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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( )
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Vc −Vd( ) = IR
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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
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1kW − hour = 1000J /s( ) 3600s /hour( ) = 3,600,000J
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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:
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−IR −QC /C = 0
Time t = 0:
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Qc = 0⇒ I =ε /R
€
⇒ I =ε /R −QC /RC
Looks like resistor & battery: uncharged cap acts like short circuit
t increases:
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Qc > 0⇒ I <ε /R VC increases, so VR decreases
Time t = :
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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
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VB −VA( ) + VD −VC( ) = 0
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Vc =Qc /C
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−IR
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⇒ I =QcRC
Charges in the current I come from capacitor:
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I = −dQcdt
Thur. Oct. 16, 2008 Physics 208 Lecture 14 20
RC discharge
RC time constant
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τ =RC
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Q =Qoe−t /τ
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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
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dQCdt
= −QCRC
Tue. Oct. 13, 2009 Physics 208 Lecture 12 26
RC analysis
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dQCdt
= −QCRC
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dQCQC
= −1
RCdt
dQCQC
= −1
RCdt
0
t
∫Qo
Q
∫
lnQC QoQ t( ) = −
t
RCQC t( ) =Qo exp −t /RC( )