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ELECTRICITY
INTRO•Intro battery/wire/light bulb activity•Use multi meter to look at battery voltage.
•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.
5.1.1 Define electric potential difference•When two charges interact there is an electric force between them.•Potential energy from this is: Electrical Potential Energy•Similar to gravitational force•Type of mechanical energy•Metotal = KE + PEg + 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 “potentially” fall down an electric field.•A positive charge is placed in a uniform electric field and is displaced.•The new position will have a ∆PE
Where does the “potential” come from•With potential gravity, an object can “potentially” fall from a height.•With electrical potential, a charge can “potentially” fall down an electric field.•A positive charge is placed in a uniform electric field and is displaced.•The new position will have a ∆PE
•Electric Potential Energy Difference between two points in an electric field is equal to the work done in moving a charge from one position to a position of higher electrical potential energy in an electric field.
•See Video•Make Drawing
Equation•∆PEelectric = -q Ed
•q = charge•E = electric field strength•d = displacement from reference point•(-) = inidcates that the electrical potential energy will INCREASE 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 “voltage”•Describes a change in energy per unit charge•Unit is volt (V)•Most batteries are 1.5V•Similar idea to a water slide***Draw diagram***
Equation•∆V = ∆PEelectric / q
•∆V = voltage•∆PEelectric = electric potential energy
•q = charge
Equation•Combine the PEelectric equations with the ∆V and you get….•∆V = -Ed***show substitution***
New Equation:∆V = -Ed
Last one•∆V = kc q/r
•kc = Coulomb constant (8.99 x 109Nm2/C2) •q = charge•r = distance separating the two charges
5.1.4 Solve problems involving electric potential difference. •Pg 599, Practice A
1) 6.4 x 10-19
2) -750V3) (charge of electron = 1.6x10-19)
2.3 x 10-16J
•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 stores electrical potential energy.•Uses• tuning the frequency of radios•Eliminating sparking in automobile ignition systems•Storing energy in electronic flash units
Capacitor•If, energized or charged has energy•Energized by connecting the plates to the terminals of a battery.•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 store energy in the form of electrically separated charges.•Is the ratio of charge to potential difference•So…•Capacitance = magnitude of charge/potential difference•C = Q/∆V
•SI unit is farad (F)
Capacitance•Depends on the size and shape 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 A/d•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 empty•Dielectric – 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 align with the field of the pates•Reduces the charge on the capacitor plates.
•Q= C∆V
•The space between isn’t always empty•Dielectric – 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 align with the field of the pates•Reduces the charge on the capacitor plates.
Discharge•Is the opposite of charging•After the two plates are charged they will stay charged until…•Connected with a conductive 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.
Discharge•Examples:•Flash in camera•Keyboards
•Size of capacitors video
Stored Energy•A charged capacitor stores electrical potential energy.•If each plate is neutral, then it takes almost no work to move charges in the circuit.•Once the plates start to build up a charge, a potential difference develops
Stored Energy•This requires more and more work to be done to move the charge through the pd. •This builds the electric potential energy.•The work done on these charges is the transfer of energy.
Stored Energy•This energy can be calculated by:
•Electric potential energy = ½ (charge on plate)(final potential difference)•PEelectric = ½ Q ∆V
•By substitution this can become•PEelectric = Q2/2C
•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 J2) a) 4.8 x 10-6 F, b) 5.4 x 10-6 J3) a) 9.00V b) 5.0 x 10-12 C4) 1.13 x 108 m2, ???
5.1.5 - Define electric current.
CURRENT• So far static electricity (charges in
electric fields)• Static electricity isn’t very useful• Moving electricity IS!!
CURRENT• The movement of electric charge• Could be positive or negative charges• Used everywhere• Lights, radios, TV, air conditioners, Cars,
computers• Bodies, discovered in mid 1700s.• Luigi Galvani experimented next to a
dissected frog• Current transmit messages between
muscles and brain
CURRENT• More specifically….• Imagine a cross section of wire• Electrons are now moving through
that wire• Current is the rate these charges
move
CURRENT• Current is the ratio of the amount of
charge to the time interval• ∆q is amount of charge• ∆t is the time it takes these charges
to pass a point
CURRENT• Current is the ratio of the amount of
charge to the time interval• ∆q is amount of charge• ∆t is the time it takes these charges to
pass a point
• I = ∆q/∆t
• I = current• Unit is Ampere, A
CURRENT***Special note***• Current is the OPPOSITE 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) 400s2) 1.00s3) 600s4) 20C5) a) 2.6 x 10-3A, b) 1.6 x 1017
electrons c) 5.1 x 10-3A
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.
Drift Velocity• 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)
Drift Velocity• 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.
Drift Velocity• 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.
Resistance5.1.6 – Define resistance5.1.7 – Apply the equation for resistance5.1.8 – State and apply Ohm’s Law
Resistance• 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 conductor.• It’s just not a good conductor.• Opposition to the motion of charge
through a conductor is RESISTANCE.• For many materials… resistance is
constant over a wide rang of applied potential differences.
Resistance• Earlier we learned that electrons do not
move in straight-lines through conductors.• This is because of collisions with the metal
atoms.• Anything that affect the number of
collisions will also affect the materials resistance.
• Primarily 4 factors.• Length• Cross-sectional Area• Material• Temperature
Resistance• Two are purely geometrical factors.• Length – the longer the wire the more
collisions which means more resistance.
• Cross-Sectional Area – a thinner wire physically restricts the path an electron can take, increasing the resistance.
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 vibrating. 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.
• ORR = ∆V / I
• 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
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
Electric current and resistance •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
•The filament lamp gets hot as there is an creasing voltage. This means the resistance increases with higher voltages.
Electric current and resistance •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
Practice•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
Practice•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
Resistance•Pg 615 Practice D # 3-6
•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 symbols to represent components of a circuit.•Anybody from around the world can read schematics
•Some basic symbols are…
Electric Circuits•A path through which charges can flow.•A schematic diagram is also called circuit 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 continous/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 un-insulated wires connect accidentally.•They heat up, could possible melt/spark/catch fire.
•Section Review Pg 645 #1, 3-5
•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.
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 conserved…. Meaning that the charge that enters one bulb is the same amount of charge that exits in a given time interval
Current•Only one path…. Meaning that charge that exits the first bulb is the same amount of charge that enters the second bulb in a given time interval
Current•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 the same!
Current•CONCLUSION: When many resistors are connected in series, the current in each resistor is the same!•OR…. Current is constant 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•Equivalent resistance of a series combination of resistors is always greater than any individual resistance. •Req = R1 + R2 + R3….
Voltage in a series•The potential difference across the battery, ∆V, must equal the potential difference across the load, ∆V1 + ∆V2…,
•So ∆V = ∆V1 + ∆V2….
CONCLUSTION: The total voltage is equal to the sum of all the individual voltages.
Voltage in a series•With a little substitution you get….
•∆V = IReq
•∆V = ∆V1 + ∆V2 + ∆V3….
•∆V = IR1 + IR2 + IR3….
•and
•∆V1 = IR1 and ∆V2 = IR2 and ∆V3 = IR3
Conclusions for a SERIES CIRCUIT•Current is constant
•∆Vtotal = ∆V1 + ∆V2 + ∆V3 …..
•Req = R1 + R2 + R3 …..
•∆V1 = IR1 and ∆V2 = IR2 and ∆V3 = IR3
PS•What happens when you unscrew one of the light bulbs?
•They all go out. Why?
•Note: all elements MUST 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.
a) Calculate the equivalent resistance.b) 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
•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 = ∆V1 = ∆V2 = ∆V3….
Current in parallel circuits•When the charges flow through the wire the reach one bulb first. Some move through the first bulb. Others move through the second bulb.
•If one bulb has less resistance, more charges move through that bulb. This is because that bulb offers less opposition to the flow of charges.
Current in parallel circuits•The sum of currents in parallel resistors is equal to the total 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) = (1/R1) + (1/R2) + (1/R3)
The total resistance must always be less than the smallest resistor!?!?
Conclusions for a PARRALLES CIRCUIT•Voltage is constant
•I = I1 + I2 + I3 ….
•(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
