Electric current
Electric Current
Chapter 27
Electric current
Current Density
Resistance, Resistivity, Conductivity
Ohm’s Law
Power Dissipated
• Movement of charge per unit time: dq/dt = I
• SI unit of current: 1 C/s = 1 Ampere (Amp)
• The direction of current is the direction that
positive charges would move.
Electric current
Current only flows in a conductor
• Movement of charge per unit time: dq/dt = I
• SI unit of current: 1 C/s = 1 Ampere (Amp)
• The direction of current is the direction that
positive charges would move.
• Conductors are made of materials (usually metal) in
which some of the electrons are free to move (not
bound to the ions). These are called conduction
electrons.
• Electrons move under the influence of an E field.
• But the individual motion of electrons is still quasirandom. However, a net average flow of charge is
set up.
Electrons move opposite
to the direction of current.
Influence of electric field on flow of electrons
E=0
Influence of electric field on flow of electrons
E=0
E=0
E
• An electric field modifies the trajectories of electrons
between collisions.
• An electric field modifies the trajectories of electrons
between collisions.
• When E is nonzero, the electrons move almost
randomly after each bounce, but gradually they drift in
the direction opposite to the electric field.
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Current density
Current density
• The current density J is defined as the current I
flowing per unit area.
• That is, if current I flows through a surface A:
J = I/A
• After an electron collides with an ion, it will
accelerate under an E field with a = eE/m.
• Suppose the average time between collisions is τ.
• Then the average velocity is vd = aτ = eEτ / m.
• This velocity is called the electron drift velocity
A Density of electrons: n
vd
Number of electrons: N=n(Avd∆t)
vd∆t
• Construct the above volume. In time ∆t all the
electrons in it move out through the right end.
• Hence the charge per time (the current) is
I = (Ne)/∆t = neAvd.
• The current density is J = I/A = nevd
= (ne2τ/m)E.
Example: What is the drift velocity of electrons in a
Example: What is the drift velocity of electrons in a
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
In Cu there is about one conduction electron per atom.
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28
The density of Cu atoms is n = 8 .49 x10 m
Find vd from J=I/A=1.3A/(π(0.0009m)2)=5.1x105 A/m2
Now use vd =
J
. ⋅105 A / m2
51
=
28
ne ( 849
. ⋅10 / m3 )(16
. x10−19 C )
v d = 3.8 ×10 −5 m s
- Much less than one millimeter per second!
Ohm’s Law
•The current density, J, is proportional to the
applied field, E, (both are vectors) or:
•
•
J=σE
- “Ohm’s Law”
• σ is the “Conductivity”, J/E.
• Units are (A/m2) divided by (V/m) = A/(Vm)
• We define the “Resistivity” as ρ = 1/σ = E/J
• Units of ρ are (V/A)m (= Ω - m).
Ohm’s Law
σ and ρ are dependent only on the material,
- not its length or area.
Consider a metal rod, resistivity ρ:
+V
ρ, area A
0 volts
L
E= ρJ
(V/L) = ρ (Ι/Α)
V = (Lρ
ρ/A) I
V=IR
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Ohm’s Law
Example:
• V = I R - the most commonly used form of Ohm’s
Law.
What is the resistance of an 0.5 m length of 22 gauge steel
wire?
• R is the “Resistance”, that depends on the material type
and shape:
•
R = ρ L/A
Units: ohms, (Ω).
• As ρ=RA/L, common units for resistivity are Ohm-meters.
• Similarly, common units for σ are (Ohm m)-1
Example:
Electrical Power Dissipation
What is the resistance of an 0.5 m length of 22 gauge steel
wire?
Note: 22-gauge has a radius of 0.321 mm
• In travelling from a to b ,
energy decrease of dq is:
dU = dq Vab
R = ρ L/A
A = πr2
πr2)
R = ρ L/(π
ρ = 10−8 Ω-m
π (0.000321 m)2)
R = 10−8 Ω-m 0.5 m /(π
= 0.015 Ω
•
•
•
•
a
dq
b
Resistance, R
Now, dq = I dt
Therefore, dU = I dt Vab
Rate of energy dissipation is dU / dt
This is the dissipated power, P. (Watts, or Joules /sec)
• Hence,
L
Vab
P = I Vab
or:
2r
Example
A heating coil of a hot water heater has a resistance of 20 Ω and
operates at 220 V.
A. How much current does this heater draw?
B. At what rate is power consumed?
C. How much energy is consumed in 15 mins?
•
P = I2 R = V2 / R
Example
A heating coil of a hot water heater has a resistance of 20 Ω and
operates at 220 V.
A. How much current does this heater draw?
I = V/R = 220 V/20 Ω = 11 Amps
B. At what rate is power consumed?
P = IV = 11 A 220 V = 2420 Watts,
or P = V2/R = (220 V)2/20 Ω = 2420 Watts,
or P = I2R = (11 A)2 20 Ω = 2420 Watts
C. How much energy is consumed in 15 mins?
dU = I dt V or U = IV ∆t = P ∆t
U = 2420 W (15 min 60 sec/min) = 2178 kJ
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