PHYSICS 222 Introduction to Classical Physics II
Prof. John Lajoie
Iowa State University
Spring 2016
LECTURE 14
Kirchhoff’s rules
The Water Circuit Analogy
Current
(water flow)
Battery  fixed potential
increase (water pump)
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
Resistors  potential
drops (ramps with
obstacles)
Ideal wires  constant
potential (flat sections)
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Changes of a Potential Around the Circuit
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchhoff’s Rules ‐ Junctions
o The algebraic sum of the currents into any junction is zero.
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchhoff’s Rules ‐ Junctions
o The algebraic sum of the currents into any junction is zero.
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchhoff’s Rules II—Loops
o The algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements, must equal zero.
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchhoff’s Rules ‐ Examples
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchhoff’s Rules ‐ Examples
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchoff’s Rules ‐ Examples
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
9/18/2016
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Kirchoff’s Rules ‐ Examples
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Kirchoff’s Rules ‐ Examples
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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I1
R
I2

I5
R
R
R I4
R
R
I3
I6
Use Kirchoff’s rules to determine which of the following statements about this circuit is FALSE. 1.
I1 = I2 + I 3
2.
I4+I5+I6 =0
3.
RI2 + RI4 = RI3
4.
ε = RI1 + RI2+ RI5
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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I1
R

I2
I5
R
R
R I4
R
R
I3
I6
Use Kirchoff’s rules to determine which of the following statements about this circuit is FALSE. 1. I1 = I2 + I3 Junction 1
2. I4+I5+I6 =0 I4-I5+I6 =0
The direction of the arrows matters!
3. RI2 + RI4 = RI3 Loop 2
4. ε = RI1 + RI2+ RI5 Loop 1
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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EXAMPLE: Two‐loop circuit
Determine the currents through the elements of
this circuit.
ε2
R1
ε1
R2
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PHYS222
- Lecture 14 - Prof. John Lajoie - Iowa State University
R3
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Step 1: How many distinct currents are there?
ε2
R1
ε1
R2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
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Step 1: How many distinct currents are there? 3
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 1: How many distinct currents are there? 3
Draw them!
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 1: How many distinct currents are there? 3
We will need 3
equations
Draw them!
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 2: How many junctions?
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 2: How many junctions?
N = 2
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 2: How many junctions?
Write N - 1
junction equations
N = 2
I2 + I3 = I1
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 3: Draw as many loops as you need to complete the
number of required equations (found in step 1)
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 3: Draw as many loops as you need to complete the
number of required equations (found in step 1)
Many possible loops. In this case we only need
2 of them.
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 4: Write the loop equations for each chosen loop
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 4: Write the loop equations for each chosen loop
ε1
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 4: Write the loop equations for each chosen loop
ε1 – I1 R1
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 4: Write the loop equations for each chosen loop
ε1 – I1 R1 – I2 R2
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Step 4: Write the loop equations for each chosen loop
ε1 – I1 R1 – I2 R2 = 0
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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ε2
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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ε2 – I2 R2
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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ε2 – I2 R2 + I3 R3
Path and current in
opposite direction
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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ε2 – I2 R2 + I3 R3 = 0
Path and current in
opposite direction
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε1
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε1 – I1 R1
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε1 – I1 R1 - ε2
Path is “against” the battery (moving from – to + ends)
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε1 – I1 R1 - ε2 – I3 R3
Path is “against” the battery (moving from – to + ends)
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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We only need two of equations, but let us do a third one just for fun…
ε1 – I1 R1 - ε2 – I3 R3 = 0
Path is “against” the battery (moving from – to + ends)
ε2
R1
ε1
R2
I1
I2
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
R3
I3
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Solve the system: 3 eqns, 3 unknowns
I 2 + I3 = I1
ε1 – I1 R1 – I2 R2 = 0
ε2 – I2 R2 + I3 R3 = 0
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Solve the system: 3 eqns, 3 unknowns
I 2 + I3 = I1
ε1 – I1 R1 – I2 R2 = 0
ε2 – I2 R2 + I3 R3 = 0
ε1 = 12 V
ε2 = 24 V
For
R1 = 5 Ω
R2 = 3 Ω
R3 = 4 Ω
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
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Solve the system: 3 eqns, 3 unknowns
I 2 + I3 = I1
ε1 – I1 R1 – I2 R2 = 0
ε2 – I2 R2 + I3 R3 = 0
ε1 = 12 V
For
ε2 = 24 V
I1 = 0.25 A
R1 = 5 Ω
I2 = 3.6 A
R2 = 3 Ω
I3 = –3.3 A
R3 = 4 Ω
PHYS222 - Lecture 14 - Prof. John Lajoie - Iowa State University
I3 flows
opposite to our
assumption
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