Plug
Earth, Green and Yellow
Live, Brown
Socket switch
Neutral,
Blue
Flex
Appliance (Electric fire)
Power
The power of the appliance can be calculated using the equations:-
P =
P = IV
E t
P = Power in watts (W),
E = Energy in joules (J), t = Time in seconds (s).
P = Power in watts (W),
I = Current in amperes, amps, (A),
V = Voltage in volts (V).

• The switch and fuse should be on the live wire. If the switch or fuse is on the neutral wire it could be possible to have an appliance which is off, but live to touch.
• The fuse protects the flex from going on fire.
• The size of the fuse should be just above the current drawn by the appliance.
• The flex rating should be above the value of the fuse.

• Mains voltage is 230 V.
• Mains frequency is 50 Hz
Example:-
Calculate:- (a) the current drawn by a 2000 watt electric kettle,
(b) suggest the size of a suitable fuse,
(c) suggest a suitable flex rating.
Answers:- (a) I = P/V = 2000/230 = 8.7
A,
(b) Suitable fuse = 10 A, (just above 8.7
A),
(c) Suitable flex = 13 A, (just above 10 A).

• The earth wire is attached to the frame of the appliance.
• If the frame becomes live:-
A large current will flow,
The fuse will blow leaving the frame safe to touch.
• Some appliances have frames made from non-conduction materials.
• They do not require an earth wire.
• They are said to be double insulated.
• The symbol for double insulation is:-

• Nobody knows exactly what charge is.
• Some particles carry charge, some do not.
• The unit of charge is the coulomb (C) .
• The symbol for charge used in equations is usually Q .
Electron
Proton
Neutron
• A neutron carries no charge.
• A proton carries a positive charge of 1.6x10 -19 coulombs.
• An electron carries a negative charge of 1.6x10 -19 coulombs.

How many electrons would be needed to carry a total charge of 1 coulomb?

How many electrons would be needed to carry a total charge of 1 coulomb?
Number of electrons =
Total charge
Charge on each electron
=
1 coulomb
1.6x10-19 coulombs
= 6.25x1018 electrons.

• A current is a flow of electrical charge.
• When electrons move along a wire they carry charge, and a current is said to be flowing.
• If one coulomb of charge passes a point, in a circuit, in one second, then the current is said to be one coulomb per second or one ampere, amp (A).
How many coulombs of charge pass a point per second if 6 coulombs of charge pass in 2 seconds?

Number of coulombs per second =
Total charge
Total time
= 6 coulombs
2 seconds
= 3

Current (I) can be calculated by using the equation:-
I =
Q t
I = Current in amperes, amps (A),
Q = Charge in coulombs (C), t = Time in seconds.

• Potential difference (p.d.) or voltage (v) is a measure of the amount of energy that is needed to move one coulomb of charge from one point to another.
• If one joule of energy is needed to move one coulomb of charge between two points, then the potential difference is said to be one joule per coulomb or volt (V).
How many joules per coulombs are needed if it takes 12 joules of energy to move 4 coulomb of charge between two points in a circuit?

Number of joules per coulomb =
=
Total energy
Charge moved
12 joules
4 coulombs
= 3

Potential difference or voltage be calculated by using the equation:-
V =
E
Q
V = p.d. or voltage in volts (V),
E = Energy in joules (J),
Q = Charge in coulombs (C).

• d.c. stands for direct current.
• Direct current flows in one direction only.
current positive direction time

• a.c. stands for alternating current.
• Alternating current flows in both the positive and negative directions.
current positive time negative

Current is measured using an ammeter .
The ammeter is connected in series with the component.
A good ammeter has a low resistance.

Measuring Potential Difference or Voltage
Voltage is measured using a voltmeter.
The voltmeter is connected in parallel with the component.
A good voltmeter has a very high resistance.

Replace Q with It
Replace E with ItV
Q = It
E = QV
E = ItV
P =
E t
P =
ItV t
Cancel ‘t’
P = IV

Ohm’s Law
Replace V with IR
P = I2R
V = IR
Replace I with V/R
P =
V2
R

AIM :
APPARATUS :
To find the relationship between the current through a resistor and the potential difference across it.
Variable voltage supply (Labpack)
A
V
METHOD :The current through the resistor was measured using an ammeter.
The potential difference was measured using a voltmeter.
The potential difference across the resistor was changed by altering the labpack setting.

RESULTS :
Table :
Potential difference in volts
0 . 0
9 . 0
1 0 . 0
1 1 . 0
1 2 . 0
1 3 . 0
1 4 . 0
1 . 0
2 . 0
3 . 0
4 . 0
5 . 0
6 . 0
7 . 0
8 . 0
Current in milliamps
1 0 . 0
1 2 . 0
1 4 . 0
1 6 . 0
1 8 . 0
2 0 . 0
2 2 . 0
2 4 . 0
2 6 . 0
2 8 . 0
0 . 0
2 . 0
4 . 0
6 . 0
8 . 0

:
Current in milliamps
0
Potential difference in volts
Conclusion : Since we have a straight line graph through the origin we can conclude that the current through a resistor is directly proportional to the potential difference across it.

α
α
=
=
=
From graph.
Reverse is true.
Replace proportion sign with equals and constant (k).
Replace k with the more usual constant R (called resistance).
Reverse Multiplication order.
This equation is called Ohm’s Law.

V = IR
V = potential difference (p.d.) or voltage in volts (V),
I = Current in amps (A),
R = Resistance in ohms (
Ω
).
Example: When a 1.5 volt cell is connected across a resistor a current of 12 milliamps flows. Calculate the resistance of the resistor.
Answer:
R =
V
I
R =
1.5
12 x 10 -3
R = 125
Ω

Used as a rheostat
+ supply
A
A
C
B
If contact C is moved towards A:-
• the total resistance of the circuit is reduced ,
• the current flowing in the circuit increases ,
• the bulb becomes brighter .
If contact C is moved towards B:-
• the total resistance of the circuit is increased ,
• the current flowing in the circuit decreases ,
• the bulb becomes dimmer .

Used as a potential divider supply
A
C
B
If contact C is moved towards A:-
• the potential difference across the bulb is reduced ,
• the bulb becomes dimmer .
If contact C is moved towards B:-
• the potential difference across the bulb is increased ,
• the bulb becomes brighter .

V
S
I
S
I
3
R
3
V
3
I
2
R
2
V
2
I
1
R
1
V
1

• The current is the same at all points in a series circuit. i.e.
‘s’ ....supply
• The sum of the potential differences across each resistor in a series circuit is equal to the supply voltage. i.e.
• The total resistance of a number of resistors connected in series is equal to the sum of the individual resistors . i.e.
‘s’ ....series

V
S
R
1
V
1
R
2
V
2
R
3
V
3
I
S
I
1
I
I
2
3

• The sum of the currents through each resistor is equal to the supply current. i.e.
‘s’ ....supply
• The potential difference across each resistor is equal to the supply voltage. i.e.
‘s’ ....supply
• The total resistance of a number of resistors connected in parallel can be calculated using the equation:-
=
+
+
‘P’ ....Parallel

X
Resistor’s in Parallel Equation
12 Ω
3 Ω
Y
Find the resistance between X and Y.

1
RP
1
=
R1
1
+
R2
Method 1
Parallel equation
1
RP
1
=
12
1
+
3
Substitute
1
RP
1
RP
1
=
12
5
=
12
4
+
12
RP
1
= 12
5
RP = 2.4 Ω
Find a common factor
Add fractions
Turn both sides up-side-down
Remember your units

1
RP
1
=
R1
1
+
R2
1
RP
1
=
12
1
+
3
1
RP
= 0.4167
RP =
1
0.4167
RP = 2.4 Ω
Method 2
Parallel equation
Substitute
Add the fractions on the calculator
To turn both sides up-side-down, use the 1/x or x -1 key on your calculator.
Remember your units

Method 3
1
RP
1
=
R1
1
+
R2
RP =
R1 x R2
R1 + R2
Parallel equation
This rearrangement is only true for two resistors.
RP =
12 x 3
12 + 3
RP = 2.4 Ω
Substitute
Remember your units

P=IV, V=IR, Q=It, P=I2R, P=V2/R
5 W = 5 joules per second
Side light
12 V, 5 W
Head light
12 V, 55 W
Head light
S
2
Side light
S
1
12 V
Car battery
Side light
Points
• The frame of the car acts as the return wire to the battery.
• Closing switch S1 will make the side lights light.
• Closing S1 and S2 will make the sidelights and head lights come on.
• It is impossible to have the headlights on and the side lights off.
• All bulbs are connected in parallel and have 12 V across them when on.
Side light

Wire
Wire
Ω
Low Resistance
(close to 0
Ω
)
Broken wire (open circuit)
Ω
High Resistance
(over 1 M
Ω
)

Bulb
Bulb
Ω
Low Resistance
(few ohms)
Blown bulb
Ω
High Resistance
(over 1 M
Ω
)

Bulb Holder
Bulb holder (no bulb)
Ω
High Resistance
(over 1 M
Ω
)
Short circuit
Ω
Low Resistance
(close to 0
Ω
)

From transformer
and power station
The earth wire is connected to a water pipe or a metal stake in the ground.
100 amp fuse.
Energy meter
1235435
J W
E = P x t s kWh kW h
1 kWh = 1 unit = 3,600,000 J
1 kWh or unit costs about 7 pence.

100 amp fuse.
Energy meter
1235435
J W
E = P x t kWh kW h s
1 kWh = 1 unit = 3,600,000 J
1 kWh or unit costs about 7 pence.
Double switch (for extra safety)
Consumer unit containing fuses or circuit breakers.

Consumer unit

Consumer unit
A
C
B
Two way switch
The lamp can be switched on and off at the top and bottom of the stairs.
D
Two way switch

The socket switch and the plug fuse should be on the live wire
The current can arrive at each socket from two directions. This means that the mains ring can be made of thinner cheaper wire.

Circuit Diagram
+
-
A x
Main points
• The MCB will trip when too much current goes through it.
• A MCB can be used instead of a mains fuse.
• A MCB can easily be reset when the fault has been fixed.

Circuit Diagram
Main points
• When a current passes through a wire a magnetic field forms around the wire.
• When a current passes through a coil of wire the magnetic field resembles that of a bar magnet .
• A core made from soft iron makes this field much stronger .
• An electromagnet can be switched on and off, but a permanent magnet cannot.

Circuit Diagram
Contact
Battery
Gong
Spring
Electromagnets
Hammer
Main points
• When a current flows through the electromagnet the hammer is attracted causing it to hit the gong.
• The current through the electromagnets stops flowing when the contact is broken.
• A spring moves the hammer back to its original position, and the whole process starts to repeat.

Circuit Diagram
S
Lamp
9 V
Relay
Circuit 2
12 V ac
Circuit 1

Main points
• When switch S is closed a current flows in circuit 1.
• This current allows the electromagnet to close the switch in circuit 2.
• Advantages
A low voltage circuit can be used to switch on a high voltage one (safer).
The low voltage circuit can be made from thinner cheaper wire.
The switching can take place at large distances (e.g. old telephone relay circuits)

Reed Switch
Alarm
N
Magnet
S
When the magnet is close to the reed switch it is closed, and the alarm will be on.
When the magnet is moved away from the reed switch the switch will open and the alarm will be off.

Force on a current carrying wire in a magnetic field.
When a current passes through a wire in a magnetic field it experiences a force (F).
The size of this force will depend on the size of the current and the strength of the magnetic field.
The direction of this force will depend on the direction of the current and the direction of the magnetic field.

F
F
S
(1) F
= Current towards
= Current away
N S N
(2)
F
The forces on the current carrying wire causes the armature to turn.

F F
S N
(3)
F
The current reverses direction every half turn to keep it rotating in the same direction.
S
(4)
F
N

shaft brushes commutator brushes

Brushes: Passes the current to the comutator . Brushes are made from soft carbon. Soft carbon conducts electricity and is soft enough to not damage the commutator, but hard enough not to wear away.
Brushes in washing machine motors usually last for about 5 years.
Armature: Turns when a current passes through it.
The armature has many loops of wire, rather than a single loop. To give the maximum turning effect. The current always flows through the loop perpendicular to the field coils.
Commutator: This is where the current enters and leaves the loops of wire in the armature. By making a sliding contact it allows the current to reverse direction every half turn.
Field coils: Makes magnetic fields . These are curved electromagnets rather than flat permanent ones.