Chp 2.[5-7] Series-Parallel Resistors Engineering 43 Chp 2.[5-7] Series-Parallel Resistors Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Series Parallel Up To Now We Have Studied Circuits That Can Be Analyzed With One Application Of KVL (Single Loop) Or KCL (Single Node-pair) We Have Also Seen That In Some Situations It Is Advantageous To Combine Resistors To Simplify The Analysis Of A Circuit Now We Examine Some More Complex Circuits Where We Can Simplify The Analysis Using Techniques: Combining Resistors Ohm’s Law

Resistor Equivalents Series Parallel Resistors Are In Series If They Carry Exactly The Same Current Parallel Resistors Are In Parallel If They have Exactly the Same Potential Across Them

Combine Resistors Example: Find RAB SERIES 6k||3k = 2k (10K,2K)SERIES (4K,2K)SERIES (3K,9K)SERIES

More Examples Step-1: Series Reduction Step-2: Parallel Reduction

Example w/o Redrawing Step-1: 4k↔8k = 12k Step-2: 12k12k = 6k Step-4: 6k(4k↔2) = 3k = RAB

Inverse Series Parallel Combo Find R: Simple Case Constraints VR = 600 mV I = 3A Only 0.1Ω R’s Available Recall R = V/I Since R>Ravail, Then Need to Run in Series

More Complex Case Find R for Constraints VR = 600 mV I = 9A Only 0.1Ω Resistors Available Either of These 0.1Ω R-Networks Will Work 33.33 mΩ 0.1║(0.1↔0.1)

Effect Of Resistor Tolerance The R Spec: 2.7k, ±10% What Are the Ranges for Current & Power? I = V/R P = V2/R For Both I & P the Tolerance.: -9.1%, +11.1% Asymmetry Due to Inverse Dependence on R

Series-Parallel Resistor Circuits Combing Components Can Reduce The Complexity Of A Circuit And Render It Suitable For Analysis Using The Basic Tools Developed So Far Combining Resistors In SERIES Eliminates One NODE From The Circuit Combining Resistors In PARALLEL Eliminates One LOOP From The Circuit

S-P Circuit Analysis Strategy Reduce Complexity Until The Circuit Becomes Simple Enough To Analyze Use Data From Simplified Circuit To Compute Desired Variables In Original Circuit Hence Must Keep Track Of Any Relationship Between Variables

Example – Ladder Network Find All I’s & V’s in Ladder Network 1st: S-P Reduction 2nd: S-P Reduction Also by Ohm’s Law

Ladder Network cont. Final Reduction; Find Calculation Starting Points Now “Back Substitute” Using KVL, KCL, and Ohm’s Law e.g.; From Before

More Examples Voltage Divider Current Divider

“BackSubbing” Example Given I4 = 0.5 mA, Find VO

BackSubbing Strategy Always ask: What More Can I Calculate? In the Previous Example Using Ohm’s Law, KVL, KCL, S-P Combinations - Calc:

Final Example Find VS Straight-Forward VB Find VS Straight-Forward IB IS Or, Recognize As Inverse V-Divider

Final Example cont VB Inverse Divider Calculation IS IB

Wye↔ Transformations This Circuit Has No Series or Parallel Resistors If We Could Make The Change Below Would Have Series- Parallel Case

Y↔ Xforms cont Then the Circuit Would Appear as Below and We Could Apply the Previous Techniques

Wye↔ Transform Eqns →Y (pg. 51) ←Y (pg. 51) →Y ←Y Rac =R1||(R2↔R3) Rab = Ra + Rb ←Y →Y (pg. 51) ←Y (pg. 51)

↔Y Application Example  Connection Find IS Use the →Y Eqns to Arrive at The Reduced Diagram Below Calc IS Req

Another ↔Y Example For this Ckt Find Vo Keep This Node-Pair Convert this Y to Delta

Another ↔Y Example cont Notice for Y→Δ in this Case Ra = Rb = Rc = 12 kΩ Only need to Calc ONE Conversion The Xformed Ckt ||-R’s Form a Current Divider

Another ↔Y Example cont The Ckt After ||-Reductions Can Easily Calc the Current That Produces Vo Then Finally Vo by Ohm’s Law

WhiteBoard Work Done Previously Let’s Use KCL to Derive the Req for N Parallel Resistors Done Previously