1. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Chp 2.2 Kirchoff’s Voltage Law

2. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Energy Conservation  One Of The Fundamental Conservation Laws In Electrical Engineering THE CONSERVATION OF ENERGY PRINCIPLE: “ENERGY CANNOT BE CREATED NOR DESTROYED”

3. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

4. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis "Gedanken" Experiment  If The Charge Comes Back To The Same Initial Point, Then The Net Energy Gain Must Be Zero (Conservative Network) Otherwise, With Repeated Loops The Charge Could End Up With Infinite Energy, Or Supply An Infinite Amount Of Energy. Mathematically

5. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Kirchoff’s Voltage Law THE ALGEBRAIC SUM OF VOLTAGE DROPS AROUND ANY CLOSED LOOP MUST BE ZERO AB  V  AB  )(V  Note That a Voltage RISE is Equivalent to a NEGATIVE DROP

6. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Examples Find V R3 Given V R1 = 18V V R2 = 12V A Start & End @ A

7. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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: V R1, V R3 Are Known  V R2, is UNknown  Determine V be  Using the “Virtual” Loop Shown

8. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

9. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

10. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Choose Loop Wisely  Find V ae, V ec Try to Find The Simplest Loop –Make Use of Virtual Loops 1 2 3 1 2 3  

11. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Dependent Sources  Dependent Sources are Just Another Voltage Drop 1 2 12

12. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Examples 1 2 1 2 12 1 2

13. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis More Examples  Find V bd First Find V R1  Remember: ALWAYS Use The Passive Sign Convention With Resistors

14. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Virtual Loop Example  Find V x, V ab Power Dissipated in the 2 kΩ Resistor  Need To Find A Closed Path With Only One Unknown Potential 1 2 1 2

15. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Combo Example  Find V 1  NO Loops With Only 1 Unknown But This is a SERIES Circuit –The Current thru the 10k & 5k Resistors is the SAME → V Across 5k is HALF that Across The 10k - Vx/2 + 1 2 1 2

16. BMayer@ChabotCollege.edu ENGR-43_Lec-02-2b_KVL.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work Problem This Nice Little Problem  Find V x & P 30kΩ