Loading presentation...

Present Remotely

Send the link below via email or IM


Present to your audience

Start remote presentation

  • Invited audience members will follow you as you navigate and present
  • People invited to a presentation do not need a Prezi account
  • This link expires 10 minutes after you close the presentation
  • A maximum of 30 users can follow your presentation
  • Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.


Electric current and electric resistance


Keith Herold

on 5 October 2016

Comments (0)

Please log in to add your comment.

Report abuse

Transcript of Electric current and electric resistance

Moving charges create electricity. The flow is the current & current is described as the amount of CHARGE "q" that moves thru the cross-sectional area of wire/unit of time.....
I = current in amps or A.
An A= 1C/sec
1 C=6.24E28 electrons
You must have a force to encourage the electrons to flow in a direction. This force is an electric field.
unfortunately collisions within the conductor of metal are not elastic. (closer to what we think of gas molecules having elastic collisions where energy is not lost in the collision...)
The conductor material is absorbing some of the energy of the collision and heating up. Not so good is the device is not a heater....
Because electron collisions are not elastic with the materials they also loose some speed. We are constantly looking for the best conductors where little loss of energy takes place...
Kinetic energy gets transferred and this is what temperature is....a record of the kinetic energy.
THREE subtopic/concepts to be familiar with:
1. HEATING EFFECTS: heating a wire through the resistance of =>electrons flowing
Current is proportional to the pushing voltage.
Resistance is directly related to the voltage and indirectly to the current...
So...therefore R=V/I
a volt/amp ó V/A= ohms
Ohm's Law: named after Georg Simon Ohm, a German physicist.
An ohm is a unit of resistance with symbol omega. 1 kg/m^2s^3A^2 just say ohms.
*superconductors: have zero resistance at extremely low temperatures {"the perfect conductors"}
In Circuitry current splits into branches if the wire has a junction and goes off in two directions.
Current tot. = Ia +Ib
I=V/R ; V is potential to do work; 1 volt can do one joule per sec of work & R, the resistance is in Ohms symbol omega & I,current is in amps=A short for amperes.
In reality the electrons are falling thru a wire having many collisions and not "elastic"(there is loss of energy). They loose their speed and once again their kinetic energy until the next collision.
This means heat will be liberated as more and more collisions occur. Materials (wires-conductors) that have better conductivity have less energy loss.
Why do the lights go on immediately upon switching on the switch and not waiting for the electrons to come from the beginning of the wire thru the wire to the light?
Electric Current Direction is weird!
By convention (accepted) it is opposite to the flow of electrons!
The elect. resist. of a conductor for a wire of a certain length is defined by the length of wire with its potential difference (Voltage) across its ends divided by the current flowing through it. R=V/I
*If you have a long-long wire, like an really long extension cord; you will have a voltage drop by the time the current goes thru...So...you must have a good quality low resistance conductor for really long wires...This mean usually thicker wire and better quality conductor materials. Otherwise you can get damage to your device plugged into the end of the cord!
What are the factors affecting Resistance?
1. material conductor is made out of...
2. the length of the wire
3. the cross sectional area of the wire.
{increase in Temp===>increase in resistance, remember superconductors work better in extreme cold. Also Ohms law refers to T=k}
R is proportional to the Length/Area
*If we double the cross sectional area of a wire we also double the amt. of current flowing thru it..as it is proportional...
Since current is also described by another equation concept we can calculate with other equations besides the Ohms Law:

current "I"=change in Q (charge)/change in t(time)

The amt. of q that passes thru the filament of a certain light bulb in 2.00s is 1.67 C. Determine the current in the light bulb.
I=*Q /*t ............ =1.67C/2.00s = 0.835 A (amps)
*Reminder: conventional current is defined in terms of a positive q movement.....
*you will need to put this link in your browser to see this page about conventional current and a useful resource... Having trouble with an active link so if you google conventional current and look for this link address. It is an interactive site.
Remember what is a Coulomb....
It is a q. It is a specific quantity. It is defined as
1C= 1amp/1 sec:::::the charge transported by a steady stream of 1 amperes in one second.
So what was an amp besides being the SI unit for current?
In practical terms, the ampere is a measure of the amount of electric charge passing a point in an electric circuit per unit time with 6.241 × 10^18 electrons, or one coulomb per second constituting one ampere.
Electric power is the conversion of electrical energy!
P=W or J/Time =*PE/*t *=delta in this Prezi
*V= *PE/q *PE=q*V substitute the expression for PE into the equation and we get P=*PE/*t=q*V/*t
Since current "I" is defined as the rate of charge movement (q/*t), we can express electric power as current multiplied by the potential difference. {i.e. the first equation P=I*V
Because *V=IR, we can express the power dissipated by a resistor in the following alternative form:
P=I*V = I (IR) = I^2R (ie current(I) squared x R)
Power is also (*V/R)*V = (*V)^2/R
Question: explain why the filament of a light bulb receives more power than the wire that supplies it?
Problem (in E-pwr)
An E heater is connected across a 120V outlet. The heater dissipated 3.5 kW of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater.
Given *V=120V P=3.5x10^3 W R=?
so R= (*V)^2/P
(120)^2/3.5x10^2 = J^2/C^2/J/s
R=4.1 V/A or ohms....
Electric companies measure energy consumed in Kilowatt-hours
1kW-h x 10^3/1kW x 60min/1minute=3.6x10^6 W-seconds = Joules.
A resistor of R value is 12 ohms has a current of 2.0 A flowing thru it. How much energy is generated in the resistor in one minute?
12x 4 W
=48watts (W) so therefore in one minute we must multiply by 60s
=2.9x10^3 J
*Electrical devices are rated by the power they consume or use...
So a lite bulb of 15 W at 220 V dissipates a different amount of energy than at 110V. What is it?
P=(*V)^2 /R 15W=(220)^2/ R since watts is the power and resistance can be assumed to be = in both cases. It is the ratio of each or (220)^2/ R over (110)^2/R =1/4 so 1/4th of 15W is = 3.75 Watts @ 110V is consumed.
Since the filament heats up, the KE of the electrons causes more resistance since their are more collisions. Since P=I^2 R , R is increased so therefore Power is directly related.
Some conductors follow Ohm's law R=V/I but some don't exactly: these are non Ohmic conductors.
e.g. a light bult filament.
As per Pearson's example page 109 SL textbook. The R at the beginning is .33 ohms and at the end of the filament we see it at .57 ohms. Once again we can explain this by the heat causing more resistance...
1. LDR: light sensor
Resistance changes with light and actually lowers the resistance allowing more current to flow or "charge-carriers"as current electrons are often called...{made of semi-conductor materials}

Sensors: thermistor--
as temp. increases resist. increases...{also semi-conductor material}
Sensors: strain gauge--
as it is stretched the length increases and the resistance increases.
Although sensors are very helpful they change only the resistance and it would be better to get a voltage change or p.d. SO ... to accomplish this we use the concept of "voltage dividers".

From Ohm's Law I=V/R (((voltage-in))
So if we know series resistors add R + R
I=V/R+R (((Voltage-out)))

So voltage divider equation becomes

V out = V in x R2/R1+R2
All batteries have cells or are called cells. There exist
an internal resistance within the battery. We put a small resister "r" above the symbol indicating the resistance of the battery. Now this must be taken into consideration with calculation the total current "I" of the system. Go back to the link on previous slide to Hyperphysics and find internal resist. of battery tutorial example.

A voltage divider is aka "Potential divider".
this is a circuit commonly used with sensors and also to produce variable potential differences.

The most basic potential divider consists of two resistors with resistances of R1 and R2 is series with a power supply.
The current in both resistors is I=total pd across both resistors/total resistance=V/R
section 5.2: Heating effects of an electric current
What is conventional current?
-it is the wrong model of how the current of charges flow in a circuit. It says the flow is one of positive charges. In reality current is the flow of electrons which is the flow of negative charges.
What is emf?
--It is about the energy changes that could occur from a device. e.g.: a battery {or cell} , photovoltaic cell (solar cell), generator (dynamo).
What is pd or potential difference?
--It is about when energy when it is transferred from one form to another. e.g.: electrical into sound, electricity into mechanical energy such as a fan.

some terms:
1. analogue
2. digital meters
3. ammeters
4. voltmeters
5. variable resistors
6. potentiometer[potential divider]: a form of
var. resistor
Resistance: the obstruction to current in a circuit caused by poor conductors and internal circuit make-up.
*familiarize yourself with these symbols.
Full transcript