Electricity 3: Ohm's Law, experimental investigations of resistance
and
IV graphs and calculations using I = V/R, Q = It and E = QV
IGCSE AQA GCSE Physics Edexcel GCSE Physics OCR GCSE
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Doc Brown's school physics revision notes: GCSE
physics, IGCSE physics, O level physics, ~US grades 8, 9 and 10
school science courses or equivalent for ~1416 year old students of
physics
What is Ohm's Law? How do you do calculations using Ohm/s Law?
What factors affect the resistance of a
circuit?
How do you construct and use a circuit to
investigate Ohm's Law?
How do you calculate the amount of electric
charge moving in a circuit?
Subindex for this page
1.
Ohm's Law,
simple investigation circuit and V= IR calculations
2.
Movement and unit of charge, the coulomb, calculations using Q = It
3.
Potential difference and
electrical energy transfer, E = QV calculations
4a.
Electrical resistance
 factors involved
4b.
Investigating the
resistance of a wire at constant temperature, varying length & width
4c.
Investigating the
current  voltage characteristics of a wire  graph explained
4d.
Investigating
the current  voltage characteristics of a metal filament lamp  graph
4e.
Investigating the
current  voltage characteristics of a diode  graph explained
See also
APPENDIX 1 for a summary of all electricity
equations you may need
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1a.
Ohm's
Law (and a mention of other units dealt with in other sections)
Ohm's law states that the current through a
conductor between two points is directly proportional to the voltage across the
two points in a circuit.
It involves a most fundamental equation
you need to know for electricity calculations.
This can be expressed mathematically as:
I
= V / R
rearrangements:
V = IR
and
R = V/I
I = current in amperes, amps,
A; a measure of the rate of flow of electric charge.
V = potential difference, p.d., volts,
V; a measure of the potential energy given to the electric charge
flowing.
The potential difference in a circuit
is the energy transferred per coulomb of electric charge that
flows between two points in an electric circuit.
The coulomb (C) is the
unit of electric charge (see
Q = It equation notes).
The energy transferred is
calculated from the p.d. and the quantity of electric charge (Q)
moved by the p.d. in V (see
E = QV
equation notes).
R = resistance of the wire, ohms,
Ω; a measure of the reluctance of a conductor to inhibit the
flow of charge.
The greater the resistance of a
resistor, the more it resists and slows down the flow of electricity.
Ohm's law means that the R in this equation
is a constant, independent of the size of the electrical current flowing.
The law correctly applies to socalled ohmic
conductors, where the current flowing is directly proportional to the
applied potential difference, but some resistors don't obey this law e.g. the heated filament of a
light bulb.
1b.
A simple
experiment to measure the resistance of a single component
If you set up circuit 31 (right diagram) you can measure the
resistance of the fixed resistor [R].
By varying the voltage from the power supply using the variable
resistor you can readily get lots of pairs of readings of p.d. (V) and current
(A).
Then use Ohm's Law equation (R = V/I) to calculate the value of
the fixed resistance.
You can then average the values of R calculated for a more
reliable result.
More details further down to get the full IV characteristic
graphs and also how to obtain the resistance by a graphical method.
This is the basic setup to investigate
the currentvoltage characteristics of any component R.
1c. Examples of
calculations using Ohm's
Law V = IR
Q1 When a p.d. of 4.5 V
is applied across a resistance, a current of 0.5 A flows.
What is the value of the resistor?
R = V/I = 4.5/0.5 =
9.0 Ω
Q2 A resistance has a
value of 50 ohms.
What p.d. must be applied across it
to cause a current of 5.0 A to flow through it?
V = IR = 5 x 50 =
250
V
Q3
A p.d. of 240 volts is applied across a heating element resistor of 30 ohms.
How much current flows through the
heater?
I = V/R = 240/30 =
8.0
A
Q4
Three 1.5 volt batteries were wired in series with three bulbs.
If the ammeter measured a current of
0.50 A, what is the resistance of each bulb?
I = V / R, so R = V / I = (3 x 1.5) /
0.50 = 9.0 Ω
Since total resistance = sum of
resistances, resistance of each bulb = 9.0 / 3 =
3.0
Ω
TOP OF PAGE
and subindex
2. Movement of charge
2a. Calculation
of the charge passing through a point in a circuit Q = It
Current (I in amps) is the rate of flow of electrical
charge around a circuit.
The greater the flow of charge in a given time the greater
the current.
The rate of flow of electric charge is
measured in coulombs/second.
You can calculate the charge passing a point in a circuit in
a given time from the formula
Q = It
where Q = electric charge in coulombs (C)
 the unit of electric charge
I = current flow in amperes (A) and t = time (s)
rearrangements from Q = It,
I = Q/t
and t = Q/I
A current flow of 1 A equals a
rate of flow of charge of 1 C/s.
2b. Examples of calculation questions involving
the equation Q = It
Q1 If
a current of 3.0 A passes through an appliance for 1 hour and 30 minutes,
how much electrical charge is transferred in the process?
Q = It, Q = 3.0 x 1.5 x 60 x 60 = 16 200 C =
1.62
x 10^{4}
C
Q2 If
9000 C of charge passes a point in an electrical circuit in 12.0 minutes,
what is the current flow?
I = Q/t = 9000/(12 x 60) = 9000/720 =
12.5
A
Q3
How long will it take, in minutes and seconds, for an electrical circuit
current of 20.0 A to transfer 5000 C of charge?
t = Q/I = 5000/20 =
250 seconds =
4
mins and 10 seconds
Q4 A laptop computer
battery charger passes a current of 1.20 A for 30 minutes with an output
p.d. of 15.0 V.
(a) Calculate how much charge is
transferred to the computer battery.
Q = It = 1.2 x 30 x 60 =
2160
C
(b) What is the resistance of the
battery charger?
V = IR, R = V/I = 15 / 1.2
=
12.5 Ω
(c) When the laptop battery is fully
charged it stores 3000 C.
How long will it take to fully
charge a flat battery?
Q = It, t = Q / I = 3000 /
1.2 =
2500 s (41 min 40 secs)
TOP OF PAGE
and subindex
3. Potential difference and
energy transfer
3a. Introduction to electrical energy
transfer
Energy transfer per unit of charge = potential difference (p.d.) and
calculations based on
E = QV
In the previous section we looked at how to calculate the
quantity of charge moving in a circuit, but said nothing about the energy
transferred.
Reminders:
Electrical circuits, terms used, circuit symbols, parallel
circuits and series circuits explained
Potential difference (p.d. in
volts, V) is the energy transferred per unit charge as electric
charge moves from one point to another in an electric circuit.
It is measured with a voltmeter,
which is always wired in parallel across a circuit component.
An electrical current transfers
energy
Just think of all the electrical
appliances you use  all need supplying with energy to work!
A power supply does work on a
charge and transfers energy to it.
Work must be done on the charge
to increase its potential energy.
Electric charge is measured in
coulombs (C)
Charge and its movement has
already been dealt with in
section 2 (Q = It).
Charges transfer energy to
components as they pass through by doing work against the resistance of
the component.
If work is done, then energy is
transferred.
If electrical charge experience a potential difference,
that charge will flow transferring energy.
The energy is supplied from the energy store of the
power supply  battery, mains electricity etc.
When charge passes through any p.d. fall it releases energy
(from a higher to a lower potential energy level).
e.g. in a thin wire
resistance, heat is released.
The potential difference between two
points is equal to the work done per unit charge.
potential difference (V) = work done (energy
transferred in J) ÷ charge (C)
i.e. 1 volt corresponds to 1 joule
per coulomb or
V = J/C
The bigger the fall in p.d., the greater the energy
transferred, because the charge starts off with a greater potential
energy.
Therefore a power supply with a bigger source
p.d. (V)
can supply more energy to the circuit per unit of electric charge (the
coulomb, C).
The bigger the p.d., the more energy the same quantity of
electric charge can carry.
3b. Another equation to calculate electrical
energy transfers
The
quantity of energy carried can be calculated from the equation:
energy transferred = charge x potential
difference.
E = QV (so Q = E/V
and V = E/Q)
E = energy transferred in joules (J)
Q = quantity of
electric charge in coulombs (C)
V = potential difference (V)
Noting that: V = E/Q =
energy transferred per unit charge
(J/C)
Just in passing and some reminders:
The more energy transferred in a given
time, the greater the power of a device or electrical appliance.
The p.d. V tells you how much
energy each unit of electrical charge transfers,
so, V = E/Q, (units
J/C), see E = QV calculations below).
The current I tells you how much
charge passes a given point in a circuit per unit time (coulombs/second,
C/s).
This means both p.d. V and current
I affect the rate at which energy is transferred to an appliance
from the electrical energy store to other energy stores.
AND some mathematical connections
based on section 2. Q = It and here in section 3 E = QV
From Q = It and E = QV, substituting
gives E = ItV,
so (i)
E = IVt (I in A,
t in s, V in volts)
Rearranging E = IVt gives IV =
E/t
This connects with the
equations for power
(ii) Power = energy
transferred / time taken =
E/t (J/s),
and
(iii) Power = current
x voltage = P (W) = I (A) x V (V),
P = IV
From (ii) and (iii) E/t =
IV, so E = IVt, which is equation (i) !!!
3c. Calculation question based on E = QV
(sometimes involving other electricity equations too)
Q1 An electric motor of a
model car is powered by a 1.5 V battery.
If 120 C of charge passes through the
motor circuit in the moving car,
(a) how much energy is transferred?
E = QV = 120 x
1.5 = 180 J
(b) Describe the likely energy store
changes when the car is running.
The chemical potential energy
store of the battery decreases and becoming electrical energy.
The kinetic energy store of the
car increases with some wasted heat from friction and sound energy
transferred to the thermal energy store of the surroundings.
Q2 What quantity of
charge is needed to transfer 500 J of energy if the p.d. of a circuit is
24.0 V?
E = QV, Q = E/V =
500/24 =
20.8 C (3
sf)
Q3 What potential
difference is required in a circuit to transfer 2000 J of energy with a
charge of 50 coulombs?
E = QV, V = E/Q = 2000/50 =
40 V
Q4
A 12.0 V battery passes a current of 2.0 A through a lamp for 5 minutes.
(a) Calculate how much charge passed
through the lamp.
Q = It = 2 x 5 x 60 =
600
C
(b) Calculate how much electrical energy
was transferred by the lamp.
Two ways:
(i) E = QV = 600 x 12 =
7200 J,
the simplest, but you can calculate it without knowing Q from:
(ii) E = IVt = 2 x 12 x 5 x 60 =
7200 J
Q5
An appliance has a power of 1.5 kW and works of a 230 V mains supply.
If the appliance is used for 15 minutes,
how much charge has flowed through the circuit?
1.5 kW ≡
1500 W ≡
1500 J/s
Total energy transferred = power x
time = 1500 x 15 x 60 = 1 350 000 J
E = QV, so Q = E / V = 1 350
000 / 230 =
5870 C (3 sf)
The answer can be calculated by
another route
P = IV, I = P / V = 1500 / 230
= 6.522 A
Q = It = 6.522 x 15 x 60 =
5870 C (3 sf)
3d. A little more on potential difference  effect of two resistors in series
The circuit 41 shows two resistors wired
in series.
On the right is shown what happens to the
p.d. going clockwise around the circuit (direction of convention current).
The potential store of the battery raises
the charge potential difference of the charge to 12 V.
As the charge passes through the 1st
resistor R_{1}, it loses energy and the p.d. falls by 8 V to a p.d.
of 4 V.
As the charge passes through the 2nd
resistor R_{1}, it loses energy again and the p.d. falls by 4 V to a
p.d. of 0 V.
As long as there is a complete circuit,
the process repeats itself.
Since E = QV, twice as much energy is released
by resistor R_{1} (p.d. 8 V) than R_{2} (p.d. 4 V) for the
same current.
TOP OF PAGE
and subindex
4. Electrical resistance  experiments to
investigate the IV characteristic of various resistances and the validity, or
otherwise, of Ohm's Law
4a.
What affects the resistance of a wire? Is resistance
constant?
and some characteristic currentvoltage graphs (IV plots)
explained
The resistance of a circuit depends on
several factors:
(i) the thickness of the resistance
wire  for a single component resistor
(ii) the length of the resistance
wire  for a single component resistor
(iii) if more than one resistance,
are they wire in series or parallel?
(iv) the temperature of a component
that acts as a resistance
Section
4. describes and explains several examples
of IV graphs  which can be investigated using circuit 31 (on the right)
The circuit diagram 31 on the right shows
how you can investigate the variation of current through a resistance (or
any component) when you vary the potential difference.
Current–potential difference graphs are used to show how the current through a component varies with the potential difference across it.
The resistance of some resistors/components does change
as the current and p.d. changes e.g. a diode or filament lamp.
See how
and why in sections 4d. and 4e.
TOP OF PAGE
and subindex
4b.
Investigating the electrical resistance of a wire  variation of length or width
Circuit 30 shows how to investigate the resistance of a wire
A relatively thin wire is fastened at
each end onto a meter ruler marked in mm using crocodile clips.
You need an ammeter to measure the
current in amps and a voltmeter to measure the p.d. across the wire in
volts.
The wire is connected in series
with a battery power supply, switch and ammeter to measure the current
flowing through the wire in amps.
The voltmeter, to measure the p.d, is
wired in parallel across the resistance wire.
Note the ammeter is always wired in
series with a component, but a voltmeter is always wired in parallel
across any component under investigation.
One end of the wire connected through the
voltmeter is fixed (on the left), but the other end has a crocodile clip that acts as a moveable
contact point to place a various distance along the resistance wire from
left to right.
Close the switch to complete the circuit
and begin taking readings.
Its best to open the switch between
readings to minimise the risk of heating up the wire.
You vary the distance d (mm) from the
left (0 mm) to a point further along to the right and take a series of pairs
of p.d and current readings e.g. every 50 mm (you can work in cm, it makes
no difference!).
Using Ohm's Law, you calculate the
resistance in ohms from the equation
R = V / I
You
can then plot a graph of resistance (Ω) versus the length of the wire d (mm)
 shown on the right.
You should find the graph is linear with
its x,y origin at 0,0.
This means the resistance is
proportional to the length of the wire.
If you don't fix the wire exactly at 0
mm, the graph should still be linear, but, the origin of the line will not
be 0,0.
If you repeat the experiment with
different diameter wires, you should find the gradient becomes lower, the
thicker the wire.
For the same length of wire the resistance is less the
thicker the wire  a good analogy is the ease with which water flows through a
thin or wider diameter pipe.
TOP OF PAGE
and subindex
4c.
Investigating the
current  voltage
characteristics of a wire
This is an Ohm's Law experiment
Circuit
31 shows you how to investigate how I varies with V for a resistance
The investigation is all about finding
out ...
... how does the current flowing
through a resistor vary with the potential difference across it?
The fixed resistor represents a
'component' in a circuit and must be at constant temperature
throughout the experiment (see later on
temperature
effects).
In this case
a simple wire resistor is
wired in series with the power supply and ammeter.
The p.d. is measured across the fixed
resistance with the voltmeter,
However,
also wired in series, a variable resistor is added, so that you can
conveniently change the potential difference and thereby change the current
flowing through the component.
This allows you to gather a whole series
of pairs of I and V readings, with which to plot suitable graphs  in this
case of V versus I, but often done as I versus V.
Using the variable resistor, you
gradually increase the potential difference across the component, taking the
matching current reading e.g. increasing at 0.5 V at a time. Repeat each
reading twice and use the average.
You can then swap the battery terminals
and repeat all the readings.
If you plot the p.d. versus current, the
graph is linear if it obeys Ohm's Law  it is then called an 'ohmic
conductor'.
This I've represented by the graph
above on the right, and the gradient is equal to the resistance in
ohms.
This corresponds to the Ohm's Law
equation V = IR, so the gradient is R.
If the graph remains linear, the
resistance is remaining constant.
This graph does not represent the
readings taken after reversing the battery terminals.
However, it does show how to get
the value of a resistance by a graphical method.
Its a linear graph and the
phrase linear component may be used.
For components like a wire that doesn't
heat up, you should get a linear plot of p.d. (V) versus I (A) with a
gradient R (Ω). (right graph).
You should make sure that the wire
doesn't heat up too much  if it starts getting warm, immediately
disconnect the resistor ('switch off') and let it cool down.
If you plot I versus V the gradient is 1/R (the reciprocal of the
resistance), linear graph.
This graph shows what you get by
plotting all the data, including the IV readings taken after reversing
the battery terminals.
The graph
(1)
is
constructed on a crosswire axis. The top right half is your first set of
results, you then reverse the terminals on the power supply and repeat the
experiment giving the bottom left part of the graph.
Note that
you will only get a linear
graph if the temperature of the wire remains constant.
When the current (A) is proportional
to the p.d (V) it is described as an
ohmic conductor
(obeys Ohm's Law!).
Using the circuit 31 you can test any
resistor or any other type of circuit component and the results are
discussed below starting with a summary of factors that affect resistance.
So,
the resistance of an ohmic conductor e.g. a circuit
component doesn't change no matter current is passed through  constant
gradient of 1/R for graph 1.
These are the expected linear graphs for
a fixed resistor using circuit 31 above.
Thinking anticlockwise on the
diagram, the different graph lines might depict a decreasing resistance
e.g. (i) a resistance wire getting shorter for the same diameter, or
(ii) the diameter getting larger for a fixed length of wire.
At a constant temperature the current flowing through an
ohmic conductor is directly proportional to the potential difference across
it  the equation is V = IR or
I = V/R.
However, this is only true, giving
a linear graph if the temperature doesn't
change.
Comments about variables in this
particular Ohm's Law experiment
Current is always determined by a
combination of the p.d. (V) and the resistance R (Ω).
The independent variable is
what we change or control in the experiment  in this case you can
consider it as the p.d. controlled by the variable resistor.
One convention is to plot the
independent variable on the xaxis, and the dependent variable on the
yaxis.
This
means the resistance R, is the reciprocal of the gradient  a bit more
awkward to calculate the resistance than from the V versus I graph plot,
where the gradient is the resistance.
Ohm's Law: I =
V / R.
The dependent variable is what
we are testing or measuring in the experiment, this is the current I
(A), which depends on the variable resistor setting, which in turn
controls the potential difference across the resistor.
The control variables are what
we keep the same during the experiment to make sure it’s a fair test
e.g. in this case the wire and temperature are kept
constant, should NOT be varied  don't change the wire or allow the wire
to heat up.
TOP OF PAGE
and subindex
4d.
Investigating the current  voltage characteristics of a metal
filament lamp
When electric charge flows through a high resistance,
like the thin metal filament of a lamp, it transfers some of the
electrical energy
to the thermal energy store of the filament. The electric charge does
work against the resistance.
Circuit 45 shows how you can
investigate the current  potential difference characteristics of a
filament
bulb.
The voltmeter is wired in parallel with
the thermistor, the p.d. V is measured in volts (V).
The variable resistor allows you to vary
the p.d. and current flow.
The ammeter, wired in series, gives you
the current I reading in amps (A).
The
passage of current heats up the filament and the rise in
temperature causes the resistance to increase. So a filament
lamp is a nonohmic conductor.
This 'heating up effect' affects
all resistors.
As the current increases, more heat energy is released and the
filament gets hotter and hotter, so further increase in temperature
further increases the resistance.
This decreases the rate at which the current increases with
increase in potential difference.
Therefore the gradient of the IV
graph curve decreases and increasingly so with increase in
temperature  graph 2. Its a nonlinear graph.
If the gradient is changing,
then the resistance is changing.
The graph
(2)
is
constructed on a crosswire axis. The top right half is your first set of
results, you then reverse the terminals on the power supply and repeat the
experiment giving the bottom left part of the graph.
The phrase nonlinear
component may be used.
When the current (A) is NOT
proportional to the p.d (V) the filament lamp is described as a
nonohmic conductor
(doesn't obey Ohm's Law!).
You get the same IV shaped graph
for a
thermistor.
Theory
 with reference to the metallic structure diagram
A metal crystal lattice consists of immobile ions and freely moving
electrons between them. As the temperature increases, the metal ions vibrate more
strongly into which the electrons collide and this inhibits the passage of electrons  reducing the flow of
charge. As the current increases, the vibrations increase causing more
of the electrical energy to be converted to heat  increasing the
temperature AND the resistance of the metallic filament, thereby
lowering the current even further.
So, an increase
in temperature increases the resistance a filament lamp (or most other
resistors) and lowers the current flowing for a given p.d.
If a resistor becomes too hot,
almost no current will flow.
There is one important exception
to this 'rule', see notes on the
thermistor where the resistance actually falls with
increase in temperature.
The filament bulb is just one of many
examples were energy is transferred usefully, BUT there is
always heat energy lost to the thermal energy store of the device and
the surroundings.
The filament is often made of the
metal tungsten that melts at >3400^{o}C, and glows brightly at
2500^{o}C, but it still evaporates very slowly. An inert gas
such as argon or nitrogen is added to reduce this evaporation  any
evaporated tungsten atoms hit the unreactive (and so nonoxidising) Ar
or N_{2} molecules and hopefully condense back on the filament.
See
Conservation of energy,
energy transfersconversions, efficiency  calculations
TOP OF PAGE
and subindex
4e.
Investigating the current  voltage characteristics of a diode
The
current through a diode flows in one direction only  see graph 3.
The resistance in the opposite direction
is very high  hence effectively a 'one way' system.
Circuit 43 shows how you can investigate
the current  potential difference characteristics of a diode.
The voltmeter is wired in parallel with
the thermistor, the p.d. V is measured in volts (V).
The variable resistor allows you to vary
the p.d. and current flow.
The ammeter, wired in series, gives you
the current I reading in amps (A).
A diode has a very high resistance in
the reverse direction.
There is also a threshold p.d.
(e.g.
1.4 V) before any current flows at all  look at the graph carefully 
there is a short horizontal portion before the current rises from zero
and eventually becomes linear.
Therefore you get a top right portion
of the graph 3 compared to graphs 1 an 2 above.
This is because when you do the
experiment using the circuit described above, on reversing the
connections, you will find no current flows as you vary the p.d.
Its a nonlinear
graph.
If the gradient is changing, then
the resistance is changing.
When the current (A) is NOT
proportional to the p.d (V), the diode is described as a
nonohmic conductor
(doesn't obey Ohm's Law!).
The phrase nonlinear component
may be used.
The graph
(3)
is
constructed on a crosswire axis. The top right half is your first set of
results, you then reverse the terminals on the power supply and repeat the
experiment giving the bottom left part of the graph.
Since the current only flows one way
through a diode, it can be used to convert an ac current into a dc
current.
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and subindex
Practical work to
help develop your skills and understanding may have included the following:
using filament bulbs and resistors to investigate potential difference
 current
characteristics,
APPENDIX 1: Important definitions, descriptions,
formulae and
units
Note: You may/may
not (but don't worry!), have come across all of these terms, it depends
on how far your studies have got. In your course, you might not need
every formula  that's up to you to find out.
V
the potential difference (p.d., commonly called
'voltage') is the driving potential that moves the electrical charge around
a circuit  usually electrons.
Potential difference is the work done in
moving a unit of charge.
It indicates how much energy is transferred
per unit charge when a charge moves between two points in a circuit
e.g. between the terminals of a battery.
The p.d. across any part of a circuit is measured in volts,
V.
I
the current is rate of flow of electrical charge in
coulombs/second (C/s), measured in amperes (amps, A).
The quantity of electric charge transferred in
a give time = current flow in amps x time elapsed in seconds
Formula connection:
Q = It,
I = Q/t, t = Q/I, Q = electrical charge moved in
coulombs (C), time t (s)
R
the resistance in a circuit, measured in ohms (Ω).
A resistance slows down the flow of electrical charge
 it opposes the flow of electrical charge.
Formula connection:
V = IR,
I = V/R, R = V/I (This is the formula for
Ohm's Law)
P
is
the power delivered by a circuit = the
rate of energy
transfer (J/s) and is measured in watts (W).
Formula connection:
P = IV,
I = P/V, V = P/I also
P = I^{2}R
(see also P = E/t below)
E = QV,
the energy transferred by the quantity of electric charge by a potential
difference of V volts.
energy transferred (joules) =
quantity of electric charge (coulombs) x potential difference
(volts)
Q =
E/V, V = E/Q, E = energy transfer in joules (J),
Q = electrical charge moved (C), V = p.d. (V)
E = Pt,
P = E/t, t = E/P, where P = power (W), E
= energy transferred (J), t = time taken (s)
Energy transferred in joules = power in watts
x time in seconds
Formula connection: Since E = Pt and P = IV,
energy transferred E =
IVt

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and subindex
What next?
Electricity and
magnetism revision
notes index
1.
Usefulness of electricity, safety, energy transfer, cost & power calculations, P = IV = I^{2}R,
E = Pt, E=IVt
2.
Electrical circuits and how to draw them, circuit symbols, parallel
circuits, series circuits explained
3. Ohm's Law, experimental investigations of
resistance, IV graphs, calculations V = IR, Q = It, E = QV
4. Circuit devices and how are they used? (e.g.
thermistor and LDR), relevant graphs gcse physics revision
5. More on series and parallel circuits,
circuit diagrams, measurements and calculations
gcse physics
6. The 'National Grid' power supply, environmental
issues, use of transformers
gcse
physics revision notes
7.
Comparison of methods of generating electricity
gcse
physics revision notes (energy 6)
8. Static electricity and electric fields, uses
and dangers of static electricity gcse
physics revision notes
9.
Magnetism
 magnetic materials  temporary (induced) and permanent magnets  uses gcse
physics
10.
Electromagnetism, solenoid coils, uses of electromagnets gcse
physics revision notes
11. Motor effect of an electric current,
electric motor, loudspeaker, Fleming's lefthand rule, F = BIL
12.
Generator effect, applications e.g. generators
generating electricity and microphone
gcse
physics
IGCSE revision
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