Unformatted text preview: Physics 212
Lecture 10
Today's Concept:
Kirchhoff’s Rules
Circuits with resistors & batteries
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0 Confused Avg = 3.3 Confident Physics 212 Lecture 10, Slide 1 Music 1
Who is the Artist?
A)
B)
C)
D)
E) Marvin Gaye
Aaron Neville
Michael Bolton
Al Green
Ray Charles 1.
2.
2.
3.
3.
4.
5.
5.
6.
7.
7.
8.
8.
9.
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10.
10. Pretty Woman
Unchained Melody
Funny How Time Slips Away
Funny
Ain't No Mountain High Enough
How Can you Mend A Broken Heart
Take Me To The River
I’m So Lonesome I Could Cry
I Can't Get Next To You
My Girl
Light My Fire Something upbeat for
Something
the day after the exam
the Different..
Different..
Mostly covers..
Very good…
Very Roy Orbison
Roy
Righteous Brothers
Righteous
Willie Nelson
Willie
Marvin Gaye
Marvin
Bee Gees
Al Green
Al
Hank Williams
Hank
Temptations
Temptations
Temptations
Temptations
Doors
Physics 212 Lecture 10, Slide 2 Physics 212
Lecture 10
Today's Concept:
Kirchhoff’s Rules
Circuits with resistors & batteries
35
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25
20
15
10
5
0 Confused Avg = 3.3 Confident Physics 212 Lecture 10, Slide 3 Key Concepts:
1) Understanding Kirchoff’s Rules
Understanding Kirchoff
2) Using Kirchoff’s Rules
Using Kirchoff Today’s Plan:
• Summary of Kirchoff’s rules
Summary
Kirchoff • Example problem • Review Preflights
Review Preflights Physics 212 Lecture 10, Slide 4 Last Time
Resistors in series:
Current through is same.
Voltage drop across is IRi Reffective = R1 + R 2 + R3 + ... Resistors in parallel:
Voltage drop across is same. 1 Current through is V/Ri Reffective 1
1
1
=+
+ + ...
R1 R2 R3 Solved Circuits
R2 R1
V = R3 V I1234 R1234 R4
Physics 212 Lecture 10, Slide 5
5 New Circuit
R1
R3
V1 V2 R2 How Can We Solve This One?
R1
R3
V1 V2 R2 = V I1234 R12 THE ANSWER: Kirchhoff’s Rules
Physics 212 Lecture 10, Slide 6
5 Kirchoff’s Voltage Rule ∑ ∆V i =0 Kirchoff's Voltage Rule states that the sum of the voltage
changes caused by any elements (like wires, batteries, and
resistors) around a circuit must be zero.
If we model voltage as height above the ground floor, see if you can come up
with the analogy to Kirchoff's Voltage Rule in terms of someone walking around
in the hallways and stairways and elevators of a highrise building.
If we start at the ground floor with a potential of 0, and walk around up and down
If
stairs, take the elevator a few flights, go to the roof, parachute off, and end up
stairs,
te
back on the ground floor, our potential is still 0. Therefore, the potential
back
he
difference is 0. We may have increased and decreased our potential as we traveled
difference
al
through the building, but we still start and end at a potential of 0. OR:
OR:
The potential difference between a point and itself is zero !
The
Physics 212 Lecture 10, Slide 7 Kirchoff’s Current Rule ∑I in = ∑ I out Kirchoff's Current Rule states that the sum of all
Current
currents entering any given point in a circuit must equal
the sum of all currents leaving the same point.
If we model electrical current as water, see if you can come up with an
analogy to Kirchoff's Current Rule in terms of household plumbing.
If you have a main water line coming into your house, it will split off to
lit
service all utilities, such as sink, toilet, shower, etc. The water in all of those
service
ter
lines must equal the amount of water coming out of the main line, and when
lines
and
all the household water drains out of the house into a main line again, all the
all
again,
smaller lines must include as much water as the drain carries out.
smaller
t. OR:
OR:
Electric Charge is Conserved
Electric
Physics 212 Lecture 10, Slide 8 Kirchhoff’s Laws
(1) Label all currents
Choose any direction (2) Label +/ for all elements R1 A + I1 + Current goes + ⇒  (for resistors) (3) Choose loop and direction
Must start on wire, not element. 3 + B
1  First sign you hit is sign to use. I2 I3  +
2 (4) Write down voltage drops + R2 R3 R5  + I4  R4 + + I5 (5) Write down node equation
Iin = Iout We’ll do calculation first today
It’s actually the easiest thing to do !! Physics 212 Lecture 10, Slide 9
17 Preflight 3
DROP GAIN BB N
AI
G A
B
C
D
E 50
40
30
20 With the current
Against the current VOLTAGE DROP 10
0 VOLTAGE GAIN Physics 212 Lecture 10, Slide 10
10 1Ω
2Ω
1Ω Calculation 2V 1V I2 In this circuit, assume Vi and Ri are known.
What is I2 ?? 1V • Conceptual Analysis:
– Circuit behavior described by Kirchhoff’s Rules:
• KVR: ΣVdrops = 0
• KCR: ΣIin = ΣIout •Strategic Analysis
–
–
– Write down Loop Equations (KVR)
Write down Node Equations (KCR)
Solve Physics 212 Lecture 10, Slide 11
11 V1 R1
++  + + R3
 I2 In this circuit, assume Vi and Ri are known.  V3
++ I1  V2 R2 Calculation What is I2 ??
I3  Label and pick directions for each current
Label the + and – side of each element
Label
This is easy for batteries
For resistors, the “upstream” side is +
For Now write down loop and node equations Physics 212 Lecture 10, Slide 12
12 V1 R1
++  + + R3
 I2  V3
++ I1  V2 R2 Calculation
BB In this circuit, assume Vi and Ri are known.
What is I2 ?? I3  • How many equations do we need to write down in order to solve for I2?
(A) 1 (B) 2
(B) (C) 3
(C) (D) 4 (E) 5 • Why??
–
– We have 3 unknowns: I1, I2, and I3
We need 3 independent equations to solve for these unknowns Physics 212 Lecture 10, Slide 13
13 V1 R1
++  + + R3
 I2  V3
++ I1  V2 R2 Calculation I3  BB In this circuit, assume Vi and Ri are known.
What is I2 ?? • Which of the following equations is NOT correct?
(A)
(B)
(C)
(D) I2 = I1 + I3
 V1 + I1R1  I3R3 + V3 = 0
 V3 + I3R3 + I2R2 + V2 = 0
 V2 – I2R2 + I1R1 + V1 = 0 • Why??
–
– (D) is an attempt to write down KVR for the top loop
Start at negative terminal of V2 and go clockwise
• Vgain (V2) then Vgain (I2R2) then Vgain (I1R1) then Vdrop (+V1)
Physics 212 Lecture 10, Slide 14
14 R1 V1 R2 V2 R3 V3 Calculation
I1
I2 BB In this circuit, assume Vi and Ri are known. I3 • We have the following 4 equations:
1.
2.
2.
3.
3.
4. I2 = I1 + I3
 V1 + I1R1  I3R3 + V3 = 0
 V3 + I3R3 + I2R2 + V2 = 0
 V2 – I2R2  I1R1 + V1 = 0 What is I2 ??
We need 3 equations:
Which 3 should we use?
A) Any 3 will do
B) 1, 2, and 4
C) 2, 3, and 4 • Why??
–
–
– We need 3 INDEPENDENT equations
Equations 2, 3, and 4 are NOT INDEPENDENT
• Eqn 2 + Eqn 3 =  Eqn 4
We must choose Equation 1 and any two of the remaining ( 2, 3, and 4)
Physics 212 Lecture 10, Slide 15
15 R1
R2
R3 V1
V2
V3 Calculation
I1
I2
I3 In this circuit, assume Vi and Ri are known.
What is I2 ??
• We have 3 equations and 3 unknowns.
I2 = I1 + I3
V1 + I1R1  I3R3 + V3 = 0
V2 – I2R2  I1R1 + V1 = 0 R
2R
R 2V V
V I1 •The solution will get very messy!
I2
I3 Simplify: assume V2 = V3 = V
V1 = 2V
R1 = R3 = R
R2 = 2R Physics 212 Lecture 10, Slide 16
16 Calculation: Simplify
In this circuit, assume V and R are known.
R
2R
R 2V V
V I1
I2 What is I2 ?? • We have 3 equations and 3 unknowns.
I2 = I1 + I3
2V + I1R  I3R + V = 0 (outside)
V – I2(2R)  I1R + 2V= 0 (top) I3 • With this simplification, you can verify:
I2 = ( 1/5) V/R
I1 = ( 3/5) V/R
I3 = (2/5) V/R Physics 212 Lecture 10, Slide 17
17 FollowUp
2V R I1 • We know:
V 2R
a • I2 = ( 1/5) V/R
I1 = ( 3/5) V/R
I3 = (2/5) V/R I2 b R V I3 Suppose we short R3: (A) Vab remains the same
remains
(B) Vab changes sign
(B)
changes
(C) Vab increases
(C)
(D) Vab goes to zero BB What happens to Vab (voltage across R2?)
R
2R Why?
Redraw:
a 2V V I1
I2 b V d I3 c Vcd = +V
Vbd = +V
Vad = Vcd = +V Vab = Vad – Vbd = V – V = 0
Physics 212 Lecture 10, Slide 18
18 a b
BB V R R Is there a current flowing between a and b ?
Is
and
A) Yes
B) No
A & B have the same potential
Current flows from battery and splits at A No current flows between A & B
Some current flows down
Some current flows right
Physics 212 Lecture 10, Slide 19
19 Preflight 5
I1 I2
BB I
I3 I4 60 I1R – I2 (2R) = 0 I1 = 2 I2 50
40 I4R – I3 (2R) = 0 I4 = 2 I3 I2 = I3 I1 = 2 I3 I = I1  I3 = 2I3 –I3 = +I3 30
20
10
0 Physics 212 Lecture 10, Slide 20
20 Prelecture What iis the same?
s Preflight Current flowing in and out of the battery
Current
2R
3
2R
3 What is different? Current flowing from a to b
Physics 212 Lecture 10, Slide 21
21 I
2/ R 3I 1/ 3I a V
2/ 2/ I
3 3I b
R 1/ 3I V/2
2R 1/
1/ 0I
3 2
1/ I
3 2R 1/ 3I 2/ 3I 3I Physics 212 Lecture 10, Slide 22
22 Preflight 7
IA IB c c BB 60 Current will flow from left to right in both cases 50
40 In both cases, Vac = V/2
In
V/2 30
20
10 I(2R) = 2I(4R)
IA = I(R) – I(2R)
I(R)
= I(R) – 2I(4R)
I(R) 2I(4R) 0 IB = I(R) – I(4R)
I(R)
Physics 212 Lecture 10, Slide 23
23 Model for Real Battery: Internal Resistance +
r
V0 r
R VL V0
R VL Usually can’t supply too much current to the
supply
load without voltage “sagging”
Physics 212 Lecture 10, Slide 24
24 ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 212 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.
 Fall '10
 Selig
 Physics

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