1. Topic 5.1 Electric potential difference, current and resistance
Electric Currents
Topic 5.1 Electric potential difference, current and resistance

2. Objectives 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 electronvolt.
5.1.4 Solve problems involving electric potential difference.

3. Electric Potential Energy
If you want to move a charge closer to a charged sphere you have to push against the repulsive force
You do work and the charge gains electric potential energy.
If you let go of the charge it will move away from the sphere, losing electric potential energy, but gaining kinetic energy.

4. When you move a charge in an electric field its potential energy changes.
This is like moving a mass in a gravitational field.

5. The electric potential V at any point in an electric field is the potential energy that each coulomb of positive charge would have if placed at that point in the field.
The unit for electric potential is the joule per coulomb (J C‑1), or the volt (V).
Like gravitational potential it is a scalar quantity.

6. In the next figure, a charge +q moves between points A and B through a distance x in a uniform electric field.
The positive plate has a high potential and the negative plate a low potential.
Positive charges of their own accord, move from a place of high electric potential to a place of low electric potential.
Electrons move the other way, from low potential to high potential.

8. In moving from point A to point B in the diagram, the positive charge +q is moving from a low electric potential to a high electric potential.
The electric potential is therefore different at both points.

9. In order to move a charge from point A to point B, a force must be applied to the charge equal to qE
(F = qE).
Since the force is applied through a distance x, then work has to be done to move the charge, and there is an electric potential difference between the two points.
Remember that the work done is equivalent to the energy gained or lost in moving the charge through the electric field.

10. Electric Potential Difference
We often need to know the difference in potential between two points in an electric field
The potential difference or p.d. is the energy transferred when one coulomb of charge passes from one point to the other point.

11. The diagram shows some values of the electric potential at points in the electric field of a positively‑charged sphere
What is the p.d. between points A and B in the diagram?

13. When one coulomb moves from A to B it gains 15 J of energy.
If 2 C move from A to B then 30 J of energy are transferred. In fact:

14. Change in Energy Energy transferred,
This could be equal to the amount of electric potential energy gained or to the amount of kinetic energy gained
W = charge, q x p.d.., V
(joules) (coulombs) (volts)

15. The Electronvolt
One electron volt (1 eV) is defined as the energy acquired by an electron as a result of moving through a potential difference of one volt.
Since W = q x V
And the charge on an electron or proton is 1.6 x 10-19C
Then W = 1.6 x 10-19C x 1V
W = 1.6 x J
Therefore 1 eV = 1.6 x J
eV  J  multiply by e-
J  eV  divide by e-

16. Objectives 5.1.5 Define electric current. 5.1.6 Define resistance.
5.1.7 Apply the equation for resistance in the form R = ρLA where ρ is the resistivity of the material of the resistor.
5.1.8 State Ohm’s law.
5.1.9 Compare ohmic and non-ohmic behaviour.
Derive and apply expressions for electrical power dissipation in resistors.
Solve problems involving potential difference, current and resistance.

17. Conduction in Metals
A copper wire consists of millions of copper atoms.
Most of the electrons are held tightly to their atoms, but each copper atom has one or two electrons which are loosely held.
Since the electrons are negatively charged, an atom that loses an electron is left with a positive charge and is called an ion.

19. The diagram shows that the copper wire is made up of a lattice of positive ions, surrounded by free' electrons:
The ions can only vibrate about their fixed positions, but the electrons are free to move randomly from one ion to another through the lattice.
All metals have a structure like this.

20. What happens when a battery is attached to the copper wire?
The free electrons are repelled by the negative terminal and attracted to the positive one.
They still have a random movement, but in addition they all now move slowly in the same direction through the wire with a steady drift velocity.
We now have a flow of charge ‑ we have electric current.

21. Electric Current Current is measured in amperes (A) using an ammeter.
The ampere is a fundamental unit.
The ammeter is placed in the circuit so that the electrons pass through it.
Therefore it is placed in series.
The more electrons that pass through the ammeter in one second, the higher the current reading in amps.

22. 1 amp is a flow of about 6 x 1018 electrons in each second!
The electron is too small to be used as the basic unit of charge, so instead we use a much bigger unit called the coulomb (C).
The charge on 1 electron is
only 1.6 x 10‑19 C.

23. In fact:
Or I = Δq/ Δt
Current is the rate of flow of charge

24. Which way do the electrons move?
At first, scientists thought that a current was made up of positive charges moving from positive to negative.
We now know that electrons really flow the opposite way, but unfortunately the convention has stuck.
Diagrams usually show the direction of `conventional current' going from positive to negative, but you must remember that the electrons are really flowing the opposite way.

25. Resistance
A tungsten filament lamp has a high resistance, but connecting wires have a low resistance.
What does this mean?
The greater the resistance of a component, the more difficult it is for charge to flow through it.

26. The electrons make many collisions with the tungsten ions as they move through the filament.
But the electrons move more easily through the copper connecting wires because they make fewer collisions with the copper ions.

27. Factors Affecting Resistance
Resistivity is the resistance per unit of a material.
Good conductors such as copper have very low resistances and poor conductors such as nichrome have higher resistances.

28. Resistivity R = ρL A R = resistance (Ω) ρ = resistivity (Ωm)
L = length of conductor (m)
A = cross sectional area (m2)

29. Resistivity
The resistivity of a material is numerically equal to the resistance between opposite faces of a cube of the material, of unit length & unit cross-sectional area.
In other words, ρ= RA/l
Resistivity also depends on temperature
Resistivity of metals increase with increase in temp
Resistivity of semiconductors decrease very rapidly with increase in temp
Look at pg. 174 chart

30. Example
Wire A has a length L and a circular cross section of diameter D. Wire B is constructed from the same material as wire A and has the same shape, but its length is 2L, and its diameter is 2D. Is the resistance of wire B the same as, twice as much, or half of wire A?
RB = RA/2

31. Example
A current of 1.82 A flows through a copper wire 1.75 m long and 1.10 mm in diameter. Find the potential difference between the ends of the wire. (The resistivity of copper is 1.68 * 10-8 Ωm.)

32. Practice Problems
A conducting wire is quadrupled in length & tripled in diameter. Find its new resistance. By what factor does the resistance change?
The tungsten filament of a lightbulb has a resistance of 0.07 Ω. If the filament is 27 cm long, what is its diameter (ρ = 5.6 * 10-8 Ωm)?

33. Resistors
Resistors are components that are made to have a certain resistance.
They can be made of a length of nichrome wire.

34. Resistance is measured in ohms (Ω) and is defined in the following way:
The resistance of a conductor is the ratio of the p.d. applied across it, to the current passing through it.
In fact:

35. Ohm’s Law
The current through a metal wire is directly proportional to the p.d. across it (providing the temperature remains constant).
This is Ohm's law.
Materials that obey Ohm's law are called ohmic conductors.

37. When X is a metal resistance wire the graph is a straight line passing through the origin: (if the temperature is constant)
This shows that: I is directly proportional to V.
If you double the voltage, the current is doubled and so the value of V/I is always the same.
Since resistance R =V/I, the wire has a constant resistance.
The gradient is the resistance on a V against I graph, and 1/resistance in an I against V graph.

39. When X is a filament lamp, the graph is a curve, as shown:

41. Doubling the voltage produces less than double the current.
This means that the value of V/I rises as the current increases.
As the current increases, the metal filament gets hotter and the resistance of the lamp rises.

42. Ohm’s law
Ohm’s law states that, for a conductor at constant temperature, the current in the conductor is proportional to the potential difference across it.

43. Power Dissipation P = W/t But W = qV and I = q/t
Therefore, P = qV/(q/I)
So, P = IV
Using Ohm’s law, we can also create 2 other forms for power
P = I2R & P = V2/R

44. Power Current through a resistance creates heat
“Energy lost” can be measured using power
Since V = IR & P = IV  P = I2R OR P = V2/R
For a given resistor, power dissipated depends on square of current
Power will be four times as great if current is double or voltage is double

45. Topic 5.2 Electric Circuits
Electric current
Topic 5.2 Electric Circuits

46. Objectives 5.2.1 Define electromotive force (emf).
5.2.2 Describe the concept of internal resistance.
5.2.3 Apply the equations for resistors in series and in parallel.
5.2.4 Draw circuit diagrams.
5.2.5 Describe the use of ideal ammeters and ideal voltmeters.
5.2.6 Describe a potential divider.
5.2.7 Explain the use of sensors in potential
divider circuits.
5.2.8 Solve problems involving electric circuits.

47. Series circuits

48. Series Circuit The current is the same throughout the circuit
Current is like water, cant split cuz only one path
All parts of the circuit are connected one after another in a loop
There is only one path for the electrons to follow
If one part goes out
The circuit goes from closed to open
Electricity will not flow
All parts go out

49. Disadvantages of series circuits
Series circuits have two disadvantages when compared with parallel circuits:
If one component in a series circuit fails, then all the components in the circuit fail because the circuit has been broken.
The more components there are in a series circuit, the greater the circuit's resistance.

50. Current in a series circuit
A series circuit has just one current.
For example:
The series circuit below connects a cell, a lamp and an ammeter. The ammeter reads that the current through the lamp is 1.5 A.

51. Parallel circuits

52. Parallel Circuit The current is split through each branch
Charge is always conserved!
There is more than one path or branch for the electrons
If a break occurs in one branch, the electrons can still flow in the other
Current is split because charges cannot be accumulate or get ‘used up’ at a junction.

53. I = I1 + I2 + I3….. Kirchhoff’s first law
The sum of the currents entering a junction in a circuit is always equal to the sum of the currents leaving it
I = I1 + I2 + I3…..

54. Advantages of parallel circuits
Parallel circuits have two advantages when compared with series circuits:
A failure of one component does not lead to the failure of the other components because a parallel circuit consists of more than one loop and has to fail in more than one place before the other components fail.
More components may be added in parallel without the need for more voltage.

55. Current in parallel circuits
The current through a parallel circuit depends upon where in the circuit the current is measured.
Ammeter 1 reads 1.5 A. Ammeter 2 also reads 1.5 A.

56. Current in parallel circuits
If we connect a third ammeter next to the cell:
What current will Ammeter 3 read?
Ammeter 3 reads 3 A.
This is because the total current flowing from a cell in a parallel circuit equals the sum of all the currents flowing in the loops.

57. Stage 1
A full current flows out from the positive terminal and reaches Junction X:

58. Stage 2
At Junction X the current splits. The current splits equally because the lamps are identical:

59. Stage 3
The two "half" currents flow through the two loops. At Junction Y the two "half" currents re-join:

60. Stage 4
A full current then flows back to the negative terminal.

61. Voltage in series circuits
Series  voltage is split at each component  sum of the voltages at each component equals the voltage across all of those components

62. Voltage in parallel circuits
Parallel  voltage is the same across each branch

63. Kirchhoff’s second law
The sum of the electromotive forces in a closed circuit is equal to the sum of the potential differences.
Energy is always conserved!

64. Example
If the supply voltage is 12 V, what voltages are across each resistor? The total resistance of the circuit below is 9 Ohms.

65. Example
If the supply voltage is 12 V, what is the current at every point?

66. Example
Using Kirchhoff’s second law, write an expression for the voltage in this circuit, in terms of the current & resistance.

67. Example
Write an expression for the current in the circuit in terms of the battery’s voltage & resistance.
DO EXAMPLE FROM WALKER TEXT, PG 741 THEN 743

68. Resistance Combinations

69. Resistors in series

71. Resistors in series V = V1 + V2 but V1 = IR1 & V2 = IR2
so if R or Req is the combined resistance then
V = IR
then, IR = IR1 + IR2
and finally, R = R1 + R2
Resistors in series  R = R1 + R2…..

72. Resistors in parallel

73. Resistors in parallel I = I1 + I2 but I1 = V/R1 & I2 = V/R2
also, I = V/R  equivalent resistance
then,
and finally,
Resistors in parallel:

74. Two resistors in parallel
= product of resistances
sum of resistances
Derive formula!

75. Two resistors in parallel
 The combined resistance of 2 resistors in parallel is less than the value of each resistor alone.
 R < R1 & R < R2
More resistors in parallel  Less combined resistance  More current (Ohm’s law!)
Example with R1 = R2

76. Example
Find the combined resistance!

77. Example
Find the combined resistance!

78. Example
The total voltage in the circuit to the right is a 12-V source and the resistor values are 6.4 Ω (R1) and 3.9 Ω (R2).
Determine the:
combined resistance of the circuit.
current in each branch resistor.
total current in the circuit.
DO GLENCOE PROBLEMS

79. Ammeters and Voltmeters
In order to measure the current, an ammeter is placed in series, in the circuit.
What effect might this have on the size of the current?
The ideal ammeter has zero resistance, so that placing it in the circuit does not make the current smaller.
Real ammeters do have very small resistances ‑ around 0.01 Ω.

80. A voltmeter is connected in parallel with a component, in order to measure the p.d. across it.
Why can this increase the current in the circuit?
Since the voltmeter is in parallel with the component, their combined resistance is less than the component's resistance.
The ideal voltmeter has infinite resistance and takes no current.
Digital voltmeters have very high resistances, around 10 MΩ, and so they have little effect on the circuit they are placed in.

81. Electromotive Force Defining potential difference
The coulombs entering a lamp have electrical potential energy;
those leaving have very little potential energy.
There is a potential difference (or p.d.) across the lamp, because the potential energy of each coulomb has been transferred to heat and light within the lamp.
p.d. is measured in volts (V) and is often called voltage.

82. The p.d. between two points is the electrical potential energy transferred to other forms, per coulomb of charge that passes between the two points.

83. Resistors and bulbs transfer electrical energy to other forms, but which components provide electrical energy?
A dry cell, a dynamo and a solar cell are some examples.
Any component that supplies electrical energy is a source of electromotive force or e.m.f.
It is measured in volts.
The e.m.f. of a dry cell is 1.5 V, that of a car battery is 12 V

84. A battery transfers chemical energy to electrical energy, so that as each coulomb moves through the battery it gains electrical potential energy.
The greater the e.m.f. of a source, the more energy is transferred per coulomb. In fact:
The e.m.f of a source is the electrical potential energy transferred from other forms, per coulomb of charge that passes through the source.
Compare this definition with the definition of p.d. and make sure you know the difference between them.

85. Internal Resistance

86. The cell gives 1.5 joules of electrical energy to each coulomb that passes through it,
but the electrical energy transferred in the resistor is less than 1.5 joules per coulomb and can vary.
The circuit seems to be losing energy ‑ can you think where?

87. The cell itself has some resistance, its internal resistance.
Each coulomb gains energy as it travels through the cell, but some of this energy is wasted or `lost' as the coulombs move against the resistance of the cell itself.
So, the energy delivered by each coulomb to the circuit is less than the energy supplied to each coulomb by the cell.

88. Very often the internal resistance is small and can be ignored.
Dry cells, however, have a significant internal resistance.
This is why a battery can become hot when supplying electric current.
The wasted energy is dissipated as heat.

89. Internal resistance All power supplies have internal resistance.
The emf delivers a current I through the resistor R.
VR is the voltage across the resistor R.
But emf has a resistance too  r
So the emf has a p.d. across it  Vr
emf = VR + Vr

90. Internal resistance VR is the terminal potential difference.
The terminal potential difference is the p.d. between the terminals of a cell when a current is being delivered.
Ideally, VR = emf, but not in real life
So emf = VR + “lost volts”
Vr = lost volts
Lost volts = current * internal resistance
E = I(R+r)
E = I (R+r) from Vr = Ir and VR = IR

91. Example
A dry cell has an internal resistance of 1.50 Ω. A resistor of 12.0 Ω is connected in series with the dry cell. If the potential difference across the 12.0 Ω resistor is 1.20 V, calculate the emf of the cell.
Emf = 1.35 V

92. Potential dividers
A potential divider is a device or a circuit that uses two (or more) resistors or a variable resistor (potentiometer) to provide a fraction of the available voltage (p.d.) from the supply.

93. The p.d. from the supply is divided across the resistors in direct proportion to their individual resistances.

94. Take the fixed resistance circuit - this is a series circuit therefore the current is the same at all points.
Isupply = I1 = I2
Where I1 = current through R1
I2 = current through R2

95. Potential divider Two resistors in series  “potential divider”
Analyzing the circuit gives:
I = V1/R1, I = V2/R2

96. Potential divider Change the value of R2 Changes the Vout
Vin R1 R2
Vout
Change the value of R2
Relative to R1
Changes the Vout
Qualitatively
R2 big… Vout big
R2 small… Vout small
So current is same at all point, then current through R2 is V2/R2 but I = Vin / Rtotal too
Then V2 = R2 Vin / R total

97. Example The voltage will be divided in the same ratio as the resistors
10kΩ
20kΩ
Vout I
…or just by inspection on the relative size of each resistor

98. Applications Replace 1 of the fixed resistors with: Variable resistor
Variable output voltage
Volume control
Sensor
Thermistor
LDR
Electronic input device
Physical quantity to electrical signal V1 Vv
Vin R1 Rv
Vout I
All these create a changing voltage level that can be processed electronically

99. Light Dependent Resistor (LDR)
A resistor whose resistance decreases when incident light intensity increases
LDR’s are used in alarm systems, street lights, camera light meters and other light-dependent circuits
Draw circuit symbol
Draw graph pg 173

100. Thermistors
A resistor whose resistance decreases as the temperature of the resistor increases
The temperature can rise by:
Heating the thermistor directly
Heating caused by the current passing through the circuit
Thermistors are used in temperature-dependent circuits (thermostats etc.)
Circuit symbol
Graph pg 172

101. Example
Calculate the readings on the meters shown below when the thermistor has a resistance of
a) 1 kΩ (warm conditions) and b) 16 kΩ. (cold conditions)