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Electric Currents
Topic 5
These notes were typed in association with Physics for use with
the IB Diploma Programme by Michael Dickinson
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.
 A man carries a bucket of water up a hill.
 Pours water down slide
 Turns water wheel
 Turns a grind stone
 Generates heat!!!
 Work done by man on the water is = heat produced
 Conservation of energy!!!!
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.
 Picture an electric field that is created by to charged plates.
 One on the left has a negative charge, one on the right has a
positive charge. (see diagram on drawn on board) This creates
an electric field.
 This is a uniform field – the strength of the field is constant no
matter where the charge moves.
 Now place a charged particle in that field. It will feel an
electric force from this field.
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.
 Now move that charge against that force.
 The charge has just done WORK!!! YAY!!!
 This work done is equal to the gain in electric potential energy.
 This is just like gravitational potential energy gained is equal to
work done on the same mass. (DO EXAMPLE)
 Work done = change in electric potential energy
W = electric potential = Fd
Fd = ΔEelec
Eqd = ΔEelec
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.
 IB Equation
ΔV = ΔE/ q
AKA
V = W/q
Where:
V= potential difference(volts)
E = Energy(Joules)
q = charge
W= work
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.
 Now lets move that charge the other direction. We’ll release it at
position B.
 The electric force “pushes” it to position A and loses electric
potential energy.
 It will also accelerate since it’s under a constant force.
 The increase of velocity gives the charge an increase of kinetic energy.
 Change in electric potential energy = change in kinetic energy
electric potential = KE
Eqd = ½ mv2
v2 = 2Eqd/m
v=
Electric Potential Difference
 As the charge moves between the platesthe “energy difference per
unit charge” remains constant (potential difference).
 Potential Difference is voltage(V)
 Potential Difference(voltage) = the energy per unit charge
V = (ΔEelec / q)
Vq = ΔEelec
Vq = KE = ½ mv2
Electric Potential Difference
 Potential Difference(voltage) = the energy per unit charge
V = (ΔEelec / q)
Unit is J/ c
Think of it as how much energy the charge is going to get as it zips
from one plate to the other.
Electric Potential Difference
5.1.3 Define the electronvolt.
 Electrical energy = potential difference x charge
 Eelec = Vq
 This gives us a unit of Joules. A Joule is such a large quantity of
energy when looking at the energy carried by electrond, that a
different unit is sometimes used.
 1 electronvolt = 1 volt x charge on 1 electron
 1eV = 1.6 x 10-19 J
Electric Potential Difference
5.1.3 Define the electronvolt.
 IB Formal Definition and Equation
 Electronvolt – the energy gained by an electron accelerated through
a potential difference of 1 volt.
 Ve = ½ mv2
Electric Potential Difference
5.1.4 Solve problems involving electric potential difference.
 An electron moves from a negative to a positive plate when a
potential difference of 50Vols applied to the plates. If the plates are
separated by a distance of 5mm, calculate:
a) the strength of the uniform field between the plates
b) the force on the electron.
c) the loss of electric potential energy as the electron moves
between the plates.
d) the gain in kinetic energy for the electron
e) the speed of the electron as it reaches the positive plate.
Electric Potential Difference
5.1.4 Solve problems involving electric potential difference.
 Solutions
 a) E = V/d
E = 50/0.005
E = 1000 V/m
 b)
F = Eq
F = 10000 x 1.6 x 10-19
F = 1.6 x 10-15N
 c)
ΔEelec = Vq
ΔEelec = 50 x 1.6 x 10-19
Δeelec = 8 x 10-18J or 50eV
Electric Potential Difference
5.1.4 Solve problems involving electric potential difference.
 Solutions
 d) Loss of Ep = gain in Ek
Δek = 8 x 10-18J or 50 eV
 e)
Δek = 8 x 10-18J
½ mv2 = 8 x 10-18J
√2 x 8 x 10-18
v = 9.11 x 10-31
v = 4.19 x 106 m/s
Electric current and resistance
5.1.5 Define electric current.
 New Diagram ***See board***
 Conducting material has a “lattice structure” – orderly rows and
columns. Because of the lattice structure electrons can freely move
from one atom to the next.
 If a potential difference(voltage) is applied to the ends of a
conducting material then the “free electrons” feel an electrical
force.
 All of them feel the same force.
 Just like F=ma says, this electrical force causes the electrons to
speed up and accelerate.
 Just like gravitational force causes you to accelerate toward earth.
Electric current and resistance
5.1.5 Define electric current.
 As they move they occasionally collided with the lattice structure.
 The more they accelerate the more collisions they have.
 Eventually, there are so many collisions that the free electrons
cannot continue to accelerate.
 This maximum velocity is called – Drift velocity.
 Drift velocity is similar in concept to terminal velocity.
Electric current and resistance
5.1.5 Define electric current.
 Which direction do the electrons move?
 From the negative terminal to the positive terminal, the source of
the potential difference.
 Electric current - is this movement of electrons.
 Convention is different!!! - electric current was originally
thought to be the movement of positively charged particles and we
are now stuck with this convention. For this reason, conventional
current moves from the positive terminal to the negative. Sorry 
Electric current and resistance
5.1.5 Define electric current.
IB Formula/Definition
 Current is defined as the amount of electric charge passing a pint in
a circuit in unit time.
 Current = charge/time
 I = Δq/ Δt
 1 Ampere = 1coulomb / 1second
Electric current and resistance
5.1.5 Define electric current.
Practice problem 2
 5 amps flow in an electric circuit. How many coulombs of charge
pass a point in the circuit in 15 seconds? How many electrons is
this equal to?
 Answer: 75 Coulombs, 4.69 x 1020 electrons
***1 coulomb = 6.25 x 1018 electrons***
Electric current and resistance
5.1.5 Define electric current.
 What kind of current flows
though your cell phone?
 Direct current(DC) is an electric current that continually flows in
one direction.
 Alternating current(AC) is a current with a regular change of
change of direction.
 Electrons oscillate back and forth in the conducting wires.
Electric current and resistance
5.1.6 Define resistance.
 Remember the concept of drift velocity
 Some of that kinetic energy is conducted to the lattice in the form
of heat
 This is called resistance, R.
 Measured in Ohms, Ω
IB Formula and Definition
 Resistance = voltage / current
 R = V/I
 Resistance – the ratio of the voltage across a conductor to the
current flowing through it.
Electric current and resistance
5.1.6 Define resistance.
 Practice 3
 When 6 volts is applied to the ends of a resistor, 3 amps of current
flows. Calculate the value of the resistor.
 Practice 4
 If 12 volts is applied to the ends of a 60Ω resistor, calculate the
current which will flow.
 Practice 5
 0.4 amps flows through a 125Ω resistor, calculate the potential
difference across the ends of the resistor.
Electric current and resistance
5.1.6 Define resistance.
 Some conductors allow electrons to flow without very much
hindrance. – good conductors
 Ex. Copper, aluminum, silver, gold.
 Resistors
 In some conductors the free flow of electrons is made difficult.
 The lattice structure is irregular or disjointed.
 Ex. Wire, light bulb, fan, ect.
 Think of conductors as a tunnel the charge has to move through.
 Resistors – have a muddy floor
 Good conductors – have paved floor
 If we make the tunnel smaller or longer it will also make it harder
for the charges to move through the tunnel.
Electric current and resistance
5.1.6 Define resistance.
 Resistance of wire can be changed.




Material
Length
Cross-sectional area
Temperature
 Which will have a lower/high resistance for each to the above
factors?
Electric current and resistance
5.1.6 Define resistance.
 Material
 Depends on each substance.
 Length
 If the length doubles then the resistance doubles R ∝ L
 Cross-sectional area
 If the CSA doubles then the resistance halves R ∝ 1/CSA
 Temp
 Higher temp, means more atomic vibrations, means more collisions
between electron and lattice.
Electric current and resistance
5.1.7 Apply the equation for resistance in the form R = ρ L/A
 Now we can take the factors, “material, length, and CSA” and
combine them we get a new equation:
IB Equation
 R = ρ L/A
Electric current and resistance
5.1.7 Apply the equation for resistance in the form R = ρ L/A
 Here is how we got there….
 1st Length - doubles then the resistance doubles R ∝ L
 2nd CSA doubles then the resistance halves R ∝ 1/CSA
 Combine those to get R ∝ L / A
 R = resistance
 L = length
 A = cross sectional area
or
R=L/A
Electric current and resistance
5.1.7 Apply the equation for resistance in the form R = ρ L/A
 Here is how we got there….
 Last we add in resistivity. (ρ) “rho”
 Resistivity is a property associated with actual material something
is made of.
 Unit of ohm meter = Ωm
 Resistance is a quantity for a specific component or part.
 So we get … R = ρ (L / A)
Electric current and resistance
5.1.8 State Ohm’s Law.
IB Formula and Definition
 Ohm’s Law sates that the current flowing through a resistor is
proportional to the voltage across it (as long as the temperature
remains constant).
 Voltage = Current x Resistance
 V = IR
 http://www.youtube.com/watch?v=zYS9kdS56l8&kw=electricity&
ad=6599373247&feature=pyv&lr=1
Electric current and resistance
5.1.9 Compare ohmic and non-ohmic behavior.
 Three different types of conductors
 Ohmic
 Filament Lamp
 Diode
Electric current and resistance
5.1.9 Compare ohmic and non-ohmic behavior.
 Resistor is “ohmic” if the current flowing is proportional to the
voltage across its ends.
 Metals are ohmic if the temperature is constant.
 Gives a straight line on I-V graph
 Ohmic conductors obey “Ohm’s Law”
Electric current and resistance
5.1.9 Compare ohmic and non-ohmic behavior.
 The filament lamp gets hot as there is an creasing voltage. This
means the resistance increases with higher voltages.
Electric current and resistance
5.1.9 Compare ohmic and non-ohmic behavior.
 Diodes
 Lets very little ore no current flow until voltage reaches a threshold.
 0.6V is very common
 Once the threshold is met, there is essentially zero risistance and a
large amount of current can flow.
 Act as a switch, only allowing current to flow in one direction.
 Can turn AC in to DC
Electric current and resistance
5.1.10 Derive and apply expressions for electrical power dissipation
in resistors
5.1.11 Solve problems involving potential difference, current and
resistance.
 Hot battery demo***
 When a current flows in a resistor electrical potential energy is
converted into heat energy.
 Where does this heat come from?
 We can solve for the amount of power, P, that is dissipated in the
resistor.
 Remember…. power = energy per second, P = ΔE/t
Electric current and resistance
5.1.10 Derive and apply expressions for electrical power dissipation
in resistors
5.1.11 Solve problems involving potential difference, current and
resistance.
 Remember…. power = energy per second, P = ΔE/t
 See work on board.
 Units for electrical power are Joules per second, J/s or Watts, W
 IB Formula
 P = VI = I2 R = V2 / R
Electric current and resistance
Practice 6
 When 20 volts is applied to the ends of a resistor, 0.5 Amps of
current flows. Calculate the power dissipated in the resistor.
 Answer: 10 Watts
Practice 7
 If 12 volts is applied to the ends of a 40Ω resistor, calculate the
power dissipated in the resistor.
 Answer: 3.6 Watts
Practice 8
 0.4 Amps flows through a 125Ω resistor, calculate the power
dissipated in the resistor.
 Answer: 20 Watts
Electric current and resistance
Practice 9
 A resistor of resistance 12Ω has a current of 2.0A flowing through
it. How much energy is generated in the resistor in one minute?
 Answer: 2.9 x 103J
Electric current and resistance
Electric devices are usually rated according to the power they use. A
light bulb rated as 60W at 220V means that it will dissipate 60W
when a potential difference of 220V is applied across its ends. If the
potential difference across its ends is anything other than 220V, the
power dissipated will be different from 60W
Practice 10
 A light bulb rated as 60W at 220V has a potential difference of
110V across its ends. Find the power dissipated in this light bulb.
 Answer: 15W
Electric current and resistance
Practice
 A resistor of resistance 12Ω has a current of 2.0A flowing through
it. How much energy is generated in the resistor in one minute?
 Answer: 2.9 x 103J