Current Electricity
Current Electricity –what have we discussed static electricity
• Properties of Conductors & Insulators.
• Transfer of charge through various methods.
• Calculation of Forces between charged objects.
• Calculation of Forces exerted by electric fields.
• Energy of charges at positions electric fields (voltage)
We’ll apply what we know to current electricity.
What does current mean?
What does electric current mean?
Flow or motion.
Flow or motion of electric charge.
What will happen if two oppositely charged metal plates touch?
Until the plates have equal charge.
How long?
Hint: RememberThe conducting spheres.
We can use E of charges in motion to do work.
If we transfer charges from high PE to low PE, the charge can do some work converting E– but not for long.
With static I can charge an object, & transfer the electric charge to do work.
Static electricity is quick transfer of charge but charge stops when equilibrium reached.
Demo with bulb
We often want continuous flow of charge because we want continuous work.
Current Electricity provides continuous motion or flow of charges.
Current still needs a potential difference/voltage to push charges to move.
Each volt does work on each coulomb of charge & each charge changing q’s Energy:
The van der Graff & sphere set up p.d.
The vdg is not continuous enough for constant work.
V = W/q. V = E/q.
Current Electricity requires constant source of potential difference or
voltage.
• Battery
• Wall outlet
• Generator
• Solar Cell
Constant Potential Dif/Voltage induces charges to move continuously.
Voltage/p.d. provides the “push” or “pressure” to charges. Think of 1 side as more + and 1 side more neg
The higher the voltage/p.d., the more push or pressure each charge gets, the more E each q
has, the more W each q can do.
+-
An electric circuit allows current to flow an to do continuous work. Circuits need the following components:
• A source of continuous voltage.
• A pathway made of conducting materials.
• Appliance(s) to convert electric E to some other form (heat, light, sound etc.)
Solids only e- flow.
Liquids any charged particles or ions can flow.
Which type of charges can move?
Measuring Current
• Rate of flow of charge.
Amperes (A) measures rate of coulombs passing a point in a wire.
1 A = 1 C/s
passing a point or cross section of wire.
# charges per sec or# Coulombs per sec.
I = q/t
I = current C/s or AQ = charge in Coulombs
t = time in seconds
Ex 1: 100 C pass a section of wire every 5 sec. How much current flows in the wire?
I = q 100C 20 C/s or 20 A. t 5 s
Ex 2: The current in a light bulb is 0.835 A. How long does it take for a total charge of 1.67 C to pass a point in the wire?
I = Q/t t = Q/I
t = 1.67C = 2.00 s
0.835 C/s
Ex 3: If, in 1 second 6.25 billion billion (6.25 x 10 18) electrons pass through a point in a wire, what is the current?
1 A
Hwk read text 19-1
• Do page 695 # 3 – 5 & pg 717 #1 – 6 neatly on lined in full sentences! Show work.
How many ways can you set up a circuit of 1 battery, 2 bulbs, 3 wires to li
• Use all the components for each circuit, but no extra components allowed.
• Sketch all the arrangements that light the bulbs.
• Sketch all the arrangements that did not light the bulbs.
• Look at your sketches. Explain what you think is happening physically. Why do the bulbs light with certain arrangements? Why won’t they light with others? Write your hypothesis under your sketches.
• For a circuit that worked, list the E transformations.
• Write down any questions you have about the circuit.
• If time allows, find the circuit symbols in your reference table. Re-sketch your circuits using proper symbols.
Electric Current Clip 9 minutes
• http://www.youtube.com/watch?v=5laTkjINHrg
http://www.youtube.com/watch?v=IpaEGhjpZgc
Battery clip old but nice. 10 min.
To get continuous flow of charge we need:
1. voltage (p.d.) to push charges.
2. Charge Pump = do work on charge which gains PE. Pumps are batteries, generators.
3. Closed Circuit – continuous pathway for charges to flow –metal wire.
Outlet has p.d. = 120V.Don’t stick a fork in it!!
How does p.d. make charge flow??
Pot. Dif. Causes electric field to spread through wire at near light speed.
e- in wire respond by moving in field & colliding with neighboring e- starting to flow.
Electric field in wire caused by voltage source induces all
e- to move.
Each e- moves slowly but all begin at once.
Circuit – Closed pathway for charges. All components need p.d. to move.
Potential Dif/ VoltageGives the push
Conducting path (wire). Easy to move current through very little p.d. needed.
Resistors – (bulbs)Difficult for current to flow. Larger p.d. needed.
Work & Energy on Charges (q) in circuit.
Battery does work on q.
Q pushed into Conducting path (wire).
Resistors – (bulbs)
PEchm – PEelc
Q in motion. PEelc - KE
Work on q, PEelc &KE – heat, light.
PEelc used up, PEchm, starts cycle again.
Voltage E transformations as skiiers. 9.5 min.
• http://www.youtube.com/watch?v=F1p3fgbDnkY
Bulbs, toasters, computers... convert KE of e- to other forms of E – heat, light etc.
Devices are called resistors or loads. They slow down the e- so they resist current flow. (Like paddle wheel in river).
A Closer Look at Resistors
As e- flow through filament, KE lost to heat & light.
What must be around the bulb to push e- through it?
p. d.
Ohm’s Law The resistance of a conductor
R = V/I
V = volts J/C
I = current A, C/s
R = resistance ohm’s or V/A.
Resistance (R) unit = ohm’s .
Ex A: A 120-V potential difference is applied to a toaster which draws a current of 4 A. What is the toaster’s resistance?
• R – V/I
• = 120-V/4 A
• =30 .
Resistance in Wires
Occurs in wires as well as appliances. Certain factors affect how much resistance a wire will offer to current flow.
Factors affecting wire resistance.
1. Length2. Area3. Temperature4. Type of material
Length – longer wire offers more resistance. More chances for friction in wire.
More resistance
Less resistance
Cross Sectional AreaThick wires offer less resistance.
TemperatureHot offers more RCold offers less R
R = resistance () = constant of resistivity (m)l = length (m) A = cross sectional area (m2) See table
At a given temperature,
Ex 1: A 9.5 cm length of copper wire has a cross sectional area of 2.5 x 10-3 m2. What is the resistance of the wire at 20oC?
From table = 1.72 x 10-8 m.
So R = (1.72 x 10-8 m)(.095 m ) 2.5 x 10-3 m2
= 6.5 x 10-7 .
Ex 2: Find the resistance of a copper wire 10 m long and 1.2 x 10-9 m2 in area.
.
For wire with known p and length use:
R = L/A
for R with known V and I use
R = V/I.
Resistance
6 V
Name _______________________________
After Film: Given the circuit connected with copper wires to a 6V battery, what should each voltage (p.d.) read at position 1 and 2? Write the voltages in the space.
• If the potential difference around the bulb is actually measured as 5.5V, what might you conclude happened ?
• _____________________________
1. _____V
2. _____V
Ex: A spool of gold wire with area 1.2 x 10-7 m2, has a resistance of 1.5 . What is its length?
• 7.4 m
Hwk Read tx. 19 – 1 & 19 - 2Do text text p 717 #15,19, 20
AND
do review book pg 225 #2 - 7
Ohm’s LawResistance, Current, Voltage
Potential DropIf a current flows in a resistor or appliance, there must be a pd across the ends of the resistor.
The voltage pushes the charge.
The resistor “drops” or lowers the PEelc of the charge. So is sometimes called potential or voltage drop.
- +R
e- current
Current (flow rate) increases w/ increased pd
& decreases with increased resistance.
• R = V/I
• I = V units V/ R
Or V = IR
Ohm’s LawWhen T across a metallic resistor is constant, the current is directly proportional to pd across it.
V = RI V/I = R = constant.
V = volts J/C
I = current A, C/s
R = total resistance
ohm’s .
What is the graph of Ohm’s Law? Usu V on y-axis, I on x-axis, R is slope of direct linear.
V = RI yield direct linear relationship.V on Y axis. I on x axis. R is slope of straight line. R = constant.
What would the slope represent if current was placed on the Y axis?
1/R
Ex 1: A 30-V battery maintains a current through a 10- resistor. Find the current in the circuit.
V = IR
I = V/R
30 V/10 = 3.0 A
Ex 2: What is the current flow through an 80- coil when it is connected to a generator supplying 120-V?
V = IR I = V/R
120V/80 = 1.5 A
Ex 3: What is the rate of electron flow through the coil above?
Need to convert coulombs of charge from current equation to electrons so use equivalence of e- in 1 C:
1.5 C/s x (6.25 x 1018 e-/C) =
9.4 x 1018 e-/s pass through a point in the coil.
Ex 4. A 1 – m long copper wire has a cross sectional area of 0.0002 m2. If the current flow is 0.5-A, what is the potential difference?
Find R using.
R = 8.6 x 10-5 .
Find V using:V = IR
V = (8.6 x 10-5 .)(0.5 A) = 4.3 x 10-5V
5. A light bulb heats up as it’s in use. Which best represents the V-I graph of a bulb filament.
Hwk read txt 19.2 do pg 703 prb #1 – 6
Power in Resistors
• Resistors convert KE & EElc to other forms.
• They do work on q.
• P rate work gets done or E used/ converted/ dissipated or supplied J/s or Watts.
• Power rating of 500 W means Eelc converted to other kinds at rate 500 J/s.
Derive some equations relating power, voltage, resistance, and
current.
P = W t
• qV q = I t t
• VI
• P = VI
• The power dissipated is thermal energy and work done in resistor.
Other Power Equations
From Ohm’s Law we use R = V/I to derive other equations for power.
P = I2R = V2.
R
Ex 1. A toaster draws a current of 2.0-A from a 120-V source. What is its power rating?
P = VI
P = (120V)(2.0 A)
240 VA or 240 W.
For bulbs, power = brightness.
• Rate bulb converts Peelc & KE to heat & light.
Work done
Since Power is Work/Time,
Power x Time = Work or Energy
Ex 2: A resistor of 12 has a current of 2.0-A flowing through it. How much energy is generated in the resistor in one minute?I = 2A R = 12 t = 60-s E = ? J
• Work = Energy
• W = I2Rt
=(2-A)2(12)(60s) = 2880 J.
Graphs P=VI
Pow
er
Current
Slope =Voltage
Power Ratings for Appliances
Devices are rated by the power they use. A bulb rated 60 W 220 V means:
the bulb will dissipate 60 W when attached to a 220 p.d.
If a different p.d. is used, then it won’t dissipate 60 W.
Fuses
As current flows, wires heat up.
Fuses designed to break circuit if current becomes to high for the wires.
Fuse should be rated just above the ideal operating current for a circuit.
Story of Electricity BBC 2 hours
• http://www.youtube.com/watch?v=mJnc79MHSs4
Finish Power Examples
Hwk Read 19.3 Text
do p 710 1 – 4,
p713 #1 – 3 and prac set
Kilowatt hours kWh.
• Power is a rate of energy use.
• Electric sold in kWh which is Pt.
• 1 kWh is energy delivered to home in 1hour.
• 1 kWh (1000 W/kW)(60 min/h)(60s/min) = 3.6 x 106 Ws = 3.6 x 106 J
