current electricity continuous motion of charged particles through a potential /voltage difference
TRANSCRIPT
Current Electricity
Continuous motion of charged particles through a potential /voltage difference.
Potential Dif/Voltage induces charges to move. Amount of work per charge.
Two models for current.
• Conventional current – positive charges in motion (IB).
• Real/electron current – electrons in motion.
Voltage gives free e- a push -charges gain lose PE and gain KE:
PE = KEqV =1/2 mv2.
Units of Current - Fundamental
Amperes (A) measures rate of charge flow in q/t.
1 A = 1 C/s passing a point or cross section of wire.
Ex 1. How many e- must pass a point in a wire every second to carry a current of 2 A?
• 2 C/s x 1.6 x 10-19 C/e- = 1.25 x 1019 e-.
To get continuous flow of charge:
1. Voltage (EMF) source = to do work on q which gains E. Batteries, generators, solar cells. The source raises the PE of charges.
2. Closed Circuit – continuous pathway for charges to flow –metal wire, ionic solutions.
4. Resistors/Load – device to convert/dissipate energy (so that source is not immediately discharged). Resistors lower the PE of charges.
P. D. Causes electric field to spread through wire at near light speed.
All e- in wire respond by moving in field & colliding with neighboring e- starting to flow.
Drift velocity is net speed in one direction. It’s slow for e- (mm/s).
How it happens
Electric field in wire caused by voltage source induces
e- to move through.
Bulbs, toasters, etc. convert Eelc to other forms– heat, light etc.
These cause the e- to lose PE. Devices are called resistors or loads. They slow down the e- through collisions so they resist current flow.
Resistors, Appliances, Loads
Direct Current (DC) circuits have current flow of e- in 1 direction fr
Envisioned as traveling from the neg to + terminal of battery.
Conventional current (pos charge) fr pos to neg..
Resistance
- Caused by internal collisions/interactions.
- Ratio of Voltage to Current
- R = V/I
- Units ohms .
- V/A
Resistance
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.
More resistance
Less resistance
Cross Sectional AreaThick wires offer less resistance.
TemperatureHot offers more RCold offers less R
R = resistance = constant of resistivityl = length A = cross sectional areaSee table
At a given temperature,
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 PE of the charge. So is sometimes called potential drop.
- +R
e- current
Ohm’s LawWhen temperature across a metallic resistor is constant, the current is directly proportional to pd across it.
V = IR V/I = R = constant.
V = volts J/C
I = current A, C/s
R = total resistance
ohm’s .
V = RI yield direct linear relationship.V on Y axis. I on x axis. R is slope of straight line. R = constant.
Switch axis 1/R is slope.
Light bulbheats up as current goes through.
Watch axis - R is 1/slope here.
I
V
Other resistors vary with temperature. See Kerr pg 133.
When a graph shows non-ohmic behavior, you can simply find R = V/I at a point – don’t use slope.
0.2 V/1.5 x 10-3 A
R = 133.3 .
• Film Clip Voltage 8.5 min• http://www.youtube.com/watch?
v=F1p3fgbDnkY&feature=relmfu
Film Clips
Resistance 10 mimhttp://www.youtube.com/watch?v=YGvu9iqjJq4
Power in Resistors
Resistors/loads convert EElc to other forms.
P is rate 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.
P = W. = Vq. = VI t t
P = VI
The power is rate thermal E dissipated & work done in
resistor.
Other Power Equations
For devices that obey Ohm’s Law we use R = V/I to derive other equations for power.
P = RI2 = V2.
R
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.
Kilowatt hours kWh.
• Power is a rate of energy use.
• Electric sold in kWh which is Pt = Energy.
• 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
Film Clip Resistance
• http://www.youtube.com/watch?v=YGvu9iqjJq4&NR=1
IB Packet and Kerr Pg 135 #3, 4, 17, 21.