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Current Electricity
Current Electricity –
what have we discussed static electricity
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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?
Flow or motion.
What does electric current mean?
Flow or motion of electric charge.
What will happen if two oppositely charged
metal plates touch?
Hint: Remember
The conducting
spheres.
How long?
Until the plates have equal charge.
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 still needs a potential
difference/voltage to push charges to move.
The pd/voltage gives a push.
Voltmeter
V = W.
q
The van der Graff & sphere set up p.d.
The vdg is not continuous enough for constant work.
Voltmeter fingers
Voltmeter battery
Current Electricity requires constant
source of potential difference or
voltage.
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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.
Outlet has p.d. = 120V.
Don’t stick a fork in it!!
An electric circuit allows current to flow an to do
continuous work. Circuits need the following
components:
• A source of continuous voltage.
• A closed pathway, circuit of conductors for charges
to flow.
• Resistor to convert electric E to some other form
(heat, light, sound etc.)
Which type of charges can move?
Solids only e- flow.
Liquids any charged
particles or ions can
flow.
Measuring Current
• Rate of flow of charge.
# charges per sec or
# Coulombs per sec.
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.
I = q/t
I = current C/s or A
Q = 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
t
100C
5s
20 C/s or 20 A.
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 = 1.67C
0.835 C/s
t = Q/I
= 2.00 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?
1A
To get continuous flow of charge we need:
1. voltage (p.d.) to push charges.
2. Charge Pump / E source = do work on charge
which gains PE. Pumps are batteries,
generators.
3. Closed Circuit – continuous pathway for
charges to flow –metal wire.
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.
• Simple Elec Circuit 5.5 min
• https://www.youtube.com/watch?v=EJeAuQ7pkpc&app=d
esktop
• Senior Physics Electric Current Clip 9 minutes
• http://www.youtube.com/watch?v=5laTkjINHrg
• Senior Physics Electric Voltage (skiers) 10 minutes
• http://www.youtube.com/watch?v=F1p3fgbDnkY
• Battery clip old but nice. 10 min.
• http://www.youtube.com/watch?v=IpaEGhjpZgc
Circuits
Circuit – Closed pathway for charges. All
components need p.d. to move.
Potential Dif/ Voltage
Gives 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.
Q pushed into Conducting
path (wire).
Q in motion.
PEelc - KE
Battery does work on q.
PEchm – PEelc
Resistors – (bulbs)
Work on q, PEelc &KE – heat, light.
PEelc used
up, PEchm,
starts cycle
again.
A Closer Look at Resistors
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).
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 Relates
Resistance, Current, Voltage
R = V/I
V = volts J/C
I = current A, C/s
R = resistance ohm’s W or V/A.
Resistance (R) unit = ohm’s W.
1. 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 W.
2. A 30-V battery maintains a current
through a 10- W resistor. Find the current
in the circuit.
• I = V/R
• 30 V/ 10 W.
• 3 -A
3. What is the current flow through an 80-W coil when
it is connected to a generator supplying 120-V?
V = IR
I = V/R
120V/80 W = 1.5 A
3a: What is the rate of electron flow through the
coil above?
9.4 x 1018 e/s
4. A resistor of 12 W has a current flow of 2-A. How
much energy is generated in the resistor in 1 minute
– or how much work does the resistor do in 1 min?
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PE = qV
V = IR
V = (2-A)(12 W )
V = 24 V.
in 1 minute,
2 C/s x 60 s = 120 C flow through.
PE = qV
(120 C )(24 V) = 2880 J.
Hwk: Read Text 19-1 and 19-2. Do pg 695 #3-5 and
pg 703 # 3-5.
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. Length
2. Area
3. Temperature
4. Type of material
Length – longer wire offers more
resistance. More chances for friction
in wire.
Less resistance
More resistance
Cross Sectional Area
Thick wires offer less resistance.
Temperature
Hot offers more R
Cold offers less R
At a given temperature,
R = resistance (W)
r = constant of resistivity (Wm)
l = length (m)
A = cross sectional area (m2)
See table
Ex 1: A 9.5 cm length of copper
wire has a cross sectional area
-3
2
of 2.5 x 10 m . What is the
o
resistance of the wire at 20 C?
From table r = 1.72 x 10-8 Wm.
So R = (1.72 x 10-8 Wm)(.095 m )
2.5 x 10-3 m2
= 6.5 x 10-7 W.
Ex 2: Find the resistance of a
copper wire 10 m long and 1.2
x 10-9 m2 in area.
143 W.
Resistance
For wire with known p and length
use:
R = rL/A
for R with known V and I use
R = V/I.
Film clip with questions. 9:30 minutes.
https://www.youtube.com/watch?v=GYJaOvkSEPw
1. _____V
Name _______________________________
6V
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.
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If the potential difference around the bulb is
actually measured as 5.5V, what might you
conclude happened ?
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_____________________________
2. _____V
Ex: A spool of gold wire with area 1.2 x 10-7
m2, has a resistance of 1.5 W. What is its
length?
• 7.4 m
Hwk Read tx. 19 – 1 & 19 - 2
Do text
717 #1, 2, 11, 17 – 21.
Potential Drop
If 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
R
Or
units V/W
V = IR
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
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 W.
Find V using:
V = IR
V = (8.6 x 10-5 W.)(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.
• qV
t
P= W
t
q = I
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 W has a current of 2.0-A
flowing through it. How much energy is
generated in the resistor in one minute?
I = 2A
R = 12W t = 60-s E = ? J
• Work = Energy
• W = I2Rt
=(2-A)2(12W)(60s) = 2880 J.
Graph
Power
• What represented by the slope?
• Slope = Voltage
• V=P
Current
Kilowatt hours kWh.
• Power is a rate of energy use.
• Electric sold in kWh which is Pt.
• 1 kWh is energy J 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
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.
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
Story of Electricity BBC 2 hours
• http://www.youtube.com/watch?v=mJnc79
MHSs4
Finish Power Examples
Hwk Read 19.3 Text
do
p 710 1 – 4,
p713 #1 – 3 and finish prac set