DC STUFF CONTINUESDC STUFF CONTINUESW7D3W7D3

THIS WEEK + PEEK INTO THE FUTURETHIS 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 EXAMREADING 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 BATTERYLAST TIME WE DISCUSSED A REAL BATTERY

E

E= enf=internal voltage

Internal resistance reducesthe effective voltage of the battery.

( )

( )

i R r emfemf

ir R

Use this in the experiment with the W wire!

THE KIRCHOFF CORPORATION.THE KIRCHOFF CORPORATION.• From the current unit:

– The current entering a node is equal to the current leaving it.

What goes in must come out!

I

Direction doesn’t matter.

THE LOOP EQUATIONTHE 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:(12) 6 (8) 24 0

Sum drops12 6 8 24 0

i i

I

I

THE REAL DEAL-SOMETIMES YOU CAN THE REAL DEAL-SOMETIMES YOU CAN REDUCE A CIRCUITREDUCE A CIRCUIT

Overview dc circuit preliminariesw7d1

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

I

-+

AI

areacurrentJ

ANOTHER DEFINITION

JAI

The total charge moving from A to B must be theSame or charge would build up at the interface.

I A B

1

A

B

B

A

BBAA

BBBAAA

A

AA

BA

AA

JJ

AJAJAJIAJI

AIJ

II

BA JJ

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!

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: V=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

IRV

Ohm’s Law

Resistivity and Resistance

L

A V

I

-+

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?

)(1 00 TT

A current I flows through a device. The difference in potentialfrom one side of the device to the other is V. How much POWERis dissipated in the device?

:

(current) x (Potential Difference)Usually written as P=IV

NoteQ

It

E Q V QP V I V

t t tP

VQ)Difference (PotentialqWorkEnergy

Remember

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 CircuitsW7D2Instructor Bindell

Course 2054 Fall 2010

Calendar Today

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. Friday

The usual quiz Next week – We should complete the chapter

with an exam to be scheduled shortly thereafter. Start studying NOW.

W8D1

DC Circuits III

Finish the chapter including the next unit.

Friday – Quiz Next Monday or Wednesday –

EXAMINATION #2

This Week

Noooo!! –

Not another one!!

Problem Review as usual at 7:30 AMMonday – 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

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!

About that experiment

From Unit 08

W WireBattery with

internal R

Length Current Measured

One Battery

Current Measured

Two Batteries

30 cm AB 455 mA 1393 mA60 cm AC 340 mA 958 mA90 cm. AD 280 mA 785 mA

A

CB

DV=1.494,

Length – 30 cm units

Currentma

1.51.5

1Batteryv

RA

Current Measurement – Keithley MeterLeast Square Estimate

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.)

=3 ohm

=2.5

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 L2

1 2 1 21 2

( )L L L LLR R R

A A A A

Parallel Resistors

R

1 2 3

1 2 3

1 2 3

(All Vs the same)

1 1 1 1

I I I IV V V VR R R R

R R R R

Bulb D is removed, what happens to the brightness of bulb B?

A BrighterB DimmerC No ChangeD It goes outE Huh??

Play it again Sam …

53 2 6

653

=0.6 why???2 5

T

B

R R RR

V VI

R RI V V

IR R

Remove D

R =2 2

0.5 0.6 B gets dimmer!2

T

B

R RR

V V V VI I

R R R R