Transcript Electric Current

Electric Currents
The Electric Battery
See howstuffworks. Another example of conservation of
energy.
Electric Current
Connecting wires (and/or lamps, etc.) to a battery permits
electric charge to flow. The current that passes any point in the
wire in a time t is defined by
OSE:
I = Q/t,
where Q is the amount of charge passing the point. One
ampere of current is one coulomb per second.
We use this symbol for a battery (the short line is negative):
+-
Here’s a really simple circuit.
Don’t try it at home! (Why not?)
Because electric charge is conserved, the
current at any point in a circuit is the same
as the current at any other point in the
same circuit at that instant in time.
+-
current
The current is in the direction of flow of positive charge.
Opposite to the flow of electrons, which are usually the charge
carriers.
+-
current electrons
An electron flowing from – to + gives rise to the same
“conventional current” as a proton flowing from + to -.
“Conventional” means our convention is always to consider the
effect of + charges.
Example: A steady current of 2.5 A flows in a wire for 4.0 min.
(a) How much charged passed through any point in the circuit?
Q
I 
t
Q  I t
ΔQ =  2.5  4.0  60 
Minutes are
not SI units!
ΔQ = 600 C
(b) How electrons would this be?
total charge
number of electrons =
charge of an electron
600 C
21
number of electrons =
=
3.8×10
electrons
-19
 1.6×10 C 


electron


“This is a piece of cake so far!”
Don’t worry, it gets “better” later.
Ohm’s Law
It is experimentally observed that the current flowing through a
wire depends on the potential difference (voltage) causing the
flow, and the resistance of the wire to the flow of electricity.
The observed relationship can be written
OSE:
V = I R,
and this is often called Ohm’s law.
Ohm’s law is not “fundamental.” It is not really a “law” in the
sense of Newton’s Laws. It only works for conductors, and
some things that conduct electricity do not obey Ohm’s law.
The unit of resistance is the ohm, and is equal to 1 Volt / 1
Ampere.
Example
A small flashlight bulb draws 300 mA from its
1.5 V battery.
(a) What is the resistance of the bulb?
V=IR
R=V/I
R = 1.5 / 300x10-3
R = 5.0 
(b) If the voltage dropped to 1.2 V, how would the current
change?
V=IR
I=V/R
I = 1.2 / 5.0
I = 0.24 A
(“If it’s this easy now, does that mean I’ll pay later?”)
Every circuit component has resistance. This is the symbol we
use for a “resistor:”
All wires have resistance. Obviously, for efficiently carrying a
current, we want a wire having a low resistance. In idealized
problems, we will consider wire resistance to be zero.
Lamps, batteries, and other devices in circuits have resistance.
Resistors are often intentionally used in
circuits. The picture shows a strip of five
resistors (you tear off the paper and
solder the resistors into circuits).
The little bands of color on the resistors have meaning. Here
are a couple of handy web links:
http://www.dannyg.com/examples/res2/resistor.htm
http://xtronics.com/kits/rcode.htm
You light me up.
http://jersey.uoregon.edu/vlab/Voltage/
http://jersey.uoregon.edu/vlab/Voltage/volt1.html
Resistivity
It is also experimentally observed that the resistance of a metal
wire is well-described by
OSE:
R = L / A,
where  is a “constant” called the resistivity of the wire
material, L is the wire length, and A its cross-sectional area.
This makes sense: a longer wire or higher-resistivity wire should
have a greater resistance. A larger area means more “space”
for electrons to get through, hence lower resistance.
Resistivities range from roughly 10-8 ·m for copper wire to
1015 ·m for hard rubber. That’s an incredible range of 23
orders of magnitude, and doesn’t even include superconductors
(we might talk about them some time).
Example
Suppose you want to connect your stereo to remote speakers.
(a) If each wire must be 20 m long, what diameter copper wire
should you use to make the resistance 0.10  per wire.
R = L / A
A = L / R
A =  (d/2)2
 (d/2)2 = L / R
geometry!
(d/2)2 = L / R
d/2= ( L / R )½
don’t skip steps!
d = 2 ( L / R )½
d = 2 [ (1.68x10-8) (20) /  (0.1) ]½
d = 0.0021 m = 2.1 mm
In the spirit of not skipping steps, you are welcome to show all units!
(b) If the current to each speaker is 4.0 A, what is the voltage
drop across each wire?
V=IR
V = (4.0) (0.10)
V = 0.4 V
Example
A wire of resistance R is stretched uniformly until it is twice its
original length. What happens to its resistance?
Hint: the volume of wire material stays the same.
Hint: R = L / A.
Resistivity depends on temperature—providing an important
research tool—but we won’t explore that.
Electric Power
Last week we defined power as the work done by a force
divided by the time it took to do the work.
PF = WF / t
We put a “bar” above the PF to indicate it is really an average
power. We had probably better do that again, hadn’t we.
OSE:
PF =
WF
.
t
We’d better use the same definition this semester! So we will.
We focus here on the interpretation that power is energy
transformed per time, instead of work by a force per time.
However, we begin with the work aspect. We know the work
done in moving a charge through a potential difference:
Wif = PEif= q Vif
The power, which is the work per time transformed by an
electric device when a charge Q moves through a potential
difference in a time t, is
P = PEif / t = q Vif / t
OSE: P = q Vif / t
Let’s get lazy and drop the  in front of the V, but keep in the
back of our heads the understanding that we are talking about
potential difference. Then
P = q V / t = (q / t) V.
But wait! A while ago, we wrote I = Q/t. The  in front of
the Q and t refer to the net flow of charge in some amount of
time. That’s just what we have here in q / t, so
P =IV.
why isn’t this an OSE…

I’m a bit bothered by the way ’s seem to fly in and out.
But “everybody else” uses P = IV so it is probably good to
follow the crowd.
Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R. It
is not clear if these should be OSE’s, because they are derived
from OSE’s. To simplify matters, I’ll make the following an OSE:
OSE: P = IV = I2R = V2/R.
Truth in advertising?* Your power company doesn’t sell you
power. It sells energy. Energy is power times time, so a
kilowatt-hour (what you buy from your energy company) is an
amount of energy.
*Just getting your attention. Not really complaining.
Example
An electric heater draws 18.0 A on a 120 V line. How much
power does it use and how much does it cost per 30 day month
if it operates 3.0 h per day and the electric company charges
10.5 cents per kWh. For simplicity assume the current flows
steadily in one direction.
What the heck is a kWh?
What’s the meaning of this assumption
about the current?
We’ll get to the second question in a minute. The current in
your household wiring doesn’t flow in one direction, but
because we haven’t talked about current other than a steady
flow of charge, we’ll make the assumption (which doesn’t wreck
the calculation.)
Remember your steps to solving problems? The first step is to
think. Maybe the step should really be titled “figure out what
kind of problem it is.”
The problem asks for power. Maybe that identifies the problem
type! Next maybe we had better go to our OSE’s and see
what they say about power.
WF
PF =
t
P = q ΔVif / Δt
P = IV = I2R = V 2 /R .
The first two equations are general and tell us power has
something to do with energy transformations per time. The last
set of three is specific to current flowing in a circuit.
We’re given current and voltage. It should be clear how to
calculate power.
P =IV
P = (15.0 A) (120 V)
P = 1800 W = 1.8 kW
How much does it cost. We are given cost per kWh and we
calculated k above. What is this kWh (and why the odd
capitalization?
(1 kW) (1 h) = (1000 W) (3600 s)
= (1000 J/s) (3600 s)
= 3.6 x 106 J
So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are
lowercase, and W (James Watt) is uppercase.
To get the cost, find the total energy and multiply by the cost
per energy “unit.”
Cost = (1.8 kW) (30 days) (used 3 h/day) ($0.105/kWh)
Cost = (1.8 kW) (30 days) (used 3 h/day) ($0.105/kWh)
Cost = $17
I wonder how much energy we actually used?
Pavg = WF /t
Energy Transformed = WF = Pavg t
Pavg and the P with the bar above it both mean average power.
Energy Transformed = (1800 J/s) (30 d) (3 h used/d) (3600 s/h)
Pavg
time t
Energy Transformed = 583,200,000 Joules used
That’s a ton of joules! Good bargain for $17. That’s about
34,000,000 joules per dollar.
OK, “used” is not an SI unit, but I stuck it in there to help me
understand. And joules don’t come by the ton.
One last quibble. You know from energy conservation that you
don’t “use up” energy. You just transform it from one form to
another. Unfortunately, in the first semester we skipped the
(important) section where you should have studied this.
Example
A typical lightning bolt can transfer 109 J
of energy across a potential difference of
perhaps 5x107 V during a time interval of
0.2 s. Estimate the total amount of
charge transferred, the current and the
average power over the 0.2 s.
learn about lightning
at howstuffworks
What kind of a problem is this?
You are given energy, potential difference, and time.
You need to calculate charge transferred, current, and average
power. OSE’s for current and power are “obvious:”
I=
ΔQ
Δt
PF =
WF
t
We could calculate power right now, but let’s do this in the
order requested. Besides, we can’t get current without Q,
charge transferred.
Can you think of anything where we have energy, potential, and
time combined?
PEif= q Vif
We need to think in terms of energy transformations rather
than work done by forces. The equation above tells us the
potential energy stored in clouds can be transferred to the
ground (at a different potential) by moving charge from cloud
to ground. We are given energy transferred and potential
difference, so we can calculate q.
Could I think of the cloud-earth system as a giant capacitor which stores energy?
You could, except our capacitor OSE U=QV/2 assumes the same charge on both plates; untrue here.
Continuing with our energy transformation idea, our OSE
becomes
Etransferred= Qtransferred Vif
Qtransferred = Etransferred / Vif
Qtransferred = 109 J / 5x107 V
Qtransferred = 20 C
That’s a lot of charge (remember, our typical charges are 10-6 C.
Once we have the charge transferred, the current is easy.
ΔQ
I=
Δt
20 C
I=
= 100 A
0.2 s
Average power is just the total energy transferred divided by
the total time.
W
PF = F
t
P=
E transferred
t
109 J
P=
0.2 s
P = 5×109 W
P = 5 GW
Maybe I should make
this into an OSE
Power in Household Circuits
I’ll do a little demo on short circuits and electrical safety…
If you want to do that at home, use the really fine steel wool. The coarse
stuff doesn’t work as well. And remember—adult supervision!
Never replace a properly-rated fuse with a higher rated one
(“higher rated” means handles more current.
Alternating Current
This topic is surprisingly complex. We will not explore it here.
We have implicitly been discussing circuits using batteries as a
power source. The current is called direct current (DC),
because it flows in one direction and does not vary with time.
Microscopic View of Electric Current
Here’s the “classical” picture of the mechanism for the
resistance of a metal:
electron “drift” velocity
E
-
+
+
+
+
+
+
+
+
+
+
+
The voltage accelerates the electron, but only until the
electron collides with a + ion. Then the electron’s velocity is
randomized and the acceleration process begins again.
Predictions made by this theory are typically off by a factor or
10 or so, but it was the best we could do before quantum
mechanics.
The free electrons move with velocities on the order of 105 or
106 m/s (fast!) but the directions are random. The “drift
velocity” represents the average velocity, or the net “drift” of
the electrons. It is on the order of 0.05 mm/s.
So how come when I flip a light switch, the light comes on “right
away?”