Unformatted text preview: Physics 212
Ph
Lecture 9 Today's Concept:
Electric Current
Ohm’s Law & resistors, Resistors in circuits, Power in circuits Physics 212 Lecture 9, Slide 1 Music
Who is the Artist?
A)
B)
C)
D)
E) Dave Brubeck
Thelonius Monk
Bill Evans
Keith Jarrett
Oscar Peterson • Now there’s an example of abstract thinking – composing in real time!
Now
• Who will be next!? Physics 212 Lecture 9 Some
Some Exam Stuff • Exam tomorrw (Wed. Sep. 21) at 7:00 – Covers material through lecture 8 (no resistors)
– Bring your ID
– Conflict exam at 5:15 – you will need to stay in the room until exam ends • Where do I go? EXAM ROOMS (by discussion section)
• Preparation
– Practice exams
– Exam Prep Exercises
– Worked examples Physics 212 Lecture 8, Slide 3 Your
Your Comments • “Thank God were are moving on to easier material. Makes my life easier :)” • “I CURRENTly have only a few concerns” • “Can we review for tommorrow's exam?” • “will we have a review section?” • “I was confused with current density” • “Can we do more practice with resistor circuits that have both parallel and
series?” • Consider the comments
•
•
• • They help us understand what you are having trouble with
Writing, even briefly, helps you to think logically
even
helps
“I bet you put typos in the powerpoints to make it look like you are
actually using what students wrote, when you are actually writing them.
Checkmate!” “Can we look at schematics for electronics we use everyday? Do they have
Can
circuits that look like these?” Physics 212 Lecture 8, Slide 4 ELECTRONICS WE USE EVERYDAY (almost) – your iClicker A BIG IDEA REVIEW
BIG
Coulomb’s Law
Force law between
point charges kq1q2
ˆ
F1,2 2 r1,2
r1,2 F
Electric Field
E
Force per unit charge
q
Gauss’ Law
Law
Flux through closed
surface is always
proportional to
charge enclosed
Electric Potential
Potential energy per
unit
unit charge 05 Capacitance
Relates charge and
potential for two
conductor system Qenc E dA 0 Va b q1 q2 Electric Field
Property of Space
Created by Charges
Superposition
Gauss’ Law
Can be used to
determine
determine E field b U a b E dl
q
a Q
C
V r1,2 F1,2 Spheres
Spheres
Cylinders
Infinite
Infinite Planes Electric Potential
Scalar Function that
can
can be used to
determine E E V Physics 212 Lecture 9, Slide 6 APPLICATIONS
APPLICATIONS OF BIG IDEAS
Conductors
Charges free to
move What Determines
How They Move?
Spheres
Cylinders
Infinite Planes They move until
E = 0 !!! Gauss’
Law E = 0 in conductor
determines charge
densities on surfaces Field Lines &
Equipotentials Capacitor Networks
Work Done By E Field b Wa b F dl qE dl
b a a Change in Potential Energy U a b Wa b qE dl
b a
05 Series:
(1/C23)=(1/C2)+(1/C3)
Parallel
C123 = C1 + C23 Physics 212 Lecture 9, Slide 7 Key
Key Concepts for Today:
1) How resistance depends on A, L, , 2) How to combine resistors in series and parallel
3) Understanding resistors in circuits Today’s Plan:
1) Review of resistance & preflights
2) Work out a circuit problem in detail Physics 212 Lecture 9, Slide 8 I A L V Conductivity – high for good conductors. Ohm’s
Ohm’s Law: J = E Observables: V = EL
I = JA
JA I/A
I/A = V/L R = Resistance
Resistance = 1/ I = V/R
V/R I = V/(L/
V/(L/
R= L
A
Physics 212 Lecture 9, Slide 9 This
This is like plumbing!
I is like flow rate of water
V is like pressure
pr
R is how hard it is for water to flow in a pipe L
R=
A To make R big, make L long or A small To make R small, make L short or A big
sm
sh Physics 212 Lecture 9, Slide 10 Checkpoint
Checkpoint 1a Checkpoint 1b
Same current through
both resistors
Compare voltages
across resistors L
R
A
L
V IR A A2 4 A1 V2 1 V1
4 L2 2 L1 V2 2V1
Physics 212 Lecture 9, Slide 11 Checkpoint
Checkpoint 3
The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same
crosssectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest?
which case is the CURRENT DENSITY through the resistor the smallest?
A. Case 1
B. Case 2
C. Case 3 “The smaller the resistor the smaller the current density”
“Has most area for current to spread out in so density is lowest”
“Case has the smallest volume of the resistor with same amount of electric current. So,
Current density is the smallest in case3.”
Physics 212 Lecture 9, Slide 12 Checkpoint
Checkpoint 3
The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same
crosssectional area as R1. In which case is the CURRENT DENSITY through the resistor the smallest?
which case is the CURRENT DENSITY through the resistor the smallest?
A. Case 1
B. Case 2
C. Case 3 I
J
A
Same Current J1 J 3 2 J 2 1
J
A Physics 212 Lecture 9, Slide 13 Resistor Summary
Series Parallel
R1 R1 R2 R2 Wiring Each resistor on the
th
same wire. Each resistor on a
different wire. Voltage Different for each
resistor.
Vtotal = V1 + V2 Same
Same for each
resistor.
Vtotal = V1 = V2 Current Same for each resistor
Itotal = I1 = I2 Different for each
resistor
Itotal = I1 + I2
Decreases
1/Req = 1/R1 + 1/R2 Resistance Increases
Req = R1 + R2 Physics 212 Lecture 9, Slide 14 Three resistors are connected to a battery with emf V as
shown. The resistances of the resistors are all the same,
i.e. R1= R2 = R3 = R. Checkpoint
Checkpoint 2a Compare the current through R2 with the current through R3:
A. I2 > I3
B. I2 = I3
C. I2 < I3 R2 in series with R3 Current through R2
and R3 is the same I 23 V
R2 R3 Physics 212 Lecture 9, Slide 15 R1 = R2 = R3 = R Checkpoint 2b
Check
Compare the current through
R1 with the current through R2 I1 I2 Checkpoint 2c
Compare the voltage
across R2 with the
voltage across R3 V2 V3 Checkpoint 2d
Compare the voltage across
R1 with the voltage across R2 V1 V2 Physics 212 Lecture 9, Slide 16 Checkpoint
Checkpoint 2b R1 = R2 = R3 = R
Compare the current through
R1 with the current through R2 A. I1/I2 = 1/2
1/2
B. I1/I2 = 1/3
C. I1/I2 = 1
D. I1/I2 = 2
E. I1/I2 = 3 “R2 and R3 make up one resistor, which means the current
R2
R3
th
would
would be multiplied by two. This means that the current
through R1 divided by R23 would equal 1/2 ”
“I2 must be equal to I3. Therefore, we can reason that I2 and
I2
I3 Th
th I2
I3
I3 are both one third of I total. Therefore, each resistor has
equal current.”
“the resistance of the series is twice as much so the current is
th
th
th
halved.”
Physics 212 Lecture 9, Slide 17 Checkpoint
Checkpoint 2b
I1
I23 R1 = R2 = R3 = R
Compare the current through
R1 with the current through R2 A. I1/I2 = 1/2
1/2
B. I1/I2 = 1/3 We know: I 23 V
R2 R3
I1 I 23 Similarly: I1 V
R1 R2 R3
R1 C. I1/I2 = 1
D. I1/I2 = 2
E. I1/I2 = 3 I1 R2 R3 2
I 23
R1
Physics 212 Lecture 9, Slide 18 Checkpoint
Checkpoint 2c V2 V23 V3 R1 = R2 = R3 = R
Compare the voltage
across R2 with the
voltage across R3
A V2 > V3 B V2 = V3 = V C V2 = V3 < V D V2 < V3 Consider loop V23 V
V23 V2 V3
R2 R3 V2 V3 V
V2 V3 2 Physics 212 Lecture 9, Slide 19 Checkpoint
Checkpoint 2d R1 = R2 = R3 = R Compare the voltage across
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 “Since V2 and V3 are in parallel with V1 and V2 and V3 are
equal, they all have the same V”
“The voltage is the same the beginning and end of the the
parallel split. Since the resistors have the same resistance,
then R1=R2+R3.”
“V1=Vbattery V2+V3=Vbattery”
V1=Vbattery V2+V3=Vbattery
Physics 212 Lecture 9, Slide 20 Checkpoint
Checkpoint 2d
V1 V23 R1 = R2 = R3 = R Compare the voltage across
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 R1 in parallel with series
combination of R2 and R3 V 1 V23
R2 = R3 , I 2 = I 3 V2 = V3 V23 V2 V3 2V2 V1 2V2 V
Physics 212 Lecture 9, Slide 21 R2 R1
V Calculation
Calculation
In the circuit shown: V = 18V,
R1 = 1R2 = 2R3 = 3and R4 = 4 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: “packing”
Evaluate voltages or currents from Ohm’s Law
Expand circuit back using knowledge of voltages and currents: “unpacking” Physics 212 Lecture 9, Slide 22 R2 R1
V Calculation
Calculation
In the circuit shown: V = 18V,
R1 = 1R2 = 2R3 = 3and R4 = 4 R3 What is V2, the voltage across R2? R4 • Combine Resistances:
R1 and R2 are connected: (A)
(A) in series (B) in parallel
ll (C) neither in series nor in parallel
ith Parallel Combination
Ra Series Combination
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
Physics 212 Lecture 9, Slide 23 R2 R1
V Calculation
Calculation
In the circuit shown: V = 18V,
R1 = 1R2 = 2R3 = 3and R4 = 4 R3
R4 What is V2, the voltage across R2? • We first will combine resistances R2 + R3 + R4:
fi
• 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 24 R2 R1
V Calculation
Calculation
In the circuit shown: V = 18V,
R1 = 1R2 = 2R3 = 3and R4 = 4 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) R1 (B) R24 (C)
Physics 212 Lecture 9, Slide 25 Calculation
Calculation R1
V In the circuit shown: V = 18V,
R3 R1 = 1R2 = 2R3 = 3and R4 = 4
R24 • Combine Resistances: What is V2, the voltage across R2?
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 (B) R234 = 2 R2 and R4 in series
(1/Rparallel) = (1/Ra) + (1/Rb) (C) R234 = 4 (D) R234 = 6 R24 = R2 + R4 = 2
1/R234 = (1/3) + (1/6) = (3/6) R234 = 2 Physics 212 Lecture 9, Slide 26 R1
V I1 = I234 Calculation
Calculation
In the circuit shown: V = 18V,
R234 R1 = 1R2 = 2R3 = 3and R4 = 4 R24 = 6R234 = 2
What is V2, the voltage across R2? R1 and R234 are in series. R1234 = 1 + 2 = 3 Our next task is to calculate the total current in the circuit V Ohm’s Law tells us
= I1234 R1234 I1234 = V/R1234
= 18 / 3
= 6 Amps Physics 212 Lecture 9, Slide 27 Calculation:
Calculation: begin unpacking
V In the circuit shown: V = 18V, I1234 R1234 R1 a V R234 = I1 = I234 R1 = 1R2 = 2R3 = 3and R4 = 4 R24 = 6R234 = 2I1234 = 6 A
What is V2, the voltage across R2?
I234 = I1234 Since R1 in series w/ R234
V234 = I234 R234
=6x2 b = 12 Volts • What is Vab, the voltage across R234 ?
(A) Vab = 1 V (B) Vab = 2 V (C) Vab = 9 V (D) Vab = 12 V (E) Vab = 16 V Physics 212 Lecture 9, Slide 28 Calculation
Calculation
R1 R1
V V R234 R3 R24 V = 18V
R1 = 1
R2 = 2
R3 = 3
R4 = 4 R24 = 6
R234 = 2
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 29 Calculation
Calculation
R1
V I1234
R3 R1 V R24 I24 R2 R3
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
R1 = 1
R2 = 2
R3 = 3
R4 = 4 R24 = 6
R234 = 2
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
Be Solved! V2 = I 2 R 2
= 2 x2
= 4 Volts!
Physics 212 Lecture 9, Slide 30 I1
V R1 Quick FollowQuick FollowUps R2 I2 R1 I3
R3 = V R4 (B) I3 = 3 A V3 = V234 = 12V R234
b • What is I3 ?
(A)
(A) I3 = 2 A a (C) I3 = 4 A I3 = V3/R3 = 12V/3 = 4A V = 18V
R1 = 1
R2 = 2
R3 = 3
R4 = 4 R24 = 6
R234 = 2
V234= 12V
V2 = 4V
I24 = 2 Amps
I1234 = 6 Amps • What is I1 ?
We
We know I1 = I1234 = 6 A
NOTE: I2 = V2/R2 = 4/2 = 2 A
4/2 I1 = I2 + I 3 Make Sense??? Physics 212 Lecture 9, Slide 31 ...
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 Fall '08
 Kim
 Physics, Current, Power, Resistor, Electrical resistance

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