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Current
+
Area
Ammeter(measures current)
Convention : Current depicts flow of positive (+) charges
Friday, July 22, 2011
Current
+
Area
Ammeter(measures current)
+
+
Convention : Current depicts flow of positive (+) charges
Friday, July 22, 2011
Current
+
Area
Ammeter(measures current)
+
+
Convention : Current depicts flow of positive (+) charges
Friday, July 22, 2011
Current
A measure of how much charge passes through an amount of time
Ammeter(measures current)
++
+
Friday, July 22, 2011
Current
++
+
Expand surface to a volume
Area = A
Count how many charges flow through
Friday, July 22, 2011
Current
++
+
Expand surface to a volume
Area = A length = !x
Count how many charges flow through
Friday, July 22, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(!x)
length = !xArea = A
Count how many charges flow through
Friday, July 22, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(!x)
length = !xArea = A
Number of charges = (charge density or charge per volume)*(volume)Number of charges = (n) * (A!x)
Count how many charges flow through
Friday, July 22, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(!x)
length = !xArea = A
Number of charges = (n) * (A!x)
Total amount of charge = (number of charges)*(charge)!Q = (n A !x)*(q)
Count how many charges flow through
Number of charges = (charge density or charge per volume)*(volume)
Friday, July 22, 2011
Current
++
+ Total volumeV = (A)(!x)
length = !x = vd !tArea = A
!Q = (n A !x)*(q)but charges have drift velocity vd = !x/!t
Friday, July 22, 2011
Current
++
+ Total volumeV = (A)(!x)
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)
Friday, July 22, 2011
Current
++
+ Total volumeV = (A)(!x)
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)
Friday, July 22, 2011
Current
This is the reason why large wires are needed to support large currents
Friday, July 22, 2011
Current
This is the reason why large wires are needed to support large currents
Friday, July 22, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
Friday, July 22, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
conductivity
Friday, July 22, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
conductivity (material property)
resistivity (material property)
Friday, July 22, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
conductivity
resistivity
Current is proportional to conductivity but inversely proportional to resistivity!
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Ohm’s LawResistancePotential difference
current
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Ohm’s Law
a much better form than ΔV = I R
ResistancePotential difference
current
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Ohm’s LawResistancePotential difference
current
a much better form than ΔV = I R
Increasing !V increases IIncreasing R decreases I
Friday, July 22, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Ohm’s LawResistancePotential difference
current
a much better form than ΔV = I R
Increasing !V increases IIncreasing R decreases I
!V = I R Increasing R does not increase !VCurrent (I) is increased because !V is increased
Friday, July 22, 2011
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
Friday, July 22, 2011
Exercise
Rank from lowest to highest amount of current
Derive the equation R = "L/Afrom V = IR, J = E/" = I/A, V = EL
Friday, July 22, 2011
Resistance and Temperature
∆T = T − T0
T0 is usually taken to be 25 °C
ρ = ρ0(1 + α∆T )
R =ρl
A
T ↑ ρ ↑
Friday, July 22, 2011
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
Friday, July 22, 2011
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.
Friday, July 22, 2011
More exercises
If the magnitude of the drift velocity of free electrons in a copper wire is 7.84 x 10-4 m/s, what is the electric field in the conductor?
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.
Friday, July 22, 2011
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
Friday, July 22, 2011
Resistors in Series
I =∆V
RRecall:
use the equation to calculate the equivalent resistance Req
Friday, July 22, 2011
Resistors in Series
Conservation of matter = Current is conserved
I1 I2
∆V1 ∆V2
I = I1 = I2
Friday, July 22, 2011
Resistors in Series
I1 I2
∆V1 ∆V2
Conservation of matter = Current is conserved
Conservation of energy∆V = ∆V1 +∆V2
I = I1 = I2
Friday, July 22, 2011
Resistors in Series
I1 I2
∆V1 ∆V2
Conservation of matter = Current is conserved
Conservation of energy∆V = ∆V1 +∆V2
I =∆V
R
Ohms LawI = I1 = I2
Friday, July 22, 2011
Conservation of matter = Current is conserved
Resistors in Series
I1 I2
∆V1 ∆V2
Conservation of energy∆V = ∆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
Friday, July 22, 2011
Conservation of matter = Current is conserved
Resistors in Series
I1 I2
∆V1 ∆V2
Conservation of energy∆V = ∆V1 +∆V2
I =∆V
R
Ohms LawI = I1 = I2
Friday, July 22, 2011
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.
Friday, July 22, 2011
Resistors in Parallel
I =∆V
RRecall:
use the equation to calculate the equivalent resistance Req
Friday, July 22, 2011
Conservation of matter = Current is conserved
I1
I2
∆V1
∆V2
I = I1 + I2
Resistors in Parallel
Friday, July 22, 2011
Conservation of matter = Current is conserved
I1
I2
∆V1
∆V2
I = I1 + I2
Conservation of energy∆V = ∆V1 = ∆V2
Resistors in Parallel
Friday, July 22, 2011
Resistors in Parallel
Conservation of matter = Current is conserved
I1
I2
∆V1
∆V2
I = I1 + I2
Conservation of energy∆V = ∆V1 = ∆V2
I =∆V
R
Ohms Law
Friday, July 22, 2011
Resistors in Parallel
Conservation of matter = Current is conserved
I1
I2
∆V1
∆V2
I = I1 + I2
Conservation of energy∆V = ∆V1 = ∆V2
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
Friday, July 22, 2011
Resistors in Parallel
Conservation of matter = Current is conserved
I1
I2
∆V1
∆V2
I = I1 + I2
Conservation of energy∆V = ∆V1 = ∆V2
I =∆V
R
Ohms Law
Friday, July 22, 2011
Exercise
Find the current passing through each resistorFind the voltage drop (potential difference) through each resistor
Friday, July 22, 2011
Kirchhoff’s Rules
Junction Rule
Loop Rule
“conservation of matter”
“conservation of energy”
Σclosed loop
∆V = 0
Friday, July 22, 2011
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!)
Friday, July 22, 2011