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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.
 Example Analogy
 A man carries a bucket of water up a hill.
 Pours water down slide
 Turns water wheel
 Turns a grind stone
 Generates heat!!!
 ______________ by man on the water is = _________ produced
 Conservation of energy!!!!
 Electrical Explanation (DRAW THIS)
 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.
 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)
 ______________ = change in electric potential energy
 W = electric potential = Fd
 _____ = ΔEelec
 Eqd = ΔEelec
 IB Equation : ______________
 AKA
V = W/q
 Where:
 V= potential difference(volts)
 E = Energy(Joules)
 q = __________
 W= work
 Now lets move that charge the other direction. We’ll release it at position B.
 The electric force “___________” it to position A and loses electric potential energy.
 It will also accelerate since it’s under a constant force.
 The increase of __________ gives the charge an increase of ________ energy.
 Change in electric potential energy = change in kinetic energy
electric potential = KE
____ = ½ mv2
v2 = 2Eqd/m
v=
 Electric Potential Difference
 As the charge moves between the plates the “energy difference per unit charge” remains constant
(________________________).
 Potential Difference is ______________(V)
Potential Difference(voltage) = the energy per unit charge
V = (ΔEelec / q)
Vq = ΔEelec
Vq = KE = ½ mv2
 Potential Difference(voltage) = the energy per unit charge
 V = (ΔEelec / q)
 Unit is ________
 Think of it as how much energy the charge is going to get as it zips from one plate to the other.
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 = _______________
IB Formal Definition and Equation
 Electronvolt – the energy gained by an electron accelerated through a potential difference of 1
volt.
 Ve = ½ mv2
5.1.4 Solve problems involving electric potential difference.
Practice 1 - 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 current and resistance
5.1.5 Define electric current.
New Diagram ***See board***
 Conducting material has a “_____________ 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 “_____________ ”
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.
 As they move they occasionally _____________ 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 is similar in concept to terminal velocity.
 New Diagram ***See board***
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Which direction do the electrons move?
From the _____________ terminal to the _____________ terminal, the source of the potential difference.
Electric _____________ - 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 
IB Formula/Definition: I = _____________
 Current is defined as the amount of electric charge passing a pint in a circuit in unit time.
 Current = charge/time
 1 _____________ = 1coulomb / 1second
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?
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What kind of current flows though your cell phone? _____________ \
Why do you have a “box” on your charger? _____________
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.
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 _____________
IB Formula and Definition: _____________
 Resistance = voltage / current
 Resistance – the ratio of the voltage across a conductor to the current flowing through it.
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. _____________ , _____________ , _____________ , gold.
 Application ____________
 Resistors
 In some conductors the free flow of electrons is made _____________ .
 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.
 _____________ – have a muddy floor
 Good _____________ – 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.
 Resistance of wire can be changed.
 Material
 Depends on each substance.
 _____________
 If the length doubles then the resistance doubles R ∝ L
 Cross-sectional area
 If the CSA doubles then the resistance halves R ∝ 1/CSA
 _____________
 Higher temp, means more atomic vibrations, means more collisions between electron and lattice.
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: _____________
 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
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Combine those to get R ∝ L / A
or R = L / A
R = resistance, L = length, A = cross sectional area
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)
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
 _____________
http://www.youtube.com/watch?v=zYS9kdS56l8&kw=electricity&ad=6599373247&feature=pyv&lr=1
5.1.9 Compare ohmic and non-ohmic behavior.
 Three different types of conductors - Draw the graphs!!!
 Ohmic
 Filament Lamp
 Diode
 Resistor is “ohmic” if the current flowing is _____________ to the voltage across its ends.
 _____________ are ohmic if the temperature is constant.
 Gives a straight line on I-V graph
 Ohmic conductors obey “Ohm’s Law”
 The filament lamp gets hot as there is an creasing voltage. This means the resistance increases with higher
voltages.
 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
 See work on board.
 Units for electrical power are Joules per second, J/s or Watts, W
IB Formula: P = _____________ _____________
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:
Practice 7
 If 12 volts is applied to the ends of a 40Ω resistor, calculate the power dissipated in the resistor.
 Answer:
Practice 8
 0.4 Amps flows through a 125Ω resistor, calculate the power dissipated in the resistor.
 Answer:
 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:
 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:
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: