Transcript Physics for Scientists & Engineers 1

PHY 184
Spring 2007
Lecture 18
Title: Resistor Circuits
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184 Lecture 18
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Announcements
 Midterm 1 will take place in class tomorrow
 Chapters 16 - 19
• Homework Sets 1 - 4
• You may bring one 8.5 x 11 inch sheet of equations, front and back,
prepared any way you prefer.
• Bring a calculator
• Bring a No. 2 pencil
• Bring your MSU student ID card
 We will post Midterm 1 as Corrections Set 1 after the exam
• You can re-do all the problems in the Exam
• You will receive 30% credit for the problems you missed
• To get credit, you must do all the problems in Corrections Set 1, not just
the ones you missed
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Seating Instructions Thursday




Please seat yourselves
alphabetically.
Sit in the row (C, D,…)
corresponding to your last
name alphabetically.
For example, Bauer would
sit in row C, Westfall in
row O.
We will pass out the exam
by rows.
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Section 2
Name
Row Name
END
BEGIN
B reen
A bron
C
k 2006 C lare
B urdic
D
Fall
Semester
F ilipiak
C oleman
E
Midterm 1
G reen
F ink
F
Section 1
H uy nh
G ruhl
G
Alphabetical Seating
Kantz
I ngham
H
Order
L antzy
Kelly
I
M iller
L egg
J
N orris
M is lik
K
P ierc e
N ov ak
L
S c afuri
P rovines
M
T ran
S c hleh
N
Work
V alentini
O
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Review - Temperature Dependence
 The temperature dependence of
the resistance of metallic
conductors is given by
R  R0  R0 T  T0 
• R is the resistance at
temperature T
• R0 is the resistance at
temperature T0
•  is the temperature coefficient
of electric resistivity for the
material under consideration
  0  0 T  T0 
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Review – Par and Ser Resistors
 We can replace n parallel resistors
with one equivalent resistor given
by
n
1
1

Req i 1 Ri
 We can replace n series resistors
with one equivalent resistor given
by
n
Req   Ri
i 1
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Example: Network of Resistors
 Consider the network of resistors shown below
 Calculate the current flowing in this circuit.
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Example: Network of Resistors (2)
 Ok, let’s look at it. R3 and R4 are in series
R34  R3  R4
 Now note that R34 and R1 are in parallel
1
1
1


R134 R1 R34
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184 Lecture 18
or R134
R1 R34

R1  R34
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Example: Network of Resistors (3)
 And now R2, R5, R6, and R134 are in series
R123456  R2  R5  R6  R134
R123456
R123456
i
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R1 R34
 R2  R5  R6 
R1  R34
R1 R3  R4 
 R2  R5  R6 
R1  R3  R4
Vemf
R123456
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Clicker Question
 Consider the circuit on the right.
 Which statement is correct?
A) R2 and R3 are in parallel
B) R1 and R3 are in series
C) R1 and R2 are in parallel
D) Several statements above are correct
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184 Lecture 18
QuickTime™ and
TIFF (Uncompressed) deco
are needed to see this p
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Clicker Question
 Consider the circuit on the right.
 Which statement is correct?
C) R1 and R2 are in parallel
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TIFF (Uncompressed) deco
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R1 and R2 have the same voltage across them. R2 and R3 do
not have the same voltage drop, so they cannot be in
parallel. R1 and R3 do not have the same current flowing
through them, so they cannot be in series.
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More resistors …
 The figure shows a circuit containing one ideal 12 V battery
(no internal resistance) and 4 resistors with R1=20 , R2=20
, R3=30, and R4=8 .
 What is the current through the battery?
Idea: Find the equivalent resistance and
use Ohm’s Law.
R2 and R3 are in parallel.
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More resistors …
 R23=12 
 What is the current through the battery?
R1, R23 and R4 are in series.
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More resistors …
 The circuit contains one ideal 12 V
battery (no internal resistance) and
4 resistors with R1=20 , R2=20 ,
R3=30, and R4=8 .
 What is the current i2 through R2?
Key Idea 1: R2 and R3 are in parallel,
so they have the same voltage
drop V2=V3=V23
Key Idea 2: R1, R23 and R4 are in
series so they have the same current
V23=iR23 =(0.3 A)(12)=3.6 V
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More resistors …
 The figure on the right shows a
circuit containing one ideal 12 V
battery (no internal resistance) and 4
resistors with R1=20 , R2=20 ,
R3=30, and R4=8 .
 What is the current i3 through R3?
Key Idea: Conservation of charge
tells us that the current i going
through R23 must be equal to the sum
of the currents through R2 and R3.
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Light Bulbs in Parallel and in Series
In parallel:
+12 V
Observation: Take
out one bulb,
nothing happens to
the others
- 12V
Assume: the bulbs are almost identical and
have the same resistance
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Clicker Question
In parallel:
+12 V
What voltage drop
will be measured
across one light
bulb?
- 12V
A) 12 V
B) 24 V
C) 36 V
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Clicker Question
In parallel:
+12 V
What voltage drop
will be measured
across one light
bulb?
- 12V
B) 24 V
Since the bulbs are wired in parallel: the voltage drop
is the same for all and equal to the voltage supplied by
the emf device
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Light Bulbs in Parallel and Series
In series:
+12 V
- 12V
Observation: Taking
one bulb out breaks
the circuit. The
more bulbs we put
in series, the
dimmer they get!
Assume: the bulbs are almost identical and
have the same resistance
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Clicker Question
In series:
+12 V
What voltage drop
will be measured
across one light
bulb?
- 12V
A) 8 V
B) 12 V
C) 24 V
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Clicker Question
In series:
+12 V
What voltage drop
will be measured
across one light
bulb?
- 12V
A) 8 V
In series: Vemf=V1+V2+V3, all resistances are the same.
We measure Vemf/3=24/3=8 V across each bulb
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Energy and Power in Electric Circuits
 Consider a simple circuit in which a source of emf with voltage V causes
a current i to flow in a circuit.
 The work required to move a differential amount of charge dq around
the circuit is equal to the differential electric potential energy dU given
by dU  dqV
 The definition of current is
i  dq / dt
 So we can rewrite the differential electric potential energy as
 The definition of power P is
 Pitting it together
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P  dU / dt
dU  idtV
dU idtV
P

 iV
dt
dt
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Energy and Power
 The power dissipated in a circuit or circuit element is given by the
product of the current times the voltage.
 Using Ohm’s Law we can write equivalent formulations of the power
2
V
P  iV  i 2 R 
R
with
 The unit of power is the watt (W).
 Electrical devices are rated by the amount of power they consume in
watts.
 Electricity bill is based on how many kilowatt-hours of electrical energy
you consume.
kW h = power times time
 The energy is converted to heat, motion, light, …
1 kW h = 1000 W X 3600 s = 3.6 x 106 joules
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Temperature Dependence of the
Resistance of a Light Bulb
 A 100 W light bulb is connected to a source of emf with Vemf
= 100 V.
 When the light bulb is operating, the temperature of its
tungsten filament is 2520 °C.
 Question:
 What is the resistance of the light bulb at room
temperature (20 °C)?
 Answer:
2
 Power when lighted
V
P
R
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Temperature Dependence of the
Resistance of a Light Bulb (2)
V 2 100 V
 … so
R

 100 
P
100 W
 The temperature dependence of the resistance
2
R  R0  R0 T  T0 
 … solve for the resistance at room temperature, R0
R  R0  R0 T  T0   R0 1   T  T0 
R
R0 
1   T  T0 
 Look up the temperature coefficient for tungsten …
R
100 
R0 

 8.2 
-3
-1
1   T  T0  1+ 4.5 10 C 2520 C  20 C

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
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Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down.
 If a battery costs US$ 0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer: $320
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