UCSD Physics 2B
Summer Session
Unit 3 Lecture Notes
Chapter 27
Section 1
Electric Current
Definition – charges in motion
Recall the analogy we made between pumping water into a storage tank and
“pumping” charge into a capacitor? There, we concentrated on the end product of
the process, namely, the resulting charge separation between the plates. Now we
focus our attention on the process of moving that charge.
When water flows through a pipe or river bed, we call that flow “current” and we
hear of dangerous currents near the beach, etc. The flow of charges is an electric
current and, in the same sense, it can flow in a particular direction, follow a
conduit or even swirl around like water.
We assign the letter I to represent the electric current and define it as the amount
of charge per unit time flowing past a particular point in space. If that flow is
uniform in time, then we define the constant current as
I=
Δq charge
=
time
Δt
Electric Current
or, if that current is changing in time, we define the instantaneous current as
I (t ) =
dq
≡ q
dt
Instantaneous Current
Since we’ll most often consider current flowing through a wire, we’ll deal with it
as a scalar rather than vector, keeping in mind the direction of the current as it
follows the path of the wire as a stream follows the path of its bed. Further, we
define the current as positive in the direction that positive charges are flowing.
Note, however, that most free charges are negative.
The MKS unit of current is called the Ampere, abbreviated A, and
1 Ampere = 1
Coulomb
second
Units of Current
1
Section 2
Conduction Mechanisms
In order for charges to move, they must be free to do so. The most common
conductors here on earth are metals which have valence (outer) electrons shared
amongst all the atoms in the otherwise fixed and solid lattice. This “sea” of
electrons can conduct current almost freely but, as we shall see, they collide with
the fixed atoms much as bumper cars collide with the posts. Other kinds of
conductors include ionic liquids like salt water, and plasmas – gases in which
some electrons have been stripped from their homes and are free to roam.
We will sometimes speak of perfect conductors, in which there are no collisions.
These are mostly mental constructs for our analyses (of circuit behavior) while
others (superconductors) actually exist in the real world.
+
+
+
+
Current
+
+
Comments: Invoking our analogy with water once again, note that the water
level of a steady state river neither rises nor falls even though water is flowing
steadily. Similarly, a current-carrying wire is neutral, even though charges are
flowing through it. Also, you might find it helpful to consider the analogy
between gallons per minute flowing through a pipe, and Coulombs per second
flowing through a conducting wire.
Section 3
Current Density
Have you ever seen a small rubber hose stretched across a highway? It’s
connected to an electronic box which counts passing cars by registering pulses as
they drive over the hose. You can determine the “current” of automobiles by
dividing the total number of cars by some time interval.
Now suppose we want to know how many cars per minute per lane are passing.
Then we divide the total car current by the number of lanes, right? Similarly, we
can define the electric current density J by dividing the total current in the wire
by the total cross-sectional area to obtain a flux, as we studied before
J=
I Amperes
=
meter 2
A
Current Density
2
There is a sort of scale correspondence between current and current density.
Current is a macroscopic measure, while current density is, in a sense,
microscopic. One way to see this is to investigate how fast individual charges
must be moving in order to comprise some total current. We consider again some
multi-lane highway and ask, can we deduce the speed of the traffic from the flux?
We can, but we must know all of the following:
1. the number of cars passing a particular point every second
2. the number of lanes in the highway
3. the density, that is, how closely apart the cars are spaced
Let’s call the total flow of cars I, the number lanes A, the flow per lane or flux
density J, the density of cars (say number of cars per meter) n, and the average
velocity v. The flux density is just J = I A and, for a given flux, the density and
the vehicle speed are inversely proportional – the faster the cars the less density is
required. To carry this over to the electrical case, we need only multiply by the
charge on each carrier, so that
J = n vD q ⇒ vD =
J
nq
Carrier Drift Velocity
This speed v is called the drift velocity of the charge carriers in a wire. Note that
we could have deduced the formula simply by dimensional analysis:
v=
⎡A ⎤ ⎡C / s ⎤
⎣⎢ m2 ⎦⎥ ⎣⎢ A ⎦⎥
J
⎡m⎤
=
=⎢ ⎥
n× q
⎡ 1 ⎤ × [C ] ⎣ s ⎦
⎣⎢ m3 ⎦⎥
In a wire, the current generally flows in only one direction, so we can treat the
current density as a scalar. However, there are some cases where we have
diverging current, or swirling currents in a conductor and we might be interested
in the total current passing through some particular surface area inside. In that
case, we must take into account the direction as well as the magnitude of the
current density and define the resulting current as an integral:
I = ∫ J • dA
Electric Current from Current Density
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Section 3
Non-perfect conductors, Collisions & Resistivity
Consider a perfect conductor. Recall that in a uniform electric field, a charge q
experiences a force F = q E and according to Mr. Newton, F = ma . So, absent
collisions, such a charge q must accelerate without bound –
E
a=
+
q
E
m
In a non-perfect conductor, otherwise “free” charges must navigate an obstacle
course, coming to a stop with each collision, and then accelerating over and over
again. The average speed of these charges during their odyssey is just the drift
velocity we derived above.
We define the resistivity ρ (rho) of a particular material as the measure of this
“friction” which limits the average velocity of charges, depending on the nature
of the material’s “obstacles” as well as the electric field strength. This is
analogous to the terminal velocity of a falling object, which depends inversely
upon the drag of the object through the air and directly as its weight. When these
two forces are balanced and cancel, terminal velocity is attained.
In conducting materials, the current density and the electric field are inversely
proportional (higher resistivity means more “friction” and thus less current.)
For certain linear materials we obtain the defining relation for resistivity
J=
1
ρ
E
Note that we can also define the conductivity σ (sigma) of a material as the
inverse of the resistivity. In other words, higher σ means a better conductor
J =σ E
or
E=ρ J
Microscopic Ohm’s Law
4
Section 3
Resistance and Ohm’s Law
Recall that we sought a means of simplifying calculations of the charge capacity
of some configuration of conductors. We wanted to find a linear expression for
charge stored at a given potential for some macroscopic configuration or device
(rather than the microscopic fields) and came up with
Q = CV
Capacitance
A current-carrying device which obeys a similar linear relation is called a resistor
and the relation is called Ohm’s Law, perhaps the most useful equation you will
learn during this course:
V = IR
Ohm’s Law (macroscopic)
Units of resistance R = Ohm ( Ω ) = Volt / Ampere
EXAMPLE A
Block or cylindrical resistor with axial current
Consider the simplest case – a rectangular solid, configured like a parallel-plate
capacitor with cross sectional area A, and plate separation d, but filled with a
(non-perfect) conducting material of resisistivity ρ . In this geometry,
V = Ed , E = J ρ and J = I A
I
⎛ ρd
V = Ed = ( J ρ ) d = ρ d = I ⎜
A
⎝ A
so that
⎞
⎟ = IR
⎠
Note that d and A are determined (and therefore constant) once the device is
made, and that ρ is an intrinsic and permanent property of the material. Hence
we’ve found the linear relationship we were seeking and, for this geometry,
R=ρ
d
A
Resistance of block (or cylinder with axial current)
This equation connects the microscopic property of the material with a
macroscopic property of a device. In the text, and hereafter, we’ll replace the
distance “d” above with the length of the resistor “L”.
Note that, from the formula above, ρ can also be expressed as Ohm · meter.
5
CAUTION: To calculate the resistance in other geometries, in particular,
concentric spheres and axial cylinders, we must integrate the electric field
through the material from source to finish (sink),
b
V = ∫ E • dr
as we did for capacitors.
a
EXAMPLE B
Spherical Resistor (radial current, concentric shells)
V = ∫ E • d r = ∫ J ( r ) ρ ( r ) • dr = ∫
b
b
b
a
a
a
I
A(r )
ρ ( r ) dr
Note that the current I is not a function of radius r since the current must be
conserved as it passes through each spherical layer. Further, if the resistivity is
uniform throughout, and not a function of radius, then
V =∫
b
a
Iρ
Iρ ⎛ 1 1 ⎞
dr =
⎜ − ⎟ ⇒
2
4π r
4π ⎝ a b ⎠
R=
ρ ⎛1 1⎞
⎜ − ⎟
4π ⎝ a b ⎠
2
Note that the surface area of each spherical shell is A ( r ) = 4π r .
EXAMPLE C
Cylindrical Resistor (radial current, coaxial shells)
Again we begin with the definitions,
V = ∫ E • d r = ∫ J ( r ) ρ ( r ) • dr = ∫
b
b
b
a
a
a
I
A(r )
ρ ( r ) dr
Here again, the current I is not a function of radius r since the current must be
conserved as it passes through each cylindrical layer. We again assume uniform
resistivity, and choose length L for the cylinder. Then
V =∫
b
a
Iρ
Iρ
⎛b⎞
dr =
ln ⎜ ⎟ ⇒
2π rL
2π L ⎝ a ⎠
R=
ρ
⎛b⎞
ln ⎜ ⎟
2π L ⎝ a ⎠
Note that the surface area of each cylindrical shell is A ( r ) = 2π rL .
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UNITS:
Note that, in each case, the resistance of the device has the units of resistivity over
length. This is most fortunate, as we don’t want the properties of the device to
change when we change the geometry. Checking our units is always a good way
to check our work!
COMMENTS:
Let’s look again at our water analogy. Consider a hose which must carry water to
quench a fire. The hose has a large cross-section in order to move a large amount
of water quickly. The reason is that there is friction between the inner wall of the
hose and the water it carries. The wider the pipe, the smaller the proportion of
water contacts the wall. By contrast, a garden hose has a small cross-section
because it must carry less “current”.
Now consider two hoses with identical cross-sections. The longer hose presents
more inside surface area to the flow and therefore more resistance.
Hence, we see that the water resistance of hoses, like that of conducting materials
made into electrical resistors, depends
• directly on the length
• inversely on the cross-sectional area
CAVEAT:
Note that, in all the examples above, the resistivity ρ (not to be confused with
the volume charge density with the same symbol) was uniform rather than a
function of position e.g., one which say varies with distance from the axis of a
cylinder. We need only, however, replace the constant function ρ with the
position-dependent one, say ρ ( r ) for a cylinder or sphere and perform the
indicated integration accordingly. That’s the only difference – the method
remains the same.
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Section 4
Resistor Networks
Now let’s see what happens when we connect some of these new devices together
in various ways. Consider the resistor whose side view is sketched below.
A
L
It is made of material with resistivity ρ and has cross-sectional area A and length
L. The electrical connections on each end are assumed to be perfect conductors
(resistivity zero). Suppose that we connect two of them in series:
A
A
L2
L1
Look at resistor R1 and R2 above. In this geometry, their resistances are
R1 = ρ
L1
A
and
R2 = ρ
L2
A
Since the “wire” in the middle connecting the two resistors together is a “perfect”
conductor, it must be an equipotential and thus must have the same voltage
everywhere. Hence it doesn’t matter in our arrangement how long we make it, so
let’s make its length zero
A
L1 + L2
Now the new single resistor created by joining R1 and R2 end-to-end must have
resistance
L + L2
L
L
RSERIES = ρ 1
= ρ 1 + ρ 2 = R1 + R2
A
A
A
Since we can keep adding resistors in the same way, we must have
RSERIES = ∑ Ri = R1 + R2 + "
Series Resistors
8
In the same way, we can construct a parallel network. This time we join two
resistor of the same length, but different areas.
A1
L
A2
Now, if we join them together side-by-side (vertically here) we’ll have one
resistor with length L and combined cross-sectional area A = A1 + A2.
The equivalent parallel resistance is thus
RPARALLEL = ρ
L
whose decomposition is not readily apparent.
A1 + A2
However, we can recast the equation (guessing the answer in advance):
1
RPARALLEL
=
A1 + A2
A
A
1
1
= 1 + 2 =
+
ρL
ρ L ρ L R1` R2
Again, we can generalize this to any number, and write
1
RPARALLEL
=∑
1
1
1
=
+
+"
RI
R1 R2
Parallel Resistors
NOTE THE SIMILARITY BETWEEN THE SERIES & PARALLEL
FORMULAS FOR RESISTOR & CAPACITOR NETWORKS.
BEWARE… THE FORMS ARE REVERSED!!!
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Alternate Derivation of Resistance
Let’s consider again the microscopic statement of Ohm’s Law
J =σ E
Suppose we dot both sides of into some area through which current flows
J • dA = σ E • dA
and integrate both sides of this new equation
∫ J • dA = ∫ σ E • dA
By definition, the left side is the macroscopic current and, if the conductivity σ
is a constant (uniform everywhere), it comes outside of the integral. Then,
I = ∫ J • dA = σ ∫ E • dA = σ Φ E = σ
Qenc
ε0
where we have used Gauss’ Law to convert the right-side of the equation.
Now, if we assume that some excess charge must be present at the boundaries of
the resistor device (as it must be for a potential, and therefore an electric field to
be present to drive free charges), then we can apply the Capacitor Equation to get
I =σ
Q
ε0
=σ
CV
ε0
⎡ ρε ⎤
⇒ V =⎢ 0⎥I
⎣ C ⎦
You can check – by plugging in the Capacitance C for our three common
geometric configurations – that we recover our previous resistance formulas!
MORAL
The lesson here is to be careful! It’s tempting to play with the math to reach a
quick solution, without considering the underlying physics which it represents.
In the special case where the conductivity σ is constant, it works. But we must
always be able to go back to the microscopic physics to obtain the right answer!
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Section 4
Electric Power
The units of potential are energy per charge, Joules per Coulomb.
The units of current are charge per time, Coulombs per second.
Let’s multiply them together
⎡ J ⎤ ⎡C ⎤ ⎡ J ⎤
V × I = ⎢ ⎥×⎢ ⎥ = ⎢ ⎥
⎣C ⎦ ⎣ s ⎦ ⎣ s ⎦
But a Joule per second is the rate of doing work or delivering energy, in other
words, Power. Hence, in electrical systems,
P = VI
EXAMPLE
Unit = Watt = Joule / second
Home Appliance
A hair dryer is rated at 1200W. If the line voltage in your house is 120V, what
current does the dryer draw?
P = VI ⇒ I =
P 1200W
=
= 10 A
120V
V
There are two other useful ways to express this. By applying Ohm’s Law,
P = ( IR ) I = I 2 R
V = IR ⇒
⎛V ⎞ V
P =V ⎜ ⎟ =
⎝ R⎠ R
2
These alternate formulas are useful when either V or I are unknown, saving us the
trouble of having to first invoke Ohm’s Law. They also can give us rapid insight
into the physical behavior of a system.
P = V I = I 2R =
V2
R
Electric Power
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EXAMPLE
Power Transmission Lines
Consider a power transmission line from a generating plant to a distant town as
shown below. Suppose we want to provide one Megawatt (1MW) to the town.
However, we are free to choose the transmission voltage which then determines
the current in the line. Two extreme possibilities are
⎧ (106 V ) × (1 A )
⎪
P = VI ⇒ 1 MW = 106 W = ⎨
6
⎪⎩ (1 V ) × (10 A )
R
1MW
Sears
Let’s consider just one “leg” of the line (wire) between two towers and, since the
wire is real and not “ideal”, let’s make it such that R = 1 Ω .
When the current is only 1 Amp, the power lost through this one leg is only
P = I 2 R = (1A ) (1Ω ) = 1 Watt , hardly noticeable. But if we send the electric
power at 1 Volt, the current through the line is a million Amps, and
2
(
P = I 2 R = 106 A
) (1Ω ) = 10
2
12
W = 1 Terawatt
In other words, all the power would be lost before it even reached the first tower.
It’s clear, then, why power companies choose very high voltages to transmit
power – it’s extremely efficient, even though the cost up front to construct the
towers may seem high.
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CAVEAT !!!
It is always of the greatest importance to apply Ohm’s Law or its corollaries
properly and not indiscriminately to the wrong part of a circuit.
It might be tempting to think, for example, that by applying the following form of
the power equation, high voltage transmission loss might be just as bad –
(
)
106 V
V2
P=
=
R
(1Ω )
2
= 1012 W = 1 Terawatt
NOT!!! The error here is that the transmission voltage 1 million Volts was used
to calculate the current to find the power lost in heating the wire along the way.
The most thorough way to look at this example is to view it as a circuit.
R
R
R
RLOAD
The battery represents the power plant, each R in the circuit above represents one
leg of the line and the Load represents the town. When using Ohm’s Law to
compute the voltage drop across one leg (or one resistor) we multiply the current
through that component by its resistance. Similarly, if we choose to compute the
power lost through a single component by using the voltage form of the power
equation, we must use the voltage drop (potential difference) across that single
circuit element. However, in the stated example, we don’t know that voltage, but
it certainly isn’t a million volts!
This is a very common mistake when doing circuit analysis, so be careful.
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