2/7/07 184 Lecture 18 1

PHY 184PHY 184PHY 184PHY 184

Spring 2007Lecture 18

Title: Resistor Circuits

2/7/07 184 Lecture 18 2

AnnouncementsAnnouncementsAnnouncementsAnnouncements 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

2/7/07 184 Lecture 18 3

Seating Instructions ThursdaySeating Instructions ThursdaySeating Instructions ThursdaySeating Instructions Thursday

Fall Semester 2006Midterm 1Section 1

Alphabetical Seating Order

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.

Section 2

2/7/07 184 Lecture 18 4

Review - Temperature DependenceReview - Temperature DependenceReview - Temperature DependenceReview - Temperature Dependence

The temperature dependence of the resistance of metallic conductors is given by

• 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

R R0 R0 T T0 R R0 R0 T T0

0 0 T T0 0 0 T T0

2/7/07 184 Lecture 18 5

Review – Par and Ser ResistorsReview – Par and Ser ResistorsReview – Par and Ser ResistorsReview – Par and Ser Resistors

We can replace n parallel resistors with one equivalent resistor given by

We can replace n series resistors with one equivalent resistor given by

1

Req

1

Rii1

n

1

Req

1

Rii1

n

Req Rii1

n

Req Rii1

n

2/7/07 184 Lecture 18 6

Example: Network of ResistorsExample: Network of ResistorsExample: Network of ResistorsExample: Network of Resistors

Consider the network of resistors shown below

Calculate the current flowing in this circuit.

2/7/07 184 Lecture 18 7

Example: Network of Resistors (2)Example: Network of Resistors (2)Example: Network of Resistors (2)Example: Network of Resistors (2)

Ok, let’s look at it. R3 and R4 are in series

Now note that R34 and R1 are in parallel

R34 R3 R4

1

R134

1

R1

1

R34

or R134 R1R34

R1 R34

2/7/07 184 Lecture 18 8

Example: Network of Resistors (3)Example: Network of Resistors (3)Example: Network of Resistors (3)Example: Network of Resistors (3)

And now R2, R5, R6, and R134 are in series

R123456 R2 R5 R6 R134

R123456 R2 R5 R6 R1R34

R1 R34

R123456 R2 R5 R6 R1 R3 R4 R1 R3 R4

i VemfR123456

2/7/07 184 Lecture 18 9

Clicker QuestionClicker QuestionClicker QuestionClicker 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

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

2/7/07 184 Lecture 18 10

Clicker QuestionClicker QuestionClicker QuestionClicker Question Consider the circuit on the right. Which statement is correct?

C) R1 and R2 are in parallel

QuickTime™ and a

TIFF (Uncompressed) decompressorare needed to see this picture.

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.

2/7/07 184 Lecture 18 11

More resistors …More resistors …More resistors …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.

2/7/07 184 Lecture 18 12

More resistors …More resistors …More resistors …More resistors …

R23=12 What is the current through the battery? R1, R23 and R4 are in series.

2/7/07 184 Lecture 18 13

More resistors …More resistors …More resistors …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 currentV23=iR23 =(0.3 A)(12)=3.6 V

2/7/07 184 Lecture 18 14

More resistors …More resistors …More resistors …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.

2/7/07 184 Lecture 18 15

Light Bulbs in Parallel and in SeriesLight Bulbs in Parallel and in SeriesLight Bulbs in Parallel and in SeriesLight Bulbs in Parallel and in Series

+12 V

- 12V

In parallel:

Observation: Take out one bulb, nothing happens to the others

Assume: the bulbs are almost identical and have the same resistance

2/7/07 184 Lecture 18 16

Clicker QuestionClicker QuestionClicker QuestionClicker Question

+12 V

- 12V

In parallel:

What voltage drop will be measured across one light bulb?

A) 12 V

B) 24 V

C) 36 V

2/7/07 184 Lecture 18 17

Clicker QuestionClicker QuestionClicker QuestionClicker Question

+12 V

- 12V

In parallel:

What voltage drop will be measured across one light bulb?

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

2/7/07 184 Lecture 18 18

Light Bulbs in Parallel and SeriesLight Bulbs in Parallel and SeriesLight Bulbs in Parallel and SeriesLight Bulbs in Parallel and Series

+12 V

- 12V

In series:

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

2/7/07 184 Lecture 18 19

Clicker QuestionClicker QuestionClicker QuestionClicker Question

+12 V

- 12V

In series:

What voltage drop will be measured across one light bulb?

A) 8 V

B) 12 V

C) 24 V

2/7/07 184 Lecture 18 20

Clicker QuestionClicker QuestionClicker QuestionClicker Question

+12 V

- 12V

In series:

What voltage drop will be measured across one light bulb?

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

2/7/07 184 Lecture 18 21

Energy and Power in Electric CircuitsEnergy and Power in Electric CircuitsEnergy and Power in Electric CircuitsEnergy 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

The definition of current is

So we can rewrite the differential electric potential energy as

The definition of power P is

Pitting it together

dU dqV

i dq / dt

dU idtVP dU / dt

P dU

dtidtV

dtiVP

dU

dtidtV

dtiV

2/7/07 184 Lecture 18 22

Energy and PowerEnergy and PowerEnergy and PowerEnergy 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

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.

The energy is converted to heat, motion, light, …

P iV i2R V 2

Rwith

kW h = power times time

1 kW h = 1000 W X 3600 s = 3.6 x 106 joules

2/7/07 184 Lecture 18 23

Temperature Dependence of the Temperature Dependence of the Resistance of a Light BulbResistance of a Light Bulb

Temperature Dependence of the Temperature Dependence of the Resistance of a Light BulbResistance 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: Power when lighted

P V 2

R

2/7/07 184 Lecture 18 24

Temperature Dependence of the Temperature Dependence of the Resistance of a Light Bulb (2)Resistance of a Light Bulb (2)

Temperature Dependence of the Temperature Dependence of the Resistance of a Light Bulb (2)Resistance of a Light Bulb (2)

… so

The temperature dependence of the resistance

… solve for the resistance at room temperature, R0

Look up the temperature coefficient for tungsten …

R V 2

P

100 V 2

100 W100

R R0 R0 T T0

R R0 R0 T T0 R0 1 T T0 R0

R

1 T T0

R0 R

1 T T0 100

1+ 4.5 10-3 C-1 2520 C 20 C 8.2

2/7/07 184 Lecture 18 25

Total Energy in a Flashlight BatteryTotal Energy in a Flashlight BatteryTotal Energy in a Flashlight BatteryTotal 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