electric current.pdf

Current and Resistance

Sunday, July 24, 2011

Current

Convention : Current depicts flow of positive (+) charges


Current

+

Area



Current

+

Area

Ammeter(measures current)



Current

+

Area


+

+



Current

A measure of how much charge passes through an amount of time


++

+


Current

++

+

Count how many charges flow through


Current

++

+

Expand surface to a volumeCount how many charges flow through


Current

++

+

Expand surface to a volume

Area = A



Current

++

+


Area = A length = !x



Current

++

+


Total volumeV = (A)(!x)

length = !xArea = A



Current

++

+



length = !xArea = A

Number of charges = (charge density or charge per volume)*(volume)Number of charges = (n) * (A!x)



Current

++

+



length = !xArea = A

Number of charges = (n) * (A!x)

Total amount of charge = (number of charges)*(charge)!Q = (n A !x)*(q)


Number of charges = (charge density or charge per volume)*(volume)


Current

++

+ Total volumeV = (A)(!x)

length = !xArea = A

!Q = (n A !x)*(q)


Current

++


length = !x = vd !tArea = A

!Q = (n A !x)*(q)but charges have drift velocity vd = !x/!t


Current

++


Area = A

!Q = (n A vd !t)*(q)

but charges have drift velocity vd = !x/!t

length = !x = vd !t

!Q = (n A !x)*(q)


Current

++


Area = A

!Q/!t = (n A vd)*(q)

I = n q vd A

but charges have drift velocity vd = !x/!t

length = !x = vd !t

!Q = (n A vd !t)*(q)

!Q = (n A !x)*(q)


Current

This is the reason why large wires are needed to support large currents


Resistance

Current density (J)current per area


Resistance


Direction of current (flow of positive charges) is same with direction of electric field


Resistance



conductivity


Resistance



conductivity (material property)

resistivity (material property)


Resistance



conductivity

resistivity

Current is proportional to conductivity but inversely proportional to resistivity!


Resistance

Current is proportional to conductivity but inversely proportional to resistivity!


Resistance

Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)


Resistance


Ohm’s LawResistancePotential difference

current


Resistance


Ohm’s Law

a much better form than ΔV = I R

ResistancePotential difference

current


Resistance



current


Increasing !V increases IIncreasing R decreases I


Resistance



current


Increasing !V increases IIncreasing R decreases I

!V = I R Increasing R does not increase !VCurrent (I) is increased because !V is increased


Resistance


Resistance

Important points:

same with capacitance, resistance does not depend on !V and I

Resistance depends on material property resistivity ", length of wire l and cross sectional area A

direction of the current I is same as direction of electric field

conventional current is flowing positive (+) charges though in reality electrons flow


Recent Equations

I =∆V

R

R =ρl

A

→J = σ

→E =

→E

ρ

→J =

→I

A

→J = nq

→v dA


Exercise

Rank from lowest to highest amount of current

Derive the equation R = "L/Afrom V = IR, J = E/" = I/A, V = EL


Resistance and Temperature

∆T = T − T0

T0 is usually taken to be 25 °C

ρ = ρ0(1 + α∆T )

R =ρl

A

T ↑ ρ ↑


Power

P =∆U

∆t

P =∆(q∆V )

∆t

P =(∆q)(∆V )

∆t

P =∆q

∆t∆V

P = I∆V


Power

P = I∆V

I =∆V

R

P =V 2

RP = I2R


An aluminum wire having a cross-sectional area of 4.00 x 10-6 m2 carries a current of 5.00 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2700 kg/m3. Assume that one conduction electron is supplied by each atom.Molar mass of Al is 27 g/mol.

The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.00 mm. (a) The beam current is 8.00 µA. Find the current density in the beam, assuming that it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as c = 3.00 x 108 m/s with negligible error. Find the electron density in the beam. (c) How long does it take for Avogadroʼs number of electrons to emerge from the accelerator?

Four wires A, B, C and D are made of the same material but of different lengths and radii. Wire A has length L but has radius R. Wire B has length 2L but with radius ½R. Wire C has length ½L but with radius 2R. Wire D has length ½L but with radius ½R.

Rank with increasing resistance

A 0.900-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.600 mm2. What is the current in the wire?resistivity of tungsten is 5.6 x 10-8 Ω-m

Exercises


Exercises

A 500-W heating coil designed to operate from 110 V is made of Nichrome wire 0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at its 20.0°C value, find the length of wire used. (b) What If? Now consider the variation of resistivity with temperature. What power will the coil of part (a) actually deliver when it is heated to 1200°C?ρ = 1.50 x 10-6 Ω-m

An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 8.00 Ω. Find the current carried by the wire and the power rating of the heater.


A 500-W heating coil designed to operate from 110 V is made of Nichrome wire 0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at its 20.0°C value, find the length of wire used. (b) What If? Now consider the variation of resistivity with temperature. What power will the coil of part (a) actually deliver when it is heated to 1200°C?ρ = 1.50 x 10-6 Ω-cm


More exercises

A certain lightbulb has a tungsten filament with a resistance of 19.0 Ω when cold and 140 Ω when hot. Assume that the resistivity of tungsten varies linearly with temperature even over the large temperature range involved here, and find the temperature of the hot filament. Assume the initial temperature is 20.0°C.4.5 x 10-3 C-1

The cost of electricity varies widely through the United States; $0.120/kWh is one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W porch light on for two weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer.


A certain lightbulb has a tungsten filament with a resistance of 19.0 Ω when cold and 140 Ω when hot. Assume that the resistivity of tungsten varies linearly with temperature even over the large temperature range involved here, and find the temperature of the hot filament. Assume the initial temperature is 20.0°C.4.5 x 10-3 C-1


The cost of electricity varies widely through the United States; $0.120/kWh is one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W porch light on for two weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer.

P =∆U

∆t=

∆U

2weeks

1week

7days

1day

24hours

1hour

3600secs=

∆U

1209600secs

40.0W =∆U

1209600s

∆U = 48384kJ

$0.120

1kWh=

$0.120

1kWh

1kW

1000W

1hour

3600secs=

$3.33× 10−8

1Joule

(a)

$3.33× 10−8

1Joule4.84× 107J = $1.61

(b) $5.82× 10−3

$0.416(c)


Electromotive Force

The electromotive force is denoted as “ε”A force that moves charges

The emf ε is the maximum possible voltagethat the battery can provide.

Direct current - current that is constant in direction and magnitude

ε = ∆V in batteries


Resistors in Series

I =∆V

RRecall:

use the equation to calculate the equivalent resistance Req


Resistors in Series

Convertto simple equivalent

circuit


Resistors in Series

Conservation of matter = Current is conserved

I1 I2

∆V1 ∆V2

I = I1 = I2


Resistors in Series

I1 I2

∆V1 ∆V2


Conservation of energy∆V = ∆V1 +∆V2

I = I1 = I2


Resistors in Series

I1 I2

∆V1 ∆V2



I =∆V

R

Ohms LawI = I1 = I2



Resistors in Series

I1 I2

∆V1 ∆V2


I =∆V

R

Ohms Law

∆V = I1R1 + I2R2

∆V = IR1 + IR2

∆V = I(R1 +R2)

∆V = IReq

Req = R1 +R2

I = I1 = I2



Resistors in Series

I1 I2

∆V1 ∆V2


I =∆V

R

Ohms LawI = I1 = I2


Resistors in ParallelI1 I2

∆V1 ∆V2

1. Imagine positive charges pass first through R1 and then through%R2. Compared to the current in R1, the current in R2 is(a) smaller(b) larger(c) the same.

2. With the switch in the circuit of closed (left), there is no current in R2, because the current has an alternate zero-resistance path through the switch. There is current in R1 and this current is measured with the ammeter (a device for measuring current) at the right side of the circuit. If the switch is opened (right), there is current in R2. What happens to the reading on the ammeter when the switch is opened?(a) the reading goes up(b) the reading goes down(c) the reading does not change.


Resistors in Parallel

I =∆V

RRecall:

use the equation to calculate the equivalent resistance Req



Convertto simple equivalent

circuit



I1

I2

∆V1

∆V2

I = I1 + I2




I1

I2

∆V1

∆V2

I = I1 + I2

Conservation of energy∆V = ∆V1 = ∆V2





I1

I2

∆V1

∆V2

I = I1 + I2


I =∆V

R

Ohms Law




I1

I2

∆V1

∆V2

I = I1 + I2


I =∆V

R

Ohms Law

I = I1 + I2

∆V

R=

∆V1

R1+

∆V2

R2

∆V

R=

∆V

R1+

∆V

R2

1

R=

1

R1+

1

R2




I1

I2

∆V1

∆V2

I = I1 + I2


I =∆V

R

Ohms Law


Recall:

Parallel

Series

Ohms Law

I =∆V

RQ = C∆V

Capacitance


Exercise

Find the current passing through each resistorFind the voltage drop (potential difference) through each resistor


Kirchhoff’s Rules

Junction Rule

Loop Rule

“conservation of matter”

“conservation of energy”

Σclosed loop

∆V = 0


Exercise

A

B

C

In solving complicated circuit problemsapply Junction rule first (conservation of current)

You may assign any direction of current as long as it is reasonable (does not violate common sense!)

Then apply the loop rule

Write down the equations for loop rules concerning loop A, B, C and the outer loop of the circuit following clockwise direction. (there must be four equations!)


electric current.pdf

Documents