Powerpoints electric circuits

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1 Electric Circuits Electric Current Electromotive Force Resistance Ohm’s Law Electric Current The flow of electric charge is called current. Charges flowing could be positive, negative or both. Electric current in wires is due to the flow of electrons in the conductor. Current flow in solutions could be positive or negative ions and electrons.

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electric circuits

Transcript of Powerpoints electric circuits

Page 1: Powerpoints electric circuits

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Electric Circuits

Electric CurrentElectromotive Force

ResistanceOhm’s Law

Electric Current

• The flow of electric charge iscalled current. Chargesflowing could be positive,negative or both.

• Electric current in wires isdue to the flow of electronsin the conductor.

• Current flow in solutionscould be positive or negativeions and electrons.

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Electric Current

Measure the total charge, ∆ q flowingpassed an area perpendicular to thecharge velocity in time, ∆ t.

t

qI

∆=

Unit of electric current:1 Ampere = 1 A = 1 C/s

In the modern SI system of units, the ampereis the fundamental quantity while theCoulomb is derived from the ampere. Theampere is defined in terms of the magneticforce between two parallel wires carryingelectric current (later in the course).

Actual Velocity of Electrons in Wiresis very Small

Copper wire has an approximate concentration of conductionelectrons of ne = 1.13 x 1029 electrons/m3. If electrons have velocityv in a wire of cross-sectional area A, the number of electrons passingthrough A in time ∆ t would be

( )

( )mm/s 0.0781 m/s 10 x 81.7

m 10 x 07.710 x 5.1 andA 10For 5-

26-23-

==

===

=∆

∆=

∆=

∆=∆

v

AI

Avent

ne

t

qI

tAvnn

e

e

π

Q: If the electrons in a wire move so slowly, how come the light turns onalmost instantaneously when you turn on a light switch?

A: The wire is full of electrons. When the switch is turned on, the electricfield moves all the charges together! The signal speed propagates at thespeed of electromagnetic waves -- about 2/3 c in wires and cables!

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Conventional Current

• Most current flow isdue to negativelycharged electrons.

• The conventionalcurrent is assumed tobe the flow of positivecharges.

• Conventional currentis opposite the actualelectron flowdirection.

Electric Circuit

• A system that allows the flow of electriccurrent by providing paths for the charges toflow in a complete loop.

• To drive an electric current there must be adriving electric field or we may say thatthere is a source of potential difference(analogous to a pump in fluid flow in pipes).

• Historically, this source is called a source ofelectromotive force (emf) although it is not aforce.

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Electric Circuit

Sources of EMF

• Sources of emf convert others forms ofenergy to electrical potential energy.

• It has two terminals at different potentials.

• Through some process (chemical, magnetic,piezoelectric, thermoelectric, photoelectric,etc.) charges are separated with one terminalwith excess positive charges and at a higherpotential than the other terminal with excessnegative charges.

• EMF is electric potential so it is in units ofvolts (V).

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Examples of Sources of EMF

• Battery

• Generator or alternator

• Fuel cells

• Photovoltaic cells

• Piezoelectric crystals

• Pyroelectric crystals

Resistance

• Electrons gain kinetic energy from the electricfield that accelerates them.

• Electrons lose most of the gained kinetic energyduring collisions with the massive atoms. Theenergy is dissipated in the form of heat.

• Thus electron velocity does not run away butremains a very low value.

• This energy dissipative resistance to flow iscalled electrical resistance or simply resistance.

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Resistance and Ohm’s Law

The relationship between voltage V, current I andthe resistance R is expressed by Ohm’s Law.

IRV =

The unit of resistance is the ohm (Ω ) (V in volts and I inamperes) in honor of Georg Simon Ohm.

Schematic symbol for resistance

Resistivity and Resistance

• Resistance depends on size, shape andcomposition of the resistor.

• Material properties such as free electronconcentration, collision frequencies, etc.are lumped together in a macroscopicquantity called resistivity, ρ (Unit: Ω ·m)

• For a material with uniform cross-sectionalarea A and length L :

A

LR ρ=

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Temperature Dependence ofResistance

For most materials resistance increases with temperature.

( )[ ]( )[ ]00

00

1

1

TTRR

TT

−+=

−+=

α

αρρ

α is the temperature coefficient of resistivity inunits of reciprocal temperature.

Some materials have a negative temperature coefficient of resistivity.Their resistance decreases with temperature and are sometimes used astemperature sensors.

Still other materials lose resistance completely below some temperaturenear absolute zero. These are the superconductors. Once a current isestablished in a superconductor it can persist indefinitely without theneed for an emf source. Superconductivity is a quantum effect.

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#8, Ch. 20, p. 619A car battery has a rating of 220 ampere-hours (A⋅h). This rating is one indication of the totalcharge that the battery can provide to a circuit before failing. (a) What is the total charge (incoulombs) that this battery can provide? (b) Determine the maximum current that the batterycan provide for 38 minutes.

( )A

t

QI

Q

347

min

s 60min 38

C 792000

C 792000h

s3600h

s

C 220hA 220

=

==

=

⋅=⋅=

#16, Ch. 20, p. 620A tungsten wire has a radius of 0.075 mm and is heated from 20.0 to 1320 °C. The temperaturecoefficient of resistivity is α = 4.5 x 10-3 (C°)-1. When 120 V is applied across the ends of thehot wire, a current of 1.5 A is produced. How long is the wire? Neglect any results due tothermal expansion.

Determine resistance at T = 1320 °C. Use Ohm’s Law.

Ω=== 805.1

120

I

VR

The textbook only has values of the resistivity at 20 °C. Hence, determine resistance at T0 = 20°C.

( )[ ]( )[ ]

Ω=

−+=

−+=−

68.13

20132010 x 5.4180

1

0

30

00

R

R

TTRR α

Finally, calculate length using table value of ρ = 5.6 x 10-8 Ω⋅m.

( )m 69.3

10 x 5.6

10 x 075.069.118-

23-

0

0

===

=

πρ

ρ

ARL

A

LR

#20, Ch 20, p. 620An aluminum wire is hung between two towers and has a length of 175 m. A current of 125 Aexists in the wire, and the potential difference between the ends of the wire is 0.300 V. Thedensity of aluminum is 2700 kg/m3. Find the mass of the wire.

We can calculate the resistance using Ohm’s Law. Then determine the area of the wire from theresistance/resistivity formula. Finally determine the volume from the area and the given lengthof wire. Mass is density times volume.

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( )( ) ( )( ) ( )

kg 9723.0

12517510 x 82.22700 28-2

2

=

===

==

==

=

=

mV

ILdVolumedensitym

V

ILALVolume

V

LI

R

LA

A

LR

I

VR

ρ

ρ

ρρ

ρ

Note that V is voltage, not volume.