current
TRANSCRIPT
Current and Resistance
Tuesday, July 19, 2011
Current
Convention : Current depicts flow of positive (+) charges
Tuesday, July 19, 2011
Current
+
Area
Convention : Current depicts flow of positive (+) charges
Tuesday, July 19, 2011
Current
+
Area
Ammeter(measures current)
Convention : Current depicts flow of positive (+) charges
Tuesday, July 19, 2011
Current
+
Area
Ammeter(measures current)
+
+
Convention : Current depicts flow of positive (+) charges
Tuesday, July 19, 2011
Current
+
Area
Ammeter(measures current)
+
+
Convention : Current depicts flow of positive (+) charges
Tuesday, July 19, 2011
Current
A measure of how much charge passes through an amount of time
Ammeter(measures current)
++
+
Tuesday, July 19, 2011
Current
++
+
Count how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volumeCount how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volume
Area = A
Count how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volume
Area = A length = Δx
Count how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(Δx)
length = ΔxArea = A
Count how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(Δx)
length = ΔxArea = A
Number of charges = (charge density or charge per volume)*(volume)Number of charges = (n) * (AΔx)
Count how many charges flow through
Tuesday, July 19, 2011
Current
++
+
Expand surface to a volume
Total volumeV = (A)(Δx)
length = ΔxArea = A
Number of charges = (n) * (AΔx)
Total amount of charge = (number of charges)*(charge)ΔQ = (n A Δx)*(q)
Count how many charges flow through
Number of charges = (charge density or charge per volume)*(volume)
Tuesday, July 19, 2011
Current
++
+ Total volumeV = (A)(Δx)
length = ΔxArea = A
ΔQ = (n A Δx)*(q)
Tuesday, July 19, 2011
Current
++
+ Total volumeV = (A)(Δx)
length = Δx = vd ΔtArea = A
ΔQ = (n A Δx)*(q)but charges have drift velocity vd = Δx/Δt
Tuesday, July 19, 2011
Current
++
+ Total volumeV = (A)(Δx)
Area = A
ΔQ = (n A vd Δt)*(q)
but charges have drift velocity vd = Δx/Δt
length = Δx = vd Δt
ΔQ = (n A Δx)*(q)
Tuesday, July 19, 2011
Current
++
+ Total volumeV = (A)(Δx)
Area = A
ΔQ/Δt = (n A vd)*(q)
I = n q vd A
but charges have drift velocity vd = Δx/Δt
length = Δx = vd Δt
ΔQ = (n A vd Δt)*(q)
ΔQ = (n A Δx)*(q)
Tuesday, July 19, 2011
Current
This is the reason why large wires are needed to support large currents
Tuesday, July 19, 2011
Current
This is the reason why large wires are needed to support large currents
Tuesday, July 19, 2011
Resistance
Current density (J)current per area
Tuesday, July 19, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
Tuesday, July 19, 2011
Resistance
Current density (J)current per area
Direction of current (flow of positive charges) is same with direction of electric field
conductivity
Tuesday, July 19, 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)
Tuesday, July 19, 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!
Tuesday, July 19, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!
Tuesday, July 19, 2011
Resistance
Current is proportional to conductivity but inversely proportional to resistivity!Current is proportional to the electric potential(specifically potential difference)
Tuesday, July 19, 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
Tuesday, July 19, 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
Tuesday, July 19, 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 ΔV increases IIncreasing R decreases I
Tuesday, July 19, 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 ΔV increases IIncreasing R decreases I
ΔV = I R Increasing R does not increase ΔVCurrent (I) is increased because ΔV is increased
Tuesday, July 19, 2011
Resistance
Tuesday, July 19, 2011
Resistance
Important points:
same with capacitance, resistance does not depend on ΔV and I
Resistance depends on material property resistivity ρ, 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
Tuesday, July 19, 2011
Recent Equations
I =∆V
R
R =ρl
A
→J = σ
→E =
→E
ρ
→J =
→I
A
→J = nq
→v dA
Tuesday, July 19, 2011
Exercise
Rank from lowest to highest amount of current
Derive the equation R = ρL/Afrom V = IR, J = E/ρ = I/A, V = EL
Tuesday, July 19, 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 ↑
Tuesday, July 19, 2011
Power
P =∆U
∆t
P =∆(q∆V )
∆t
P =(∆q)(∆V )
∆t
P =∆q
∆t∆V
P = I∆V
Tuesday, July 19, 2011
Power
P = I∆V
I =∆V
R
P =V 2
RP = I2R
Tuesday, July 19, 2011
End of Chapter Exercises
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?
Three 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
Tuesday, July 19, 2011