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Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
If you have ever looked inside an
electronic device you have no doubt
seen what a resistor looks like.
A resistor’s working part is usually
made of carbon, which is a
semiconductor.
The less carbon
there is, the harder it is for
current to flow through the resistor.
As the animation shows, carbon is
spiraled away to cut down the
cross-sectional area, thereby increasing the resistance
to whatever value is desired.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
Some very precise resistors are
made of wire and are called
wire-wound resistors.
And some resistors can be made to vary their
resistance by tapping them at various places. These are
called variable resistors and potentiometers.
Thermistors are temperaturedependent resistors, changing
their resistance in response
to their temperature.
Light-dependent resistors (LDRs)
change their resistance in
response to light intensity.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
Electrical resistance R is a measure of how hard it is
for current to flow through a material. Resistance is
measured in ohms () using an ohm-meter.
0.L
0.0
330.4
This resistor
has a
resistance of
330.4 .
FYI
A reading of 0.L on an
ohmeter means “overload”.
The resistance is too high
to record with the meter.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
The different types of resistors have different
schematic symbols.
fixed-value resistor
2 leads
potentiometer
3 leads
variable resistor
2 leads
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
The different types of resistors have different
schematic symbols.
light-dependent
thermister
resistor (LDR)
2 leads
2 leads
As temperature increases
resistance decreases
As brightness increases
resistance decreases
Electrical resistance (symbol R)
• Why is it necessary to keep pushing the charges to make
them move?
• The electrons do not move unimpeded through a
conductor. As they move they keep bumping into the ions
of crystal lattice which either slows them down or bring
them to rest.
path
atoms
(actually
positive ions)
free electron
.
The resistance is a measure of how hard it is to
pass something through a material.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
The resistance R of a material is the ratio of the
potential difference V across the material to the current I
flowing through the material.
R=V/I
electric resistance
The units from the formula are (V A-1) which are called
ohms ().
PRACTICE:
Orange = 3
A fixed resistor has a current of 18.2 mA
Orange = 3
when it has a 6.0 V potential difference
Brown = 1
across it. What is its resistance?
SOLUTION: Last color is number of zeros.
 R = V / I = 6.0 / 18.210-3 = 330 .
Factors affecting resistance
Conductors, semiconductors and insulators differ
in their resistance to current flow.
DEF: The electrical resistance of a piece of material is
defined by the ratio of the potential difference across
the material to the current that flows through it.
V
R=
I
The units of resistance are volts per ampere (VA-1).
However, a separate SI unit called the ohm Ω is defined
as the resistance through which a current of 1 A flows
when a potential difference of 1 V is applied.
Wires, wires, wires
As you are going to see, the resistance of a wire can be
completely ignored – if it is a thin wire connecting two,
three or more resistors, or becoming very important if it is
a long, long wire as in the case of iron, washing machine,
toaster ….., where it becomes resistor itself.
The resistance of a conducting wire depends on four main
factors: • length • cross-sectional area • resistivity • temperature
Cross Sectional Area (A)
The cross-sectional area of a conductor (thickness) is similar to the cross section
of a hallway. If the hall is very wide, it will allow a high current through it, while a
narrow hall would be difficult to get through. Notice that the electrons seem to be
moving at the same speed in each one but there are many more electrons in the
larger wire. This results in a larger current which leads us to say that the resistance
is less in a wire with a larger cross sectional area.
Length of the Conductor (L)
The length of a conductor is similar to the length of a hallway.
A shorter hallway will result in less collisions than a longer one.
Temperature
To understand the effect of temperature you must picture what happens in a
conductor as it is heated. Heat on the atomic or molecular scale is a direct
representation of the vibration of the atoms or molecules. Higher temperature
means more vibrations. In a cold wire ions in crystal lattice are not vibrating much so
the electrons can run between them fairly rapidly. As the conductor heats up, the
ions start vibrating. As their motion becomes more erratic they are more likely to get
in the way and disrupt the flow of the electrons. As a result, the higher the
temperature, the higher the resistance.
At extremely low temperatures, some materials, known as superconductors, have no
measurable resistance. This is called superconductivity. Gradually, we are creating
materials that become superconductors at higher temperatures and the race is on to
find or create materials that superconduct at room temperature. We are painfully far
away from the finish line.
Resistance also depends on temperature, usually
increasing as the temperature increases.
At low temperatures some materials, known as superconductors,
have no resistance at all. Resistance in wires produces a loss of
energy (usually in the form of heat), so materials with no resistance
produce no energy loss when currents pass through them.
And that means, once set up in motion (current) you don’t
need to add additional energy in oder to keep them going.
The dream: current without cost!!!!!!!!! Both in money and
damage to environment!!!!!!!!
Of course, resistance depends on the material being used.
Resistance of a wire when the temperature is kept constant is:
L
R=ρ
A
The resistivity, ρ (the Greek letter rho), is a value that only
depends on the material being used. It is tabulated and you can
find it in the books. For example, gold would have a lower value
than lead or zinc, because it is a better conductor than they are.
The unit is Ω•m.
In conclusion, we could say that a short fat cold wire makes
the best conductor.
If you double the length of a wire, you will double the
resistance of the wire.
If you double the cross sectional area of a wire you will cut
its resistance in half.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
The Greek  is the resistivity of the particular material
the resistor is made from. It is measured in m.
Resistivities and Temperature Coefficients for Various Materials at 20C
Material
 (m)
 (C -1)
Conductors
Material
 (m)
 (C -1)
360010-8
-5.010-4
Semiconductors
Aluminum
2.8210-8
4.2910-3
Carbon
Copper
1.7010-8
6.8010-3
Germanium
4.610-1
-5.010-2
1010-8
6.5110-3
Silicon
2.5102
-7.010-2
Mercury
98.410-8
0.8910-3
Nichrome
10010-8
0.4010-3
Nickel
7.810-8
6.010-3
Platinum
1010-8
3.9310-3
1.5910-8
6.110-3
5.610-8
4.510-3
Iron
Silver
Tungsten
Nonconductors
Glass
1012
Rubber
1015
Wood
1010
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Resistance
Note that resistance depends on temperature. The IBO
does not require us to explore this facet of resistivity.
PRACTICE: What is the resistance
of a 0.00200 meter long carbon
core resistor having a core
diameter of 0.000100 m? Assume
the temperature is 20 C.
r = d / 2 = 0.0001 / 2 = 0.00005 m.
A = r2 = (0.00005)2 = 7.85410-9 m2.
From the table  = 360010-8  m.
R = L / A
= (360010-8)(0.002) / 7.85410-9 = 9.17 .
L
A
Ohm’s law
The German Ohm studied resistance of
materials in the 1800s and in 1826 stated:
“Provided the temperature is kept constant, the
resistance of very many materials is constant
over a wide range of applied potential
differences, and therefore the current is
proportional to the potential difference .”
DEF: Current through resistor (conductor) is
proportional to potential difference on the resistor if
the temperature/resistance of a resistor is constant.
 In formula form Ohm’s law looks like this:
𝑉
or 𝐼 =
𝑅
𝑉
Ohm’s law
= 𝑐𝑜𝑛𝑠𝑡.
IV
𝐼
 Ohm’s law applies to components with constant R.
or
Examples
• If a 3 volt flashlight bulb has a resistance of 9 ohms, how
much current will it draw?
• I = V / R = 3 V / 9  = 0.33 A
• If a light bulb draws 2 A of current when connected to a
120 volt circuit, what is the resistance of the light bulb?
• R = V / I = 120 V / 2 A = 60 
Effects of electric current on the BODY- electric shock
Current (A)
Effect
0.001
can be felt
0.005
painful
0.010
involuntary muscle contractions (spasms)
0.015
loss of muscle control
0.070
if through the heart, serious disruption; probably
fatal if current lasts for more than 1 second
questionable circuits: live (hot) wire ? how to avoid being electrified?
1. keep one hand behind the body (no hand to hand current through the body)
2. touch the wire with the back of the hand. Shock causing muscular contraction
will not cause their hands to grip the wire.
human body resistance varies:
100 ohms if soaked with salt water;
moist skin - 1000 ohms;
normal dry skin – 100 000 ohms,
extra dry skin – 500 000 ohms.
What would be the current in your body if you touch the
terminals of a 12-V battery with dry hands?
I = V/R = 12 V/100 000  = 0.000 12 A
quite harmless
But if your hands are moist (fear of AP test?) and you
touch 24 V battery, how much current would you draw?
I = V/R = 24 V/1000  = 0.024 A
a dangerous amount of current.
Ohmic and Non-Ohmic behaviour
How does the current varies with potential difference for some typical devices?
current
potential
difference
𝐼
𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝑉
𝑉
𝑠𝑜 𝑖𝑠 𝑅 =
𝐼
diode
current
filament lamp
current
metal at const. temp.
potential
difference
potential
difference
𝑎𝑠
devices are non-ohmic if
resistance changes
Devices for which current through them is directly
proportional to the potential difference across device are
said to be ‘ohmic devices’ or ‘ohmic conductors’ or
simply resistors. In other words the resistance stays
constant as the voltage changes.
There are very few devices that are trully ohmic.
However, many useful devices obey the law at least
over a reasonable range.
Example
A copper wire has a length of 1.60 m and a diameter of 1.00 mm. If the wire is
connected to a 1.5-volt battery, how much current flows through the wire?
The current can be found from Ohm's Law, V = IR. The V is the battery
voltage, so if R can be determined then the current can be calculated.
The first step, then, is to find the resistance of the wire:
L = 1.60 m.
r = 0.5 mm
 = 1.72x10-8 m, copper - books
The resistance of the wire is then:
R =  L/A = (1.72x10-8 m)(1.60)/(7.9x10-7m2 ) = 3.50 
The current can now be found from Ohm's Law:
I = V / R = 1.5 / 3.5 = 0.428 A
Do Now
Answer the following
1) What is power? What is unit for power?
2) Can you figure out 2 reasons for this year Nobel prize in physics?
It is an invention of greatest benefit to mankind
“for the invention of efficient blue light-emitting diodes which has
enabled bright and energy-saving white light sources"
The main advantage is efficiency. In conventional tungsten/wolfram
bulbs, the light-production process involves generating a lot of heat
(the filament must be warmed). This is completely wasted energy,
unless you're using the lamp as a heater, because a huge portion of
the available electricity isn't going toward producing visible light. LEDs
generate very little heat, relatively speaking. A much higher percentage
of the electrical power is going directly to generating light, which cuts
down on the electricity demands considerably.
https://www.youtube.com/watch?v=oCEKMEeZXug
Ohmic and Non-Ohmic behaviour
EXAMPLE: The graph shows the applied voltage V vs resulting current I
through a tungsten filament lamp.
a. Find R when I = 0.5 mA and 1.5 mA. Is this filament ohmic or non-ohmic?
 At 0.5 mA: V = 0.08 V
R = V / I = 0.08 / 0.510-3 = 160 .
 At 1.5 mA: V = 0.6 V
R = V / I = 0.6 / 1.510-3 = 400 .
b. Explain why a lamp filament
might be non-ohmic.
 tungsten is a conductor.
 Therefore, the hotter the filament the
higher R.
But the more current, the hotter a lamp
filament burns.
Thus, the bigger the I the bigger the R.
Since R is not constant
the filament is non-ohmic.
Ohmic and Non-Ohmic behaviour
EXAMPLE: The I-V characteristic
is shown for a non-ohmic component. Sketch in the I-V characteristic for a 40  ohmic component
in the range of 0.0 V to 6.0 V.
SOLUTION:
”Ohmic” means V = IR and R is
constant (and the graph is linear).
Thus V = I40 or I = V / 40.
If V = 0, I = 0 / 40 = 0.0.
If V = 6, I = 6 / 40 = 0.15 A.
But 0.15 A = 150 mA.
Power dissipation
Power is the rate at which electric energy is converted
into another form such as mechanical energy, heat, or light.
It is rate at which the work is done.
▪
▪
▪
▪
P=W/t
P=W/t
P = qV / t
P = (q / t)V
P=IV
1W =
1J
= 1A 1V
1s
P = IV = V2/R = I2 R
This power represents the
energy per unit time delivered
to, or consumed by, an electrical
component having a current I
and a potential difference V.
Power dissipation
PRACTICE: Use the definition of resistance R = V / I.
together with the one we just derived (P = VI) to derive
the following two formulas:
(a) P = I 2R
(b) P = V 2/ R.
SOLUTION:
(a) From R = V / I we get V = IR.
P = IV = I ( IR ) = I 2R.
(b) From R = V / I we get I = V / R.
P = IV = (V / R) (V) = V 2/ R.
P = VI = I 2R = V 2/ R
electrical power
Power dissipation
PRACTICE:
The graph shows the V-I
characteristics of a tungsten
filament lamp.
What is its power consumption
at I = 0.5 mA and at I = 1.5 mA?
SOLUTION:
At 0.5 mA, V = 0.08 V.
P = IV = (0.510-3)(0.08) = 4.010-5 W.
At 1.5 mA, V = 0.6 V.
P = IV = (1.510-3)(0.6) = 9.010-4 W.
Electric circuits
solder joints
An electric circuit is a set of conductors (like wires)
and components (like resistors, lights, etc.) connected
to an electrical voltage source (like a cell or a battery) in
such a way that current can flow in complete loops.
Here are two circuits consisting of cells, resistors, and
wires.
Note current flowing from (+) to (-) in each circuit.
triple-loop circuit
single-loop
circuit
Circuits diagrams
A complete circuit will always
contain a cell or a battery.
The schematic diagram of a cell is this:
this is a
battery…
this is really
a cell…
A battery is just a group
this is the same
of cells connected in series:
battery…
If each cell is 1.5 V, then the battery above is 3(1.5) =
4.5 V. What is the voltage of your calculator battery?
A fixed-value resistor looks like this:
this is a
The schematic of a fixed-value
resistor…
resistor looks like this:
Drawing and interpreting circuit diagrams
EXAMPLE: Draw schematic diagrams of each of the
following circuits:
SOLUTION:
Resistors in Series
• connected in such a way that all components
have the same current through them.
• Burning out of one of the lamp filaments or simply
opening the switch could cause such a break.
▪ Conservation of energy: q = qV1 + qV2 + qV3.
▪  = IR1 + IR2 + IR3 = I(R1 + R2 + R3)
▪  = I(R), where R = R1 + R2 + R3 equivalent resistance in series
the one that could replace all resistors
resulting in the same current
logic: the total or effective resistance would have length L1+ L2+ L3
and resistance is proportional to the length
Resistors in Series
EXAMPLE: Three resistors
of 330  each are connected to
a 6.0 V battery in series
R1
R2
(a) What is the circuit’s equivalent resistance?
(b) What is the current in the circuit?
(a) R = R1 + R2 + R3
R = 330 + 330 + 330 = 990 .
(b) I = V / R = 6 / 990 = 0.0061 A.
R3

(c) What is the voltage on each resistor?
(c) The current I we just found is the same everywhere. Thus each
resistor has a current of I = 0.0061 A.
From Ohm’s law, each resistor has a voltage given by
V = IR = (0.0061)(330) = 2.0 V.
In series the V’s are different if the R’s are different.
Resistors in Parallel
• Electric devices connected in parallel are connected to the
same two points of an electric circuit, so all components have
the same potential difference across them.
• A break in any one path does not interrupt the flow of charge in the other
paths. Each device operates independently of the other devices. The greater
resistance, the smaller current.
• The current flowing into the point of splitting is equal to the sum of the
currents flowing out at that point:
I = I1 + I2 + I3
𝑉
𝑉
𝑉
𝑉
=
+
+
𝑅 𝑅1 𝑅2 𝑅3
1
1
1
1
=
+
+
𝑅 𝑅1 𝑅2 𝑅3
equivelent resistance
is smaller than the
smallest resistance.
equivalent resistance in parallel
the one that could replace all resistors
resulting in the same current
RESISTORS IN COMPOUND CIRCUITS
Now you can calculate current, potential drop and
power dissipated through each resistor
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Investigating combinations of resistors in parallel
EXAMPLE: Three resistors of
330  each are connected to a
R2 R3
R1
6.0 V cell in parallel as shown.
(a) What is the circuit’s resistance?
(b) What is the voltage on each resistor?
SOLUTION:
(a) In parallel, 1 / R = 1 / R1 + 1 / R2 + 1 / R3 so that
1 / R = 1 / 330 + 1 / 330 + 1 / 330 = 0.00909.
Thus R = 1 / 0.00909 = 110 .
(b) The voltage on each resistor is 6.0 V, since the
resistors are in parallel. (Each resistor is clearly directly
connected to the battery).

Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Investigating combinations of resistors in parallel
EXAMPLE: Three resistors of
330  each are connected to a
R2 R3
R1
6.0 V cell in parallel as shown.
(c) What is the current in each
resistor?
SOLUTION:
(c) Using Ohm’s law (I = V / R):
I1 = V1 / R1 = 6 / 330 = 0.018 A.
I2 = V2 / R2 = 6 / 330 = 0.018 A.
I3 = V3 / R3 = 6 / 330 = 0.018 A.
FYI
In parallel the I’s are different if the R’s are different.

Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - voltmeters are connected in parallel
PRACTICE: Draw a schematic diagram
for this circuit:
1.06
SOLUTION:
FYI
Be sure to position the
voltmeter across the desired
resistor in parallel.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - voltmeters are connected in parallel
EXAMPLE:
09.4
00.0
A battery’s voltage is measured
as shown.
(a) What is the uncertainty in
it’s measurement?
SOLUTION:
For digital devices always use
the place value of the least
significant digit as your raw uncertainty.
For this voltmeter the voltage is measured
to the tenths place so we give the raw
uncertainty a value of ∆V = 0.1 V.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - voltmeters are connected in parallel
EXAMPLE:
09.4
A battery’s voltage is measured
as shown.
(b) What is the fractional error in
this measurement?
SOLUTION: Fractional error is
just V / V. For this particular
measurement we then have
V / V = 0.1 / 9.4 = 0.011 (or 1.1%).
FYI
When using a voltmeter the red lead is
placed at the point of highest potential.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - voltmeters are connected in parallel
Consider the simple circuit of battery, lamp, and wire.
To measure the
voltage of the
circuit we merely
01.6
00.0
connect the
voltmeter while
the circuit is
in operation.
lamp cell
voltmeter
in parallel
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams - ammeters are connected in series
To measure the current of the circuit we must break
the circuit and insert the ammeter so that it intercepts all
of the electrons
that normally
00.2
00.0
travel through
the circuit.
ammeter
in series
lamp
cell
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
the circuit must
be temporarily
broken to insert
the ammeter
Circuit diagrams - ammeters are connected in series
PRACTICE: Draw a schematic diagram
for this circuit:
.003
SOLUTION:
FYI
Be sure to position the
ammeter between the desired
resistors in series.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Circuit diagrams
PRACTICE: Draw a schematic diagram for this circuit:
SOLUTION:
FYI
This circuit is a combination series-parallel. In a later
slide you will learn how to find the equivalent resistance
of the combo circuit.
Ammeters and voltmeters
In practical use, we need to be able to measure currents through
components and voltages across various components in electrical
circuits. To do this, we use AMMETERS and VOLTMETERS.
An ammeter – measures current passing through it
• is always connected in series with a component we want to
measure in order that whatever current passes through the
component also passes the ammeter.
• has a very low resistance compared with the
resistance of the circuit so that it will not alter the
current the current being measured.
• In order to not alter the original properties of the circuit ammetrs
would ideally have no resistance, to minimize the effect on the
current that is being measured. Potential difference across it
would be almost zero so no energy would be dissipated in it.
A voltmeter – measures voltage drop between two points
• is always connected across a device (in parallel).
• has a very high resistance so that it takes very little
current from the device whose potential difference
is being measured.
• In order to not alter the original properties of the circuit an ideal
voltmeter would have infinite resistance with no current passing
through it and no energy would be dissipated in it.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
potential
Consider a battery of  = 6 V.
R1 divider
Suppose we have a light bulb
that can only use three volts.
How do we obtain 3 V from a
R2
6 V battery?
A potential divider is a circuit
made of two (or more) series resistors that allows us to
tap off any voltage we want that is less than the battery
voltage.
The input voltage is the emf of the battery.
The output voltage is the voltage drop across R2.
Since the resistors are in series R = R1 + R2.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
potential
Consider a battery of  = 6 V.
R1 divider
Suppose we have a light bulb
that can only use three volts.
How do we obtain 3 V from a
R2
6 V battery?
From Ohm’s law the current I
of the divider is given by I = VIN / R = VIN / (R1 + R2).
But VOUT = V2 = IR2 so that
VOUT = IR2
= R2VIN / (R1 + R2).
VOUT = VIN [ R2 / (R1 + R2) ]
potential divider
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE:
Find the output voltage if the
battery has an emf of 9.0 V, R1
is a 2200  resistor, and R2 is a
330  resistor.
SOLUTION:
Use VOUT = VIN [ R2 / (R1 + R2) ]
VOUT = 9 [ 330 / (2200 + 330) ]
VOUT = 9 [ 330 / 2530 ] = 1.2 V.
FYI
The bigger R2 is in comparison to R1, the closer VOUT is
in proportion to the total voltage.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE:
Find the value of R2 if the battery
has an emf of 9.0 V, R1 is a
2200  resistor, and we want an
output voltage of 6 V.
SOLUTION:
Use the formula VOUT = VIN [ R2 / (R1 + R2) ]. Thus
6 = 9 [ R2 / (2200 + R2) ]
6(2200 + R2) = 9R2
13200 + 6R2 = 9R2
13200 = 3R2
R2 = 4400 
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE: A light-dependent
resistor (LDR) has R = 25  in
bright light and R = 22000  in low
light. An electronic switch will turn
on a light when its p.d. is above 7.0 V.
What should the value of R1 be?
SOLUTION: Use VOUT = VIN [ R2 / (R1 + R2) ]. Thus
7 = 9 [ 22000 / (R1 + 22000) ]
7(R1 + 22000) = 9(22000)
7R1 + 154000 = 198000
7R1 = 44000
R1 = 6300  (6286)
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE: A thermistor has a
resistance of 250  when it is in
the heat of a fire and a resistance
of 65000  when at room
temperature. An electronic switch
will turn on a sprinkler system when
its p.d. is above 7.0 V.
(a) Should the thermistor be R1 or R2?
SOLUTION:
Because we want a high voltage at a high temperature,
and because the thermistor’s resistance decreases with
temperature, it should be placed at the R1 position.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE: A thermistor has a
resistance of 250  when it is in
the heat of a fire and a resistance
of 65000  when at room
temperature. An electronic switch
will turn on a sprinkler system
when its p.d. is above 7.0 V.
(b) What should R2 be?
SOLUTION: In fire the thermistor is R1 = 250 .
7 = 9 [ R2 / (250 + R2) ]
7(250 + R2) = 9R2
1750 + 7R2 = 9R2  R2 = 880  (875)
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
PRACTICE: A filament lamp is rated
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(a) Sketch the variation of the p.d. V vs. the current I for
a typical filament lamp. Is it ohmic? ohmic means linear
SOLUTION: Since the temperature increases with
the current, so does the resistance.
V
But from V = IR we see that R = V / I,
non-ohmic
which is the slope.
Thus the slope should increase with I.
I
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
R1
PRACTICE: A filament lamp is rated
R2
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(b) The potentiometer is adjusted so that the meter
shows 4.0 V. Will it’s contact be above Y, below Y, or
exactly on Y?
SOLUTION: The circuit is acting like a potential divider
with R1 being the resistance between X and Y and R2
being the resistance between Y and Z.
Since we need VOUT = 4 V, and since VIN = 6 V, the
contact must be adjusted above the Y.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
R1
PRACTICE: A filament lamp is rated
R2
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(c) The potentiometer is adjusted so that the meter
shows 4.0 V. What are the current and the resistance of
the lamp at this instant?
SOLUTION: P = 0.80 W and V = 4.0 V.
From P = IV we get 0.8 = I(4) so that I = 0.20 A.
From V = IR we get 4 = 0.2R so that R = 20. .
You could also use P = I 2R for this last one.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
R1
PRACTICE: A filament lamp is rated
R2
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(d) The potentiometer is adjusted so that the meter
shows 4.0 V. What is the resistance of the Y-Z portion
of the potentiometer?
SOLUTION: Let R1 = X to Y and R2 = Y to Z resistance.
Then R1 + R2 = 24 so that R1 = 24 – R2.
From VOUT = VIN [ R2 / (R1 + R2) ] we get
4 = 7 [ R2 / (24 – R2 + R2) ]  R2 = 14  (13.71).
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
R1
PRACTICE: A filament lamp is rated
R2
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(e) The potentiometer is adjusted so that the meter
shows 4.0 V. What is the current in the Y-Z
portion of the potentiometer?
SOLUTION:
V2 = 4.0 V because it is in parallel with the lamp.
I2 = V2 / R2
= 4 / 13.71 = 0.29 A
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Potential divider circuits
R1
PRACTICE: A filament lamp is rated
R2
at “4.0 V, 0.80 W” on its package.
The potentiometer has a resistance
from X to Z of 24  and has linear
variation.
(f) The potentiometer is adjusted so that the meter
shows 4.0 V. What is the current in the ammeter?
SOLUTION: The battery supplies two currents.
The red current is 0.29 A because it is the I2 we just
calculated in (e).
The green current is 0.20 A found in (c).
The ammeter has both so I = 0.29 + 0.20 = 0.49 A.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Solving problems involving circuits
PRACTICE: A battery is connected to a 25-W lamp as
shown.
What is the lamp’s
01.4
00.0
resistance?
SOLUTION:
Suppose we connect
a voltmeter to the
circuit.
We know P = 25 W.
We know V = 1.4 V.
From P = V 2 / R we get
R = V 2/ P = 1.4 2 / 25 = 0.078 .
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Solving problems involving circuits
PRACTICE: Which circuit shows the correct setup to
find the V-I characteristics of a filament lamp?
SOLUTION:
lamp
two
The voltmeter must
current
currents
be in parallel with the
lamp.
It IS, in ALL cases.
short
no
The ammeter must
circuit!
currents
be in series with the
lamp and must read
only the lamp’s current.
The correct response is B.
Topic 5: Electricity and magnetismequivalent ckt
5.2 – Heating effect of electric currents
Solving problems involving circuits
PRACTICE: A non-ideal voltmeter is
used to measure the p.d. of the
20 k resistor as shown. What will
its reading be?
SOLUTION: There are two currents in the circuit
because the voltmeter does not have a high enough
resistance to prevent the green one from flowing.
The 20 k resistor is in parallel with the 20 k so that
1 / R = 1 / 20000 + 1 / 20000 = 2 / 20000.
R = 20000 / 2 = 10 k.
But then we have two 10 k resistors in series and
each takes half the battery voltage, or 3 V.
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Solving problems involving circuits
PRACTICE: All three circuits use the same resistors
and the same cells.
Highest I
0.5R
Lowest I
2R
parallel
series
Which one of the following
shows the correct ranking
for the currents passing
through the cells?
SOLUTION: The bigger the
R the smaller the I.
Middle I
1.5R
R
0.5R
combo
Topic 5: Electricity and magnetism
5.2 – Heating effect of electric currents
Solving problems involving circuits
PRACTICE: The voltmeter has
infinite resistance.
What are the readings on the
voltmeter when the switch is
open and closed?
SOLUTION:
With the switch open the
green R is not part of the
circuit. Red and orange
split the battery emf.
With the switch closed the
red and green are in parallel and are (1/2)R.
E
Kirchhoff’s rules – junction, branch, and loop
”Solving” a circuit consists of finding the
voltages and currents of all of its components.
Consider the following circuit containing a
few
batteries and resistors:
A junction is a point in a circuit where three or more wires are
connected together.
A branch is all the wire and all the components connecting
one junction to another.
A loop is all the wire and all the components in a
complete circle.
Kirchhoff’s rules – the rule for current I
”Solving” a circuit consists of finding the
voltages and currents of all of its components.
From conservation of charge
the sum of the currents at each
junction is zero.
I = 0 (junction)
Kirchhoff’s rule for I
From conservation of energy
the sum of the voltages in each
loop is zero.
V = 0 (loop)
Kirchhoff’s rule for V
Kirchhoff’s rules – the rule for current I
 ”Solving” a circuit consists of finding the voltages and currents of
all of its components.
 STEP 1: Assign a current to each branch.
 If you have a good idea which way it flows, choose that direction.
 If you don’t, an arbitrary direction will do just fine.
 If a current turns out to have a negative
solution, you will interpret that as
meaning that it flows in the opposite
direction.
I3
I1
 If a current enters a junction it is a “gain” and
is assigned a POSITIVE value.
 If a current leaves a junction it is a “loss” and
is assigned a NEGATIVE value.
I2
For the TOP junction, I1 and I3 are both POSITIVE and I2 is NEGATIVE.
I1 – I2 + I3 = 0
For the BOTTOM junction, I1 and I3 are both NEGATIVE and I2 is POSITIVE.
I2 – I1 – I3 = 0
Kirchhoff’s rules – the rule for voltage V
 ”Solving” a circuit consists of finding the voltages and currents
of all of its components.
 Give each resistor a voltage V and each cell or battery an emf .
 Cells and batteries increase the energy of the current,
resistors decrease the energy.
 Since the energy is qV (or q), if we sum up all the energy gains
and losses in each LOOP we must get zero.
 We can go either CW or CCW – it doesn’t matter.
V2
1
V1
I1
I3
V4
V3
2
I2
If our loop goes in the direction of the
branch current assigned earlier, the
resistor energy change is negative.
negative
 If our loop goes against the branch
current, the sign is reversed for the resistor. positive
 For our loop we have
–qV1 and –qV2 and –qV4.
V2
I1
I3
V4
1
 If our loop goes through a cell from
negative to positive, the cell energy positive
V3

2
change is positive.
V1
I2
 If our loop goes from positive to
negative through a cell the energy
change is negative.
negative
 For our loop we have
+q1 and –q2.
Use Kirchhoff’s rule for V for each loop.
For our loop we have
–V +  + –V + –V + – = 0
1
1
2
4
2
q cancels
Kirchhoff’s rules – the rule for voltage V
PRACTICE: Using the voltage rule
write the equation for the other
loop.
SOLUTION:
negative
–
2
+
–V + –V
3
4
= 0.
negative
V2
1
V1
I1
I3
V4
2
V3
I2
Kirchhoff’s rules – solving the circuit
EXAMPLE: Suppose each of the resistors is R = 2.0 , and the emfs are
1 = 12 V and 2 = 6.0 V. Find the voltages and the currents of the circuit.
▪ I1 – I2 + I3 = 0 (1)
+ –V2 + –V4 + 1 + – 2 = 0,
– – V – V = 0
2
3
4
▪ –V
▪
–V
1
1
–
2
+ –V2 + –V4 + 1 + – 2 = 0 & V=IR  –2I1 + –2I1 + –2I2 + 12 + –6 = 0 (2)
– V3 – V4 = 0 & V=IR 
–– 6
We now have three equations in I:
(1)  I3 = I2 – I1.
I1 = 1.8 A
(2)  3 = 2I1 + I2.
I2 = -0.6 A
I3 = -2.4 A
(3)  3 = -I2 + -I3
– 2I3 – 2I2 = 0 (3)
Finally, we can redraw our currents:
resistor voltages: V = IR
V1 = 1.8(2) = 3.6 V.
V2 = 1.8(2) = 3.6 V.
V3 = 2.4(2) = 4.8 V.
V4 = 0.6(2) = 1.2 V