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Upcoming Schedule
Oct. 8
boardwork
Oct. 10
19.3-19.5
Quiz 4
Oct. 13
19.6-19.9
Oct. 15
boardwork
Oct. 17
boardwork
Quiz 5
Oct. 20
review
Oct. 22
Exam 2
Ch. 18, 19
Oct. 24
20.1-20.2
Oct. 6
19.1-19.2
“An expert is someone who has made all the mistakes in his field.”—
Anonymous. “When studying physics, new make mistakes as fast as you
can.”—UMR Prof. Emeritus D. M. Sparlin
Ohm’s Law and Quiz 3
Here’s a secret that really bothered me when I first learned it:
V=IR
is not Ohm’s law. These are the different, correct, ways of
writing Ohm’s law (I’ll explain the symbols in class):
E=J
J=E
If the resistivity of a conductor is constant under the conditions
of whatever problem you are solving, so that R does not
change, then V=IR can be derived from E=J.
What does this have to do with Physics 35?
Answer #1: V=IR is an approximate form of Ohm’s law, valid
under limited conditions (but still very useful).
Answer #2: “All” textbooks write V=IR, but that’s not what they
mean.
My college physics textbook, copyright Dark
Ages (1964) wrote Vab=IR. It was clear that Vab
was the potential difference between points a
and b, and the text always used the words
“potential difference,” not “voltage.”
Vab=Vb-Va
R
a
b
I
Your textbooks writes V=IR, but calls V a “potential
difference.”
You have been thinking that the V in V=IR is a
voltage.
Physicists know that the V in V=IR is a potential difference, or
difference in voltage.
You are not physicists!
I need to change my OSE’s, and be more careful in the words
I use.
V = IR
P = IV = I2R = (V)2/R
There are also signs and current directions to worry about. For now, let’s
consider everything in the equations above to be magnitudes.
The electric utility company supplies
your house with electricity from the
main power lines at 120 V. The wire
from the pole to your house has a
resistance of 0.03 . Suppose your
house is drawing 110 A of current.
I
VT
VH
R
(a) Find the voltage at the point where the power wire enters
your house. One of you got this right. - for 3.3 V.
(b) How much power is being dissipated in the wire from the
pole to your house? My pre-quiz mini-lecture was intended to help you
get this right. Five of you got this right. Shows how much my lectures are
worth! - if you used P=IV with the wrong V.
(c) How much power are you using inside your house? Two of
you got this, one “almost” got it (math errors). - for 363 W. Hint: your
“incoming” power wire is at VH and your “outgoing” power wire is at 0 V.
The electric utility company supplies
your house with electricity from the
main power lines at 120 V. The wire
from the pole to your house has a
resistance of 0.03 . Suppose your
house is drawing 110 A of current.
I
VT
VH
R
(a) Find the voltage at the point where the power wire enters
your house.
VHT = IR
VT-VH = IR
VH = VT-IR
VH = (120 V) – (110 A) (0.03) = 116.7 V
(b) How much power is being dissipated in the wire from the
pole to your house?
I
VT
VH
R
Three different ways
to solve; all will give
the correct answer.
P = IV = I2R = (V)2/R
P = I(VT-VH) = I2R = (VT-VH)2/R
P = (110 A) (120 V -116.7 V) = 363 W
or P = (110 A)2 (0.03) = 363 W
or P = (120 V – 116.7 V)2 / (0.03) = 363 W
(c) How much power are you using inside your house?
I
VT
VH
R
This is cruel because you need to understand that your
household voltage represents the potential difference between
the “incoming” and “outgoing” power lines, and the “outgoing”
is at ground (0 V in this case)…except…
you can get this correct if you simply multiply the current by the
voltage at the point where the power wire enters your house…
and I was expecting most of you to do this and get your 4
points…
points…except
most of you used the wrong voltage (3.3 instead
of 120-3.3). Shows how in touch I am with what you are actually learning!
(c) How much power are you using inside your house?
I
VT
VH
R
P = IV
P = (110 A) (116.7 V – 0 V)
P = 12840 W
You don’t want to use the P=I2R=V2/R equations because you
don’t know the effective resistance of your house (although you
could calculate it).
P = (110 A) (120 V) – (110 A)(3.3 V) is also a reasonable way to work this part.
I have one day on the schedule to do a lecture which has
grown to 53 slides.
What are the chances I’ll finish today’s lecture?
If I do finish today’s lecture, what are the chances you’ll
remember any of it?
Exam 2 is scheduled for Wednesday, October 22. We’ll have to
keep a close watch on our schedule.
Chapter 19
DC Circuits
Resistors in Series and Parallel
R
resistor:
A resistor is any circuit element that has
electrical resistance (heater, light bulb, etc.).
Usually we assume wires have no resistance.
+ -
battery:
V
resistors connected in series:
A
B
Put your finger on the wire at A. If you can move along the
wires to B without ever having a choice of which wire to
follow, the circuit components are connected in series.
Here’s a circuit with three resistors and a battery:
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
current flows
in the steady state, the same current flows through all
resistors
there is a potential difference (voltage drop) across each
resistor
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
An electric charge q is given a potential energy qV by the
battery.
As it moves through the circuit, the charge loses potential
energy qV1 as it passes through R1, etc.
The charge ends up where it started, so the total energy lost
must equal the initial potential energy input:
qV = qV1 + qV2 + qV3 .
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
qV = qV1 + qV2 + qV3
V = V1 + V2 + V3
V = IR1 + IR2 + IR3
Now imagine replacing the three resistors by a single resistor,
having a resistance R such that it draws the same current as
the three resistors in series.
I
Req
V
+ -
I
As above:
From before:
Combining:
V
V = IReq
V = IR1 + IR2 + IR3
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
For resistors in series, the total resistance is the sum of the
separate resistances.
We can generalize this to make an OSE:
OSE:
Req = Ri
(resistors in series)
a consequence of
conservation of energy
resistors connected in parallel:
A
B
Put your finger on the wire at A. If in moving along the
wires to B you ever having a choice of which wire to
follow, the circuit components are connected in parallel.
I1
R1
current flows
different currents flows through
different resistors
V
I2
R2
V
R3
the voltage drop across each
resistor is the same
I3
V
+ -
I
V
This figure may look different than Giancoli’s figure 19-2a, but
it is “identical.” Circuits which are drawn to appear very
different may be electrically equivalent.
I1
In the steady state, the
current I “splits” into I1, I2,
and I3 at point A.
V
A
I1, I2, and I3 “recombine” to
make a current I at point B.
R2
B
R3
I3
I2 =
V
+ V
I
Because the voltage drop across
each resistor is V:
V
R1
I2
V
Therefore, the net current
flowing out of A and into B is I
= I1 + I2 + I3 .
I1 =
R1
V
R2
I3 =
V
R3
I
I
Now imagine replacing the
three resistors by a single
resistor, having a resistance R
such that it draws the same
current as the three resistors in
parallel.
Req
A
B
V
+ -
From above, I = I1 + I2 + I3, and
V
I1 =
R1
So that
V
I2 =
R2
V
V
V
V
= + + .
R eq R1 R 2 R 3
I
V
V
I3 =
.
R3
I
Dividing both sides by V gives
1
1
1
1
= + + .
R eq R1 R 2 R 3
We can generalize this to make an OSE:
OSE:
1
=
R eq

i
1
Ri
(resistors in parallel)
a consequence of
conservation of charge
Example 19-2
Two 100  resistors are connected (a) in series and (b) in
parallel to a 24 V battery. What is the current through each
resistor and what is the equivalent resistance of each circuit?
(a) Series combination.
OSE: Req = Ri
R1
Req = R1 + R2
V = I Req
R2
+ -
I
V = 24 V
V = I (R1 + R2)
I = V / (R1 + R2) = 24 V / (100  + 100 ) = 0.12 A
The same current of 0.12 A flows through each resistor.
The equivalent resistance is
Req = R1 + R2
Req = 100  + 100  = 200  .
(b) Parallel combination.
OSE:
1
=
R eq

i
1
Ri
I1
R1
V
I2
R2
V
1
1
1
=
+
R eq
R1
R2
+ -
I
V = 24 V
I
V = I Req
V=
I
1
1
+
R1
R2
1
1 
I= V  +

R
R
2 
 1
1 
 1
I = 24V 
+

100
Ω
100
Ω


 200 
I = 24 
 = 0.48 A
 10000 
The equivalent resistance is
1
1   200 Ω 
 1
= 
+
=
2 
R eq
100
Ω
100
Ω
10000
Ω

 

R eq = 50 Ω
Conceptual Example 19-1. For which circuit would the bulb(s)
be brighter…
R1
R2
Series?
+ -
I
V = 24 V
I1
R1
V
Parallel?
Please lock in your
votes now.
I2
R2
V
+ -
I
V = 24 V
I
To answer the question, we must calculate the power dissipated
in the bulbs for each circuit. The more power “consumed,” the
brighter the bulb.
In both circuits, the bulbs are identical and have identical
currents passing through them. We pick either bulb for the
calculation.
Series circuit: we know the resistance and current through
each bulb, so we use:
OSE:
P = I2 R
P = (0.12 A)2 (100 )
P = 1.44 W
Parallel circuit: we know the resistance and voltage drop across
each bulb, so we use:
OSE:
P = V2 / R
P = (24 V)2 / ( 100 )
P = 5.76 W
Compare:
Pseries = 1.44 W
Pparallel = 5.76 W
The bulbs in parallel are brighter.
The hands-on activity we did failed to “prove” to all groups that
the lamps are brighter in parallel than in series.
Why?
The calculation assumes the voltage of the battery remains
constant.
Some of the batteries I have have been used many times over
the last 5 years (or more). They last a very long time when
used with the little light bulbs. Nevertheless, the do “wear out.”
I should have replaced them with new ones.
The calculation above shows 4 times as much current flows in
the parallel circuit compared to the series circuit.
You may have also noticed that when you first connected the
parallel circuit, the lights were dim, and then they got brighter
after a while.
This is another sympton of dying batteries.
The reason for the misbehaving circuits is that the batteries, as
they near the end of their lifetime, are not able to deliver their
full rated voltage, especially under heavy load conditions.
Hence the current and bulb brightness is noticeably reduced in
the parallel circuit.
Don’t believe me? Then I’ll have to bring in fresh batteries
some time and demonstrate.
This is what you see if you connect 40 W bulbs directly to a 120
V outlet. (DO NOT TRY AT HOME.)
Off
On
Here are some on-line toys:
Light bulbs in series and parallel. (seems to be dead)
The Ohm Zone. You can build your own circuits.
Science Joy Wagon. Lessons on electricity. (now a pay site)
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