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DC STUFF CONTINUES
W7D3
THIS WEEK + PEEK INTO THE FUTURE
• Quiz on DC Circuits
• Complete Unit #8
– Be sure to download the next unit from the website.
• Best Guess about Exam #2
– Monday, October 18th
– Yes, that’s in about one week!
• Review session on Monday morning @ 7:30AM
• Next major topic will be MAGNETISM.
READING SUMMARY /NEXT EXAM
• Chapter 18
– Parallel Plate Capacitor (548)
• Chapter 19
– Sections 19.1-19.4
– Section 19.5 – Only what we cover in class
• Chapter 20
– Section 20.1-20.4
– Section 20.6-20.11
– Section 20.12 – A covered in clss
LAST TIME WE DISCUSSED A REAL BATTERY
E= enf=internal voltage
E
Internal resistance reduces
the effective voltage of the
battery.
i ( R  r )  emf
emf
i
(r  R)
Use this in the experiment with the W wire!
THE KIRCHOFF CORPORATION.
• From the current unit:
– The current entering a node is equal to the current
leaving it.
I
What goes in must come out!
Direction doesn’t matter.
THE LOOP EQUATION
If you start at a point in a circuit and go around the loop and
return to the same place, the change in potential is zero. Or: The
sum of the voltage rises = the sum of the voltage drops=0
Sum Rises:
i (12)  6  i (8)  24  0
Sum drops
12 I  6  8I 24  0
THE REAL DEAL-SOMETIMES YOU CAN
REDUCE A CIRCUIT
Overview dc circuit preliminaries
w7d1
October 4, 2010
This week

We continue our exploration or DC circuits.
There is a WebAssign which you should be able to
answer later in the week. Feel free to actually read the
textbook and start sooner!
There will be a QUIZ on Friday.

Examinations have been returned.


Be aware



The “exploration” approach leaves much out of the
classroom discussion.
Some of this required material may be found in the
textbook.
WebAssign can serve as a guide to some of this.
Current
L
A
+
V
ANOTHER DEFINITION
current I
J

area
A
I  JA
The total charge moving from A to B must be the
Same or charge would build up at the interface.
I
A
B
I A  IB
JA 
IA
AA
I A  J A AA  I B  J B AB
J A AA  J B AB
J A AB

1
J B AA
J A  JB
NOTE


Electric Current is DEFINED as the flow of POSITIVE
CHARGE.
It is really the electrons that move, so the current is actually
in the opposite direction to the actual flow of charge.
(Thank Franklin!)
Charge is moving so there must be an E in the metal
conductor!
Ohm





A particular object will
resist the flow of current.
It is found that for any
conducting object, the
current is proportional to
the applied voltage.
STATEMENT: DV=IR
R is called the resistance of
the object.
An object that allows a
current flow of one ampere
when one volt is applied to
it has a resistance of one
OHM.
Ohm’s Law
DV  IR
Resistivity and Resistance
L
A
+
V
How?
A wire has a resistance of 20 Ω. It is melted down,
and from the same volume of metal a new wire is
made that is three times longer than the original
wire.What is the resistance of the new wire?
  0 1   (T  T0 )
A current I flows through a device. The difference in potential
from one side of the device to the other is V. How much POWER
is dissipated in the device?
Note :
Remember
Energy  Work
DQ
I
 q  (Potential Difference )  DQDV
Dt
DE DQDV DQ
P


DV  I DV
Dt
Dt
Dt
P  (current) x (Potential Difference)
Usually written as P=IV
Reading materials



Sections: 20.1-20.4
Sections: 20.6-20.9
Watch for a new WebAssign that will be due on
TUESDAY evening so we can get back on schedule.
DC Circuits
W7D2
Instructor Bindell
Course
2054 Fall 2010
Calendar

Today





Friday


Some DC Issues not yet covered
The usual review stuff
Continue on Units 7 & 8
There is a new WebAssign on DC. Watch for more, so don’t
wait to get started.
The usual quiz
Next week – We should complete the chapter with an
exam to be scheduled shortly thereafter.

Start studying NOW.
DC Circuits III
W8D1
This Week
 Finish the chapter including the next unit.
 Friday – Quiz
 Next Monday or Wednesday –
EXAMINATION #2
Noooo!! –
Not another
one!!
Problem Review as usual at 7:30 AM
Monday – Rm 218 CLA I
Reading Summary /next exam
• Chapter 18
– Parallel Plate Capacitor (548)
• Chapter 19
– Sections 19.1-19.4
– Section 19.5 – Only what we cover in class
• Chapter 20
– Section 20.1-20.4
– Section 20.6-20.11
– Section 20.12 – A covered in clss
About that experiment





The meter doesn’t seem to function correctly for this
experiment
Significant contact resistance … the pressure on the
contact is important. The meter works better at higher
currents.
Here is data from another meter.
Use this data and bypass the experiment.
Sorry about that!
From Unit 08
A
Battery with
internal R
W Wire
B
C
D
V=1.494,
Length
Current
Measured
One Battery
Current
Measured
Two Batteries
30 cm AB
60 cm AC
90 cm. AD
455 mA
340 mA
280 mA
1393 mA
958 mA
785 mA
Current Measurement – Keithley Meter
Least Square Estimate
2500
Current
ma
Axis Title
2000
1500
Series1
Series2
1000
Linear (Series1)
Linear (Series2)
500
0
0
1
2
3
4
Axis Title
RBattery
1.5v

 1.5
1A
Length – 30 cm units
PREVIOUS SLIDES ARE POSTED
ON THE WEBSITE
Find the magnitude and direction of the current in
the 2.0-Ω resistor in the drawing. (Let R = 3.0 Ω
and V = 2.5 V.)
=2.5
=3 ohm
A portable CD player operates with a voltage of 4.5
V, and its power usage is 0.21 W. What is the current
in the player?
Back to WORK!
An especially violent lightning bolt has an
average current of 1.15 103 A lasting 0.146 s.
How much charge is delivered to the ground by
the lightning bolt?
Two wires are identical, except that one is aluminum
and one is iron. The aluminum wire has a resistance of
0.20 Ω. What is the resistance of the iron wire?
0.688Ω
The filament in an incandescent light bulb is made from
tungsten. The light bulb is plugged into a 60 V outlet and
draws a current of 0.96 A. If the radius of the tungsten
wire is 0.0050 mm, how long must the wire be?
0.0877m
Adding R’s
L
L1
R
L
A

 ( L1  L2 )
A
L2

 L1
A

 L2
A
 R1  R2
Parallel Resistors
R
I  I1  I 2  I 3
V V
V
V



R R1 R2 R3
(All Vs the same)
1
1
1
1



R R1 R2 R3
Bulb D is removed, what happens to the
brightness of bulb B?
A
B
C
D
E
Brighter
Dimmer
No Change
It goes out
Huh??
Play it again Sam …
R R 5R
 
3 2
6
V 6V
I 
R 5R
I 3V
V
IB  
=0.6 why???
2 5R
R
RT 
Remove D
R R
RT =   R
2 2
V
V
V
V
I
IB 
 0.5   0.6 
B gets dimmer!
R
2R
R
R