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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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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 12 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 13 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 14 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 15 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 16 Bruce Mayer, PE [email protected] • ENGR-43_Lec-02-2b_KVL.ppt