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Engineering 43
Chp 2.2
Kirchoff’s Voltage
Law
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Energy Conservation
 One Of The Fundamental Conservation
Laws In Electrical Engineering
THE CONSERVATION OF
ENERGY PRINCIPLE:
“ENERGY CANNOT BE
CREATED NOR DESTROYED”
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Kirchoff’s Voltage Law (KVL)
 KVL Is A Conservation Of Energy
Principle
• A Positive Charge Gains Energy As It
Moves To A Point With Higher Voltage And
Releases Energy If It Moves To A Point
With Lower Voltage
W  q (VB  VA )
B
VB
q
q
q

VA
Engineering-43: Engineering Circuit Analysis
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 Vab 

a
b
 Vcd 

c
d
LOSES W  qVab
GAINS W  qVcd
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
"Gedanken" Experiment

AB
V
C
V B

 If The Charge Comes
Back To The Same
Initial Point, Then The
Net Energy Gain Must
Be Zero (Conservative
Network)
B VB
q

VA  VCA 
• Otherwise, With Repeated Loops The Charge
Could End Up With Infinite Energy, Or Supply An
Infinite Amount Of Energy. Mathematically
q(VAB  VBC  VCA )  0
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
VC
Kirchoff’s Voltage Law
THE ALGEBRAIC SUM OF VOLTAGE
DROPS AROUND ANY CLOSED LOOP
MUST BE ZERO
 Note That a Voltage RISE is Equivalent
to a NEGATIVE DROP

V 
A
Engineering-43: Engineering Circuit Analysis
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B
( V ) 
A
B
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Examples
Find VR3 Given
• VR1 = 18V
• VR2 = 12V
A
VR  18V
Start & End @ A
1
VR  12V
2
 Vs  VR1  VR2  VR3  0
VR3  20V
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
KVL Problem Solving
 KVL Is Useful To Determine Voltages
• Find A Loop Including The Unknown Voltage
– The Loop Does NOT Have To Be Physical
 Example: VR1, VR3
Are Known
 VR2, is UNknown
 Determine Vbe

Vbe

 Using the “Virtual”
Loop Shown
VR  Vbe  VR  30[V ]  0
1
Engineering-43: Engineering Circuit Analysis
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3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Linear Dependence
 Background: For KCL We Saw That Not All
Possible KCL Equations Are Independent
• The Same Situation Arises When Using KVL
• The Third Equation
Is The Sum Of The
Other Two
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Dependence Quantified
 For a Given Circuit Define
• N  Number of Nodes
• B  Number of Braches
 Then
• N-1  No. of Linearly Independent KCL Eqns
• B-(N-1)  No. of Linearly Independent KVL
Eqns
– Example: For Previous Circuit We Have
N = 6, B = 7.
 Hence There Are Only Two Independent KVL Equations
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Choose Loop Wisely
 Find Vae, Vec
• Try to Find The
Simplest Loop
1
2
3
– Make Use of
Virtual Loops
1

2
3
Engineering-43: Engineering Circuit Analysis
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
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Dependent Sources
 Dependent
Sources are Just
Another Voltage
Drop
1
2
1
2
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Examples
Vac  4V  6V  0
1
1
Vac  10V
Vbd  2V  4V  0
2
2
Vbd  6V
Vad  12V  8V  6V  0
1
Vad  26V
Veb  4V  6V  12V  0
1
2
2
Veb  10V
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
More Examples
 Find Vbd
• First Find VR1
 12  VR1  1  10VR1  0  VR1  1V
Vbd  VR2  10VR1  11V
 Remember: ALWAYS Use The Passive
Sign Convention With Resistors

V

Engineering-43: Engineering Circuit Analysis
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 V 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Virtual Loop Example
 4V 
 Find

• Vx, Vab
b  Vx 
2
R  2k
V1
• Power Dissipated

in the 2 kΩ Resistor
+
-
+
-
1
V1  12V , V2  4V a
 Need To Find A Closed Path With
Only One Unknown Potential
VX  V2  Vab  0
VX  V2  V1  4V  0
VX  4V  12V  4V  0
Vab  VX  V2
1
V22k 4V 


 8mW
R
2k
2
2
P2 k
Engineering-43: Engineering Circuit Analysis
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
Vx 
Vab
Pow
the
P2k
Vab  4V  4V  8V
VX  4V
V2
DET
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Combo Example  Find V1
 NO Loops
With Only
1 Unknown
5k
10k

25V
• But This
is a SERIES
Circuit
 Vx 
+
-

1
-
Vx/2 +
+
-
V1

2
– The Current thru the 10k & 5k Resistors is the SAME
→ V Across 5k is HALF that Across The 10k
VX VX
 25[V ]  VX 

0
2
4
1
VX  20[V ]
Engineering-43: Engineering Circuit Analysis
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VX VX
V1 

0
4
2
2
VX
V1  
 5[V ]
4
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt
Vx
4
WhiteBoard Work
 Let’s Work Problem
This Nice Little Problem
30 k
2 Vx
12 V
10 k
+
Vx
_
 Find Vx & P30kΩ
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt