Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Electric potential difference
5.1.1 Define electric potential difference.
5.1.2 Determine the change in potential energy
when a charge moves between two points at
different potentials.
5.1.3 Define the electron-volt.
5.1.4 Solve problems involving electric potential
difference.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
Recall from previous classes that there are two
types of charge: (+) and (-).
You may also recall the old charge rule that says
like charges repel and unlike charges attract.
The basic charge that we will work with is the
electron. The electron is a negative charge.
The algebraic symbol for charge is q and it is
measured in coulombs (C).
An electron and a proton have the following
charges:
electron qe = -1.610-19 C
electron, proton
and elementary
proton qp = +1.610-19 C
charge
elementary e = +1.610-19 C
Thus qe = -e and qp = e.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
EXAMPLE:
How many electrons are in a charge of q = -15 C?
SOLUTION:

q = (-1510-6 C)(1 electron/-1.610-19 C)

q = 9.41013 electrons
EXAMPLE:
What is the net charge of a mole of sulfur S+2
ions?
SOLUTION:
(1 mol S+2)(2 e/S+2)(6.021023/mol)(1.610-19 C/e)

q = 1.9105 C.
FYI
The word electron/s is often abbreviated e-.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Cathode
+
+
+
+
+
+
+
+
The chemical cell
Define electric potential difference.
We begin our study of electric potential
difference by looking at a chemical cell.
To make a chemical cell, or a battery, you can
begin with a container of weak acid, and two
electrodes made of different metals.
Different metals dissolve in
Anode
acids at different rates.
- +
When a metal dissolves, it
- +
enters the acid as a positive
- +
ion, leaving behind an electron.
- +
We call the weak acid the
electrolyte. We call the least
- +
negative metal the anode. We
call the most negative metal the
Weak Acid
cathode.
Electrolyte
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
The chemical cell
Define electric potential difference.
Think of a chemical cell as a
device that converts chemical
energy into electrical energy.
+
If we connect conductors and a
light bulb to the anode and the
cathode we see that electrons
Anode
Cathode
begin to flow in what’s called
(+)
(-)
- +
an electric current.
- +
- +
FYI
- +
- +
- +
Why do the electrons run from
- +
- +
cathode to anode in the external
- +
- +
circuit instead of the reverse?
- +
+
They try to escape the more
- +
highly negative cathode for the
Electrolyte
less negative anode.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
-
Cathode (-)
+
+
+
+
+
-
Electrolyte
+
+
+
+
+
+
+
+
The chemical cell
Define electric potential difference.
Each time an electron leaves
the cathode, the acid creates
another in the cathode.
+
Each time an electron enters
the anode, the acid neutralizes
an electron in the anode.
Anode (+)
This process is maintained
until one of the metals or
the electrolyte is used up
completely.
Thus, the battery will continue
to supply electrons to the
external circuit until its
chemical energy runs out.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
The chemical cell
PRACTICE: A current isn’t really
just one electron moving through
a circuit. It is in reality more
+
like a chain of them, each one
shoving the next through the
circuit. Recalling the rule of
Anode (+) B
A Cathode (-)
charges which says “like
charges repel, and unlike
- +
- +
- +
charges attract,” explain why
- +
- +
the current flows from cathode
- +
to anode in terms of the force
- +
- +
of repulsion.
- +
- +
Consider electrons at A and B.
- +
+
Both electrons feel a repulsive
- +
force at their respective
Electrolyte
electrodes.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
The chemical cell
PRACTICE: A current isn’t really
just one electron moving through
a circuit. It is in reality more
+
like a chain of them, each one
shoving the next through the
circuit. Recalling the rule of
Anode (+) B
A Cathode (-)
charges which says “like
charges repel, and unlike
- +
- +
- +
charges attract,” explain why
- +
- +
the current flows from cathode
- +
to anode in terms of force of
- +
- +
repulsion.
- +
- +
But A’s is bigger than B’s
- +
+
because there are more negatives
- +
repelling at A than B.
Electrolyte
A wins the “push of war.”
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
The chemical cell
PRACTICE: Consider an electron
at the point A and an electron
at the point B. Which one has
+
more potential energy?
The electron at B has already
traveled through the circuit
Anode (+) B
A Cathode (-)
and is returning to the
battery so it can do no more
- +
- +
- +
work on the circuit (like
- +
- +
lighting up the bulb).
- +
The electron at A still has the
- +
- +
full circuit to travel and can
- +
- +
thus still do work (like
- +
+
lighting up the bulb).
- +
Thus the electron at A has
Electrolyte
more potential energy.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
-
Cathode (-)
+
+
+
+
+
-
Electrolyte
+
+
+
+
+
+
+
+
The chemical cell
Define electric potential difference.
Back in the days of Ben
Franklin scientists understood
that there were two types of
+
electricity: positive and
negative.
What they didn’t know was
Anode (+)
which one was actually free
to travel through a circuit.
The influential Ben Franklin
guessed wrongly that it was the
positive charge.
Thus, his vision of our
chemical cell looked like the
one on the next slide, not the
one on this slide.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
+ + + + -
Electrolyte
FYI
Positive charge
flow is called
conventional
current and is
still used in
circuit
analysis.
- +
- +
- +
- +
- +
more negative
+ -
= less positive
= more positive
Franklin’s chemical cell
-
Cathode (-)
Anode (+)
less negative
Cathode (-)
Anode (+)
+
+
+
+
+
+
+
+
-
+
-
Electrolyte
+
+
+
+
+
+
+
+
The chemical cell
-
+
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
Franklin’s chemical cell
PRACTICE: Consider positive
charges at points A and B.
Which one has more potential
+
energy?
The charge at B has already
traveled through the circuit
B Cathode (-)
Anode (+) A
and is returning to the
battery so it can do no more
+ + + work on the circuit (like
+ + lighting up the bulb).
+ The charge at A still has the
+ + full circuit to travel and can
+ + thus still do work (like
+ + lighting up the bulb).
+ Thus the positive charge at A
Electrolyte
has more potential energy.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
We define the electric
potential difference ∆V as the
amount of work done in moving
+
a positive charge q from a point
of lower potential energy to a
point of higher potential
Anode (+)
energy.
+
∆V = ∆EP/q
potential difference
+
The units for electric
+
potential difference are volts
+
(V) or, as can be seen from
+
-1
the formula, (J C ).
+
FYI
Electric potential difference
is often abbreviated p.d.
-
Cathode (-)
+ + -
+ -
+ + + + -
Electrolyte
BLACK = low
potential
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
The instrument used to measure
electric potential difference
is called the voltmeter.
01.6
00.0
This cell has
a potential
difference
of 1.6 V.
Cathode (-)
Anode (+)
+
+
+
+
+
+
+
+
-
+ -
+ + + + -
Electrolyte
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Solve problems involving electric potential
difference.
EXAMPLE:
A battery’s voltage is
measured as shown.
(a) What is the uncertainty
in it’s measurement?
SOLUTION:
For measuring 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.
09.4
00.0
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Solve problems involving electric potential
difference.
EXAMPLE:
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%).
09.4
FYI
When using a voltmeter the red lead is
placed at the point of highest potential.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
(+) (-)
-
Electrolyte
low potential
Cathode (-)
Chemical
energy
high potential
Define electric potential difference.
Think of a battery as an engine
that uses chemical energy to
take positive charges and move
+
them from low to high potential
within the battery so that they
can do work outside the battery
Anode (+)
in the external circuit.
To simplify the drawing of
circuits schematic symbols
have been
developed.
lamp
switch
For this
circuit we
cell
have:
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric potential difference.
As an analogy to gravitational potential energy,
think of the chemical cell as an elevator,
powered by a chemical engine.
Inside the cell, positive
(+)
charges at low potential
are raised through chemical
energy to high potential.
Outside the cell, positive
charges at high potential
are released into the
(-)
external circuit to do
chemical
external
their electrical work.
cell
circuit
FYI
Remember that there is a whole chain of
electrons moving at once. One pushes the next…
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
EXAMPLE:
200 C of charge is brought from an electric
potential of 2.0 V to an electric potential of 14
V through use of a car battery. What is the
change in potential energy of the charge?
SOLUTION:
From ∆V = ∆EP/q we see that ∆EP = q∆V. Thus
∆EP = q(V – V0)
∆EP = (20010-6)(14 – 2)
∆EP = 0.0024 J.
FYI
Note the increase rather than the decrease.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
PRACTICE:
An elementary charge is carried from the cathode
to the anode within a chemical cell whose
potential difference is 1.64 V. What is the
change in potential energy of the charge?
SOLUTION:
We see that q = +1.610-19 C and ∆V = 1.64 V.
From ∆V = ∆EP/q we see that ∆EP = q∆V so that
∆EP = (1.610-19)(1.64)
∆EP = 2.610-19 J.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
EXAMPLE: An evacuated glass tube is shown
with a potential difference of 1000 V between the two plates. A thin-wire filament that emits electrons when heated up is
located behind a hole in the negative plate.
(a) Find the change in potential energy of the
electron as it passes through the hole and
travels between the plates.
SOLUTION:
We see that q = -1.610-19 C and ∆V = +1000 V.
From ∆EP = q∆V we see that
∆EP = (-1.610-19)(+1000)
∆E = -1.610-16 J.
+
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
EXAMPLE: An evacuated glass tube is shown
with a potential difference of 1000 V between the two plates. A thin-wire filament that emits electrons when heated up is
located behind a hole in the negative plate.
(b) Find the change in kinetic energy of the
electron as it passes through the hole and
travels between the plates.
SOLUTION:
From ∆EK + ∆EP = 0 we get

∆EK = -∆EP = +1.610-16 J.
Note that the loss in potential energy is equal
to the gain in kinetic energy.
+
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
EXAMPLE: An evacuated glass tube is shown
with a potential difference of 1000 V between the two plates. A thin-wire filament that emits electrons when heated up is
located behind a hole in the negative plate.
(c) Given that the mass of an electron is
9.1110-31 kg, find the speed of the electron if
its speed at the negative plate is close to zero.
SOLUTION:
From the previous problem ∆EK = +1.610-16 J.
Thus 1.610-16 = EK,f – EK,0 = (1/2)mv2 - (1/2)mu2

1.610-16 = (1/2)(9.1110-31)v2

v = 1.9107 m s-1.
+
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
PRACTICE: Show that for a general free
charge q having mass m and an initial
speed of zero, qV = (1/2)mv2, where V is the
potential difference the charge accelerates
through as it loses potential energy.
From the previous problems recall ∆EK = -∆EP and
∆EP = -qV.
Thus
∆EK = -∆EP = -(-qV)

qV = EK,f – EK,0

qV = (1/2)mv2 - (1/2)mu2

qV = (1/2)mv2.
FYI
A loss of EP is always a gain in EK.
+
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Determine the change in potential energy when a
charge moves between two points at different
potentials.
We can tailor the previous formula to fit the
case where q = e, the elementary charge. This
formula is in the physics data booklet.
Ve = (1/2)mv2
kinetic energy of an accelerated
elementary charge
PRACTICE: Find the final velocity of a proton
that has been accelerated from rest through a
potential difference of 125 V. mp = 1.6710-27 kg.

Ve = (1/2)mv2
125(1.610-19) = (1/2)(1.6710-27)v2
v = 1.5105 m s-1.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define the electron-volt.
We define a new unit of energy called the
electron volt (eV) as the energy gained by a
single electron in moving through a potential
difference of exactly 1 V.
From ∆EP = q∆V we see that ∆EP = (1.610-19)(1) or
1 eV = 1.610-19 J
electron volt
PRACTICE:
From the last example we discovered that ∆EP =
2.610-19 J. Convert this energy into electronvolts.
SOLUTION:
Since 1 eV = 1.610-19 J we have

∆EP = (2.610-19 J)(1 eV/1.610-19 J)
∆EP = 1.6 eV.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Electric current and resistance
5.1.5 Define electric current.
5.1.6 Define resistance.
5.1.7 Apply the resistance equation.
5.1.8 State Ohm’s law.
5.1.9 Compare ohmic and non-ohmic behavior.
5.1.10 Derive and apply expressions for
electrical power dissipation in resistors.
5.1.11 Solve problems involving potential
difference, current and resistance.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
Electric current I is the time rate ∆t at which
charge ∆q moves past a particular point in a
circuit.
I = ∆q/∆t or I = q/t
electric current
From the formula it should be clear that current
is measured in coulombs per second (C s-1) which
is called an ampere (A).
PRACTICE:
Many houses have 20-amp(ere) service. How many
electrons per second is this?
SOLUTION:
20 A is 20 C per s so we only need to know how
many electrons are in 1 C.

(20 C)(1 e- / 1.610-19 C) = 1.31020 e-.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
Electric current I is the time rate ∆t at which
charge ∆q moves past a particular point in a
circuit.
A simple model may help clarify how electrons
move through a circuit.
Think of conductors as pipes that hold electrons.
The chemical cell pushes an electron out of the
(-) side. This electron in turn pushes the next,
and so on, because like charges repel.
This electromotive force is transfered
simultaneously to every charge in the circuit.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
Electric current I is the time rate ∆t at which
charge ∆q moves past a particular point in a
circuit.
EXAMPLE: Suppose each (-) represents an electron.
Find the current in electrons per second and then
convert it to amperes.
SOLUTION:
Choose a reference point and start a timer.
The rate is about 7 electrons each 12 s or
I = 7/12 = 0.58 e- s-1.
I = (0.58 e- s-1)(1.610-19 C/1 e-) = 9.310-20 A.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
EXAMPLE: Explain why when a wire is cut current
stops everywhere and does not leak into the air.
SOLUTION:
In order to free an e- from a conductor takes
quite a bit of energy. This is why when a metal
is conducting electricity and you stand near it
you don’t get electrocuted by e- jumping off of
the metal. This is also why when you cut the wire
e- do not leak out into the surrounding
environment.
Finally, if the chain is broken the push stops,
so the current stops everywhere.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
EXAMPLE: Sketch in the electron current and the
conventional current in each of the circuits.
SOLUTION:
Electron current originates at
lamp
switch
the cathode (-).
electron
Conventional current originates at
current
the anode (+).
Finally, as we just discussed,
(+) (-)
the current is not just a single
charge. Rather it is a whole
lamp
chain of charge that is moving
switch
as one stream.
conventional
FYI
current
All currents start and stop as
(+) (-)
one in a circuit.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
Consider the simple circuit of battery, lamp, and
wire. As we have just learned, in order to make
the bulb light the circuit has to be complete.
To measure the
voltage of the
circuit we merely connect the
voltmeter while
the circuit is
in operation.
01.6
00.0
lamp
cell
voltmeter
in parallel
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define electric current.
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
travel through
the circuit.
00.2
00.0
ammeter
in series
lamp
cell
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define 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: Electric currents
5.1 Electric potential difference,
current and resistance
Define 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: Electric currents
5.1 Electric potential difference,
current and resistance
Define 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 ohmmeter.
This resistor has a resistance
of 330.4 .
0.L
0.0
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: Electric currents
5.1 Electric potential difference,
current and resistance
Define resistance.
The different types of resistors have different
symbols.
fixed-value
resistor
2 leads
variable
resistor
2 leads
potentiometer
3 leads
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define resistance.
The different types of resistors have different
symbols.
thermister
2 leads
As temperature
increases
resistance
decreases
Lightdependent
resistor
(LDR) 2 leads
As brightness
increases
resistance
decreases
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Define resistance.
The precise definition of resistance is as
follows:
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:
A fixed resistor has a current of 2.5 mA when it
has a 6.0 V potential difference across it. What
is its resistance?
 R = V/I = 6/2.510-3 = 2400  (or 2.4 k).
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Apply the resistance equation.
In order to better understand electrical
resistance, consider two identical milk shakes.
In the first
Resistance is a
experiment the
measure of how
straws have the
hard it is to
same diameter, but
pass something
different lengths.
through a
material.
In the second
experiment the
straws have the
R  L
R  1/A
same length, but
different
diameters.
Thus we see
that R  L/A.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Apply the resistance equation.
Of course conductors and resistors are not hollow
like straws. And instead of milk shake current we
have electrical current.
A
Even through solids R  L/A.
CARBON
But R also depends on the material
L
through which the electricity is
flowing.
A
For example the exact same size
COPPER
of copper will have much less
L
resistance than the carbon.
It turns out that the complete resistance
equation looks like this:
R = L/A
resistance equation
The Greek rho () is the resistivity of the
particular material the resistor is made from.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Apply the resistance equation.
Here is a table of resistivities and thermal
coefficients. Note that from the resistance
equation resistivity  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: Electric currents
5.1 Electric potential difference,
current and resistance
Apply the resistance equation.
From the previous table it should be clear 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
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
State 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
potential difference is proportional
to the current.”
In formula form Ohm’s law looks like this:
V  I or V/I = CONST or V = IR
Ohm’s law
FYI
Don’t write Ohm’s as R = V/I because this just
the definition of resistance.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Compare ohmic and non-ohmic behavior.
A material is considered ohmic if it behaves
according to Ohm’s law. In other words the
resistance stays constant as the voltage changes.
EXAMPLE: Three graphs for three electrical
components are shown. Label appropriate graphs
with the following labels: ohmic, non-ohmic, R
increasing, R decreasing, and R constant.
SOLUTION:
V
V
V
nonR
R
First label the
ohmic
resistance
nonR
dependence.
ohmic
ohmic
R = V/I so R is
I
I
I
just the slope of the V vs. I graph.
Ohm’s law states the R is constant. Thus only one
graph is ohmic.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Compare ohmic and non-ohmic behavior.
EXAMPLE: The graph shows
the applied voltage V vs.
resulting current I through
a tungsten filament lamp.
Find R when I = 0.5 mA and
1.5 mA. Is this filament
ohmic or non-ohmic?
SOLUTION:
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 .
Since R is not constant the filament is nonohmic. (Or observe that V vs. I is non-linear.)
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Compare ohmic and non-ohmic behavior.
EXAMPLE: The graph shows
the applied voltage V vs.
resulting current I through
a tungsten filament lamp.
Give a plausible explanation
as to why the lamp filament
is non-ohmic.
SOLUTION:
The temperature coefficient
for tungsten is positive, as
it is for all of the conductors in the table.
Therefore, the hotter the filament the higher R.
But the more current, the hotter and brighter a
lamp filament burns.
Thus, the bigger the I the bigger the R.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Solve problems involving potential difference,
current and resistance.
EXAMPLE: The graph shows the
I-V characteristic for a nonohmic component. In the graph
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 graph is linear).
Thus V = 40I 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.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Derive and apply expressions for electrical power
dissipation in resistors.
Way back when, we defined the electrical
potential difference as V = ∆EP/q.
We rewrote this as ∆EP = Vq.
Recall that power P is energy production (or
usage or delivery) per unit time. Thus for
electrical power we have:
P = ∆EP/t
= Vq/t
= V(q/t)
P = VI.
FYI
The IBO expects you to be able to derive P = VI
just like this.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Derive and apply expressions for electrical power
dissipation in resistors.
PRACTICE: Recall that the definition of
resistance is R = V/I. Using this relationship
together with the one we just derived (P = VI)
derive the following two formulas:
(a) P = I2R and (b) P = V2/R.
SOLUTION:
From R = V/I we get V = IR.
For (a): P = IV = I(IR) = I2R.
From R = V/I we get I = V/R.
For (b): P = IV = (V/R)(V) = V2/R.
Thus we can put it all together.
P = VI = I2R = V2/R
electrical power
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Derive and apply expressions for electrical power
dissipation in resistors.
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.
Topic 5: Electric currents
5.1 Electric potential difference,
current and resistance
Derive and apply expressions for electrical power
dissipation in resistors.
PRACTICE: A battery is connected to a 25-W lamp
as shown.
What is the
lamp’s
resistance?
SOLUTION:
Suppose we connect
a voltmeter to the
circuit.
We know P = 25 W.
We know V = 1.4 V.
From P = V2/R we get
R = V2/P = 1.42/25 = 0.078 .
01.4
00.0