A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
SECTION V
Electricity and Magnetism
CIE A-Level [AS and A2]
________________________
Course Notes
DIPONT Educational Resource - Science
1
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Syllabus Details______________________
DIPONT Educational Resource - Science
2
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
17. Electric Fields [AS &A2]__________________
Content
17.1 Concept of an electric field
17.2 Uniform electric fields
17.3 Force between point charges
17.4 Electric field of a point charge
17.5 Electric potential
[AS]
[AS]
[A2]
[A2]
[A2]
Learning outcomes_____________________________________
Candidates should be able to:
(a) show an understanding of the concept of an electric field as an example of a field
of force and define electric field strength as force per unit positive charge acting
on a stationary point charge
Electric Field
Force
E=F/q
Electric field =
Force per unit test charge
From Coulombs law
E = q1
q2
Force
q2
4 0r
2
q1




The electric field produced by a charge (q1) is defined in terms of a test
charge (q2)
This test charge will experience a force dependent only on the test
charge
The test charge must be so small as to not disturb the charge being
considered
The test charge is positive
DIPONT Educational Resource - Science
3
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(b) represent an electric field by means of field lines
Electric Fields
Charged Sphere
Parallel Plates
+ + + + + + + + +
+
+
------------
Point Charge
+
-
+
+
Two Point Charges
(c) recall and use E = V / d to calculate the field strength of the uniform field
between charged parallel plates in terms of potential difference and separation
Energy Difference in an Electric Field
+
+
+
+
+
+
+
+
+
+
Lower electrical
potential energy
Higher electrical
potential energy
Force
+ q
+ q
d
-
Change in electrical potential energy from A to B = force x distance
Potential difference = electrical potential energy / charge
Potential difference x charge = force x distance
Electric field strength = force / charge = potential difference / distance
E=V/d
DIPONT Educational Resource - Science
4
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(d) calculate the forces on charges in uniform electric fields
(e) describe the effect of a uniform electric field on the motion of charged particles
(f) recall and use Coulomb’s law in the form F = Q1Q2/4πε0r 2 for the force between
two point charges in free space or air
Coulombs Law
q1
force
q2
force
r
F = kq1q 2
r
OR
2
F = q1 q 2
2

4 0r
 0 = permittivity of free space
q = Charge
r = Distance between charges
k = coulombs constant
q = Charge
r = Distance between charges
(g) recall and use E = Q/ 4πε0r 2 for the field strength of a point charge in free space
or air
Electric Field
Force
E=F/q
Electric field =
Force per unit test charge
From Coulombs law
E = q1
q2
Force
q2
4 0r
2
q1
DIPONT Educational Resource - Science
5
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(h) define potential at a point in terms of the work done in bringing unit positive
charge from infinity to the point
zero potential
at infinity
potential increase
F2
+Q
F1
q
q
Force on q increases
Electric Potential Energy of charge q =
Qq
40r
Electric potential energy: work done in moving a charge from infinity to a point



Zero of potential energy is at infinity
Potential energy taken as a negative value
The work done in moving a charge between two points in an electril field
is independent of the path taken
Electric potential: Energy per unit test charge.
(i) state that the field strength of the field at a point is equal to the negative of
potential gradient at that point
Electric potential
Velec =
Q
40r
Gradient of this line = E
r
DIPONT Educational Resource - Science
6
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
F
E=
=
q
Q
40r 2
E = V
r
Electric Field = - potential gradient
(j) use the equation V =Q/4πε0r for the potential in the field of a point charge
Electric potential: Energy per unit test charge.
Velec =
V =
Velec =
work done
test charge
E
q
Q
40r
V = electric potential (JC-1)
Q = charge producing field
This formula is given at the start of the test paper
DIPONT Educational Resource - Science
7
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(k) recognise the analogy between certain qualitative and quantitative aspects of
electric fields and gravitational fields.
potential energy decreases as
gravitational force does work
zero potential energy
at infinity
F1
F2
M
m
m
Force on m increases
Gravitational potential energy of mass m= -
GMm
r
Gravitational potential energy: work done in moving a mass from infinity to a
point



Zero of potential energy is at infinity
Potential energy taken as a negative value
The work done in moving a mass between two points in a gravitational
field is independent of the path taken
Gravitational potential: Energy per unit test mass.
18. Capacitance [A2]_________________________
Content
18.1 Capacitors and capacitance
18.2 Energy stored in a capacitor
Learning outcomes_____________________________________
Candidates should be able to:
(a) show an understanding of the function of capacitors in simple circuits
Function of capacitors:
 Storing charge for rapid discharge (e.g. in a camera flash)
DIPONT Educational Resource - Science
8
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(b) define capacitance and the farad
(c) recall and solve problems using C= Q/V
Capacitance: The charged stored (in coulombs) per volt of potential difference
Farad (F): Unit of capacitance. 1 coulomb per volt.
C= Q
q/V
(d) derive, using the formula C= Q/V , conservation of charge and the addition of
p.d.s, formulae for capacitors in series and in parallel
+q
-q +q
C1
V
V2
V1
-q
=
+q
-q
C
C2
V = V1 + V2
V1 = q / C1
and V2 = q / C2
(q = charge induced on one plate)
q / C = q / C1 + q / C2
1 / C = 1 / C1 + 1 / C2
1/Ctotal = 1/C1 + 1/C2
DIPONT Educational Resource - Science
9
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
V
V
+q1
-q1
C1
=
+q
-q
C
+q2
-q2
C2
q = q1 + q2
q1 = C1V
q2 = C2V
CV = C1V + C2V
C = C1 + C2
Ctotal = C1 + C2
These formula are given at the start of the test paper
(e) solve problems using formulae for capacitors in series and in parallel
DIPONT Educational Resource - Science
10
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(f) deduce, from the area under a potential-charge graph, the equation W=1/2 QV
and hence W = ½ CV2.
Potential Difference /V
V0
q
Charge /C
Q = CV
W= Ep = VQ
Ep = V0q
W = Ep = ½ QV
W = Ep = ½ CV2
This formula is given at the start of the test paper
DIPONT Educational Resource - Science
11
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
19. Current of Electricity [AS]_________________
Content
19.1 Electric current
19.2 Potential difference
19.3 Resistance and resistivity
19.4 Sources of electromotive force
Learning outcomes_____________________________________
Candidates should be able to:
(a) show an understanding that electric current is the flow of charged particles
Electrical Conduction
Conventional Current
Metal Wire
Metal Ions
Electrons
Drift Velocity
Current = Charge flowed
time
I = dQ
dt
1 amp = 1 coulomb
1 second
DIPONT Educational Resource - Science
12
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(b) define charge and the coulomb
Charge: a fundamental property of some subatomic particles, which determines their
electromagnetic interactions
The Coulomb: The charge passing a point in a circuit when there is a current of one
ampere for one second.
(c) recall and solve problems using the equation Q = It
Q = It
(d) define potential difference and the volt
(e) recall and solve problems using V = W/Q
Potential difference: Energy difference between two points per unit
charge for a charge in an electric field
p.d. = Energy difference
Charge
Units of p.d. = JC-1 = Volt (V)
V = W/Q
(f) recall and solve problems using P = VI, P = I 2R
Potential difference = energy difference / charged flowed
Current = charged flowed / time taken
Potential difference x current = (energy difference/charged flowed) x (charged
flowed/time taken)
Potential difference x current = energy difference / time taken = Power dissipated
P = VI
DIPONT Educational Resource - Science
P = V2
R
13
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(g) define resistance and the ohm
(h) recall and solve problems using V = IR
Resistance: The ratio of the potential difference across an object to the current flowing
through it.
Ohm: The resistance when 1 Volt of potential difference gives a current of 1 Amp
Resistance )  Potential Difference (V)
Current (A)
SEE PHET SIM
R = V
I
(i) sketch and explain the I-V characteristics of a metallic conductor at constant
temperature, a semiconductor diode and a filament lamp
Ohmic and Non-ohmic behavior
Filament Lamp
Potential
Difference
Ohmic Behavior
DIPONT Educational Resource - Science
Current
Diode
Current
Current
Metal at
constant temperature
Potential
Difference
Potential
Difference
Non - ohmic Behavior
14
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Current
(j) sketch the temperature characteristic of a thermistor (thermistors will be
assumed to be of the negative temperature coefficient type)
Hot
Cold
Potential
Difference
(k) state Ohm’s law
Ohm’s Law: The current flowing through a piece of metal is proportional to
the potential difference across it providing the temperature remains constant
(l) recall and solve problems using R = L / A
R=
L
A
R = Resistance
 = Resistivity of material
L = Length of conductor
A = Area
SEE PHET SIM
DIPONT Educational Resource - Science
15
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(m) define e.m.f. in terms of the energy transferred by a source in driving unit
charge round a complete circuit
Electromotive force (e.m.f.): The total energy difference per unit charge around a
circuit
(n) distinguish between e.m.f. and p.d. in terms of energy considerations
e.m.f. = (energy converted from other forms to electrical) / charge
p.d. = (energy converted from electrical to other forms) / charge
(o) show an understanding of the effects of the internal resistance of a source of
e.m.f. on the terminal potential difference and output power.
Internal Resistance
Internal Resistance
Perfect Battery
e.m.f
ri
e.m.f
Terminals of Battery
R
External Resistance
= I x Rtotal
= I(ri + R)
= Iri + IR
IR = e.m.f. - Iri
Terminal p.d.
‘Lost’ volts
SEE PHET SIM
DIPONT Educational Resource - Science
16
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
20. DC Circuits [AS]_________________________
Content
20.1 Practical circuits
20.2 Conservation of charge and energy
20.3 Balanced potentials
Learning outcomes_____________________________________
Candidates should be able to:
(a) recall and use appropriate circuit symbols as set out in the ASE publication
Signs, Symbols and Systematics
(b) draw and interpret circuit diagrams containing sources, switches, resistors,
ammeters, voltmeters, and/or any other type of component referred to in the
syllabus
Cell
Bell
Variable
resistor
Fuse
Resistor
Transformer
Lamp
V
Voltmeter
A
Ammeter
Magnetizing Coils
Relay
Switch
(c) recall Kirchhoff’s first law and appreciate the link to conservation of charge
Kirchhoff’s First Law: The sum of the currents entering a junction is always equal to
the sum of the currents leaving it
DIPONT Educational Resource - Science
17
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(d) recall Kirchhoff’s second law and appreciate the link to conservation of energy
Kirchhoff’s Second Law: The sum of the electromotive forces in a closed circuit is
equal to the sum of the potential differences
(e) derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in series
R1
R2
V1
V2
For a series circuit…
Emf = V1 + V2
IRtotal = IR1 + IR2
Rtotal = R1 + R2
(f) solve problems using the formula for the combined resistance of two or more
resistors in series
Series Circuit
24V
Example at R1
I=2A
V = IR
V=2x3
V = 6V
R1 = 3
R2 = 4
6V
8V
R3 = 5
10V
Rtotal = R1 + R2 + R3
This formula is given at the start of the test paper
DIPONT Educational Resource - Science
18
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(g) derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in parallel
V
A
R1
A1
V1
R2
A2
V2
I = I1 + I2
V / R = V1 / R1 + V2 / R2
V = V1 = V2
1 / R = 1 / R1 + 1 / R2
(h) solve problems using the formula for the combined resistance of two or more
resistors in parallel
Parallel Circuit
24V
Itotal = 18.8 A
24V
R1 = 3
I1 = 8 A
24V
R2 = 4
I2 = 6 A
24V
R3 = 5
I2 = 4.8 A
1/Rtotal = 1/R1 + 1/R2 + 1/R3
Itotal = I1 + I2 + I3
Itotal = V/R1 + V/R2 + V/R3
This formula is given at the start of the test paper
DIPONT Educational Resource - Science
19
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(i) apply Kirchhoff’s laws to solve simple circuit problems
(j) show an understanding of the use of a potential divider circuit as a source of
variable p.d.
Potential Divider
Output voltage
A Potentiometer: A variable potential divider.
(k) explain the use of thermistors and light-dependent resistors in potential dividers
to provide a potential difference that is dependent on temperature and illumination
respectively
Light Dependent Resistor (LDR)
DIPONT Educational Resource - Science
LDR
When light shines on the
LDR the resistance
decreases
10 k
When light shines on the
LDR the p.d. across the
fixed resistor increases
20
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Negative Temperature Coefficient (NTC) Resistors
NTC
Resistors
10 k
When the temperature of
the NTC increases the
resistance decreases
When the temperature of
the NTC increases the
p.d. across the fixed
resistor increases
(l) recall and solve problems using the principle of the potentiometer as a means of
comparing potential differences.
21. Magnetic Fields [A2]______________________
Content
21.1 Concept of Magnetic Field
Learning outcomes_____________________________________
Candidates should be able to:
(a) show an understanding that a magnetic field is an example of a field of force
produced either by current-carrying conductors or by permanent magnets
(b) represent a magnetic field by field lines.
Magnetic field lines
Bar magnet
DIPONT Educational Resource - Science
21
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Magnetic field around a current carrying wire
22. Electromagnetism [A2]____________________
Content
22.1 Force on a current carrying conductor
22.2 Force on a moving charge
22.3 Magnetic fields due to currents
22.4 Force between current-carrying conductors
Learning outcomes_____________________________________
Candidates should be able to:
(a) show an appreciation that a force might act on a current-carrying conductor
placed in a magnetic field
(b) recall, and solve problems using, the equation F = BIl sinθ, with directions as
interpreted by Fleming’s left-hand rule
The Left Hand Rule
Direction
of Force
Direction
of Magnetic Field
Direction
of Current
DIPONT Educational Resource - Science
22
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
A current carrying wire in a magnetic field
Magnetic Field
Current Carrying Wire
Direction of Force
For a current carrying wire in a magnetic field…




Magnetic Field, B
Current, I
Length of conductor in magnetic field, l
Angle between field and current, 
F = BIl sin
SEE PHET SIM
(f) define magnetic flux density and the tesla
Magnetic Flux Density: a measure of the strength of a magnetic field at a given point,
expressed by the force per unit length on a conductor carrying unit current at that point.
Tesla: The uniform magnetic flux density which, acting normally to a long straight wire
carrying a current of 1 amp, causes a force per unit length of 1 Nm-1 on the conductor.
DIPONT Educational Resource - Science
23
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(d) show an understanding of how the force on a current-carrying conductor can be
used to measure the flux density of a magnetic field using a current balance
A
Magnetic field
00250.0g
(e) predict the direction of the force on a charge moving in a magnetic field
A moving charge in a magnetic field
S
Electron
N
Force
(f) recall and solve problems using F = BQv sinθ
For a moving charge in a magnetic field…




Magnetic Field, B
Charge, q
Velocity of the charge, v
Angle between field and velocity of charge, 
F = Bqv sin 
DIPONT Educational Resource - Science
24
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(g) sketch flux patterns due to a long straight wire, a flat circular coil and a long
solenoid
Solenoid
Solenoid
Magnetic Field
Magnetic field around a current carrying wire
DIPONT Educational Resource - Science
25
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Magnetic field lines
A Flat Coil
Circular current carrying coil
SEE PHET SIM
(h) show an understanding that the field due to a solenoid may be influenced by the
presence of a ferrous core
Magnetic field inside a Solenoid
DIPONT Educational Resource - Science
26
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Current carrying
conductors
Current carrying
conductors
(i) explain the forces between current-carrying conductors and predict the direction
of the forces
Force
Force
Force
Force
(j) describe and compare the forces on mass, charge and current in gravitational,
electric and magnetic fields, as appropriate.
Substance
Field
Effect
Uncharged mass
Gravitational field
Uncharged mass
Uncharged mass
Charged mass
Electric field
Magnetic field
Gravitational field
Positive charge
Electric field
Positive charge moving
perpendicular to field
Negative charge
Magnetic field
Negative charge moving
perpendicular to field
Current
Current perpendicular to
field
Magnetic field
Attracted in direction of
field line
No effect
No effect
Attracted in direction of
field line
Attracted in direction of
field line
Repelled perpendicular to
field line
Repelled in opposite
direction to field line
Repelled perpendicular to
field line
No significant effect
Repelled perpendicular to
field line
DIPONT Educational Resource - Science
Electric field
Gravitational field
Magnetic field
27
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
23. Electromagnetic Induction [A2]_____________
Content
23.1 Laws of electromagnetic induction
Learning outcomes_____________________________________
Candidates should be able to:
(a) define magnetic flux and the weber
(b) (b) recall and solve problems using Φ = BA
Magnetic Flux ( )
a
Are
A
B

 = BA cos
Weber (Wb) = 1 Tm-2
Magnetic Flux: The product of the magnetic flux density and the area normal to the lines
of flux
Weber: A magnetic flux of 1 Tesla meter-squared = 1 Tm2
(c) define magnetic flux linkage
Flux linkage = 
B
N

S
Rotating coil
Flux linkage = NBA cos
N = number of turns on of coil
DIPONT Educational Resource - Science
28
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(d) infer from appropriate experiments on electromagnetic induction:
• that a changing magnetic flux can induce an e.m.f. in a circuit
• that the direction of the induced e.m.f. opposes the change producing it
• the factors affecting the magnitude of the induced e.m.f.
(e) recall and solve problems using Faraday’s law of electromagnetic induction and
Lenz’s law
Electron moving in a metal conductor
in a magnetic field
Electron moving
in magnetic field
-
X
X
X
B
X
X
X
X
X
X
X
X
X
X
X
X
-
X
X
B
X
X
X
X
X
X
X
F = Bqv sin
Induced
Potential difference
Velocity
+
X
X
X
X
X
X
X
X
X
X
X
X
-
Electron at equilibrium
V
Wire length = l
Magnetic field = B
di f
fe
r
en
ce
=
Fm
Po
te
n
tia
l
Fe
Velocity = v
Electric force due to emf
Magnetic force due to movement
Fe = Eq = (V/l)q
Fm = Bqv
Bqv = (V/l)q
V = Blv
Induced emf = Blv (as no current is flowing V=emf)
DIPONT Educational Resource - Science
29
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Wire length = l
Magnetic field = B
In time t …. Area A = lx
velocity = v
= Blv
x
v = x/t
 = Bl(x/t)
 = BA/t
 = BA
 =  / t
Area swept out by wire
Faraday’s law: The magnitude of an induced emf is proportional to the rate of
change of flux linkage.
 = N / t
Lenz’s law: The direction of the induced emf is such that if an induced current
were able to flow, it would oppose the change which caused it.
Lenz’s law
Force from induced current opposes original motion
In
d
uc
ed
cu
r
re
nt
Magnetic field = B
Velocity = v
DIPONT Educational Resource - Science
30
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(g) explain simple applications of electromagnetic induction.
Rotation
AC Generator
Induced emf
Sinusoidal output
Time
24. Alternating Currents [A2]_________________
Content
24.1 Characteristics of alternating currents
24.2 The transformer
24.3 Transmission of electrical energy
24.4 Rectification
Learning outcomes_____________________________________
DIPONT Educational Resource - Science
31
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Candidates should be able to:
(a) show an understanding of and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage
(b) deduce that the mean power in a resistive load is half the maximum power for a
sinusoidal alternating current
p.d.
+
AC Supply
current
Time
Power
+
Mean power
Time
-
Period: The time for one complete cycle of the a.c.
Root mean squared current (or voltage) is the value of the direct current (or
voltage) that dissipates power in a resistor at the same rate.
Peak Power Pmax = I02R
Io = peak current
Energy = power x time
Energy supplied = area under the power-time graph
Over one cycle the area under the graph = area under the midway level
Mean power Pav= 1/2 I02R
(Irms)2R = 1/2 I02R
Irms = (1/√2) I0
R = Vo / Io = Vrms / Irms
DIPONT Educational Resource - Science
32
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(c) represent a sinusoidally alternating current or voltage by an equation of the
form x = x0sinωt
p.d.
V0
+
current
AC Supply
I0
Time
I = I0 sin wt
-
V = V0 sin wt
t
This formula is given at the start of the test paper
(d) distinguish between r.m.s. and peak values and recall and solve
problems using the relationship I rms = I0 /sqrt2 for the sinusoidal case
Irms = (1/√2)Io
Vrms = (1/√2)Vo
(e) show an understanding of the principle of operation of a simple laminated ironcored transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an
ideal transformer
AC Generator
Primary coil
NP turns
Secondary coil
NS turns
IP
IS
VP
VS
VP / VS = NP / NS = IS / IP
DIPONT Educational Resource - Science
33
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(f) show an appreciation of the scientific and economic advantages of alternating
current and of high voltages for the transmission of electrical energy
Power losses in Electricity Transmission….








Resistance in the power lines causes heating
Resistance in the coils of transformers causing them to heat up
Eddy currents in the iron core. These can be reduced by making the
core out of laminated sheets of metal connected with electrically
insulating layers
Flux losses caused by not all the magnetic flux produced by the
primary coil linking with the secondary coil
Electricity transmission lines have resistance, therefore, energy will be lost
through heating in the wires
Power dissipates in the wire as I2R. If a low current is used there will be less
power dissipated.
Transformers are used to step-up the voltage and so reduce the current used.
High voltages also have risks (electrocution). The supply is passed through a
step-down transformer before reaching the users.
Full-wave rectification
Potential Difference Output(V)
Potential Difference Input (V)
(h) distinguish graphically between half-wave and full-wave rectification
DIPONT Educational Resource - Science
Time
Time
34
Half-wave rectification
Time
Potential Difference Output(V)
Potential Difference Input (V)
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Time
(h) explain the use of a single diode for the half-wave rectification of an alternating
current
Input AC
Current
Half-wave rectifier
DIPONT Educational Resource - Science
Output Rectified
Current
35
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
(i) explain the use of four diodes (bridge rectifier) for the full-wave rectification of
an alternating current
Input AC
Current
Full-wave rectifier
Output Rectified
Current
(j) analyse the effect of a single capacitor in smoothing, including the effect of the
value of capacitance in relation to the load resistance.
Potential Difference Output(V)
Full-wave
rectified input
DIPONT Educational Resource - Science
Output
Time
36
A-Level Course Notes: PHYSICS
SECTION V: Electricity and Magnetism
Background Reading_________________
PHYSICS, Giancoli 6th edition, Chapter 16 - 21
Useful Websites______________________
http://phet.colorado.edu/en/simulations/category/new
http://www.s-cool.co.uk/alevel/physics.html
http://www.physicsclassroom.com/mmedia/index.cfm
http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html
http://www.colorado.edu/physics/2000/index.pl
Constants___________________________
[These are given on each test paper]
DIPONT Educational Resource - Science
37