Last time…

Resistors
Circuits
Begin circuits
Today…


Physical layout
Resistor circuits
Start resistor-capacitor circuits
Schematic layout
Tue. Oct. 13, 2009
Physics 208 Lecture 12
1
Tue. Oct. 13, 2009
Physics 208 Lecture 12
2
Question
Quick Quiz
Which bulb is brighter?
When 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?
I
A. A
I
B. B
C. Both the same
A.  qVd – qVc
I
Current through each must be same
B.  qVc – qVd
C.  qVd + qVc
I
D.  zero
Conservation of current (Kirchoff’s current law)
Charge that goes in must come out
Tue. Oct. 13, 2009
Physics 208 Lecture 12
3
Tue. Oct. 13, 2009
Physics 208 Lecture 12
Power dissipation (Joule heating)
Quick Quiz

What is the change in kinetic energy
as it moves from c to d?
Charge loses energy from c to d.
E lost = −ΔE = −(ΔKE + ΔU ) = 0 − q(Vd − Vc )

Ohm’s law:

€
A.  qVd – qVc

B.  qVc – qVd
C.  qVd + qVc
€
D.  zero
€


(Vc − Vd ) = IR
E lost = qIR
Energy dissipated in resistor as
  Heat (& light) in bulb
Power dissipated in resistor =
dE lost dq
=
IR = I 2 R
dt
dt
Tue. Oct. 13, 2009
4
Physics 208 Lecture 12
5
Tue. Oct. 13, 2009
Physics 208 Lecture 12
Joules / s = Watts
6
€
1
Light bulbs and power
I
Household voltage is 120V
  60W = 60J /s = I 2 R = I IR = VI
( )



Two different bulbs
60 Watt
b
P1 = I 2 R1
I = 60W /120V = 0.5A
R = V /I = 120V /0.5A = 240Ω
a
c
P2 = I R2
€
€
€
Cost



€
(60J /s)(24hour)(3600s /hour) = 5,184,000J

Current same through each
Power dissipated different


€
e
€
24 hours on requires
I
R1
2
I
R2
d
I
Brightness different
MG&E ~ 13¢ / kWatt-hour
1kW − hour = (1000J /s)( 3600s /hour) = 3,600,000J
19¢ / day
Tue. Oct. 13, 2009
Physics 208 Lecture 12
7
Tue. Oct. 13, 2009
Physics 208 Lecture 12
8
€
Kirchoff’s junction law

Charge conservation
I2
What happens to the brightness of the
bulb A when the switch is closed?
I1
Iin
I3
I1=I2+I3
A.  Gets dimmer
B.  Gets brighter
I1
C.  Stays same
I3
Iout = Iin
D.  Something else
I2
Iout
I1+I2=I3
Tue. Oct. 13, 2009
Physics 208 Lecture 12
9
Tue. Oct. 13, 2009
10 A
I
Toaster
R2
R3
R4
5A
12 A
Coffee Pot
Microwave
….
Everything works great until you plug in your space heater,
then you smell smoke. This is because
Increases if each Ri getting bigger
Increases if each Ri getting smaller
Always increases
Each resistor added
Always decreases
adds ΔV/Ri to the
Stays the same
total current I
Tue. Oct. 13, 2009
10
You use one power strip to plug in your toaster,
coffee pot, microwave.
As more and more resistors are added to the
parallel resistor circuit shown here the total
current flowing I…
R1
Physics 208 Lecture 12
Question
Question
A.
B.
C.
D.
E.
Quick Quiz
Physics 208 Lecture 12
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
11
Tue. Oct. 13, 2009
Physics 208 Lecture 12
12
2
More complicated
circuits


Determine equivalent
resistance
Replace combinations
with equivalent
resistance
Tue. Oct. 13, 2009

The circuit below contains three 100W light bulbs.
The emf ε = 110 V. Which light bulb(s) is(are)
brightest ?
Both series & parallel


Quick Quiz
A. A
B. B
C. C
D. B and C
E. All three are equally bright.
Physics 208 Lecture 12
13
Tue. Oct. 13, 2009
Physics 208 Lecture 12
Measurements in a circuit
Kirchoff’s loop law
A multimeter can measure currents (as an ammeter), potential
difference (as a voltmeter)
Electrical measuring devices must have minimal impact in the
circuit

ΔVV
ΔV
I
V
R
ε
A
IA
Ammeter
The internal resistance of the
ammeter must be very small
I = IA
ε = ΔV+ΔVA = RI + rAI → RI
for rA →0
Tue. Oct. 13, 2009
IV
I
ε
R2
R3
I2
I3
Voltmeter
ε
The internal resistance of the
voltmeter must be very large
I = Iv+IR
ε ε
ε
→∞
I = + rV
→
ΔVV = ε
Physics 208 Lecture 12
R1
I1
R
IR
ΔVA
Conservation of energy
14
rV
R
15
R
Tue. Oct. 13, 2009
Physics 208 Lecture 12
16
€
Resistorcapacitor circuit

Charging a capacitor

What happens to charges on
the capacitor after switch is
closed?
Again
Kirchoff’s loop law:
ε − IR − Q /C = 0
⇒ I = ε /R − Q
C

Why does the charge flow
through the resistor?
C
€

Why does the charge on the
capacitor change in time?
/RC
Time t = 0: Qc = 0 ⇒ I = ε /R
€
t increases: Qc > 0 ⇒ I < ε /R
Looks like resistor & battery:
uncharged cap acts like short circuit
VC increases, so VR decreases
€
Time t = ∞: VC = ε ⇒ VR = 0 ⇒ I = 0
Thur. Oct. 16, 2008
Physics 208 Lecture 14
17
€
Thur. Oct. 16, 2008
Fully charged capacitor acts
like open circuit
Physics 208 Lecture 14
18
€
3
Discharging the capacitor

Kirchoff’s loop law
(VB − VA ) + (VD − VC ) = 0
ΔVc = Qc /C
€
€
€
−IR
RC discharge
B
C
A
D
Q
⇒I= c
RC

RC time constant
τ = RC
Charges in the current I come from capacitor:
Q = Qoe−t / τ
dQ
I=− c
dt
€
Thur. Oct. 16, 2008
Physics 208 Lecture 14
I = Ioe−t / τ
€
19
Thur. Oct. 16, 2008
Physics 208 Lecture 14
20
€
€
€
Charging a capacitor
Question
The circuit contains three identical light
bulbs and a fully-charged capacitor.
Which is brightest?
A.  A
B.  B
Q = Qmax (1− e−t / τ )
C.  C
D.  A & B
E.  All equally bright
€
Thur. Oct. 16, 2008
Physics 208 Lecture 14
21
Question
Thur. Oct. 16, 2008
Physics 208 Lecture 14
RC discharge

The circuit contains three identical light
bulbs and an uncharged capacitor.
Which is brightest?
22
RC time constant
τ = RC
€
A.  A
B.  B
C.  C
D.  A & B
time t
E.  All equally bright
Thur. Oct. 16, 2008
Physics 208 Lecture 14
23
Tue. Oct. 13, 2009
Physics 208 Lecture 12
24
4
RC analysis

RC analysis
dQC
Q
=− C
dt
RC
Kirchoff loop law:
ΔVC + ΔVR = 0
Q
⇒ C − IR = 0
C

€
€
I related to QC
dQ
I=− C
dt
€
Tue. Oct. 13, 2009
€
dQC
1
=−
dt
QC
RC
Q
∫
Qo
t
∫ dt
0
t
o
RC
QC ( t ) = Qo exp(−t /RC )
Q( t )
lnQC Q = −
QC
dQ
+R C =0
C
dt
dQC
Q
=− C
dt
RC
Physics 208 Lecture 12
dQC
1
=−
QC
RC
25
Tue. Oct. 13, 2009
Physics 208 Lecture 12
26
€
€
5