R = L / A.

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Transcript R = L / A.

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“Life is uncertain. Eat dessert first.”—UMR Professor Emeritus D. M. Sparlin
Chapter 18 Electric Currents
18.1 The Electric Battery
Read this section. There will not be exam problems on it.
There could be exam questions on it.
Howstuffworks shows how batteries work.
So does energizer.com.
Looks like lots of nasty
chemical things going
Ugh! Chemistry!
18.2 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
I = Q/t,
where Q is the amount of charge passing the point.
One ampere of current is one coulomb per second. 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.
We use this symbol for a battery (the short line is negative):
Here’s a really simple circuit:
Don’t try that at home! (Why not?)
The current is in the direction of flow of positive charge.
Opposite to the flow of electrons, which are usually the charge
+current electrons
An electron flowing from – to + gives rise to the same
“conventional current” as a proton flowing from + to -.
“Conventional” refers to our convention, which is always to
consider the effect of + charges.
“Hey, that figure you just showed me is confusing. Why don’t
electrons flow like this?”
+current electrons
Good question.
Electrons “want” to get away from - and go to +.
Chemical reactions (or whatever energy mechanism the battery
uses) “force” electrons to the negative terminal. The battery
won’t “let” electrons flow the wrong way inside it. So electrons
pick the easiest path—through the external wires towards the +
Of course, real electrons don’t “want” anything.
Example 18-1
A steady current of 2.5 A flows in a wire for 4.0 min.
(a) How much charge passed through any point in the circuit?
ΔQ = I Δt
60 s 
ΔQ =  2.5   4.0 min×
ΔQ = 600 C
This is the symbolic answer.
Minutes are not SI units! So
convert minutes to seconds.
(b) How many electrons would this be?
total charge
number of electrons =
charge of an electron
600 C
number of electrons =
 1.6×10 C 
 electron 
“This is a piece of cake so far!”
Don’t worry, it gets “better,” especially in chapter 19.
18.3 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
V = I R,
Is this V the same “thing”
that we saw in Chapter 17?
and this is often called Ohm’s law.
Ohm’s law is not “fundamental,” so it is not really a “law” in the
sense of Newton’s Laws. It only works for conductors, and
things that conduct electricity do not necessarily obey Ohm’s
The unit of resistance is the ohm, and is equal to 1 Volt / 1
Example 18-3
A small flashlight bulb draws 300 mA from its
1.5 V battery.
(a) What is the resistance of the bulb?
R = 1.5 / 300x10-3
R = 5.0 
(b) If the voltage dropped to 1.2 V, how would the current
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 efficiency in 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:
You light me up.
18.4 Resistivity
It is also experimentally observed that the resistance of a metal
wire is well-described by
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.
R = L / A,
The longer a wire, the “harder” it is to push electrons through
The greater the resistivity, the “harder” it is to push electrons
through it.
The greater the cross-sectional area, the “easier” it is to push
electrons through it.
Resistivity is a useful tool in physics because it depends only on
the properties of the wire material, and not the wire geometry.
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 18-4
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
(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 = (4.0) (0.10)
V = 0.4 V
Example 18-5
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 (see equation 18-4), but I
don’t plan on testing you on this part of the section (all of page
A couple more links. Sorry about the shockwave you have to
go through on the first one. Too bad they both seem to be
dead now.