In order to get the G.C.S.E. grade you are capable of, you must make
your own revision notes using your Physics notebook.
When summarising notes, use different colours and draw
diagrams/pictures. If you do, you will find them easier to remember.
Once you have made revision notes for a topic, re-visit these regularly
(on the day of your examination you will not remember something you
revised 4 weeks previously). Each time you re-visit a note, tick the top
of the page/card. This will allow you to identify any notes you have
neglected.
WARNING: DO NOT RELY SOLELY ON THE REVISION
POWERPOINTS!
ELECTRON
NEUTRON
Negative
No Charge
PROTON
Positive
Two positive protons and two negative electrons
Gain an electron to become negative
Lose an electron to become positive
It’s all about moving electrons
Acetate
positive
-
+
-
+
-
Polythene
Negative
Cloth
Negative
+
+
-
+
-
+
-
+
-
+
Cloth
Positive
+
-
+
Like charges repel:
Opposite charges repel:
1. Painting a Car
Positive Car
Negative
spray gun
1.
2.
The paint spreads out as each negative
drop repels
No paint is wasted as the positive car
attracts the negative paint
2. Removing particles from waste gases
Electricity
Contents
Static charge can be dangerous when a charged object discharges (loses
its extra charge).
If an object discharges through a person the person will get an electric
shock.
If a charged object discharges and produces a spark it can cause a fire.
To prevent these, objects are connected, usually to the ground, so that
they cannot become charged, e.g. a petrol tanker.
Maxol
Chain connects the tanker to the
ground, preventing the build up of
charge.
An electric current is a flow of electric charge. In a wire, an electric
current is due to the movement of electrons.
Every circuit will have a power source, and this is where the
electrons gain their energy. This energy is then taken around the
circuit and given to the different circuit components
In circuit diagrams, components are represented by the
following symbols:
cell
battery
diode
variable
resistor
switch
lamp
resistor
voltmeter
ammeter
Current
Contents
All sources have a positive terminal and a negative terminal. Electric
current flows from positive to negative, even though the electrons,
which are negatively charged, move from negative to positive
Current is measured in amps. The
current at a point in a circuit is equal
to the amount of charge (in Coulombs)
that passes that point in one second.
charge = current x time
Coulombs
amps
Where:
Q
IXt
Q =charge
I = current
T = time
seconds
+
-
current
electron
flow
x
Current is measured using an ammeter.
SERIES CIRCUIT
2A
• current is the same
at all points in the
circuit.
2A
2A
PARALLEL CIRCUIT
• current is shared
between the
components
4A
4A
2A
2A
Voltage
Contents
The potential difference (or voltage) between two points in a circuit is the
amount of electrical energy converted into other forms when one coulomb
moves between those points. (click to see the potential difference between
points c and d)
The voltage of a battery/source tells you how much electrical energy one
coulomb gains when it moves through the battery/source. (click to see the
battery voltage)
a+
- b
Voltage = 5 V  when 1
coulomb moves through the
battery it gains 5 joules of
electrical energy.
5V
5V
c
x
d
Voltage = 5 V  when 1
coulomb moves between c
and d, 5 joules of electrical
energy is converted into
heat and light.
Voltage is measured using
a voltmeter.
3V
SERIES CIRCUIT
Supply voltage is
shared between all
components.
PARALLEL CIRCUIT
The voltages across
parallel sections are
identical
1.5V
1.5V
3V
3V
3V
What are the unknown voltages and currents in the following circuits?
(1)
(2)
1A
?
5V
3A
3V
?V
3A
?A
1A
?
1V
?A
1A
1V
?A
1.5A
0.5A
?
(3)
4.5V
3A
1.5V
?V
?A
1.5A
?
3V
?A
1.5A
Resistance
Contents
To investigate how the voltage across a component affects the current through it you
should use the following circuit. LEARN THIS!
12 V
12 V
Method:
1. Set up the circuit opposite, with
the variable resistor set at its
maximum value.
2. Turn on the power supply and
record the current and voltage.
3. Change the setting on the
variable resistor and record the
new voltage and current values
Component
symbol
A
6
18 
V
Remember:
X
diode
filament
bulb
resistor
Current
Whatever happens to the gradient
the opposite happens the resistance.
Voltage
Current
Current
Voltage
Metal at constant temp.
 constant gradient
 constant resistance
Current
Voltage
Filament bulb
 decreasing gradient
 increasing resistance
Voltage
Diode
 only allows current to
flow in one direction (–ve
voltage  no current)
Anything placed in an electric circuit will offer some opposition to the
flow of current (charge). This opposition is called resistance.
A Resistor is anything which is specially constructed to have resistance.
There are two types: a fixed resistor and a variable resistor.
Fixed resistor symbol:
Conductors  low resistance
Variable resistor symbol:
Insulators  high resistance
The resistance of a conductor depends on:
1. Length  resistance is directly proportional to length – if you double
length you double resistance
2. Area  the greater the area the smaller the resistance
3. Material
1. Series:
R1
R2
R3
RT
RT = R1 + R2 + R3
2. Parallel:
R1
R2
RT
R3
1 = 1 +1 +1
RT = R1 R2 R3
Ohm’s law states that the current (amps) flowing through
a conductor is directly proportional to the potential
difference (volts) across its ends provided the
temperature remains constant.
V = IR
V
IXR
Calculate:
(a) the current leaving the battery (click for solution)
(b) the current through each component (click for solution)
(C) the voltage across each component (Click for solution)
12 V
1
(a) I = V/R
I = supply voltage / total resistance
I = 12 / (18 + 6) = 12 / 24 = 0.5 A
18 
6
(b) Series circuit  current is constant
 current = 0.5 A
(c) V = IR
18 : V = 0.5 x 18 = 9 V
6 : V = 0.5 x 6 = 3 V
Calculate:
(a) the current leaving the battery (click for solution)
(b) the current through each component (click for solution)
(C) the voltage across each component (Click for solution)
2
12 V
(a) I = V/R
I = supply voltage / total resistance
Total resistance: 1/RT = 1/R1 + 1/R2
Total resistance: 1/RT = 1/12 + 1/12
Total resistance: 1/RT = 2/12 = 1/6
12 
12 
RT = 6 
I = 12 / 6 = 2 A
(b) Parallel circuit  V across each section = 12 V
 V across each 12  resistor = 12 V
I = V/R = 12/12 = 1 A
(c) V = 12 V (see above)
Calculate:
(a) the current leaving the battery (click for solution)
(b) the current through each component (click for solution)
(C) the voltage across each component (Click for solution)
(a) I = V/R
3
10 V
I = supply voltage / total resistance
Total resistance: 1/RT = 1/R1 + 1/R2
of parallel section
Total resistance: 1/RT = 1/20 + 1/20
20 
Total resistance: 1/RT = 2/20 = 1/10
10 
RT = 10 
Total resistance: RT = 10 + 10 = 20 
I = 10 / 20 = 0.5 A
20 
(b) Current through 10  = current leaving battery
= 0.5 A
V across 10  = IR = 0.5 x 10 = 5V
 V across 20  = (10 – 5) = 5 V
Current through 20  = V/R = 5/20 = 0.25 A
(c) V across 20  = 5 V; V across 10  = 5 V (see above)
Graphs
Contents
An electrical device converts electrical energy into other forms. The amount of
energy it converts per second is the power of the device. Power is measured in
watts (W).
If the current through a device and the voltage across it are I and V
respectively, the power is given by:
P
P =VI
VXI
Since V = IR, the power equation can also be written as:
P = I 2R
P = V2/R
Questions
Contents
Lightning
Fire/damage to the building
Electrons transfer from balloon to cloth
Balloon now has a deficiency of electrons
Balloon is positively charged
Click for solutions
Same
The two balloons are repelling one another
Like charges repel
Therefore, both balloons are positively charged
V
I = V/R
I = 1/5
I = 0.2 A
Series:
Rt = R1 + R2
Rt =5 + 10 = 15
Click for solutions
V=IxR
Battery voltage = current leaving the battery x total resistance
V = 0.2 x 15
V=3V
Click for solutions
Q=Ixt
Q = 0.2 x 600
Q = 120 C
1/Rt = 1/R1 + 1/R2
1/Rt = 1/60 + 1/40
1/Rt = 1/24
Rt = 24
I = V/R
I = 24/40
I = 0.6 A
Click for solutions
It operates at a voltage of 240 volts and has a power rating of 80 watts
Click for
solutions
I =P/V
I = 80/240
I = 0.33 A
Contents