Download (v). - Rosshall Academy

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=
E
t
P = IV
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).
Switch, flex and fuse
• 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 electricity
• 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).
Earth Wire
• 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:-
Charge (Q)
• 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.
Atom
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.
Question:How many electrons would be needed to carry a total charge of 1 coulomb?
Question:How many electrons would be needed to carry a total charge of 1 coulomb?
Answer:-
Number of electrons =
=
=
Total charge
Charge on each electron
1 coulomb
1.6x10-19 coulombs
6.25x1018 electrons.
Current (I).
• 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).
Question:How many coulombs of charge pass a point per second if 6
coulombs of charge pass in 2 seconds?
Answer:-
Number of coulombs per second =
Total charge
Total time
= 6 coulombs
2 seconds
= 3
Equation: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 (v).
• 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).
Question: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?
Answer:Number of joules per coulomb = Total energy
Charge moved
= 12 joules
4 coulombs
= 3
Equation: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).
a.c. & d.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
Measuring Current
Current is measured using an ammeter.
A
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.
V
The voltmeter is connected in parallel with the component.
A good voltmeter has a very high resistance.
Equations
Q = It
E = QV
Replace Q with It
E = ItV
P=
E
t
Replace E with ItV
P=
ItV
t
Cancel ‘t’
P = IV
Ohm’s Law
Replace V with IR
P = I2 R
V = IR
Replace I with V/R
P=
V2
R
Ohm’s Law
AIM:
To find the relationship between the current through a
resistor and the potential difference across it.
APPARATUS:
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
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
Current in
milliamps
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
16.0
18.0
20.0
22.0
24.0
26.0
28.0
Graph:
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.
IαV
From graph.
VαI
Reverse is true.
V = kI
Replace proportion sign with equals
and constant (k).
V = RI
Replace k with the more usual
constant R (called resistance).
V = IR
Reverse Multiplication order.
This equation is called Ohm’s Law.
Ohm’s Law
V = potential difference (p.d.) or voltage in volts (V),
V = IR
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
Variable resistors
Used as a rheostat
-
+
supply
A
C
A
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.
Variable resistors
Used as a potential divider
supply
B
A
C
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.
Resistors in Series
IS
VS
R3
I2
I3
V3
R1
R2
I1
V2
V1
Experiments show:
• The current is the same at all points in a series circuit. i.e.
Is = I1 = I2 = I3
‘s’ ....supply
• The sum of the potential differences across each resistor in a
series circuit is equal to the supply voltage. i.e.
Vs = V1 + V2 + V3
• The total resistance of a number of resistors connected in series
is equal to the sum of the individual resistors . i.e.
Rs = R1 + R2 + R3
‘s’ ....series
Resistors in Parallel
VS
IS
R1
V1
I1
R2
V2
R3
V3
I2
I3
Experiments show:
• The sum of the currents through each resistor is equal to the supply
current. i.e.
Is = I1 + I2 + I3
‘s’ ....supply
• The potential difference across each resistor is equal to the supply
voltage. i.e.
Vs = V1 = V2 = V3
‘s’ ....supply
• The total resistance of a number of resistors connected in
parallel can be calculated using the equation:-
1
1
=
+
RP
R1
1
+
R2
1
R3
‘P’ ....Parallel
Resistor’s in Parallel Equation
12 Ω
X
3Ω
Y
Find the resistance between X and Y.
1
=
RP
1
=
RP
1
=
1
+
Method 1
Parallel equation
1
Substitute
3
+
12
=
1
R2
12
RP
1
+
R1
RP
1
1
4
12
Find a common factor
5
Add fractions
12
RP
= 12
1
5
RP = 2.4 Ω
Turn both sides up-side-down
Remember your units
Method 2
1
=
RP
1
=
1
1
+
R1
RP
1
1
Parallel equation
R2
1
+
12
Substitute
3
= 0.4167
Add the fractions on the calculator
RP
RP =
1
0.4167
RP = 2.4 Ω
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
Parallel equation
R2
RP = R1 x R2
R1 + R2
This rearrangement is only true for
two resistors.
RP = 12 x 3
12 + 3
Substitute
RP = 2.4 Ω
Remember your units
Car lights
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
Side light
S2
S1
+
-
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
Detecting Faults
Wire
Wire
Ω
Low Resistance
(close to 0 Ω)
Broken wire (open circuit)
Ω
High Resistance
(over 1 MΩ)
Detecting Faults
Bulb
Bulb
Blown bulb
Ω
Ω
Low Resistance
(few ohms)
High Resistance
(over 1 MΩ)
Detecting Faults
Bulb Holder
Bulb holder (no bulb)
Short circuit
Ω
High Resistance
(over 1 MΩ)
Ω
Low Resistance
(close to 0 Ω)
Mains Electricity
From transformer
and power station
The earth wire is connected to a
water pipe or a metal stake in the
ground.
100 amp fuse.
J
W
s
E=Pxt
1235435
Energy meter
kWh
kW
h
1 kWh = 1 unit = 3,600,000 J
1 kWh or unit costs about 7 pence.
J
100 amp fuse.
W
s
E=Pxt
1235435
Energy meter
kWh
kW
h
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.
Lighting Circuit
Consumer unit
Consumer unit
Two way switch
A
Stairs
Lighting
Circuit
C
B
D
The lamp can be switched on
and off at the top and bottom
of the stairs.
Two way switch
The socket switch and the plug
fuse should be on the live wire
Mains
Ring
Circuit
The current can arrive at each socket
from two directions. This means that the
mains ring can be made of thinner
cheaper wire.
Miniature Circuit Breaker (MCB)
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.
Electromagnet
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.
Electric Bell
Circuit Diagram
Spring
Contact
Battery
Electromagnets
Main points
• When a current flows through
the electromagnet the hammer
is attracted causing it to hit the
gong.
Gong
• The current through the
electromagnets stops flowing
when the contact is broken.
Hammer
• A spring moves the hammer
back to its original position, and
the whole process starts to
repeat.
Relays
Circuit Diagram
S
Lamp
9V
12 V ac
Relay
Circuit 1
Circuit 2
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
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.
N
F
S
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.
Electric Motor
F
F
S
N
(1)
F
= Current towards
= Current away
N
S
(2)
F
The forces on the current
carrying wire causes the
armature to turn.
F
F
S
N
F
S
N
(3)
The current reverses
direction every half turn
to keep it rotating in the
same direction.
(4)
F
Electric Motor
shaft
rotating
Rotating
coilscoils
or
armature
or armature
Field
Field
Coils
coils
brushes
commutator
brushes
Parts of a real Electric Motor.
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.