Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1. Engineering 43 Series/Parallel,Dividers,Nodes & Meshes Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Up To Now We Have Studied Circuits That Can Be Analyzed With One Application Of KVL Or KCL We will see 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 Series Parallel

3. Series Resistors Are In Series If TheyCarry Exactly The Same Current Resistor Equivalents • Parallel • Resistors Are In Parallel IfThey have Exactly the Same Potential Across Them

4. Conductance Equivalents • ReCall: G = 1/R • For SERIES Connection • For PARALLELConnection GS = 1.479 S GP = 15 S

5. SERIES Combine ResistorsExample:Find RAB 6k||3k = 2k (10K,2K)SERIES (4K,2K)SERIES (3K,9K)SERIES

6. More Examples • Step-1: Series Reduction • Step-2: Parallel Reduction 9 kΩ

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

8. 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

9. 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

10. 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

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

12. The Voltage Divider • Ohm’s Law KVL ON THIS LOOP • Ohm’s Law in KVL • Find i(t) by 

13. Voltage Divider cont. • Now Sub i(t) Into Ohm’sLaw to Arrive at TheVoltage Divider Eqns KVL ON THIS LOOP • Quick Chk → In Turn, Set R1, R2 to 0  

14. V-Divider Summary • Governing Equations • The Larger the R, The Larger the V-drop • Example • Gain/Volume Control • R1 is a VariableResistor Called aPotentiometer, or “Pot” for Short

15. Volume Control Example • Case-I → R1 = 90 kΩ 9 V • Case-II → R1 = 20 kΩ 30kΩ

16. Practical Example  Power Line • Using Voltage Divider Also • Power Dissipated by the Line is a LOSS 8.25% of Pwr Generated is Lost to Line Resistance!* How to Reduce Losses?

17. º + R R R R 1 2 1 2 Equivalent Circuit • The Equivalent Circuit Concept Can Simplify The Analysis Of Circuits • For Example, Consider A Simple Voltage Divider • As Far As TheCurrent IsConcerned BothCircuits AreEquivalent • The One On The Right Has Only One Resistor SERIES Resistors →

18. 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 Schematic vs. Physical

19. PHYSICAL NODE PHYSICAL NODE CONNECTOR SIDE ILLUSTRATING THE DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS SECTION OF 14.4 KB VOICE/DATA MODEM CORRESPONDING POINTS COMPONENT SIDE

20. Voltage Sources In SeriesCan Be AlgebraicallyAdded To Form AnEquivalent Source We Select The ReferenceDirection To Move AlongThe Path i(t) Generalization  Multiple v-Sources • Voltage Rises AreSubtracted From Drops • Apply KVL

21. Multiple v-Source Equivalent • Collect All SOURCES On One Side • The Equivalent Circuit: • V-source in SeriesADD directly

22. Apply KVL (rise = Σdrops) Generalization Mult. Resistors • [Rk/RS] is the Divider RATIO KVL • Now by Ohm’s Law • And Define RS • Then Voltage Division For Multiple Resistors

23. Find: I, Vbd, P30kΩ Apply KVL & Ohm Example APPLY KVL TO THIS LOOP • Solving for I • Now Vbd • Finally, The 30 kΩ Resistor Power Dissipation

24. Find: I, Vbd Use KVL and Ohm’s Law APPLY KVL TO THIS LOOP Examples • Find VS by V-Divider • The V20k Divider Eqn • Solving for VS

25. When In Doubt → ReDraw • From The Last DiagramIt Was Not ImmediatelyObvious That This Wasa V-Divider Situation • UnTangle/Redraw at Right

26. SNP Circuits Are Characterized By ALL the Elements Having The SAME VOLTAGE Across Them → They Are In PARALLEL Single Node-Pair (SNP) Circuits • SNP Example This Element is INACTIVE • The Inactive Element Has NO Potential Across it → SHORT Circuited

27. UnTangling Reminder • Nodes Can Take STRANGE Shapes NODE → A region of Constant Electrical Potentiale.g.; a group of connected WIRES is ONE Node LowDistortionPowerAmplifier

28. LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW SOME PHYSICAL NODES COMPONENT SIDE CONNECTION SIDE

29. Basic Circuit APPLY KCL The Current Divider • Apply KCL at Top Node • Use Ohm’s Law to Replace Currents • The Current i(t) Enters The Top Node then Splits, or DIVIDES, into the the Currents i1(t) and i2(t)

30. Basic Circuit The Current Divider cont. • The Current Division • By KCL & Ohm • Define PARALLEL Resistance

31. Current Divider Example • For This Ckt Find: I1, I2, Vo • When in doubt… REDRAW the circuit to Better Visualize the Connections • By I-Divider 2-Legged Divider is more Evident

32. Car Stereo and Circuit Model Real World Example • Thus the Speaker Power • Use I-Divider to Find Current thru the 4Ω Speakers • Power Per Speaker by Joule

33. Current & Power Example KCL • For This Ckt Find: • I1, I2, • P40k Power ABSORBED by 40 kΩ Resistor • By I-Divider • The 40k Power by RI2 • Find I2 by I-Divider OR KCL • Choose KCL

34. Generalization: Multi i-Sources • Given Single Node-Pair Ckt w/ Multiple Srcs KCL • KCL on Top Node: • Combine Src Terms To Form Equivalent Source • The Equivalent Ckt

35. Generalization: Multi i-Sources • By Analysis and Electrical Physics of KCL  • Thus CURRENT Sources in PARALLEL ADD directly • Compare to VOLTAGE Sources in SERIES which also ADD Directly =

36. + W k 6 V O mA 10 W k 3 - mA 15 i-Source Example • For This Ckt Find Vo, and the Power Supplied by the I-Srcs • Combine Srcs to Yield Equivalent Ckt • Vo by Ohm’s Law • Use PASSIVE SIGN Convention for Power

37. Generalization: Multi Resistors • Given Single Node-Pair Ckt w/ Multiple R’s KCL • KCL on Top Node: • The EquivalentResistance & v(t) Ohm’s Law atEach Resistor

38. i 1 k k 4 20 k 5 mA 8 Multi-R Example • For This Ckt Find i1, and the Power Supplied by the I-Source • Find Rp • Recall the General Current Divider Eqn

39. i 1 k k 4 20 k 5 mA 8 Multi-R Example cont. • Find i1, by Divider • Take Care with Passive Sign Conv • Find v for Single-Node-Pair by Ohm • Note: this time For Passive Sign Convention CURRENT Direction assigned as POSITIVE • Find Psrc by v•i

40. i 1 k k 4 20 k 5 20k||5k mA 8 i 1 k 4 k 4 mA 8 Multi-R: Alternative Approach • Start by Combining R’s NOT associated with i1 • The Ckt After the R-Combination • Now Have 1:1 Current Divider so

41. Example: Multi-R, Multi-Isrc SNP • Given Single Node-Pair Ckt: Find IL • Soln Game Plan: Convert The Problem Into A Basic CURRENT DIVIDER By Combining Sources And Resistors • Combine Sources • Assume DOWN = POSITIVE

42. Multi-R, Multi-Isrc SNP cont. • Given Single Node-Pair Ckt: Find IL • Next Combine Parallel Resistors • IL by 3:1 I-Divider • Then the Equivalent Circuit → Note MINUS Sign

43. The SAME Ckt Can Look Quite Different I2 I1 9mA I1 I1 I2 I2

44. UnTangling Utility • Redrawing A Circuit May, Sometimes, Help To Better Visualize The Electrical Connections I1 I2 I1 I2 • Be FAITHFUL to the Node-Connections

45. k 2 k 3 k 4 mA 20 Another Example + V _ • For This Ckt Find the I-SrcPower, P20 • Alternatives for P • By vi & passive sign: • Use ||-Resistance • By Joule and Energy Balance

46. Nodal Analysis (based on KCL) • A Systematic Technique To Determine Every Voltage and Current in a Circuit • The variables used to describe the circuit will be “Node Voltages” • The voltages of each node Will Be Determined With Respect To a Pre-selected REFERENCE Node • The Reference Node is Often Referred to as • Ground (GND) • Or • COMMON

47. Consider Resistor Ladder • Goal: Determine All Currents & Potentials in this “Ladder” Network • Plan • Use Series/Parallel Transformation to Find I1 • Back-Substitute Using KVL, KCL, Ohm to Find Rest

48. Xform1 Combine 3 Resistors at End of Network Series-Parallel Transformations • Xform2 • Combine 3 Resistors at End of Network • Note By Ohm’s Law

49. Xform3 To Single-Loop Ckt Xform cont. • Now Back Substitute • Recall • By KCL

50. Recall Xform2 In Summary I1 = 1 mA I2 = I3 = 0.5 mA I4 = 0.375 mA I5 =0.125 mA Va = 3 V Vb = 1.5 V Vc = 0.375 V Xform cont. • Finally by KCL