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• When the ends of an electric conductor are at different electric
potential, charge flows from one end to the other. Voltage is what
causes charge to move in a conductor. Charge moves toward lower
potential energy the same way as you would fall from a tree.
• Voltage plays a role similar to pressure in a pipe; to get water to flow
there must be a pressure difference between the ends, this pressure
difference is produced by a pump
• A battery is like a pump for charge, it provides the energy for pushing
the charges around a circuit
• You can have voltage, but without a
path (connection) there is no current.
An
electrical
outlet
voltage
Current– flow of electric charge
If I connect a battery to the ends of the copper bar the
electrons in the copper will be pulled toward the positive
side of the battery and will flow around and around.
 this is called current – flow of charge
copper
An electric circuit!
Duracell
+
Electric current (symbol I)
◊ the flow of electric charge q that can occur in solids, liquids and gases.
q
• DEF: the rate at which charge flows
past a given cross-section.
• measured in amperes (A)
q
I=
t
1C
1A =
1s
Solids – electrons in metals and graphite, and holes in semiconductors
Liquids – positive and negative ions in molten and aqueous electrolytes
Gases – electrons and positive ions stripped from gaseous molecules
by large potential differences.
Electrical resistance (symbol R)
• 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.
atoms
(actually
positive ions)
free electron
.
The resistance is a measure of how hard it is to
pass something through a material.
path
Electrical Resistance
▪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.
▪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 temperature- dependent resistors, changing their
resistance in response to their temperature.
▪Light-dependent resistors (LDRs) change their resistance in
response to light intensity.
Electrical Resistance
▪The different types of resistors have different schematic symbols.
fixed-value resistor
2 leads
potentiometer
3 leads
variable resistor
2 leads
light-dependent
resistor (LDR)
2 leads
thermister
2 leads
As temperature increases
resistance decreases
As brightness increases
resistance decreases
– photoconductivity
Electrical 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 .
▪A reading of 0.L on an ohmeter means
“overload”. The resistance (of the air) is
too high to record with the meter.
Definition of 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
Factors affecting resistance
▪ we ignore the resistance of a connecting wire calculations
▪ but resistance of a wire can not be neglected
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 low temperatures some materials, 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 order to keep them going.
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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.
Electrical 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
Tungsten
5.610-8
4.510-3
Gold
2.310-8
Iron
Silver
Nonconductors
Glass
1012
Rubber
1015
Wood
1010
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 .
A
L
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:
IV
or 𝐼 =
𝑉
𝑅
or
𝑉
= 𝑐𝑜𝑛𝑠𝑡.
𝐼
Ohm’s law
▪ Ohm’s law applies to components with constant R.
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
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
shown for a non-ohmic compoSketch in the I-V character40  ohmic component
of 0.0 V to 6.0 V.
▪”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.
▪ 0.15 A = 150 mA.
is
nent.
istic for a
in the range
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 = 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:
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?
▪ 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.
this is really a
cell…
▪The schematic diagram of a cell is this:
this is a
battery…
▪A battery is just a group
of cells connected in series:
this is the same
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?
▪The schematic of a fixed-value resistor :
Drawing and interpreting circuit diagrams
Schematic diagrams of each of the following circuits:
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 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
equivalent resistance in parallel
equivalent resistance is
smaller than the
smallest resistance.
the one that could replace all resistors
resulting in the same current
RESISTORS IN COMPOUND CIRCUITS
Now you can calculate current, potential
drop/potential difference/voltage and power
dissipated through each resistor
Resistors in series
Three resistors of 330  each are connected to a 6.0 V battery in series
R1
R2
R3

What is the voltage and current on each resistor?
▪ R = R1 + R2 + R3
R = 330 + 330 + 330 = 990 
▪ I = V / R = 6 / 990 = 0.0061 A
▪ The current I = 0.0061 A is the same in each resistor.
▪ voltage/potential difference across each resistor:
V = I R1 = I R2 = I R3 = (0.0061)(330)
= 2.0 V
▪ In series the V’s are different if the R’s are different.
Resistors in series and parallel
Three resistors of 330  each are connected to a 6.0 V cell in
parallel.
What is the voltage and current on each resistor?
▪ 1/R = 1/R1 + 1/R2 + 1/R3
R = 110 
1/R = 1/330 + 1/330 + 1/330 = 0.00909
▪ The voltage on each resistor is 6.0 V, since the resistors are in
parallel. (Each resistor is clearly directly connected to the battery).
▪ I1 = V1 / R1 = 0.018 A
▪ I2 = V2 / R2 = 6 / 330 = 0.018 A
▪ I3 = V3 / R3 = 6 / 330 = 0.018 A
▪ In parallel the I’s are different if the R’s are different.
Find power of the source, current in each resistor, terminal potential, potential drop across each resistor
and power dissipated in each resistor.
1. step: find total/equivalent resistance
Req = 120 Ω
2. step: find current in main circuit: 𝐼1
=
𝜀
𝑅
I1 = ε ∕ Req = 0.3 A
3. step: to find currents in parallel branches use the Kirchhoff's laws:
voltage is the same across all resistors in parallel:
𝐼2 𝑅2 = 𝐼3 𝑅3 , and 𝐼1 = 𝐼2 +𝐼3
one could use 𝜀 = 𝐼1 r + 𝑉𝐴𝐵
𝑉𝐴𝐵 = 𝐼2 𝑅2 = 𝐼3 𝑅3
100𝐼2 = 50𝐼3 → 𝐼3 = 2𝐼2
0.3 = 𝐼2 +𝐼3 → 0.3 = 3𝐼2 → 𝐼2 = 0.1 A
80 Ω
100 Ω
50 Ω
6.7 Ω
potential drops
V = IR
0.3x80 = 24 V
0.1x100 = 10 V
0.2x50 = 10 V
0.3x6.7 = 2 V
power dissipated
P = IV
0.3x24 = 7.2 W
0.1x10 = 1 W
0.2x10 = 2 W
0.3x2 = 0.6 W
𝐼3 = 0.2 A
ε = Σ all potential drops: (Kirchhoff's
law)
36 V = 2 V + 24 V + 10 V
power dissipated in the circuit =
power of the source
0.6 + 2 + 1 + 7.2 = 0.3x36
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.
To measure the potential difference across resistor, we use a VOLTMETER.
• Voltmeter is in PARALLEL with the resistor we are measuring.
Rvoltmeter >> R 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 (1/ Rvoltmeter ≈ 0) with no current passing
through it and no energy would be dissipated in it.
Circuit diagrams - voltmeters are connected in parallel
Draw a schematic diagram for this circuit
1.06
▪Be sure to position the voltmeter
across the desired resistor in parallel.
Circuit diagrams - voltmeters are connected in parallel
EXAMPLE:
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.
09.4
00.0
(b) What is the fractional error in this
measurement?
SOLUTION: Fractional error is just V / V.
For this particular measurement :
V / V = 0.1 / 9.4 = 0.011 (or 1.1%).
▪When using a voltmeter the red lead is placed at the point of highest potential.
Circuit diagrams - voltmeters are connected in parallel
▪Consider the simple circuit of battery, lamp, and wire.
▪To measure the voltage in the circuit we merely connect the
voltmeter while the circuit is in operation.
01.6
00.0
lamp
cell
voltmeter
in parallel
To measure the current, we use an AMMETER.
• Ammeter is in SERIES with resistor R in order that whatever current
passes through the resistor also passes through the ammeter.
Ram << R, so it doesn’t change the current being measured.
• Req = R+ Rammeter ≈ 𝑅 No energy would be dissipated in ammeter.
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 travel through the circuit.
ammeter
in series
lamp
cell
00.2
00.0
Circuit diagrams - ammeters are connected in series
PRACTICE: Draw a schematic diagram for this circuit:
SOLUTION:
the circuit must
be temporarily
broken to insert
the ammeter
.003
▪Be sure to position the ammeter
between the desired resistors in series.
Potential divider circuits
▪Consider a battery of  = 6 V.
Suppose we have a light bulb that can only use three volts.
How do we obtain 3 V from a 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.
▪ R = R1 + R2.
▪ I = VIN / R = VIN / (R1 + R2).
▪ VOUT = V2 = IR2
𝑉𝑜𝑢𝑡 =

𝑅2
𝑉
𝑅1 + 𝑅2 𝑖𝑛
𝑉1 =
𝑅1
𝑉
𝑅1 + 𝑅2 𝑖𝑛
R1
R2
potential
divider
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:
▪ VOUT = VIN [ R2 / (R1 + R2) ]
VOUT = 9 [ 330 / (2200 + 330) ]
VOUT = 9 [ 330 / 2530 ] = 1.2 V.
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:
▪ VOUT = VIN [ R2 / (R1 + R2) ] 
6 = 9 [ R2 / (2200 + R2) ]
6(2200 + R2) = 9R2  13200 = 3R2
R2 = 4400 
▪The bigger R2 is in comparison to R1, the closer VOUT
is in proportion to the total voltage.
Using a potential divider to give a variable pd
variable resistor circuit
a power supply, an ammeter, a variable resistor and a resistor.
When the variable resistor is set to its minimum value, 0 Ω, pd across the
resistor is 2 V and a current of 0.2 A in the circuit.
When the variable resistor is set to its maximum value, 10 Ω, pd across the
resistor is 1 V and a current of 0.1 A in the circuit.
Therefore the range of pd across the fixed resistor can only vary from 1 V to 2 V.
The limited range is a significant limitation in the use of the variable resistor.
potential divider
The same variable resistor can be used but the set up is different and involves the
use of the three terminals on the variable resistor (sometimes called a rheostat.)
Terminals of rheostat resistor are connected to the terminals of the cell.
The potential at any point along the resistance winding depends on the position of
the slider (or wiper) that can be swept across the windings from one end to the other.
Typical values for the potentials at various points on the windings are shown for the
three blue slider positions.
The component that is under test (again, a resistor in this case) is connected in a secondary
circuit between one terminal of the resistance winding and the slider on the rheostat. When the
slider is positioned at one end, the full 2 V from the cell is available to the resistor under test.
When at the other end, the pd between the ends of the resistor is 0 V (the two leads to the
resistor are effectively connected directly to each other at the variable resistor). You should
know how to set this arrangement up and also how to draw the circuit and explain its use.
• Potentiometer – rheostat – variable resistor
is a resistor whose resistance decreases with increasing incident light intensity;
in other words, it exhibits photoconductivity
PRACTICE: A light sensor consists of a 6.0 V battery, a 1800 Ω resistor and a light-dependent
resistor in series. When the LDR is in darkness the pd across the resistor is 1.2 V.
(a) Calculate the resistance of the LDR when it is in darkness.
(b) When the sensor is in the light, its resistance falls to 2400 Ω. Calculate the pd across the
LDR.
(a) As the pd across the resistor is 1.2 V, the pd across the
LDR must be 6-1.2=4.8.
The current in the circuit is
The resistance of the LDR is
(b)
For the ratio of pds to be 1.33, the pds must be
2.6 V and 3.4 V with the 3.4 V across the LDR.
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: VOUT = VIN [ R2 / (R1 + R2) ]
7
= 9 [ 22000 / (R1 + 22000) ]
7(R1 + 22000) = 9(22000)
7R1 + 154000 = 198000
R1 = 6300  (6286)
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.
(b) What should R2 be?
SOLUTION: In fire the thermistor is R1 = 250 .
7
= 9 [ R2 / (250 + R2) ]
7(250 + R2) = 9R2
R2 = 880  (875)
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.
▪But from V = IR we see that R = V / I, which is the slope.
▪Thus the slope should increase with I.
R1
R2
(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.
V
non-ohmic
▪Since we need VOUT = 4 V, and since VIN = 6 V, the
contact must be adjusted above the Y.
I
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.
(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.
▪ P = IV  0.8 = I(4)  I = 0.20 A.
▪ V = IR  4 = 0.2R  R = 20. .
▪You could also use P = I 2R for this last one.
(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).
R1
R2
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.
R1
R2
(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
(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.
Solving problems involving circuits
PRACTICE: A battery is connected to a 25-W lamp as shown.
What is the lamp’s resistance?
SOLUTION:
01.4
00.0
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 .
Solving problems involving circuits
PRACTICE: Which circuit shows the correct
setup to find the V-I characteristics of a
filament lamp?
SOLUTION:
▪The voltmeter must be in
parallel with the lamp.
▪It IS, in ALL cases.
▪The ammeter must be in series
with the lamp and must read
only the lamp’s current.
two
currents
lamp
current
no
currents
short
circuit!
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:
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.
equivalent ckt
Solving problems involving circuits
PRACTICE: All three circuits use the same resistors and the same cells.
Highest I
0.5R
parallel
Lowest I
2R
series
Middle I
1.5R
R
0.5R
combo
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.