ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 43 Chp 2.3 Single-Loop Circuits

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Single Loop Ckts - Background  Using KVL And KCL We Can Write Enough Equations To Analyze ANY Linear Circuit  Begin The Study Of Systematic And Efficient Ways Of Using The Fundamental KCL & KVL Circuit Laws This Time → Single LOOP Circuits

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Power of Loop Analysis  Consider this Circuit  Alternative Analyses Write N-1 (5) KCL Equations Determine Only the SINGLE Loop Current –An Easy Choice  The Plan for Loop Analysis Begin With The Simplest One-Loop Circuit Extend Results To Multiple-Source and Multiple-Resistor Circuits

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Voltage Divider KVL ON THIS LOOP  Ohm’s Law in KVL  Find i(t)  Ohm’s Law

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Voltage Divider cont.  Now Sub i(t) Into Ohm’s Law to Arrive at The Voltage Divider Eqns KVL ON THIS LOOP    Quick Chk → In Turn, Set R 1, R 2 to 0

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis V-Divider Summary  Governing Equations The Larger the R, The Larger the V-drop  Example Gain/Volume Control –R 1 is a Variable Resistor Called A Potentiometer, or “Pot” for Short

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Volume Control Example Case-I → R 1 = 90 kΩ 30kΩ 9 V Case-II → R 1 = 20 kΩ

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Practical Example  Power Line 8.25% of Pwr Generated is Lost to Line Resistance! * How to Reduce Losses?  Power Dissipated by the Line is a LOSS  Using Voltage Divider

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Equivalent Circuit  The Equivalent Circuit Concept Can Simplify The Analysis Of Circuits For Example, Consider A Simple Voltage Divider SERIES Resistors → 1 R 2 R  21 RR  –As Far As The Current Is Concerned Both Circuits Are Equivalent  The One On The Right Has Only One Resistor

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Schematic vs. Physical  Sometimes, For Practical Construction Reasons, Components That Are Electrically Connected May Be Physically Quite Apart Each Resistor Pair Below Has the SAME Node-to-Node Series-Equivalent Circuit

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis COMPONENT SIDE CONNECTOR SIDE ILLUSTRATING THE DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS PHYSICAL NODE SECTION OF 14.4 KB VOICE/DATA MODEM CORRESPONDING POINTS

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Generalization – Multiple Sources  Voltage Sources In Series Can Be Algebraically Added To Form An Equivalent Source We Select The Reference Direction To Move Along The Path i(t) –Voltage Rises Are Subtracted From Drops  Apply KVL

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Multiple-Source Equivalent  Collect All SOURCES On One Side  The Equivalent Circuit

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Generalization – Mult. Resistors  Apply KVL (rise = Σdrops)  Now by Ohm’s Law  And Define R S  Then Voltage Division For Multiple Resistors

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Find: I, V bd, P(30kΩ)  Apply KVL & Ohm APPLY KVL TO THIS LOOP  Solving for I  Now V bd  Finally, The 30 kΩ Resistor Power Dissipation

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis “Inverse” Voltage Divider  The Std V-Divider 1 R R S V   O V Inverse Divider Use Inverse Divider  Example Find V S

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Student Exercise  Lets Turn on the Lights for 5-7 min  Students are invited to Analyze the following Ckt  Determine the CURRENT, I V bd + -

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Examples  Find: I, V bd from Exercise Use KVL and Ohm’s Law APPLY KVL TO THIS LOOP  Find V S V 20k Divider The V S INVERSE Divider

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis When In Doubt → ReDraw  From The Last Diagram It Was Not Immediately Obvious That This Was a V-Divider Situation UnTangle/Redraw at Right

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Find V x, V ab P 3Vx –The Power Absorbed or Supplied By the Dependent Source  Apply KVL + - x V 3  V 4 k 4   X V a b S V VV S 12   Find DS Power Passive Sign Conven.  Ohm’s Law Then Pwr ABSORBED

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Find: V DA, V CD, I Apply KVL & Ohm k V 9 k 20 k 10 A B C D E V Ohm’s Law

ENGR-43_Lec-02-4_Single_Loop_Ckts.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work This Problem