Tue. Oct. 13, 2009 Physics 208 Lecture 12 1

Last time…

Begin circuits

Resistor circuits Start resistor-capacitor circuits

Today…

Tue. Oct. 13, 2009 Physics 208 Lecture 12 2

Resistors

Schematic layout

Circuits

Physical layout

Tue. Oct. 13, 2009 Physics 208 Lecture 12 3

Quick Quiz

Which bulb is brighter?

A. A

B. B

C.Both the same

Current through each must be same

Conservation of current (Kirchoff’s current law)

Charge that goes in must come out

I

I

I

I

QuestionWhen current flows, charge moves around the circuit.

Suppose that positive charge carriers flow around the circuit. What is the change in potential energy of a positive charge as moves from c to d?

Tue. Oct. 13, 2009 Physics 208 Lecture 12 4

A. qVd – qVc

B. qVc – qVd

C. qVd + qVc

D. zero

Quick Quiz

What is the change in kinetic energy as it moves from c to d?

Tue. Oct. 13, 2009 Physics 208 Lecture 12 5

A. qVd – qVc

B. qVc – qVd

C. qVd + qVc

D. zero

Tue. Oct. 13, 2009 Physics 208 Lecture 12 6

Power dissipation (Joule heating)

Charge loses energy from c to d.

Ohm’s law:

Energy dissipated in resistor as Heat (& light) in bulb

Power dissipated in resistor =

€

E lost = −ΔE = − ΔKE + ΔU( ) = 0 −q Vd −Vc( )

€

Vc −Vd( ) = IR

€

dE lostdt

=dq

dtIR = I2R Joules / s = Watts

€

E lost = qIR

Tue. Oct. 13, 2009 Physics 208 Lecture 12 7

Light bulbs and power

Household voltage is 120V

Cost 24 hours on requires

MG&E ~ 13¢ / kWatt-hour

60 Watt

€

60W = 60J /s = I2R = I IR( ) =VI

€

I = 60W /120V = 0.5A

€

60J /s( ) 24hour( ) 3600s /hour( ) = 5,184,000J

€

1kW − hour = 1000J /s( ) 3600s /hour( ) = 3,600,000J

€

R =V /I =120V /0.5A = 240Ω

19¢ / day

Tue. Oct. 13, 2009 Physics 208 Lecture 12 8

Two different bulbs

Current same through each Power dissipated different

Brightness different

R1

R2

a

b

c

d

e

I

I

I

I€

P1 = I2R1

€

P2 = I2R2

Tue. Oct. 13, 2009 Physics 208 Lecture 12 9

Kirchoff’s junction law

Charge conservation

Iin

Iout

Iout = Iin

I1

I2

I3I1=I2+I3

I2

I3

I1

I1+I2=I3

Tue. Oct. 13, 2009 Physics 208 Lecture 12 10

Quick QuizWhat happens to the brightness of the

bulb A when the switch is closed?

A. Gets dimmer

B. Gets brighter

C. Stays same

D. Something else

Tue. Oct. 13, 2009 Physics 208 Lecture 12 11

QuestionAs more and more resistors are added to the

parallel resistor circuit shown here the total current flowing I…

….R1

R2

R3

R4

I

A. Increases if each Ri getting bigger

B. Increases if each Ri getting smallerC. Always increasesD. Always decreasesE. Stays the same

Each resistor added adds V/Ri to the total current I

Tue. Oct. 13, 2009 Physics 208 Lecture 12 12

You use one power strip to plug in your toaster, coffee pot, microwave.

Toaster Coffee Pot Microwave

10 A 5 A 12 A

Everything works great until you plug in your space heater, then you smell smoke. This is because

Question

A. The resistance of the circuit is too high

B. The voltage in the circuit is too high

C. The current in the circuit is too high

Tue. Oct. 13, 2009 Physics 208 Lecture 12 13

More complicated circuits

Both series & parallel Determine equivalent

resistance Replace combinations

with equivalent resistance

Tue. Oct. 13, 2009 Physics 208 Lecture 12 14

Quick Quiz

The circuit below contains three 100W light bulbs. The emf = 110 V. Which light bulb(s) is(are) brightest ?

A. AB. BC. CD. B and CE. All three are equally bright.

Tue. Oct. 13, 2009 Physics 208 Lecture 12 15

Measurements in a circuit A multimeter can measure currents (as an ammeter), potential

difference (as a voltmeter) Electrical measuring devices must have minimal impact in the

circuit

R

VVoltmeter

The internal resistance of the ammeter must be very smallI = IA= V+VA = RI + rAI RIfor rA 0

VA

A

R

Ammeter

I

IA

V

The internal resistance of the voltmeter must be very largeI = Iv+IR VV =

VV

IV

IR

I

€

I =ε

rV+ε

RrV →∞ ⏐ → ⏐ ⏐ ε

R

Tue. Oct. 13, 2009 Physics 208 Lecture 12 16

Kirchoff’s loop law

Conservation of energy

R1

R2 R3

I1

I2 I3

Thur. Oct. 16, 2008 Physics 208 Lecture 14 17

Resistor-capacitor circuit What happens to charges on

the capacitor after switch is closed?

Why does the charge on the capacitor change in time?

Why does the charge flow through the resistor?

Thur. Oct. 16, 2008 Physics 208 Lecture 14 18

Charging a capacitor

Again Kirchoff’s loop law:

€

−IR −QC /C = 0

Time t = 0:

€

Qc = 0⇒ I =ε /R

€

⇒ I =ε /R −QC /RC

Looks like resistor & battery: uncharged cap acts like short circuit

t increases:

€

Qc > 0⇒ I <ε /R VC increases, so VR decreases

Time t = :

€

VC =ε ⇒ VR = 0⇒ I = 0 Fully charged capacitor acts like open circuit

Thur. Oct. 16, 2008 Physics 208 Lecture 14 19

Discharging the capacitor

Kirchoff’s loop law

A

B C

D

€

VB −VA( ) + VD −VC( ) = 0

€

Vc =Qc /C

€

−IR

€

⇒ I =QcRC

Charges in the current I come from capacitor:

€

I = −dQcdt

Thur. Oct. 16, 2008 Physics 208 Lecture 14 20

RC discharge

RC time constant

€

τ =RC

€

Q =Qoe−t /τ

€

I = Ioe−t /τ

Thur. Oct. 16, 2008 Physics 208 Lecture 14 21

Charging a capacitor

€

Q =Qmax 1− e−t /τ( )

Thur. Oct. 16, 2008 Physics 208 Lecture 14 22

Question

The circuit contains three identical light bulbs and a fully-charged capacitor. Which is brightest?

A. A

B. B

C. C

D. A & B

E. All equally bright

Thur. Oct. 16, 2008 Physics 208 Lecture 14 23

Question

The circuit contains three identical light bulbs and an uncharged capacitor. Which is brightest?

A. A

B. B

C. C

D. A & B

E. All equally bright

Tue. Oct. 13, 2009 Physics 208 Lecture 12 24

RC discharge RC time constant

time t

€

τ =RC

Tue. Oct. 13, 2009 Physics 208 Lecture 12 25

RC analysis

Kirchoff loop law:

€

VC + ΔVR = 0

⇒QCC

− IR = 0

I related to QC

€

I = −dQCdt

€

QCC

+ RdQCdt

= 0

€

dQCdt

= −QCRC

Tue. Oct. 13, 2009 Physics 208 Lecture 12 26

RC analysis

€

dQCdt

= −QCRC

€

dQCQC

= −1

RCdt

dQCQC

= −1

RCdt

0

t

∫Qo

Q

∫

lnQC QoQ t( ) = −

t

RCQC t( ) =Qo exp −t /RC( )