[email protected] • ENGR-43_Lec-02-2b_KVL.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
Chp 2.2Chp 2.2Kirchoff’s Kirchoff’s VoltageVoltage
LawLaw
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Energy ConservationEnergy Conservation One Of The Fundamental Conservation
Laws In Electrical Engineering
THE CONSERVATION OF THE CONSERVATION OF ENERGY PRINCIPLE:ENERGY PRINCIPLE:
““ENERGY CANNOT BE ENERGY CANNOT BE CREATED NOR CREATED NOR DESTROYED”DESTROYED”
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Kirchoff’s Voltage Law (KVL)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
AV
BBV)( AB VVqW
q
abVa b
q
abqVW LOSES
cdVc d
q
cdqVW GAINS
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
"Gedanken" Experiment "Gedanken" Experiment If The Charge Comes
Back To The SameInitial Point, Then TheNet Energy Gain MustBe Zero (ConservativeNetwork)• Otherwise, With Repeated Loops The Charge
Could End Up With Infinite Energy, Or Supply An Infinite Amount Of Energy. Mathematically
AV
BBV
q
CV
ABV
BCV
CAV
0)( CABCAB VVVq
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Kirchoff’s Voltage LawKirchoff’s Voltage Law
THE ALGEBRAIC SUM OF THE ALGEBRAIC SUM OF VOLTAGEVOLTAGE
DROPSDROPS AROUND ANY CLOSED AROUND ANY CLOSED LOOP MUST BE ZEROLOOP MUST BE ZERO
A B V
A B )( V
Note That a Voltage RISE is Equivalent to a NEGATIVE DROP
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExamplesExamples
0321 RRRs VVVV
VVR 181
VVR 122
VVR 203
Find VR3 Given• VR1 = 18V
• VR2 = 12V
A
Start & End @ A
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
KVL Problem SolvingKVL 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 KnownVR2, is UNknownDetermine Vbe
beV
0][3031
VVVV RbeR
Using the “Virtual”Loop Shown
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Linear DependenceLinear 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
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Dependence QuantifiedDependence 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
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Choose Loop WiselyChoose Loop Wisely Find Vae, Vec
• Try to Find TheSimplest Loop– Make Use of
Virtual Loops
1
2
3
12
3
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Dependent SourcesDependent Sources Dependent
Sources are JustAnother VoltageDrop
1
2
1 2
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExamplesExamples
VVVVV
ac
ac
10064
VVVVV
bd
bd
6042
1
2
1
2
VVVVVV
ad
ad
2606812
VVVVVV
eb
eb
1001264
1 2
1
2
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
More ExamplesMore Examples Find Vbd
• First Find VR1
VVVV
VVVV
RRbd
RRR
1110
1010112
12
111
V V
Remember: ALWAYS Use The Passive Sign Convention With Resistors
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Virtual Loop ExampleVirtual Loop Example Find
• Vx, Vab
• Power Dissipatedin the 2 kΩ Resistor
Need To Find A Closed Path With Only One Unknown Potential
VVVVVV
VVVV
X
X
X
404124
0412
VVVVVVV
VVV
ab
Xab
abX
844
0
2
2
1 2
1
2
mWkV
RVP k
k 824 22
22
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Combo Example Combo Example Find Find VV11
NO LoopsWith Only1 Unknown• But This
is a SERIESCircuit– The Current thru the 10k & 5k Resistors is the SAME
→ V Across 5k is HALF that Across The 10k
+-
+-
k10 k5
xV
V25 4
xV
1V- Vx/2 +
][20
042
][25
VV
VVVV
X
XXX
][54
024
1
1
VVV
VVV
X
XX
1 2
1
2
[email protected] • ENGR-43_Lec-02-2b_KVL.ppt16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard WorkWhiteBoard Work Let’s Work Problem
This Nice Little Problem
3 0 k
1 2 V2 V x
1 0 k
+
V x_
Find Vx & P30kΩ