Electricity 3: Ohm's Law, experimental investigations of resistance
and
IV graphs and calculations using I = V/R, Q = It and E = QV
Doc Brown's Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
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?
See
APPENDIX 1 for a summary of all electricity
equations you may need.
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.
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
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
curve 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.
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.
Tthe 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.
Examples of questions involving 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
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: 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.
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)
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.
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 charge in coulombs (C)
V = potential difference (V)
Noting: 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.
A little more on potential difference
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.
Question based on E = QV
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
Investigating the 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 is 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 water flows through a
thin or wider diameter pipe.
An
experiment to investigate the IV characteristic of a resistance
and the validity, or
otherwise, of Ohm's Law
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).
As with the wire experiment above, it 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.
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.
This corresponds to the Ohm's Law
equation V = IR, so the gradient is R.
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.
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.
What affects the resistance of a wire? Is resistance
constant?
and some characteristic currentvoltage graphs (IV plots)
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
Examples
of IV graphs
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.
The experiment is
described
in detail above.
Current–potential difference graphs are used to show how the current through a component varies with the potential difference across it.
The following three graphs are
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 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.
This is the expected linear graph for
a fixed resistor using circuit 31 above.
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.
The resistance of some resistors/components does change
as the current and p.d. changes e.g. a diode or filament lamp. So how
and why?
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.
The phrase nonlinear
component may be used.
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 lower the resistance of a
component,
See
Conservation of energy,
energy transfersconversions, efficiency  calculations
The
current through a diode flows in one direction only  see graph 3.
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.
Therefore you get a top right portion
of the graph 3 compared to graphs 1 an 2 above. Its a nonlinear
graph.
The phrase nonlinear component
may be used.
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.
There is also a threshold p.d. (e.g.
1.4 V) before and current flows at all  look at the graph carefully 
there is a short horizontal portion before the current rises from zero.
Since the current only flows one way
through a diode, it can be used to convert an ac current into a dc
current.
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 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 force that moves the electrical charge around a
circuit. p.d. is measured in volts, V.
I
the current = rate of flow of electrical charge in
coulombs/second (C/s), measured in amperes (amps, A).
Formula connection:
Q = It, I =
Q/t, t = Q/I, Q = electrical charge moved in coulombs (C),
time t (s)
R
a resistance in a circuit, measured in ohms (Ω). A
resistance slows down the flow of electrical charge.
Formula connection:
V = IR,
I = V/R, R = V/I (This is the formula for
Ohm's
Law)
P
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, 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)
Formula connection: Since E = Pt and P = IV,
energy transferred E = IVt
TOP OF PAGE
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 revision
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 revision
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 gcse physics
12.
Generator effect, applications e.g. generators
generating electricity and microphone
gcse
physics revision
IGCSE revision
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