.60
8.00
0.75
0.55
6.00
4.40
Plot
a
graph
of
ln I
against
x
and
draw
a
line
of
best fit.
[4]
The following
per
[8]
b
Determine
c
State
d
Write
the
the
gradient
absorption
down
the
and
y-intercept.
coefficient
equation
of
the
μ
of
line
[4]
lead.
of
[1]
best fit.
[1]
0.
15
in
V
/V
out
1.20
181
List
of
physical
constants
−11
Universal
gravitational
constant
G
=
6.67
×
10
g
=
9.80 m s
=
6380 km
=
5.98
×
10
=
7.35
×
10
2
N m
–2
kg
–2
Acceleration
Radius
of
due
the
to
gravity
Earth
R
E
24
Mass
of
the
Earth
M
kg
E
22
Mass
of
the
Moon
M
kg
M
5
Atmosphere
atm
=
1.00
×
10
–2
N m
–23
Boltzmann’s
constant
k
=
1.38
×
10
–1
J K
9
Coulomb
constant
=
9.00
×
10
2
N m
–2
C
–31
Mass
of
the
electron
m
=
9.11
×
10
kg
e
–19
Electron
charge
e
=
1.60
×
10
C
3
Density
of
water
=
1.00
×
10
Resistivity
of
steel
=
1.98
×
10
Resistivity
of
copper
=
1.80
×
10
=
400 W m
–3
kg m
–7
Ω m
–8
Ω m
–1
Thermal
conductivity
of
copper
–1
K
–1
–1
Specific
heat
capacity
of
aluminium
=
910 J kg
K
Specific
heat
capacity
of
copper
=
387 J kg
Specific
heat
capacity
of
water
=
4200 J kg
Specific
latent
heat
of
fusion
=
3.34
×
10
Specific
latent
heat
of
vaporisation
=
2.26
×
10
–1
–1
K
–1
–1
K
5
of
ice
–1
J kg
6
of
water
–1
J kg
23
Avogadro
constant
N
=
6.02
×
10
per
mole
A
8
Speed
of
light
in
free
space
c
=
3.00
×
10
=
4π
=
8.85
×
10
h
=
6.63
×
10
u
=
1.66
×
10
–1
m s
–7
Permeability
of
free
space
μ
×
10
–1
H m
0
12
Permittivity
of
free
space
ε
–1
F m
0
–34
The
Planck
constant
J s
–27
Unified
atomic
mass
constant
kg
–27
Rest
mass
of
proton
m
=
1.67
×
10
R
=
8.31 J K
σ
=
5.67
×
10
=
1.67
×
10
kg
p
–1
Molar
gas
constant
–1
mol
–8
Stefan–Boltzmann
constant
–2
W m
–27
Mass
of
neutron
m
n
182
kg
–4
K
Glossary
Coulomb’s
law
The force
acting
Electromotive
force
(e.m.f.)
The
A
between
Activity
per
The
number
nuclei
decaying
second.
Ammeter
measure
circuits
instrument
electric
in
by
used
to
the
current.
circuit
which
driven
proportional
charges
An
Asynchronous
are
of
the
circuit
in
and
square
point
to
charges
the
product
inversely
of
the
is
energy
of
their
proportional
distance
digital
operations
the
input
to
between
its
signals.
A
register
number
clock
Cut-off
energy
per
through
Electron
Counter
the
converted
mechanical)
them.
Sequential
changes
two
of
that
clock
is
able
pulses
to
count
arriving
at
input.
of
the
The
maximum
incident
unit
A
1
charge
the
(or
electrical
fl owing
1
charged
nucleus
of
an
electronvolt
transformed
moves
of
into
negatively
Electronvolt
it
energy
chemical
it.
orbits
energy
wavelength
wavelength
that
from
through
a
by
particle
atom.
(eV)
an
is
the
electron
potential
as
difference
volt.
B
electromagnetic
Background
radiation
radioactivity
The
electrons
random
detected from
to
be
radiation
required for
emitted.
Energy
level diagram
shows
an
the
the
various
A
diagram
energy
that
levels
within
atom.
D
Equipotiential
surroundings.
Bandwidth
which
The
the
range
gain
of
of frequencies for
an
amplifier
remains
Decay
constant
decay
of
a
The
nucleus
probability
per
unit
of
of
equal
line
A
line
joining
points
potential.
time.
F
constant.
Barrier
Depletion
potential
difference
a
p-type
n-type
Binding
The
across
a
material
potential
p-n
is
on
junction
placed
when
against
an
material.
energy
completely
energy
separate
the
required
nucleons
to
of
a
nucleus.
Binding
by
the
per
energy
total
Bistable
side
are
Dielectric
no
The
plates
Diffusion
occurs
A
nucleon
of
the
number
device
nucleus
of
that
The
in
divided
nucleons.
has
two
a
of
Doping
of
stable
to
a
a
The
region
junction
where
carriers.
placed
between
capacitor.
A
of
of
that
difference
holes
The
and
process
capacitor
in
the
(F)
coulomb
1
volt
Faraday’s
electrons
by
which
element)
semiconductor
to
an
is
is
if
a
The
e.m.f.
the
of
is
a
capacitance
charge
stored
potential
applied
induced
rate
has
the
when
law
Feedback
added
across
it.
of
proportional
of
is
difference
magnitude
change
enhance
current
its
adding
into
that flows
The
fraction
properties.
A
A
1 farad
the
to
magnetic flux
linkage.
(another
current
1
of
current
a
Farad
of
junction.
conductive
Drift
p-n
charge
current
impurity
states.
a
layer)
material
because
p-n
(or
net
concentration
energy
binding
either
there
the
The
region
when
of
it
an
process
the
to
of
output
the
input
taking
signal
signal
a
and
being fed
amplifier.
Ferromagnetic
Substances
such
as
iron,
C
Capacitance
potential
Capacitor
stores
The
charge
difference
An
in
electrical
stored
a
per
unit
that
The
a
p-n
arrangement
difference
is
applied
across
junction.
electrons
potential
charge.
Cascaded
potential
Drift velocity
capacitor.
component
a
steel
and
nickel
magnetic
The
along
net
a
velocity
metal
difference
is
of
when
applied
to
a
a
it.
of
to
properties
show
when
strong
subjected
magnetising force.
Fleming’s
across
which
left
predict
hand
the
rule
A
direction
rule
of
used
a force
current-carrying
conductor
right
magnetic field.
on
placed
a
at
E
connecting
that
the
several
output
connected
to
of
the
amplifiers
one
such
amplifier
input
of
is
Eddy
another
in
amplifier.
Charge
which
the
determined
Conductor
electrons
a
A
circuit
output
by
the
digital
the
current
material
to flow
A
of
that
through
circuit
circuit
is
inputs.
it
easily,
e.g.
the
same
that
throughout
an
experiment.
Coulomb
of
1
charge
coulomb
a
(C)
passes
section
of
when
current
a
current
conductor
of
1
is
the
quantity
through
in
1
any
second
ampere
is flowing.
in
a
See
A flow
of
unit
Electric
point
A
region
a force
positive
the
angles
in
to
Flip-flop
charged
a
charged
experienced.
The force
acting
the
a
a
The
potential
done
charge from
in
at
moving
infinity
to
a
unit
that
the
an
induction
electric
The
effect
current
using
conductor
the
basic
rule
the
used
induced
moving
memory
When
across
p-type
n-type
A
of
at
right
element
in
circuits.
positive
negative
a
a
p-n
cell
terminal
material
terminal
is
is
junction
is
and
such
connected
the
connected
to
the
material.
Frequency
cycles
point.
producing
that
rule
direction
magnetic field.
The
connected
a
hand
Forward-biased
to
charge.
work
Electromagnetic
of
predict
current
resistance
around
is
strength
potential
is
to
to
right
sequential
where
positive
that
situated
magnetic field.
Electric field
per
Variables
is
angles
Fleming’s
that flow
particles.
body
metal.
kept
Electric
that
resistance
Electric field
allows
Currents
conductor
Electrical
Controlled variable
are
a
changing
Q = I × t
Combinational
in
currents
The
number
generated
Full-adder
one-bit
A
per
circuit
of
complete
second.
that
adds
three
numbers.
magnetism.
183
Glossary
Logic
gates
The
building
blocks
provided
of
that
there
is
no
change
in
the
G
digital
Gain
The
the
ratio
input
of
the
output
voltage
physical
circuits.
to
conditions
Open-loop
voltage.
M
Magnetic field
A
gain
The
An
arithmetic
circuit
used
where
region
a
around
one-bit
to
The
average
time
p-type
number
of
undecayed
nuclei
to
half
of
its
product
density
initial
is
and
the
area
The
across
conductor when
the force
straight
magnetic field
is
conductor
normal
A
covalent
bond
that
is
length
Period
on
to
carrying
unit
linkage
The
linking
or
passing
numerically
Magnetic
pole
equal
of
magnetic
through
One
to
of
a
coil.
It
two
in
carriers
which
are
the
‘holes’.
maximum
value
of
time
an
of
an
taken for
alternating
a
the
ability
of
a
alternating
ends
of
one
complete
current.
space
A
medium
measure
to
transmit
magnetic field.
Permittivity of free
Nφ
the
material
Permeability of free
the field.
electron.
is
A
charge
The
The
cycle
missing
flux
one
unit
applied.
Magnetic flux
Hole
per
a
current
perpendicular
amplifier
current.
Numerically
a
a
current-carrying
to
of
instantaneous
potential difference
up transversely
material
majority
Peak value
through
value.
equal
set
magnetic
passes.
Magnetic flux density
Hall voltage
of
to
which
decrease
The
taken for
flux
the
an
is
numbers.
Magnetic flux
Half-life
of
a
magnetic force
experienced.
two
gain
conductor.
P
magnet
add
the
without feedback.
H
Half-adder
of
of
how
easy
it
is
space
to
A
measure
transmit
an
electric
I
a
Induced
flows
current
in
a
A
conductor
electromagnetic
Insulator
an
A
current
e.g.
as
a
result
that
changed
of
in
an
Mass defect
does
to flow
magnet.
Manipulated variable
that
induction.
material
electric
easily,
current
field
not
allow
through
it
difference
mass
of
of
constituent
its
variable
that
is
nucleus
and
between
the
total
the
mass
area
The
on
a
power
incident
per
Photon
The
a
material
A
material
in
surface
A
to
The
rate
at
emitted from
when
exposed
to
radiation.
effect
a
The
metal
emission
surface
electromagnetic
quantum
of
of
when
radiation.
electromagnetic
which
radiation.
resistance
the
to
metal
are
electromagnetic
exposed
unit
surface.
resistance
internal
current
electrons from
n-type
Internal
a
space.
electrons
Photoelectric
nucleons.
rubber.
N
Intensity
Photoelectric
which
experiment.
The
the
A
through
majority
of
charge
carriers
are
Positive
cell.
charge
A
deficiency
of
electrons.
Intrinsic
semiconductor
Pure
electrons.
elements
Negative
such
as
silicon
and
charge
An
A
charged
atom
or
Atoms
of
have
the
the
same
a
same
different
mass
negative fraction
atomic
of
positive fraction
of
an
amplifier
to
to
the
input
signal
the
the
of
an
amplifier
input
signal
Potential difference
whereby
The
An
induced
sum
any
of
an
the
bombarded
in
nucleus
unstable
by
a
nucleus
difference
V
circuit
is
equal
to
the
a
the
circuit
is
point
splits
out
sum
of
the
of
fragments
the
as
that
Nuclear force
point.
The
between
potential
two
work
done
points
in
(energy
neutron. The
into
two
or
more
well
as
several
Very
strong
electrical
energy
to
stable
of
energy)
in
moving
unit
neutrons.
positive
currents flowing
it
into
is
other forms
a
adding
process
converted from
into
and
being fed
amplifier.
K
law
taking
amplifier.
into
currents flowing
of
output
being fed
number.
Nuclear fission
Kirchhoff ’s first
process
the
and
number
it
of
the
signal
signal
The
of
element
adding
but
electrons.
process
a
output
that
The
molecule.
taking
Isotopes
of
Positive feedback
Negative feedback
Ion
excess
germanium.
charge from
one
point
to
the
short-range
other.
Kirchhoff ’s
second
law
The
force
algebraic
inside
the
nucleus
of
an
atom.
Power
sum
of
the
e.m.f.s
around
any
loop
in
a
Nuclear fusion
A
process
The
rate
at
which
energy
is
whereby
converted.
circuit
the
is
p.d.s
equal
to
around
the
the
algebraic
sum
of
light
nuclei
combining
loop.
become
with
more
other
stable
light
by
nuclei
Q
to form
L
law
The
induced
e.m.f.
acts
in
such
a
direction
effects
to
oppose
by
the
nucleus,
release
protons
and
of
energy.
Quantised
can
neutrons
only
Refers
take
to
energy
discrete
levels
that
values.
the
the
nucleus
of
an
Quantum
atom.
A
packet
of
energy.
to
Nucleus
produce
The
stable
(or
inside
current)
heavier
accompanied
Nucleons
Lenz’s
a
The
central
part
of
an
atom.
change
R
causing
it.
O
Line
absorption
continuous
dark
Line
of
spectrum
bright
Radioactive decay
A
spectrum
crossed
by
emission
spectrum
bright
lines
A
on
series
a
dark
of
spectrum
discrete
184
of
1
ohm
A
spectrum
lines.
that
consists
potential
a
(Ω)
the
through
law
The
the
is
resistance
which
when
difference
conductor
to
is
ampere flows
Ohm’s
background.
Line
1
conductor
lines.
discrete
Ohm
of
1
a
potential
there
volt
is
stable
across
it.
through
across
random
unstable
a
proportional
difference
and
a
current
current flowing
directly
of
it
by
nucleus
of
nucleus
spontaneous
whereby
attempts
disintegrating
and
emitting
the following:
particles,
The
process
gamma
into
any
alpha
rays.
to
an
become
another
one
or
particles,
more
beta
Glossary
Rectification
The
alternating
direct
process
current
is
by
which
converted
an
into
Slip
a
Several flip-flops
in
connected
of
an
ratio
The
electric
of
d.c.
an
together.
Resistance
the
A
opposition
current.
potential
to
the flow
Resistance
difference
is
the
(V)
device
used
in
a.c.
Transformer
motors
the
generators.
Split-ring
current.
Register
ring
and
commutator
motors
electric
or
A
device
generators
current
to flow
to
in
used
allow
one
the
conductor
to
the
current
(I)
device
of
an
used
to
alternating
change
power
supply.
Truth table
A
table
logic function
of
representing
a
digital
the
circuit.
direction.
Stopping
potential
difference
that
current
reduce
The
causes
U
potential
a
photoelectric
Unified
across
A
voltage
to
to
atomic
mass
unit
(u)
zero.
1/12
the
12
mass
of
the
carbon
atom
C.
6
flowing
through
it.
Synchronous
digital
RA
Resistivity
ρ
circuits
circuit
in
A
which
sequential
a
clock
input
is
V
=
used
l
to
drive
all
the
circuit
operations.
Virtual
Responding variable
A
variable
terminal
T
changes
when
some
action
is
earth
taken
of
experiment.
Temperature
coefficient
The
When
a
cell
is
change
of
a
physical
property
across
a
p-n
junction
such
the
temperature
is
changed
by
1
to
the
the
positive
n-type
negative
terminal
material
terminal
is
is
and
connected
the
Terminal
potential difference
potential
connected
to
1
p-type
the
material.
even
volt
Tesla
1
difference
tesla
(T)
is
across
the
a
flux
square
(value of
an
acts
density
on
a
if
of
(V)
1
joule
of
direct
average
the
current)
current
which
power
alternating
to
a
That
steady
delivers
resistive
the
same
load
as
carrying
a
of
1
newton
length
1
metre,
current
perpendicular
Thermionic
to
of
the
emission
1
coulomb
1
which
electrons
are
ampere
The
process
surface
by
two
The
emitted from
output
energy
of
an
frequency
whereby
amplifier
possible
heating
The
of
the
value
is
at
electromagnetic
its
(limited
by
electrons
to
be
power
supply
1
A
of
converted
charge flows
An
instrument
potential
watt
between
used
to
difference.
(W)
of
1
is
a
joule
rate
per
of
conversion
second.
and
matter
have
The
idea
both
that
wave
and
incident
radiation
properties.
required for
emitted from
a
1
weber
(Wb)
is
the
magnetic
metal
when
a flux
density
of
1
tesla
surface.
voltage).
material
when
when
it.
passes
Semiconductor
difference
circuit
minimum
flux
the
is
Wave–particle duality
condition
a
points.
energy
Weber
maximum
in
a
particle
the
potential
W
light
Saturated
the
points
by
S
Threshold frequency
not
placed
of
metal
is
magnetic field.
Watt
current.
is
earth.
cell.
measure
alternating
it
to
magnetic
a force
wire
though
connected
two
Voltmeter
mean
a
zero
The
the
Root
at
kelvin.
between
that
is
when
Volt
connected
whereby
op-amp
relative
physically
Reverse-biased
concept
in
potential,
an
A
an
that
that
Time
is
constant
The
time
taken for
perpendicularly
through
an
area
the
2
of
neither
a
good
conductor
nor
a
good
charge
on
a
capacitor
to fall
to
1 m
.
1/e
Work function
(0.368)
insulator.
of
its
initial
needed
Sequential
circuits
Digital
Timing diagram
circuits
A
diagram
used
which
the
determined
previous
output
by
the
outputs.
of
the
current
circuit
inputs
is
and
illustrate
inputs
the
and
logic
states
outputs
over
of
a
to free
an
minimum
energy
electron from
a
to
metal
in
The
value.
surface.
various
period
of
time.
185
Index
Headings
in
bold
indicate
glossary
terms.
circuits
6–7
,
and
12–23
in
capacitors
38–43
Kirchhoff ’s first
photoelectric
law
14,
current
15
124
A
digital
100–9
op-amp
absorption
spectra
132,
clipped
a.c.
78–81,
cloud
and
generators
64–5,
160,
agricultural
161,
signals
99
chambers
49,
51,
155,
60,
62,
154,
in
motors
in
transformers
decay
156,
Geiger–Marsden
157
,
64–5,
experiment
gates
ionising
67
electric
potential
capacitors
circuits
100
electrical
90–1
3,
100,
number
139
42–3
energy
7
,
12,
79,
125
induction
6
electromagnetic
radiation
60–9
144–5
48–51,
3,
8,
56–7
particulate
82–3
see also
nature
gamma
120–7
,
rays;
136
light;
circuits
106–7
146,
ultraviolet; X-rays
100
147
,
176
electromagnets
58–9,
89
160
current
6,
52
electromotive force
(e.m.f.)
12–13,
142–6
energy
levels
132–5,
139,
6
22–3
145
Coulomb’s
attenuation
of X-rays
B
background
law
29
radiation
109
in
current
see
current
balance
electric
90,
see
electronics
electrons
conductors
wavelength
e.m.f.
14–15
88–109
2–3,
48–51,
56–7
in
atoms
10
3,
122,
138–40
as
beta
132–5,
particles
144–5,
as
cathode
156,
146
160
83
160
rays
142
D
diffraction
154,
156,
157
,
158,
of
136–7
160
in fields
d.c.
counters
64,
65,
67
,
33
84–5
109
free
circuits
logic
law
93
potential
particles
induced
Kirchhoff ’s
155
cut-off
becquerel
current
55
current-carrying
bandwidth
induced
139–40
counters
see free
electrons
14–23
100
in
decay
binary
energy
102
circuits
coulomb
binary
157
30–1
88–99
conventional
binary
7
,
store
electromagnetic
controlled variable
beta
radiation
potential
142–4
control
barrier
30–5
53
6
electronics
asynchronous
3,
and
electric
80–1
semiconductors
atoms
156–7
68–9
current-carrying
atomic
28,
28–35,
158–9
conductors
AND
8,
160
comparators
analogue
7
,
4
combinational
alpha
ammeter
electric fields
162–3
spraying
particles
in
10–11
strength
66–7
coils
alpha
resistance
electric field
84–5
motors
activity
and
98–9
133
numbers,
adding
see
radioactive
the
photoelectric
effect
120–3,
decay
110
124
decay
binding
energy
constant
160,
161–3
148–51
in
delocalised
binding
energy
per
nucleon
electrons
semiconductors
148–9
in X-ray
depletion
bistable
element
region/layer
production
detectors
voltage
of
radiation
122
154–5
83
electrostatics
dielectric
36,
diffraction
of
electrons
see
36–7
,
38–41,
electronics
with
85
binding
amplifiers
96
11,
83,
6,
7
,
10,
charge
9
of
capacitors
and
electric fields
36,
28,
reactions
levels
29
E
130–1
currents
of
of
subatomic
66–7
,
carriers
8,
particles
8–9,
38,
34–5,
see also
electrons
nuclear
in
the
energy
current
energy
currents
energy
levels
9,
caused
147
,
effect
42–3,
121,
7
,
132–5,
81,
133,
145
63,
by
line
30
48–9
EXNOR
characteristics
10–11,
60,
61,
gates
101,
102
83
EXOR
current
139
81
62–3,
65,
66
gates
101,
102
148–51
121–2
139
133
63,
spectra
equipotential
induced
146,
83
52–3
I–V
of
level diagrams
and X-ray
fields
186
reactions
photoelectric
83
energy
current
eddy
33,
energy
levels
138
39
138
in fields
149–51
158
84–5
circuits
82–3
6,
148,
156,
52
146
drift
charged
78–81,
capacitor
diffusion
charge
kinetic
energy
in
thermal
145,
mass
81
6–7
,
131
particles
with
quantisation
in
14
125
radiations
see
see
potential
current
30–1
a.c.
quantisation
63,
147
electric
2,
12,
79,
4
38–41
measuring
charges
12,
8–9
extraction
eddy
nuclear
point
ionising
kinetic
and
148–51
9,
83
53
dust
7
,
82
current
drift velocity
elementary,
7
,
equivalence
52,
6–7
,
6,
88–9
12–13
centripetal force
in
(LEDs)
circuits
electrical
drift
2–3,
energy
84–5
in
doping
cells
133
100–9
42
light-emitting
cascaded
132,
energy
diodes
rectification
spectra
83
36–43
digital
capacitors
current
electromotive force
136–7
emission
diffusion
2–5
37
e.m.f.
C
capacitance
138–40
83
108
electronvolt
breakdown
82–3
3
134
Index
induction,
F
charging
insulators
farad
disc
Faraday’s
law
92,
62,
139–40,
63
photoelectric
materials
48,
hand
right
resistance
rule
hand
semiconductors
NOT
48–9,
square
law
61
radiation
see also
34–5,
alpha
particles;
gamma
147
,
100,
nuclei
K
148
147
,
of
148,
149
146–51
146,
154,
number
nucleons
electrons
120–1,
122,
148
160–2
146,
147
,
148–9,
160
nuclides
production
148
125
83
8,
149
energy
in X-ray
6,
144–5,
nucleon
of
11,
148,
148
reactions
decay
electromotive force
forward-biased
102
146–7
,
168–9
conductors
kinetic
160
102–3
rays
nuclear
56–7
nuclear forces
93
8–9
isotopes
3,
gates
nuclear fusion
6,
92,
147
,
154–9
50–3
current-carrying
electrons
101,
nuclear force
ions
see also
gates
nuclear fission
magnetic flux
between
146,
158
108–9
28–33,
145,
82
50
rule
103
12–13
NOR
ionising
see
102,
2
124
neutrons
inverse
Fleming’s
charge
120–1,
58
intrinsic
flip-flops
effect
nuclear fission
left
82–3
101,
negative feedback
122,
Fleming’s
gates
negative
the
internal
see
material
100
158
93
ferromagnetic
fission
90,
NAND
in
feedback
circuits
68
intensity
free
N
n-type
integrated
forces
3
36
Faraday’s
flux
by
3
138,
146
139–40
10
fission
releases
146
O
frequency
Kirchhoff ’s first
of
a.c.
law
14,
15
78
ohm
Kirchhoff ’s
and
op-amps
90,
93,
second
law
10
14–15
97
ohmic
in
the
photoelectric
122,
effect
120–1,
L
law
spectra
leakage
133
current
drop
experiment
full-adders
left
111
hand
rule
gain
Lenz’s
nuclear fusion
law
62–3
and
line
spectra
gates
resistors
(LDRs)
voltage
generators
light-emitting
of
cascaded
of
summing
amplifiers
open-loop
rays
154,
(LEDs)
2,
line
absorption
line
emission
156–8,
line
spectra
spectra
spectra
132,
132,
133
P
133
132–5,
139,
junctions
material
gates
parallel
components
parallel
plates
176
plotting
M
particles
magnetic fields
33
in
H
electromagnetic
and
ionising
induction
radiation
in
transformers
nature
161,
period
162–7
51,
58,
81
54–5
permeability of free
electromagnets
electricity
linkage
62,
permittivity of free
poles
48,
photocopiers
48,
mass
148,
V
characteristics
photoelectric
current
photoelectric
effect
atoms
nuclear
reactions
147
,
121–2,
132–5
148
Planck
140–1,
subatomic
particles
122,
op-amp
charges
2,
30–1
90
mass defect
148–9,
150–1
positive
current
60,
61,
charge
2
62–3
mass
number
146,
147
,
148–9,
160
positive feedback
generators
produce
65,
corkscrew
rule
48
potential
61,
62,
63,
92
66
Maxwell’s
e.m.f.
125
145
point
impedance,
constant
169
of
induced
140
146
photons
induced
136
149–51
83
in
imaging
124
120–7
,
10–11
of
diodes
37
176
photography, X-ray
I
29,
4–5
68–9
manipulated variable
I
space
62–3
82–3
magnets
51
37
63
5
magnetic
space
58
permittivity
magnetic flux
holes
50,
54
static
78
permeability
in
120–7
80–1
magnetic flux density
of
radiation
78
61–3
54–5
Hall voltage
particles
of
157
110
48,
charged
60–3
peak value
magnetic flux
144–5
48–59
particulate
hazards
42
32–5
136–7
atoms
see also
half-adders
28,
176–9
gravitational fields
effect
16,
100–9
in
Hall
82–3
177
154
65–6
gradient/slope
half-life
82–3
145
142–4
logic
graphs,
97
160
logarithms
generators
94–5,
gain
experiment
tube
66
93,
5
p-type
Geiger–Muller
91,
94–5
p-n
Geiger–Marsden
65,
88–9
op-amps
lightning
see also
diodes
96
amplifiers
90–9
102
88
92–3
gamma
amplifiers
101,
132–5
output
light-dependent
gain
97
136
OR
G
91,
gain
operational
light
90,
50
see also
see
130–1
83
open-loop
fusion
11
10–11
125
oil
and
conductors
Ohm’s
see
electric
motors
64–5,
67
potential difference
generators
produce
65,
motors
in
transformers
7
,
9,
12
66
barrier
in
potential
68–9
potential
83
67
80
187
Index
and
capacitors
and
electric field
in
Kirchhoff ’s
36–7
,
38–41
strength
second
law
reverse-biased
32–3
right
14–15
root
hand
rule
mean
11,
83
48–9,
square
61
transmission
of
truth tables
100–1
electrical
energy
79
78
U
and
potential
dividers
20–1
S
potentiometers for
22–3
ultraviolet
and
resistance
salts
stopping
radiation
potential
6,
8–9
unified
atomic
mass
saturated output
91,
97
V
search
dividers
potential
energy
coils
69
20–1
semiconductor
diodes
11,
83
variables
9,
42–3,
diodes
virtual
gradient
176
139
see also
potential
124–5
149
124–5
see also voltage
potential
120,
10–11
earth
92
32
semiconductors
3,
8,
82–3
voltage
potentiometers
22–3
and
Hall
effect
55
and
power
9,
78–9,
a.c.
78,
sequential
circuits
100,
108–11
breakdown
proton
number
146,
147
,
voltage
components
16,
42
comparing,
145,
146,
electron
3,
6,
slip
rings
64,
65
solenoids
49,
51,
digital
Hall
Q
131,
in
133
electromagnets
V
voltage
90–1
and
induced
counters
10–11,
83
58–9
current
see
output
voltage
62–3
rectification
spark
100
54
characteristics
output
121
electronics
69
I
quantum
op-amps
145
in
121,
with
160
shells,
quantisation
83
160
series
protons
79
81
of
84–5
155
transformers
change
80–1
R
spectra
132–5,
138–9,
145
see also
split-ring
radiation
see
background
64,
radiation;
voltage followers
potential
voltmeter
amplifiers
96
124–5
ionising
summing
difference
65
radiation;
stopping
electromagnetic
commutators
potential
94–5,
7
97
radiation
symbols,
radioactive decay
154,
circuit
and fission
W
88–9
160–7
synchronous
radioactivity
9,
circuits
100
146
watt
9
wave
theory
154–69
T
radioisotopes
121,
136
168–9
wave-particle duality
radiotherapy
141
temperature
coefficient
11,
wavelength
rectification
84–5
terminal
potential difference
109
tesla
59,
89
thermal
10–11,
16–19
wavelength
energy
7
,
63,
81,
particles
thermionic
emission
61
138
Wheatstone
internal
resistance
12–13
thermistors
11,
of
22,
23
bridge
threshold frequency
120–1,
done
7
,
30,
thermal
resistivity
11,
resistors
9,
7
,
energy
losses
81
82
time
constant
39–40,
timing diagrams
16–19
transducers
99,
capacitor
circuits
38–41
transfer
light-dependent
(LDRs)
188
176
88
X
104–5
characteristics
transformers
transistors
122,
88–9
spectra
83
79,
80–1
138–9
97
X-rays
responding variable
121,
85
X-ray
in
42–3
122
work function
and
22
88
work
measurements
122,
136–7
138
weber
resistance
134–5
50
of
relays
132–3,
12
cut-off
registers
136–7
88
138–41
125
138–40
Physics
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