Name ________________ Date ________ Block ___
GAC Physics
Topic 5: Electric Currents
Ch 17
5.1 Electric potential difference, current and resistance
Electric Potential difference and Capacitance
5.1.1
Sec 1,
Define electric potential difference.
Practice
A,
5.1.2
Determine the change in potential energy when a charge moves between two points at different
SR pg 601
potentials.
5.1.4
Solve problems involving electric potential difference.
5.1.12 Sec 2,
Relate capacitance to the storage of electrical potential energy in the form of separated charges.
5.1.13 Practice B,
Calculate the capacitance of various devices.
5.1.14 SR pg 607
Calculate the energy stored in a capacitor.
Electric Current and resistance
Assessment statement
5.1.5 Sec 3,
Define electric current.
Practice C,
SR pg 616
5.1.6
Sec 3,
Practice D,
SR pg 616
5.1.7
Define resistance.
Apply the equation for
resistance in the form:
where ρ is the resistivity
of the material of the resistor.
State Ohm’s law
Solve problems involving potential
difference, current and resistance.
5.1.8
5.1.11
Teacher’s notes
It is sufficient for students to know that current is
defined in terms of the force per unit length
between parallel current-carrying conductors.
Students should be aware that R = V/I is a
general definition of resistance. It is not a
statement of Ohm’s law. Students should
understand what is meant by resistor.
For example, students should be able to draw I–
V characteristics of an ohmic resistor and a
filament lamp.
5.2 Electric Circuits
5.2.1
5.2.2
5.2.3
Ch 18, Sec 2, Practice A, SR pg656
Ch 18, Sec 2, Practice B, SR pg 656
Ch 18 Sec 3, Practice C & D, SR pg 663
5.2.4
Ch 18, Sec 1, SR pg 645
5.2.5
Apply the equations for resistors in series.
Apply the equations for resistors in parallel.
Apply the equations for combinations of resistors in
complete circuits.
Draw circuit diagrams. Students should be able to
recognize and use the accepted circuits symbols.
Describe the use of ideal ammeters and ideal voltmeters.
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Electricity – Topic 5
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.
 When two charges interact there is an electric ____________ between them.
 Potential energy from this is: Electrical Potential Energy
 Similar to gravitational force
 Type of _____________ energy
 Metotal = KE + _______ + PEspring + PEelectric
Where does the “potential” come from
 With potential gravity, an object can “potentially” fall from a height.
 With electrical potential, a charge can “_______________” fall down an electric field.
 A positive charge is placed in a uniform electric field and is _______________.
 The new position will have a ∆PE
 Electric Potential Energy Difference between two points in an electric field is equal to the ________ done
in moving a charge from one position to a position of higher electrical potential energy in an electric field.
 See Video/ Make Drawing
2
 Equations
 ∆PEelectric = _____________
 q = charge
 E = ________________
 d = displacement from reference point
 (-) = inidcates that the electrical potential energy will _____________ if the charge is NEGATIVE
 Unit is a Joule(J)
 Potential Difference
 A measure of the difference in the electrical potential energy between two positions in space.
 AKA “_______________”
 Describes a change in energy per unit charge
 Unit is volt (V)
 Most batteries are ________
 Similar idea to a water slide
 ***Draw diagram***
 Equation
 ∆V = _____________
 ∆V = voltage
 ∆PEelectric = electric potential energy
 q = _____________
 Equation
 Combine the PEelectric equations with the ∆V and you get….
 ∆V = -Ed
 ***show substitution***
 New Equation:
 ∆V = -Ed
 Last one
 ∆V = ____________
 kc = Coulomb constant (_____________Nm2/C2)
 q = charge
 r = distance separating the two charges
5.1.4 Solve problems involving electric potential difference.
 Pg 599, Practice A
 Section Review pg 601
CAPACITANCE!!!
5.1.12 - Relate capacitance to the storage of electrical potential energy in the form of separated charges.
5.1.13 - Calculate the capacitance of various devices.
5.1.14 - Calculate the energy stored in a capacitor.
 Capacitor
 Device that _______________ electrical potential energy.
 Uses
 tuning the frequency of radios
 Eliminating sparking in automobile ignition systems
 Storing energy in electronic flash units
 If, _______________ or charged has energy
 Energized by connecting the plates to the terminals of a __________________.
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 Charge transfer between the plates stops when the potential difference between the plates is equal to the
pd of the battery.
 Capacitance
 Is the ability of a conductor to ___________ energy in the form of electrically separated charges.
 Is the ratio of charge to potential difference
 So…
 Capacitance = magnitude of charge/potential difference
 C=
 SI unit is farad (F)
 Capacitance
 Depends on the _________ and _________ of the capacitor.
 Many capacitors have parallel plates with no material
between them
 The equation shows that as the area of increases the
capacitance increases
 C=
 ε = permittivity of a vacuum (8.85 x10-12 C2/Nm2)
 A = area of one of the plates
 d = distance between the plates
 Dielectric
 The space between isn’t always ____________
 __________ – insulating material that fills the space between two capacitor’s plates
 Ex – air, rubber, glass, waxed paper, ____________
 Capacitance increases with use of dielectric.
 Molecules in a dielectric can __________ with the field of the pates
 Reduces the charge on the capacitor plates.
 Q=
 Discharge
 Is the opposite of ____________
 After the two plates are charged they will stay
charged until…
 Connected with a ______________ material.
 Charges move back from one plate to the other
until both are balanced and uncharged.
 This is because this is the lowest potential
energy.
 Examples:
 _________ in camera
 ___________
 Stored Energy
 A charged capacitor stores ______________________.
 If each plate is neutral, then it takes almost ______________ to move charges in the circuit.
 Once the plates start to build up a charge, a potential difference develops
 This requires more and more work to be done to move the charge through the pd.
 This builds the __________________ energy.
 The work done on these charges is the transfer of energy.
 Stored Energy
 This energy can be calculated by:
4








Electric potential energy = ½ (charge on plate)(final potential difference)
PEelectric =
By substitution this can become
PEelectric =
 C= capacitance
Example
 A capacitor, connected to a 12V battery, holds 36 μC of charge on each plate. What is the capacitance of
the capacitor? How much electrical potential energy is stored in the capacitor?
 Identify your variables. Identify your equations. Watch your units.
Answer: C = 3.0 x 10-6 μF, PE = 2.2 x 10-4 J
Practice
Pg 607 Practice B #1-4
 1) a) 4.8 x 10-5 C,
b) 4.5 x 10-6 J
-6
 2) a) 4.8 x 10 F,
b) 5.4 x 10-6 J
 3) a) 9.00V
b) 5.0 x 10-12 C
8
2
 4) 1.13 x 10 m , ???
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5.1.5 - Define electric current.
CURRENT
 So far static electricity (charges in _____________________)
 Static electricity isn’t very useful
 Moving electricity IS!!
 CURRENT
 The _________________ of electric charge
 Could be ________________ or negative charges
 Used everywhere
 Lights, radios, TV, air conditioners, Cars, computers ____________________
 Bodies, discovered in mid 1700s.
 Luigi Galvani experimented next to a dissected __________________
 Current transmit messages between muscles and brain
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



More specifically….
Imagine a cross section of wire
Electrons are now moving through that wire
Current is the ____________ these charges move
 CURRENT
 Current is the ratio of the amount of charge to the
time interval
 _________ is amount of charge
 ∆t is the time it takes these charges to pass a point
 Current is the ratio of the amount of charge to the time interval
 ___________ is amount of charge
 ∆t is the time it takes these charges to pass a point
 I = ∆q/∆t
 I = current
 Unit is ___________, A
***Special note***
 Current is the _________________ direction of the movement of the negative charges.
Example
 The current in a light bulb is 0.835A. How long does it take for a total charge of 1.67C to pass through the
filament of the bulb?
 Identify what you know. Identify what you want to know. Identify your correct equation.
 Answer: 2.00s
Practice
 Pg 609 Practice C #2-5
 Answers:
1) 400s
2) 1.00s
3) 600s
4) 20C
5) a) 2.6 x 10-3A, b) 1.6 x 1017 electrons,
c) 5.1 x 10-3A
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Drift Velocity
 When you flip the switch how fast do the lights come on? Instantaneously?
 Does this mean the electrons are moving at the speed of light?
 Not quite. 
Drift Velocity
 When you turn on the light the electron motion near the switch changes the
electric field there.
 That change in the electric field then propagates throughout the wire at nearly
the speed of light.
 The charges move much more slowly.
 When these charges do move do they move in a straight line?
 Not quite. Sorry. 
 Remember that electrons in a metallic bond jump from nucleus to nucleus.
 These jumps are random and sporadic.
Drift Velocity
 What causes these crazy movements?
 As they move the individual atoms are vibrating. (see Brownian Movement)
 In these collisions energy is transferred, which causes more vibration and an increase in temperature. (think
friction)
 Electrons want to accelerate in this electrical field but can’t because of these collisions.
 Eventually things will balance out.
 Drift velocity is the net velocity of these individual movements.
 So how fast does an electron really move???
 In a wire with current 10A, the drift speed would be 2.46 x 10-4m/s.
 Meaning it would take about 68 minutes for it to travel 1 meter.
Resistance
5.1.6 – Define resistance
5.1.7 – Apply the equation for resistance
5.1.8 – State and apply Ohm’s Law
 What is a conductor?
 What is an insulator?
 What is a resistor?
 Is there a difference between an insulator and a resistor?
 Is there a difference between a conductor and a resistor?
 Resistance
 A resistor is a ___________________.
 It’s just not a good conductor.
 _________________ to the motion of charge through a conductor is RESISTANCE.
 For many materials… resistance is _______________over a wide rang of applied potential differences.
 Earlier we learned that electrons do not move in ______________________through conductors.
 This is because of collisions with the metal atoms.
 Anything that affect the number of collisions will also affect the materials _________________.
 Primarily 4 factors.
 Length
 Cross-sectional Area
 Material
 Temperature
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 Resistance
 Two are purely geometrical factors.
 _________________ – the longer the wire the more collisions which means more resistance.
 Cross-Sectional Area – a _____________ wire physically restricts the path an electron can take,
increasing the resistance.
 Material – different material have different structures of atoms. Some are more conductive than others.
 Examples….____________
 Temperature – Temperature is a measure of how much an object’s atoms are _____________. More
vibrations means it’s harder for atoms to travel, which means more resistance.
 Resistance
 The relationship between resistance, voltage and current can be expressed by saying resistance is the
ratio of potential difference to current.
 OR
 R=
 SI unit for resistance is ohm, Ω
Ohm’s Law
 Ohm’s law states that voltage will cause a current to flow through a
resistor.
 Commonly stated as V = IR
 Watch Video: https://www.youtube.com/watch?v=zYS9kdS56l8
Electric current and resistance
 Three different types of conductors
 Ohmic
 Filament Lamp
 Diode
 Resistor is “_______” if the current flowing is proportional to the
voltage across its ends.
 Metals are ohmic if the temperature is constant.
 Gives a _____________ line on I-V graph
 Ohmic conductors obey “Ohm’s Law”
 The filament lamp gets hot as there is an increasing voltage. This
means the resistance _____________ with higher voltages.
 Diodes
 Lets very little or no current flow until voltage reaches a threshold.
 0.6V is very common
 Once the threshold is met, there is essentially ________ resistance
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
Example
 The resistance of a steam iron is 19.0 Ω. What is the current in the iron when it is connected across a
potential difference of 120V?
 Identify what you know. Identify what you are looking for. Identify your equations. Solve.
 Answer: 6.32Ω
Example
 A 1.5V battery is connected to a small light bulb with a resistance of 3.5Ω. What is the current in the
bulb?
 Identify what you know. Identify what you are looking for. Identify your equations. Solve.
 Answer: 0.43A
 Practice
10
 Pg 615 Practice D # 3-6
 Answers:
1) 0.43A
2) 1.8 A
3) a)2.5A
b) 6.0A
4) 110 V
5) 46Ω
6)a 0.41A
b) 0.59A
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 Pg 616 Section Review
 Answers:
1) yes; When charge carriers are negative
2) a) 0.60A
b) 2.2 x 1020 electrons
3) 12 A
4) 3.6Ω
5) The diode is non-ohmic
6) Resistors regulate the magnitude of the current. Diodes regulate the direction of current.
7) 1.5A, 2.4A
8) Although electrons undergo a force that moves them across the wire, collisions with atoms
continually randomize their motion. As a result, the drift speed is much less than the average speed
between collisions.
9) You could decrease current by making the wire as long as possible, thereby increasing its resistance.
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Circuits
5.2.4 - Draw circuit diagrams. Students should be able to recognize and use the accepted circuits symbols.
Ch 18, Sec 1, SR pg 645
 Remember when we made the light bulb light up at the beginning of this unit.
 Draw a picture of what we did.
 Volunteers to put their drawing on the board.
Schematic Diagrams
 A diagram that depicts the construction
of an electrical apparatus
 Uses ____________ to represent
components of a circuit.
 Anybody from around the ________
can read schematics
 Some basic symbols are…
Electric Circuits
 A path through which charges can
_______________.
 A schematic diagram is also called
______________ diagram.
 All circuits must have a source of potential difference (ex_____________________ ), resistance, also called
a “load” (ex… __________________), and some thing connecting them (ex… _____________________)
 It must be a continuous/complete loop for the charges to flow and work.
Short circuits
 With out a load, or resistance, the circuit has very little resistance to the movement of electrons.
 Happens when ______-insulated wires connect accidentally.
 They heat up, could possible melt/spark/catch fire.
Section Review Pg 645 #1, 3-5
Series Circuits
5.2.1 - Apply the equations for resistors in series. - Ch 18, Sec 2, Practice A, SR pg656
Current in series
 Break up into groups of two.
 Each group gets a battery/wire/bulb
 Build a circuit.
 Combine and make both bulbs turn on.
 What happened to the second bulb?
Current
 Charge is ______________…. Meaning that the charge that enters one bulb is the same amount of charge
that exits in a given time interval
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 Only one path…. Meaning that charge that exits the first bulb is the ___________ amount of charge that
enters the second bulb in a given time interval
 Current is the amount of charge per time…. Meaning the current in bulb 1 is the same current in bulb 2.
 CONCLUSION: When many resistors are connected in series, the current in each resistor is ____________!
 OR…. Current is __________________ in a series circuit
Resistance
 The total current in a series circuit depends on how many resistors there are.
 To find the total current you must find the total resistance,
 Total resistance is called ______________________
 Equivalent resistance of a series combination of resistors is always greater than any individual resistance.
 Req =
Voltage in a series
 The potential difference across the battery, ∆V, must equal the potential difference across the load, ∆V1 +
∆V2…,
 So ∆V =
 CONCLUSTION: The total voltage is equal to the ____________ of all the individual voltages.
 With a little substitution you get….
 ∆V = IReq
 ∆V = IR1 + IR2 + IR3….
And
 ∆V1 = IR1 and ∆V2 = IR2 and ∆V3 = IR3
Conclusions for a SERIES CIRCUIT
 Current is _______________
 ∆Vtotal = ∆V1 + ∆V2 + ∆V3 …..
 Req = R1 + R2 + R3 …..
 ∆V1 = IR1 and ∆V2 = IR2 and ∆V3 = IR3
 P.S. What happens when you unscrew one of the light bulbs?
 They all go out. Why?
 Note: all elements ____________be connected for a series circuit to work.
Sample Problem A
 A 9.0V battery is connected to four light bulbs. They have 2.0Ω, 4.0Ω, 5.0Ω and 7.0Ω resistance
respectively. Find the equivalent resistance for the circuit and the current in the circuit.
 Draw a schematic. Label it. Identify what you need. Identify what you know. Solve.
 Answer: Req = 18.0Ω, I = 0.50A
Practice as class
 A 12.0V storage battery is connected to three resistors, 6.75Ω, 15.3Ω, and 21.6Ω respectively. The resistors
are joined in series.
 Calculate the equivalent resistance. What is the current in the circuit?
 Draw a schematic. Label it. Identify what you need. Identify what you know. Solve.
 Answer: Req = 43.65Ω, I = 0.27A
Practice: Pg 650, Practice A # 3-6
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Parallel Circuits
5.2.2 - Apply the equations for resistors in parallel. - Ch 18, Sec 2, Practice B, SR pg 656
 Get back into those groups.
 Now make both bulbs turn on in a parallel circuit. Meaning one can be unscrewed and the other still stay on.
 What happens to the brightness of the bulbs as one is connected/disconnected?
 Resistors in parallel have the same potential differences across them.
 Meaning….
 ∆Vtotal = _____________________
Current in parallel circuits
 When the ___________ flow through the wire they reach one bulb first. Some move through the first bulb.
Others move through the second bulb.
 If one bulb has ___________resistance, more charges move through that bulb. This is because that bulb
offers less opposition to the flow of charges.
 The sum of currents in parallel resistors is equal to the _____________ current.
 I = I1 + I2 + I3 ….
 Based upon that and ∆V = IR…. I = ∆V/R
So …. (∆V/Req) = (∆V/R1) + (∆V/R2) + (∆V/R3) …
Resistors in parallel
 Since
 ∆Vtotal = ∆V1 = ∆V2 = ∆V3….
 And
 (∆V/Req) = (∆V/R1) + (∆V/R2) + (∆V/R3)
 We can then simplify to….
 (1/Req) =
 The total resistance must always be__________________ than the smallest resistor!?!?
Conclusions for a PARRALLES CIRCUIT
 ____________________ is constant
 I=
 (1/Req) = (1/R1) + (1/R2) + (1/R3)
Sample Problem B
 A 9.0V battery is connected to four resistors in a parallel fashion. They include 7.0Ω, 5.0Ω, 4.0Ω, and 2.0Ω.
Find the equivalent resistance for the circuit and the total current in the circuit.
 Draw a schematic. Label it. Identify what you need. Identify what you know. Solve.
 Answer: Req = 0.917Ω, I = 9.8A
Class Practice
 The potential difference across the equivalent resistance in sample problem B equals the potential difference
across each of the individual parallel resistors. Calculate the value for the current in each resistor.
 Draw a schematic. Label it. Identify what you need. Identify what you know. Solve.
 Answer: 1.3A, 1.8A, 2.25A, 4.5A
Practice - Pg 655, Practice B # 2-4
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Schematic Practice Questions
Answers:
1) one battery, one closed switch, two resistors, and three bulbs
2) ….
3) Figure 6 and Figure 8 will have no current I them.
4) 3.0 V
5) It is converted to thermal energy and light energy.
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Series Circuits Practice - Pg 650
Answers:
1) a. 43.6 Ω b. 0.275 A
2) a. 24Ω
b. 1.00A
c. 1.00A
3)1V, 2V, 2.5V, 3.5V
4) a. 11.28Ω, 0.798A
b. 5.79V, 3.22V
5) 0.5 Ω
6) a. 67.6 Ω b. 45 bulbs
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Parallel Circuit Practice – pg 655
Answers:
1) 4.5A, 2.2A, 1.8A, 1.3A
2) 50.0Ω
3) a. 2.2Ω
b. 6.0A, 3.0A, 2.0A
4) a. 2.99Ω
b. 36.0V
c. 2.00A, 6.00A
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Series and Parallel Circuits - pg 656
Answers:
1) in the series circuit
2) a. All have equal I
b. 15Ω
3) a. 5Ω
b. All have equal ∆V
4) a. 2Ω: 2A, 4V
b. 2Ω: 6A, 12V
4Ω: 2A, 8V
4Ω: 3A, 12V
5) Because the resistance of each bulb is the same, the brightness depends only on the potential
difference. Bulbs a, d, and e have equal potential difference across them and thus equal
brightnesses. Because bulbs b and c have the same resistance and re in series, they have equal but
lesser potential differences and thus are equally bright but less bright than bulbs a, d, and e.
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20
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Practice C – pg 659
Answers
1) a. 27.8Ω b. 26.6Ω
2) a. 50.9Ω b. 57.6Ω
c. 23.4Ω
Practice D – pg 662
Answers
Ra = 0.5A, 2.5V
Rb = 0.5A, 3.5V
Rf = 2.0A, 4.0V
Rc = 1.5A, 6.0V
Rd = 1.0A, 4.0V
Re = 1.0A, 4.0V
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Combination Circuits Section Review – pg 663
Answers: 1) 9.8Ω 2) 1.8A
3) 2.7V
4) 175Ω
5)0.229A, 3.44V
6) 0.235A, 3.52V
7) a. No, the current is the toaster is less than the current in the microwave,
b. Yes, the potential differences are equal because they are in parallel.
c. Yes, because it is in series with the rest of the circuit
d. 3.6Ω
e. 7.1 A
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