1. Topic 5: Electricity and magnetism
5.2 – HEATING EFFECTS OF ELECTRIC CURRENTS

2. 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

3. Current– flow of electric charge
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 +

4. Electric current (symbol I)
◊ the flow of electric charge q that can occur in solids, liquids and gases.
DEF: the rate at which charge flows past a given cross-section.
measured in amperes (A) q
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.

5. 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. .
path
atoms
(actually positive ions)
free electron
The resistance is a measure of how hard it is to pass something through a material.

6. 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.
© 2006 By Timothy K. Lund 6

7. 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)
© 2006 By Timothy K. Lund
2 leads
thermister
2 leads
As brightness increases resistance decreases
– photoconductivity
As temperature increases resistance decreases 7

8. This resistor has a resistance of 330.4 .
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.
This resistor has a resistance of .
330.4
0.L
0.0
© 2006 By Timothy K. Lund
▪A reading of 0.L on an ohmeter means “overload”. The resistance (of the air) is too high to record with the meter. 8

9. 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.
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.

10. Factors affecting resistance
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

11. 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.
The dream: current without cost!!!!!!!!! Both in money and damage to environment!!!!!!!!
WHICH WE JUST MIGHT HAVE DISCOVERED
Hydrogen turned into metal in stunning act of alchemy that could revolutionise technology and spaceflight

12. Of course, resistance depends on the material being used.
Resistance of a wire when the temperature is kept constant is:
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.

13. 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)
Material
 (m)
 (C -1)
Conductors
Semiconductors
Aluminum
2.8210-8
4.2910-3
Carbon
360010-8
-5.010-4
Copper
1.7010-8
6.8010-3
Germanium
4.610-1
-5.010-2
Iron
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
Nonconductors
© 2006 By Timothy K. Lund
Nickel
7.810-8
6.010-3
Glass
1012
Platinum
1010-8
3.9310-3
Rubber
1015
Silver
1.5910-8
6.110-3
Wood
1010
Tungsten
5.610-8
4.510-3
Gold
2.310-8 13

14. 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 meter long carbon core resistor having a core diameter of m? Assume the temperature is 20 C.
▪r = d / 2 = / 2 = m.
▪A = r2 = ( )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
© 2006 By Timothy K. Lund L 14

15. 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 .”
▪ In formula form Ohm’s law looks like this:
DEF: Current through resistor (conductor) is proportional to potential difference on the resistor if the temperature/resistance of a resistor is constant.
© 2006 By Timothy K. Lund
I  V
Ohm’s law
or 𝐼= 𝑉 𝑅
𝑉 𝐼 or
=𝑐𝑜𝑛𝑠𝑡.
▪ Ohm’s law applies to components with constant R. 15

16. 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 

17. 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?
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.

18. human body resistance varies:
100 ohms if soaked with salt water;
moist skin ohms;
normal dry skin – ohms,
extra dry skin – 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/ W = 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 W = A
a dangerous amount of current.

19. Ohmic and Non-Ohmic behaviour
How does the current varies with potential difference for some typical devices?
metal at const. temp filament lamp diode
current
current
current
potential
difference
potential
difference
potential
difference
𝑎𝑠 𝐼 𝑉 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝑠𝑜 𝑖𝑠 𝑅= 𝑉 𝐼
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.

20. 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
r = 1.72x10-8 Wm, copper - books
                         
The resistance of the wire is then:
R = r L/A = (1.72x10-8 Wm)(1.60)/(7.9x10-7m2 ) = 3.50 W
The current can now be found from Ohm's Law:
I = V / R = 1.5 / 3.5 = A

21. 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.
© 2006 By Timothy K. Lund
▪ 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. 21

22. Ohmic and Non-Ohmic behaviour
EXAMPLE: The I-V characteristic is shown for a non-ohmic compo nent. Sketch in the I-V character istic for a 40  ohmic component in the range 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.
© 2006 By Timothy K. Lund 22

23. 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 = I V
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.

24. 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.
© 2006 By Timothy K. Lund 24

25. Electric circuits
▪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
© 2006 By Timothy K. Lund
single-loop circuit

26. Circuits diagrams
▪A complete circuit will always contain a cell or a battery.
▪The schematic diagram of a cell is this:
▪A battery is just a group of cells connected in series:
▪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 :
this is really a cell…
this is a battery…
this is the same battery…
© 2006 By Timothy K. Lund

27. Primary and secondary cells
primary cells – non rechargeable cells
The cells are used until they are exhausted and then thrown away. The original chemicals
have completely reacted and been used up, and they cannot be recharged. Examples include AA cells (properly called dry cells) and button mercury cells as used in clocks and other small low current devices.
secondary cells – rechargeable cells
When the chemical reactions have finished, the cells can be connected to a charger. Then the chemical reaction is reversed and the original chemicals form again. When as much of the re-conversion as is possible has been achieved, the cell is again available as a chemical energy store.
To reverse the chemical processes we need to return energy to the cell using electrons as the agents, so that the chemical action can be reversed. When charging, the electrons need to travel in the reverse direction to that of the discharge current and you can imagine that the charger has to force the electrons the “wrong” way through the cell.
capacity
of a cell is the quantity used to measure the ability of a cell to release charge: if a cell can supply a constant current of 2 A for 20 hours then it said to have a capacity of 40 amp-hours (40 A h).
The implication is that this cell could supply 1 A for 40 hours, or 0.1 A for 400 hours, or 10 A for 4 hours. However, practical cells do not necessarily discharge in such a linear way and this cell may be
able to provide a small discharge current of a few milliamps for much

28. Drawing and interpreting circuit diagrams
Schematic diagrams of each of the following circuits:
© 2006 By Timothy K. Lund

29. Electromotive force: emf or e
The voltage generated by a cell/battery has a special name. It is energy per unit charge and is measured in volts, not newtons, and thus, is not actually a force.
Although, do not be surprised if we still write V instead of e for cell voltage.
▪ Electromotive force (ε) is the energy per unit charge made
available by an electrical source.
ε = U/q
▪ Electromotive force, (ε) is the power supplied to the circuit per unit current
P = I V ⟹ V = P/ I
▪ Power supplied by the source will be dissipated in the circuit:
In short, emf is voltage generated by cell

30. Resistors in Series R = R1 + R2 + R3 equivalent resistance 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: q𝜀 = qV1 + qV2 + qV3.
▪ 𝜀 = IR1 + IR2 + IR3 = I(R1 + R2 + R3)
▪ 𝜀 = I(R)
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

31. 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 𝑅 = 𝑉 𝑅 1 + 𝑉 𝑅 2 + 𝑉 𝑅 3
1 𝑅 = 1 𝑅 𝑅 𝑅 3
equivalent resistance in parallel
equivalent resistance is smaller than the smallest resistance.
the one that could replace all resistors resulting in the same current

33. RESISTORS IN COMPOUND CIRCUITS
Now you can calculate current, potential drop/potential difference/voltage and power dissipated through each resistor

34. In the mid-nineteenth century, G. R
In the mid-nineteenth century, G.R. Kirchoff ( ) stated two simple rules using the laws of conservation of energy and charge to help in the analysis of direct current circuits.
These rules are called Kirchoff’s rules.

35. 𝜀 = V1 + V2 + V3 1. Junction rule – conservation of charge.
‘The sum of the currents flowing into a point in a circuit equals the sum of the currents flowing out at that point’.
I1 + I2 = I3 + I4 + I5
2. loop rule – conservation of energy.
‘In a closed loop, the sum of the emfs equals the sum of the potential differences’.
𝜀 = V1 + V2 + V3
Energy supplied by cell equals the energy released in this closed path
energy: qε = qV1 + qV2 + qV3

36. 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 + R R = = 990 
▪ I = V / R = 6 / 990 = A
▪ The current I = A is the same in each resistor.
© 2006 By Timothy K. Lund
▪ 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.

37. 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/R /R = 1/ / /330 =
R = 110 
𝐼= 𝜀 𝑅 𝑒𝑞 =0.054 𝐴
▪ The voltage on each resistor is 6.0 V, since the resistors are in
parallel. (Each resistor is clearly directly connected to the battery).
© 2006 By Timothy K. Lund
▪ I1 = V1 / R1 = A
▪ I2 = V2 / R2 = 6 / 330 = A
▪ I3 = V3 / R3 = 6 / 330 = A
𝐼=3×0.018𝐴=0.054 𝐴
▪ In parallel the I’s are different if the R’s are different.

38. Internal resistance, emf and terminal voltage of a cell
The materials from which the cells are made have electrical resistance in just the same way as the resistors in the external circuit. This INTERNAL RESISTANCE (r ) has an important effect on the total resistance and current in the circuit. When electrons flow around a circuit, they gain potential energy in the cell and then lose the energy in the resistors and in the cell as well. Thus the TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is less then cell’s emf.

39. Internal resistance, emf and terminal voltage of a cell
Assumptions:
▪ internal resistance is constant
(for real cell it varies with the state of discharge)
▪ emf is constant
(which also varies with discharge current).
Kirchhoff’s loop rule: emf of the cell supplying energy to the circuit = the sum of the pds
ε = IR + Ir
If the pd across the external resistance is V=IR, then:
V = ε – Ir
V, which is the pd across the external resistance, is equal to the terminal pd across real cell
(in other words between A and B).
TERMINAL VOLTAGE (the actual voltage delivered to the circuit) is: V = ε - Ir
The emf is the open circuit pd across the terminals of a power source –
in other words, the terminal pd when no current is supplied.

40. Emf and internal resistance of the battery (made up of two cells)
When voltage sources are connected in series in the same polarity, their emfs and internal resistances are add.
Two voltage sources with identical emfs connected in parallel have a net emf equivalent to one emf source, however, the net internal resistance is less, and therefore produces a higher current.

41. power of the source = power dissipated in the circuit
Summary:
I – current through resistor,
V – potential difference across R
𝐼= 𝑉 𝑅
Power dissipated in resistor R: P = IV
𝐼= 𝜀 𝑅 𝑒𝑞 = 𝜀 𝑅+𝑟 = 𝑉 𝑅
𝑉= 𝜀−𝐼𝑟
1 𝑅 𝑒𝑞 = 1 𝑅 𝑅 𝑅 3
I1 + I2 = I3 + I4 + I5
𝑉 𝐴𝐵 = 𝐼 1 𝑅 1 = 𝐼 2 𝑅 2 = 𝐼 3 𝑅 3
𝑅 𝑒𝑞 = R1 + R2 + R3
𝐼= 𝜀 𝑅
𝜀 = V1 + V2 + V3
Any circuit: 𝐼𝜀= Ʃ 𝐼𝑉
power of the source = power dissipated in the circuit

42. 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: 𝐼 1 = 𝜀 𝑅
I1 = ε ∕ Req = 0.3 A
3. step: for currents in parallel branches use the Kirchhoff's rules:
𝐼 2 𝑅 2 = 𝐼 3 𝑅 ( 𝑉 𝐴𝐵 )
𝐼 1 = 𝐼 2 + 𝐼 3
100 𝐼 2 = 50𝐼 3 → 𝐼 3 = 2 𝐼 2
0.3= 𝐼 2 + 𝐼 → 0.3=3 𝐼 2 → 𝐼 2 =0.1 A 𝐼 3 =0.2 A
potential drops
V = IR
power dissipated
P = IV
80 Ω
0.3x80 = 24 V
0.3x24 = 7.2 W
100 Ω
0.1x100 = 10 V
0.1x10 = 1 W
50 Ω
0.2x50 = 10 V
0.2x10 = 2 W
6.7 Ω
0.3x6.7 = 2 V
0.3x2 = 0.6 W
control:
ε = Σ all potential drops:
36 V = 2 V + 24 V + 10 V
power dissipated in the circuit =
power of the source
= 0.3x36

43. 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.

44. 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.

45. 1.06 Circuit diagrams - voltmeters are connected in parallel
Draw a schematic diagram for this circuit
1.06
© 2006 By Timothy K. Lund
▪Be sure to position the voltmeter across the desired resistor in parallel.

46. Circuit diagrams - voltmeters are connected in parallel
tenths place
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 = (or 1.1%).
© 2006 By Timothy K. Lund
▪When using a voltmeter the red lead is placed at the point of highest potential. 46

47. 00.0 01.6 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.
00.0
01.6
© 2006 By Timothy K. Lund
voltmeter
lamp
cell
in parallel 47

48. • Ammeter is in SERIES with resistor R in order that whatever current
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.

49. 00.0 00.2 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.
00.0
00.2
ammeter
in series
© 2006 By Timothy K. Lund
lamp
cell 49

50. the circuit must be temporarily broken to insert the ammeter
Circuit diagrams - ammeters are connected in series
the circuit must be temporarily broken to insert the ammeter
PRACTICE: Draw a schematic diagram for this circuit:
SOLUTION:
.003
© 2006 By Timothy K. Lund
▪Be sure to position the ammeter
between the desired resistors in series.

51. 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?
potential divider
▪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. R1
▪The input voltage is the emf of the battery. R2
▪The output voltage is the voltage drop across R2.
▪ R = R1 + R2.
▪ I = VIN / R = VIN / (R1 + R2).
▪ VOUT = V2 = IR 
𝑉 𝑜𝑢𝑡 = 𝑅 2 𝑅 1 + 𝑅 2 𝑉 𝑖𝑛
𝑉 1 = 𝑅 1 𝑅 1 + 𝑅 2 𝑉 𝑖𝑛

52. 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 / ( ) ]
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  resistor, and we want an output voltage of 6 V.
SOLUTION:
▪ VOUT = VIN [ R2 / (R1 + R2) ]  = 9 [ R2 / ( R2) ]
6( R2) = 9R2  = 3R2
R2 = 4400 
▪The bigger R2 is in comparison to R1, the closer VOUT is in proportion to the total voltage.

53. 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.

54. • Potentiometer – rheostat – variable resistor
is a resistor whose resistance decreases with increasing incident light intensity;
in other words, it exhibits photoconductivity

55. PRACTICE: A light sensor consists of a 6
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.
Calculate the resistance of the LDR when it is in darkness.
When the sensor is in the light, its resistance falls to 2400 Ω. Calculate the pd across the LDR.
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.

56. Potential divider circuits
PRACTICE: A light-dependent resistor (LDR) has R = 25  in bright light and R =  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 [ / (R ) ]
7(R ) = 9(22000)
7R =
R1 = 6300  (6286)

57. Potential divider circuits
PRACTICE: A thermistor has a resistance of 250  when it is in the heat of a fire and a resistance of  when at room temperature.
An electronic switch will turn on a sprinkler system
when its p.d. is above 7.0 V.
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)

58. 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?
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
ohmic means linear 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?
© 2006 By Timothy K. Lund
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 I
non-ohmic
▪Since we need VOUT = 4 V, and since VIN = 6 V, the contact must be adjusted above the Y.

59. R1
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. R2
(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).

60. 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 / = 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 = = 0.49 A.

61. 01.4 00.0 Solving problems involving circuits
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 = V 2 / R we get
▪R = V 2/ P = / 25
= .
01.4
00.0
© 2006 By Timothy K. Lund 61

62. 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
no currents
short circuit!
lamp current

63. Solving problems involving circuits
equivalent ckt
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 / / = 2 /
R = / 2 = 10 k.
▪But then we have two 10 k resistors in series and each takes half the battery voltage, or 3 V.
© 2006 By Timothy K. Lund

64. Solving problems involving circuits
PRACTICE: All three circuits use the same resistors and the same cells.
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.
Highest I
Lowest I
Middle I
1.5R
0.5R 2R R
0.5R
parallel
series
combo
© 2006 By Timothy K. Lund

66. Kirchhoff’s rules – solving the circuit (less lucky)
If a circuit has MORE than ONE cell you are less lucky.
But you are seniors now.
You have to use Kirchhoff's laws in purest form.
EXCELLENT EXPLANATION of KIRCHOFF’s LAWS – thank you Kiran

67. 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.
© 2006 By Timothy K. Lund 67

68. Kirchhoff’s first law \ˈkir-ˌkȯf\ 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 ∑𝐼 = 0
the sum of the currents into a junction equals the sum of the currents away from a junction Or
total charge flowing into a junction equals the total charge flowing away from the junction.
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 ∑𝐼 = 0
𝐼 1 + 𝐼 2 + 𝐼 3 = 𝐼 4 + 𝐼 5
It is equivalent to a statement of conservation of charge.

69. Kirchhoff’s rules – solving the circuit
PRACTICE: Calculate the unknown branch current in the following junctions.
I = – 3 = 11 A out of the junction.
I = 12 – 14 = –2 A directed away from the junction.

70. Kirchhoff’s second law
in a complete circuit loop, the sum of the emfs in the loop is equal to the sum of the potential differences in the loop or the sum of all variations of potential in a closed loop equals zero
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑐𝑙𝑜𝑠𝑒𝑑 𝑙𝑜𝑜𝑝 𝑖𝑛 𝑎 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ∑𝜀 = ∑𝐼𝑅
Equivalent to conservation of energy.
Loop GABCDEG travelling anticlockwise round the loop
The direction of loop travel and the current direction are in all cases the same. We give a positive sign to the currents when this is the case. The emf of the cell is driving in the same direction as the loop travel direction; it gets a positive sign as well. If the loop direction and the current or emf were to be opposed then they would be given a negative sign.
𝜀=𝐼 1 𝑅 1 + 𝐼 2 𝑅 2
loop EFGE travelling clockwise round the loop
resistor R3: loop direction is in the same direction as the conventional current. resistor R1: loop direction is in opposite direction to the conventional current,
so there has to be a negative sign.
There is no source of emf in the loop so the Kirchhoff equation becomes
0=𝐼 3 𝑅 3 − 𝐼 2 𝑅 2

71. Kirchhoff’s rules – solving the circuit
PRACTICE: Calculate the currents in the circuit shown.
For loop 1
Current directions have been assigned and two
loops 1 and 2 and junction A defined in the diagram.
3 = 3I1 + 9I2 [equation 1]
(the emf in the loop is 3 V)
3 equations with three unknowns:
For loop 2
0 = 6I3 – 9I2 [equation 2]
(there is no source of emf in this loop, current I2 is
in the opposite direction to the loop direction 0.
I1 = 0.45 A
I2 = 0.18 A
For junction A
I3 = 0.27 A
I1 = I2 + I3

72. 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)
▪ –V1 + –V2 + –V4 + 1 + – 2 = 0,
▪ – 2 – V3 – V4 = 0
–V1 + –V2 + –V4 + 1 + – 2 = 0 & V=IR  –2I1 + –2I1 + –2I –6 = 0 (2)
– 2 – V3 – V4 = 0 & V=IR  –– 6 – 2I3 – 2I2 = 0 (3)
We now have three equations in I:
(1)  I3 = I2 – I1.
(2)  3 = 2I1 + I2.
(3)  3 = -I2 + -I3
I1 = 1.8 A
I2 = -0.6 A
I3 = -2.4 A
© 2006 By Timothy K. Lund 72

73. 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
© 2006 By Timothy K. Lund 73