Unformatted text preview: Physics 212
Lecture 9
Today's Concept:
Electric Current
Ohm’s Law & resistors, Resistors in circuits, Power in circuits
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0 Confused Confident Physics 212 Lecture 9, Slide 1 Music
Who is the Artist?
BB A)
B)
C)
D)
E) Stephane Grappelli
Pearl Django
Pearl Django
Mark O’Connor’s Hot Swing Trio
Bob Wills
Hot Club of Cowtown
Hot
Cowtown
Why?? THEY ARE COMING TO TOWN !!
THEY
SATURDAY, SEPT 25, 8:45PM
SATURDAY,
http://www.folkandroots.org/site/ (THU – SAT)
(THU
Catch Prof Kwiat (your 213/214 lecturer next
Catch
spring) do a swing dance demo before the show..
spring) Great musicians… wide variety of styles
Western Swing meets Gypsy Jazz
Great Live Show !!!! Physics 212 Lecture 9, Slide 2 Some Exam Stuff • Exam Tomorrow at 7:00
–
–
–
– Covers material through lecture 8
Bring your ID: Rooms determined by discussion section (see link)
Conflict exam at 5:15 for those who signed up in gradebook
If you have conflicts with both of these, you should have heard from
Prof Beck about scheduling https://my.physics.illinois.edu/undergrad/examprep/index.asp Physics 212 Lecture 9, Slide 3 Thanks for your feedback about the course
There were, of course, a mixed bag of opinions on what would make the
the
course better.
course
However, here’s our initial take on the comments
1) Make sure we have time to devote to the problem for the day
2) Reduce the votes on preflight questions that most students get
3) The homeworks are challenging; it would be good to give more help
The homeworks
We will probably construct a survey, based on these suggestions from
We
you, to be given after the exam to get a better quantitative
measure of your opinions.
measure Physics 212 Lecture 9, Slide 4 A BIG IDEA REVIEW
BIG
Coulomb’s Law
Force law between
Force
point charges
point
Electric Field
Force per unit charge
Gauss’ Law
Flux through closed
Flux
surface is always
proportional to
charge enclosed
charge
Electric Potential
Potential energy per
Potential
unit charge 05 Capacitance
Capacitance
Relates charge and
Relates
potential for two
conductor system kq1q2
ˆ
F1,2 = 2 r1,2
r1,2 q1 Electric Field
Electric
Property of Space
Created by Charges
Superposition F
E≡
q ∫ E idA = Gauss’ Law
Can be used to
Can
determine E field Qenc ε0 ∆U a →b
≡
= − ∫ E ⋅ dl
q
a
b ∆Va →b Q
C≡
V q2 r1,2 F1,2 Spheres
Spheres
Cylinders
Infinite Planes Electric Potential
Electric
Scalar Function that
Scalar
can be used to
determine E
determine E = −∇V Physics 212 Lecture 9, Slide 5 APPLICATIONS OF BIG IDEAS
APPLICATIONS
Conductors
Charges free to
Charges
move
move What Determines
What
How They Move?
How
Spheres
Cylinders
Infinite Planes They move until
They
E = 0 !!!
!!! Gauss’
Law E = 0 in conductor
in
determines charge
densities on surfaces
densities s
lls
t
tiia
en
en
ot
po
uiip
Eq
Eq Fiel
d
L ine
s Field Lines &
Field
Equipotentials
Equipotentials Capacitor Networks
Work Done By E Field
b b a a Wa →b = ∫ F ⋅ dl = ∫ qE ⋅ dl Change in Potential Energy
Change
b ∆U a →b = −Wa →b = − ∫ qE ⋅ dl
a
05 Series:
(1/C23)=(1/C2)+(1/C3)
Parallel
C123 = C1 + C23 Physics 212 Lecture 9, Slide 6 Key Concepts:
1) How resistance depends on A, L, σ, ρ
How
2) How to combine resistors in series and parallel
3) Understanding resistors in circuits Today’s Plan:
1) Review of resistance & preflights
Review
preflights
2) Work out a circuit problem in detail Physics 212 Lecture 9, Slide 7 σ I A L V Conductivity – high for good conductors.
Conductivity Ohm’s Law: J = σ E
Observables: V = EL
I = JA I/A = σV/L
I/A R = Resistance
ρ = 1/σ I = V/R I = V/(L/σΑ)
R= L
σA
Physics 212 Lecture 9, Slide 8 This is just like plumbing!
I is like flow rate of water
V is like pressure
R is how hard it is for water to flow in a pipe
is L
R=
σA To make R big, make L long or A small
big,
long To make R small, make L short or A big
To
small,
short Physics 212 Lecture 9, Slide 9 Preflight 2 Preflight 4
Same current through
Same
both resistors
both
Compare voltages
Compare
across resistors
across L
R∝
A 70
60 V = IR ∝ 50
40
30 L
A 60
50
40
30
20 20 10 10
0 70 0 A2 = 4 A1 ⇒ V2 = 1 V1
4 L2 = 2 L1 ⇒ V2 = 2V1
Physics 212 Lecture 9, Slide 10
10 Preflight 12 BB 70 I
J≡
A
Same Current 60 J1 = J 3 = 2 J 2 50
40
30 1
J∝
A 20
10
0 Physics 212 Lecture 9, Slide 11
11 Resistor Summary
Series Parallel
R1 R1 R2
R2 Wiring
Voltage
Current
Resistance Each resistor on the
same wire. Each resistor on a
different wire. Different for each
resistor.
Vtotal = V1 + V2 Same for each
resistor.
Vtotal = V1 = V2 Same for each resistor
Itotal = I1 = I2 Different for each
resistor
Itotal = I1 + I2
Decreases
1/Req = 1/R1 + 1/R2 Increases
Req = R1 + R2 Physics 212 Lecture 9, Slide 12
12 Preflight 6 100
80
60
40
20
0 R2 in series with R3 Current through R2
and R3 is the same I 23 = V
R2 + R3 Physics 212 Lecture 9, Slide 13
13 R1 = R2 = R3 = R Preflight 7
Compare the current through
Compare
R1 with the current through R2 I1 I2 Preflight 9
Compare the voltage
Compare
across R2 with the
across
with
voltage across R3
voltage V2 V3 Preflight 10
Compare the voltage across
Compare
R1 with the voltage across R2 V1 V2 Physics 212 Lecture 9, Slide 14
14 I1 50
40 I23 30
20
10
0 R1 = R2 = R3 = R Preflight 7
Compare the current through
Compare
R1 with the current through R2
I1/I2 = 1/2
I1/I2 = 1/3 We know: I 23 = V
R2 + R3
I1 = I 23 Similarly: I1 = V
R1 R2 + R3
R1 I1/I2 = 1
I1/I2 = 2
I1/I2 = 3 I1 R2 + R3
=
=2
I 23
R1
Physics 212 Lecture 9, Slide 15
15 BB V2 V3
V23 50
40
30
20 R1 = R2 = R3 = R 10 Preflight 9
Compare the voltage
Compare
across R2 with the
across
with
voltage across R3
voltage
A V2 > V3 B V2 = V3 = V C V2 = V3 < V D 0 Consider loop V23 = V V2 < V3 V23 = V2 + V3
R2 = R3 ⇒ V2 = V3 V
V2 = V3 =
2 Physics 212 Lecture 9, Slide 16
16 V1
BB
50
40 V23 30
20 R1 = R2 = R3 = R 0 Preflight 10
Compare the voltage across
Compare
R1 with the voltage across R2
A V1 = V2 = V
B V1 = ½ V2 = V C V1 = 2V2 = V D V1 = ½ V2 = 1/5 V
1/5 E V1 = ½ V2 = ½ V 10 R1 in parallel with series
in
combination of R2 and R3
combination V 1= V23
R2 = R3 ⇒ V2 = V3 V1 = 2V2 = V V23 = V2 + V3 = 2V2
Physics 212 Lecture 9, Slide 17
17 R2 R1 Calculation
In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V R3
R4 What is V2, the voltage across R2? • Conceptual Analysis:
– – Ohm’s Law: when current I flows through resistance R, the potential
drop V is given by: V = IR.
Resistances are combined in series and parallel combinations
• Rseries = Ra + Rb
• (1/Rparallel) = (1/Ra) + (1/Rb) • Strategic Analysis
–
–
– Combine resistances to form equivalent resistances
Evaluate voltages or currents from Ohm’s Law
Expand circuit back using knowledge of voltages and currents Physics 212 Lecture 9, Slide 18
18 R2 R1
V Calculation
In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R3 What is V2, the voltage across R2? R4 • Combine Resistances:
R1 and R2 are connected: (A) in series (B) in parallel
(B) (C) neither in series nor in parallel
(C) Parallel Combination
Parallel
Ra Series Combination
Series
Ra Rb Parallel: Can make a loop that
contains only those two resistors Rb Series : Every loop with resistor 1 also
has resistor 2.
Physics 212 Lecture 9, Slide 19
19 R2 R1
V Calculation
In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R3
R4 What is V2, the voltage across R2? • We first will combine resistances R2 + R3 + R4:
• Which of the following is true?
(A) R2, R3 and R4 are connected in series
(B) R2, R3, and R4 are connected in parallel
(C) R3 and R4 are connected in series (R34) which is connected in parallel with R2
(D) R2 and R4 are connected in series (R24) which is connected in parallel with R3
(E) R2 and R4 are connected in parallel (R24) which is connected in parallel with R3 Physics 212 Lecture 9, Slide 20
20 R2 R1 Calculation
In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V BB R3 What is V2, the voltage across R2? R4 R2 and R4 are connected in series (R24) which is connected in parallel with R3
Redraw the circuit using the equivalent resistor R24 = series combination of R2
and R4.
R1
V R1 R3 R3 V V R3 R24 R24 (A)
(A) R1 (B) R24 (C)
Physics 212 Lecture 9, Slide 21
21 Calculation R1
V In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω.
R3 BB R24 What is V2, the voltage across R2?
• Combine Resistances: R2 and R4 are connected in series = R24
R3 and R24 are connected in parallel = R234 What is the value of R234? (A) R234 = 1 Ω
(A) (B) R234 = 2 Ω
(B) R2 and R4 in series
(1/Rparallel) = (1/Ra) + (1/Rb) (C) R234 = 4 Ω
(C) (D) R234 = 6 Ω
(D) R24 = R2 + R4 = 2Ω + 4Ω = 6Ω
1/R234 = (1/3) + (1/6) = (3/6) Ω−1 R234 = 2 Ω Physics 212 Lecture 9, Slide 22
22 R1 Calculation
In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. V
I1 = I234 R234 R24 = 6Ω
R234 = 2Ω
What is V2, the voltage across R2? R1 and R234 are in series. R1234 = 1 + 2 = 3 Ω
Our next task is to calculate the total current in the circuit Ohm’s Law tells us V
= I1234 R1234 I1234 = V/R1234
= 18 / 3
= 6 Amps Physics 212 Lecture 9, Slide 23
23 Calculation
V I1234 R1234 In the circuit shown: V = 18V,
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. BB R24 = 6Ω
R234 = 2Ω I1234 = 6 A
What is V2, the voltage across R2?
R1 a I234 = I1234 Since R1 in series w/ R234 V
R234 = I1 = I234 V234 = I234 R234
=6x2 b = 12 Volts • What is Vab, the voltage across R234 ?
(A) Vab = 1 V (B) Vab = 2 V
(B) (C) Vab = 9 V (D) Vab = 12 V
(C)
(D) (E) Vab = 16 V
(E) Physics 212 Lecture 9, Slide 24
24 Calculation
R1 R1
V V
R234 R3 R24 V = 18V
R1 = 1Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω. BB R24 = 6Ω
R234 = 2Ω
I1234 = 6 Amps
I234 = 6 Amps
V234 = 12V
What is V2? Which of the following are true?
A) V234 = V24 B) I234 = I24 C) Both A+B R3 and R24 were combined in parallel to get R234 D) None
Voltages are same! Ohm’s Law
I24 = V24 / R24
= 12 / 6
= 2 Amps
Physics 212 Lecture 9, Slide 25
25 Calculation
R1
V I1234 R3 R1 V I24 R2 R3 R24 R4
Which of the following are true?
A) V24 = V2 B) I24 = I2 C) Both A+B R2 and R4 where combined in series to get R24 D) None V = 18V
R 1 = 1Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω. BB R24 = 6Ω
R234 = 2Ω
I1234 = 6 Amps
I234 = 6 Amps
V234 = 12V
V24 = 12V
I24 = 2 Amps
What is V2? Currents are same! Ohm’s Law
The Problem Can Now
The
Be Solved!
Be V2 = I2 R2
= 2 x2
= 4 Volts!
Physics 212 Lecture 9, Slide 26
26 I1 Quick FollowUps R1 R2 I2 R1 I3
V = R3 a V
R234 R4
b • What is I3 ?
(A) I3 = 2 A (B) I3 = 3 A
(B) V3 = V234 = 12V (C) I3 = 4 A
(C) V = 18V
R1 = 1Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω. BB R24 = 6Ω
R234 = 2Ω
V234= 12V
V2 = 4V
I1234 = 6 Amps I3 = V3/R3 = 12V/3Ω = 4A • What is I1 ?
We know I1 = I1234 = 6 A
We
NOTE: I2 = V2/R2 = 4/2 = 2 A I1 = I2 + I3 Make Sense??? Physics 212 Lecture 9, Slide 27
27 ...
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 Spring '11
 MESTRE
 Physics, Current, Power, Resistor, Electrical resistance

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