# Lecture 5 - PowerPoint 2

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```					    Physics 212
Lecture 10
Today's Concept:
Kirchhoff’s Rules
Circuits with resistors & batteries

Physics 212 Lecture 10, Slide 1
Music
Who is the Artist?

A)   Norah Jones
B)   Diana Krall
C)   Jane Monheit
D)   Nina Simone
E)   Marcia Ball

• New album with Paul McCartney (and Eric Clapton and Stevie Wonder)
• Anyone know the connection between Oscar Peterson and Diana Krall?

• Both great Canadian jazz pianists – Peterson was one of Krall’s
mentors

Physics 212 Lecture 10
“This stuff is loopy”

“The conventions for signs of voltage drops and gains make absolutely no
sense. Why is a voltage drop considered positive and a gain negative?”

“determining which way current flows and
application of Kirchoff’s rules will need a ton of         We’ll start simply, with
practice since this is all new to me.”                    the checkpoints, then to
a calculation
“Discuss multiple currents more. It is pretty
confusing to understand.”

“I don't really understand how Kirchhoff's Rules           This is a great example
apply to the blue wire problem. Could we please go          – we’ll do it at the end
over these problems in lecture? That would be a              to make sure of the
huge help!”                                                        concepts

"Fluke" is not a good brand name for a multimeter. Advertisement: "If the
data is good, it must have been a Fluke."           Physics 212 Lecture 10, Slide      3
Today’s Plan:

• Summary of Kirchoff’s rules – these are the key
concepts

• Example problem

• Review Checkpoints

Physics 212 Lecture 10, Slide 4
Last Time
Resistors in series:
Current through is same.
Voltage drop across is IRi      Reffective  R1  R2  R3  ...

Resistors in parallel:
Voltage drop across is same.             1         1 1   1
    ...
Current through is V/Ri            Reffective    R1 R2 R3

Solved Circuits
R1            R2

V
V                                               I1234
R1234
R3             =
R4
Physics 212 Lecture 10, Slide 5
5
Last Time

Physics 212 Lecture 10, Slide 6
5
New Circuit
R1

R3
V1          V2            R2

How Can We Solve This One?
R1

R3                    V
V1        V2            R2                      I1234
R12
=

THE ANSWER: Kirchhoff’s Rules

Physics 212 Lecture 10, Slide 7
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.

WHY?
The potential difference between a point and itself is zero !

Physics 212 Lecture 10, Slide 8
Kirchoff’s Current Rule

I     in     I out
Kirchoff's Current Rule states that the sum of all
currents entering any given point in a circuit must equal
the sum of all currents leaving the same point.

WHY?
Electric Charge is Conserved

Physics 212 Lecture 10, Slide 9
Checkpoint 1
How many potentially different currents are there in the circuit shown?

A. 3       B. 4       C. 5          D. 6   E. 7

“two for the parallel branch and then one for the first
and last resistors .”

“There are four different series connections”
A
B
C
“there is D
potentially one between every pair of
resistors.”

Physics 212 Lecture 10, Slide 10
Checkpoint 1
How many potentially different currents are there in the circuit shown?

I1      I3

I2

I1      I3

A. 3       B. 4      C. 5       D. 6      E. 7

Look at the nodes!
Top node: I1 flows in, I2 and I3 flow out
A
Bottom
B
C   node: I2 and I3 flow in, I1 flows out
D
That’s all of them!

Physics 212 Lecture 10, Slide 11
Checkpoint 2
In the following circuit, consider the loop abc. The direction of the current
through each resistor is indicated by black arrows.

If we are to write Kirchoff's voltage equation for this loop in the clockwise direction
starting from point a, what is the correct order of voltage gains/drops that we will
encounter for resistors R1, R2 and R3?
A
A. drop, drop, drop               B. gain, gain, gain              C. drop, gain, gain
B gain, drop, drop
D.                                E. drop, drop, gain
C
D
“going with current is drop, against current is gain”
E
“The voltage gains when the current is flowing with the voltage”
“drops 2 times at the split then gains when merging”
Physics 212 Lecture 10, Slide 12
Checkpoint 2
In the following circuit, consider the loop abc. The direction of the current
through each resistor is indicated by black arrows.         DROP

GAIN
If we are to write Kirchoff's voltage equation for this loop in the clockwise direction
starting from point a, what is the correct order of voltage gains/drops that we will
encounter for resistors R1, R2 and R3?
A
A. drop, drop, drop               B. gain, gain, gain              C. drop, gain, gain
B gain, drop, drop
D.                                E. drop, drop, gain
C
D
E
With the current                       VOLTAGE DROP

Against the current                     VOLTAGE GAIN Physics 212        Lecture 10, Slide 13
2V
Calculation
1
In this circuit, assume Vi and Ri are known.
2     1V    I2
What is I2 ??
1      1V

• Conceptual Analysis:
–    Circuit behavior described by Kirchhoff’s Rules:
• KVR: SVdrops = 0
• KCR: SIin = SIout
•Strategic Analysis
–    Write down Loop Equations (KVR)
–    Write down Node Equations (KCR)
–    Solve

Physics 212 Lecture 10, Slide 14
V1
Calculation
R1               I1
-        + +    -
In this circuit, assume Vi and Ri are known.
R2         V2    I2
+         - +    -                              What is I2 ??
R3          V3   I3
-        + +    -

(1) Label and pick directions for each current
(2) Label the + and – side of each element
This is easy for batteries
For resistors, the “upstream” side is +

Now write down loop and node equations

Physics 212 Lecture 10, Slide 15
V1
Calculation
R1               I1
-        + +    -

R2         V2    I2
In this circuit, assume Vi and Ri are known.
+        - +    -

R3          V3   I3                    What is I2 ??
-        + +    -

• How many equations do we need to write down in order to solve for I2?
(A) 1           (B) 2           (C) 3       (D) 4            (E) 5

• Why??
–    We have 3 unknowns: I1, I2, and I3
–    We need 3 independent equations to solve for these unknowns

(3) Choose loops and directions

Physics 212 Lecture 10, Slide 16
V1
Calculation
R1               I1
-        + +    -

R2         V2    I2
+        - +    -               In this circuit, assume Vi and Ri are known.
R3          V3   I3                         What is I2 ??
-        + +    -

• Which of the following equations is NOT correct?
(A)   I2 = I1 + I3                               (4) Write down voltage drops
(B)   - V1 + I1R1 - I3R3 + V3 = 0                (5) Write down node equation
(C)   - V3 + I3R3 + I2R2 + V2 = 0
(D)   - 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 17
V1
Calculation
R1            I1

R2      V2    I2
In this circuit, assume Vi and Ri are known.

R3       V3   I3                          What is I2 ??

• We have the following 4 equations:                 We need 3 equations:
1.   I2 = I1 + I3                                  Which 3 should we use?
2.   - V1 + I1R1 - I3R3 + V3 = 0               A) Any 3 will do
3.   - V3 + I3R3 + I2R2 + V2 = 0               B) 1, 2, and 4
4.   - V2 – I2R2 - I1R1 + V1 = 0               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 18
V1
Calculation
R1         I1
In this circuit, assume Vi and Ri are known.
R2   V2    I2
What is I2 ??

R3    V3   I3     • We have 3 equations and 3 unknowns.
I2 = I 1 + I 3
V1 + I1R1 - I3R3 + V3 = 0
V2 – I2R2 - I1R1 + V1 = 0

R    2V           (6) Solve the equations
I1
•The solution will get very messy!
2R   V     I2          Simplify: assume V2 = V3 = V
V1 = 2V
R1 = R3 = R
R     V    I3
R2 = 2R

Physics 212 Lecture 10, Slide 19
Calculation: Simplify
In this circuit, assume V and R are known.          What is I2 ??

R      2V
I1
• We have 3 equations and 3 unknowns.
I2 = I1 + I3
2R       V      I2             -2V + I1R - I3R + V = 0 (outside)
-V – I2(2R) - I1R + 2V= 0 (top)

R       V      I3

current     • With this simplification, you can verify:
direction          I2 = ( 1/5) V/R
I1 = ( 3/5) V/R
I3 = (-2/5) V/R

Physics 212 Lecture 10, Slide 20
Follow-Up
R       2V
I1
• We know:
2R       V       I2                           I2 = ( 1/5) V/R
a                                                 I1 = ( 3/5) V/R
b
I3 = (-2/5) V/R
R        V      I3

•       Suppose we short R3:            What happens to Vab (voltage across R2?)
R            2V
(A)     Vab remains the same                                                        I1
(B)     Vab changes sign
Why?
(C)     Vab increases                           Redraw:          2R           V    I2
(D)     Vab goes to zero                                                                   d
a        b

Bottom Loop Equation:                                    V   I3
c
Vab + V – V = 0

Vab = 0
Physics 212 Lecture 10, Slide 21
A              B

V                            R              R

Is there a current flowing between A and B ?

A) Yes
B) No

A & B have the same potential              No current flows between A & B
Some current flows down
Current flows from battery and splits at A
Some current flows right
Physics 212 Lecture 10, Slide 22
Checkpoint 3a
Consider the circuit shown below. Note that this question is not identical to the similar looking one
I1           I2
you answered in the prelecture.

Which of the following best describes the current flowing in the blue wire connecting points a and b?
A. Positive current flows from a to b    B. Positive current flows from b to a
C. No current flows between a and b

“Energy flows toward least resistance, which is 1R.”

“because b has higher voltage than a”

“The top half is the same as the bottom half.”
Physics 212 Lecture 10, Slide 23
Checkpoint 3a
Consider the circuit shown below. Note that this question is not identical to the similar looking one
I1           I2
you answered in the prelecture.

I1             I2
I

I3             I4

Which of the following best describes the current flowing in the blue wire connecting points a and b?
current flowing in the blue wire to a
Which of the following best describes theB. Positive current flows from bconnecting points a and b?
A. Positive current flows from a to b
C. No current flows between a and b

I1R – I2 (2R) = 0           I2 = ½ I 1

I4R – I3 (2R) = 0           I4 = 2 I 3

a: I1 = I + I3
I1 - I3 + ½ I1 = 2I3           I1 = 2I3           I = +I3
b: I + I2 = I4                                                       Physics 212 Lecture 10, Slide 24
Prelecture                          Checkpoint

What is the same?   Current flowing in and out of the battery

2R
3

2R
3

What is different?       Current flowing from a to b
Physics 212 Lecture 10, Slide 25
I

1/
3I
2/            R                 2R
3I

V                             a                 b
V/2
R   1/            2R
2/
3I                    3I

2/ I                                                      1/
3                                                            3I
1/
0I
3
1/
3I
2
1/ I                                                      2/
3                                                            3I

Physics 212 Lecture 10, Slide 26
Consider the circuit shown below.
Checkpoint 3b

In which case is the current flowing in the blue wire connecting points a and b the largest?
A. Case A                       B. Case B                      C. They are both the same

“Case A has a lower Req”

“The resistors have a value of 4R in Case B, so the current will be more apt
to bypass this section of the circuit”

“The total resistance in the path taken is the same for both”

Physics 212 Lecture 10, Slide 27
Consider the circuit shown below.
Checkpoint 3b

IA                                            IB

c                                         c

In which case is the current flowing in the blue wire connecting points a and b the largest?
A. Case A                       B. Case B                      C. They are both the same

Current will flow from left to right in both cases

In both cases, Vac = V/2

I2R = 2I4R
IA = IR – I2R               IB = IR – I4R
= IR – 2I4R
Physics 212 Lecture 10, Slide 28
Model for Real Battery: Internal Resistance

+
r
r                                V0
R      VL
V0                                             R            VL

Usually can’t supply too much current to the
load without voltage “sagging”
Physics 212 Lecture 10, Slide 29
Kirchhoff’s Laws
(1) Label all currents
Choose any direction
R1         I1
(2) Label +/- for all elements          A                -
+                                      I6
Current goes +  - (for resistors)                                 +                          +
R2                V3
Battery signs fixed!                     +                         -                          -
B
(3) Choose loops and directions              V
- 1
I2             I3
-
I4
-
+
Must start on wire, not element.                          V2
R3             R4
-
(4) Write down voltage drops                                                   R5
+           +
First sign you hit is sign to use.                                      -             + I
5
(5) Write down node equations
Iin = Iout
(6) Solve set of equations

Physics 212 Lecture 10, Slide 30
17

```
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