ENGR-43_Lec-04a_1st_Order_Ckts.pptx 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical.

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 RC & RL 1st Order Ckts

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis C&L Summary

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Introduction  Transient Circuits  In Circuits Which Contain Inductors & Capacitors, Currents & Voltages CanNOT Change Instantaneously  Even The Application, Or Removal, Of Constant Sources Creates Transient (Time- Dependent) Behavior

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Basic Concept  Inductors & Capacitors Can Store Energy  Under Certain Conditions This Energy Can Be Released  RATE of Energy Storage/Release Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st & 2 nd Order Circuits  FIRST ORDER CIRCUITS Circuits That Contain ONE Energy Storing Element –Either a Capacitor or an Inductor  SECOND ORDER CIRCUITS Circuits With TWO Energy Storing Elements in ANY Combination

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Circuits with L’s and/or C’s  Up to this point we have considered DC circuits which generate a System of Linear Equations.  We will now consider Circuits where the Power Supply can Switched IN or OUT

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Graphical Example  Given  The Ckt Before Switching  The Ckt After Switching

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis White Board Example

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Solutions Graphically

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis MATLAB Code for Plots function CartPlot = TempPlotXY(xU,yV) % Bruce Mayer, PE % ENGR25 * 10Sep12 % Function file = plotXY.m clc % workspace window % % Make a nicely formatted XY plot plot(xU,yV, 'LineWidth', 3) xlabel('t (mS)') ylabel('v_C (volt)') title('ENGR43 RC Transient RC Circuit') grid function CartPlot = TempPlotXY(xU,yV) % Bruce Mayer, PE % ENGR25 * 10Sep12 % Function file = plotXY.m clc % workspace window % % Make a nicely formatted XY plot plot(xU,yV, 'LineWidth', 3) xlabel('t (µS)') ylabel('i_L (µAmp)') title('ENGR43 RL Transient RC Circuit') grid

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Example  Slide-35  Students should CAREFULLY Study:

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Circuits with L’s and/or C’s  Conventional DC Analysis Using Mathematical Models Requires The Determination of (a Set of) Equations That Represent the Circuit Response  Example; In Node Or Loop Analysis Of Resistive Circuits One Represents The Circuit By A Set Of Algebraic Equations The Ckt The DC Math Model Analysis

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Ckt w/ L’s & C’s cont.  When The Circuit Includes Inductors Or Capacitors The Models Become Linear Ordinary Differential Equations (ODEs)  Thus Need ODE Tools In Order To Analyze Circuits With Energy Storing Elements Recall ODEs from ENGR25 See Math-4 for More Info on ODEs

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis First Order Circuit Analysis  A Method Based On Thévenin Will Be Developed To Derive Mathematical Models For Any Arbitrary Linear Circuit With One Energy Storing Element  This General Approach Can Be Simplified In Some Special Cases When The Form Of The Solution Can Be Known BeforeHand Straight-Forward ParaMetric Solution

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Basic Concept  Inductors & Capacitors Can Store Energy  Under Certain Conditions This Energy Can Be Released  RATE of Energy Storage/Release Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example: Flash Circuit  The Battery, V S, Charges the Cap To Prepare for a Flash  Moving the Switch to the Right “Triggers” The Flash i.e., The Cap Releases its Stored Energy to the Lamp Say “Cheese”

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Flash Ckt Transient Response  The Voltage Across the Flash-Ckt Storage Cap as a Function of TIME  Note That the Discharge Time (the Flash) is Much Less Than the Charge-Time

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis General Form of the Response  Including the initial conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form  x p (t)  ANY Solution to the General ODE Called the “Particular” Solution  x c (t)  The Solution to the General Eqn with f(t) =0 Called the “Complementary Solution” or the “Natural” (unforced) Response i.e., x c is the Soln to the “Homogenous” Eqn  This is the General Eqn  Now By Linear Differential Eqn Theorem (SuperPosition) Let

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Response Eqns  Given x p and x c the Total Solution to the ODE  Consider the Case Where the Forcing Function is a Constant f(t) = A  Now Solve the ODE in Two Parts  For the Particular Soln, Notice that a CONSTANT Fits the Eqn:

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Response Eqns cont  Sub Into the General (Particular) Eqn x p and dx p /dt  Next, Divide the Homogeneous (RHS=0) Eqn by  ∙x c (t) to yield  Next Separate the Variables & Integrate  Recognize LHS as a Natural Log; so  Next Take “e” to The Power of the LHS & RHS

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 28 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Response Eqns cont  Then  Note that Units of TIME CONSTANT, , are Sec  Thus the Solution for a Constant Forcing Fcn  For This Solution Examine Extreme Cases t =0 t → ∞  The Latter Case (K 1 ) is Called the Steady-State Response All Time-Dependent Behavior has dissipated

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Effect of the Time Constant Tangent reaches x-axis in one time constant Decreases 63.2% after One Time Constant time Drops to 1.8% after 4 Time Constants

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Large vs Small Time Constants  Larger Time Constants Result in Longer Decay Times The Circuit has a Sluggish Response Quick to Steady-State Slow to Steady-State

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Ckt Solution Plan

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 32 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Ckt Solution Plan

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 33 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 1 st 1 st Order Ckt Solution Plan  The Solution Should include a Negative Exponential with an UnKnown PreFactor 7.Use the initial condition in the total solution to determine the Prefactor which completes the total solution 8.Check the total Solution for extreme cases: t = 0+t → ∞

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 34 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  1 st Order Ckt Soln

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 43 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Time Constant Example  Charging a Cap  Use KCL at node-a  Now let v C (t = 0 sec) = 0 V v S (t)= V S (a const)  ReArrange the KCL Eqn For the Homogenous Case where V s = 0  Thus the Time Constant

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 44 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Time Constant Example cont  Charging a Cap  The Solution Can be shown to be  “Fully” Charged Criteria v C >0.99V S OR

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 45 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Differential Eqn Approach  Conditions for Using This Technique Circuit Contains ONE Energy Storing Device The Circuit Has Only CONSTANT, INDEPENDENT Sources The Differential Equation For The Variable Of Interest is SIMPLE To Obtain –Normally by Using Basic Analysis Tools; e.g., KCL, KVL, Thevenin, Norton, etc. The INITIAL CONDITION For The Differential Equation is Known, Or Can Be Obtained Using STEADY STATE Analysis Prior to Switching –Based On: Cap is OPEN, Ind is SHORT

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 46 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Given the RC Ckt At Right with Initial Condition (IC): –v(0 − ) = V S /2  Find v(t) for t>0  Looks Like a Single E-Storage Ckt w/ a Constant Forcing Fcn Assume Solution of Form  Model t>0 using KCL at v(t) after switch is made  Find Time Constant; Put Eqn into Std Form Multiply ODE by R

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 47 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example cont  Compare Std-Form with Model Const Force Fcn Model  Note: the SS condition is Often Called the “Final” Condition (FC)  In This Case the FC  Now Use IC to Find K 2 In This Case  Next Check Steady- State (SS) Condition In SS the Time Derivative goes to ZERO

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 48 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example cont.2  At t = 0 + The Model Solution  The Total/General Soln  Recall The IC: v(0 − ) = V S /2 = v(0 + ) for a Cap Then  Check  

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 49 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Exmpl  Find i(t) Given i(0 − ) = 0  Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn Assume Solution of Form  In This Case x→i  To Find the ODE Use KVL for Single-Loop ckt KVL  Now Consider IC By Physics, The Current Thru an Inductor Can NOT Change instantaneously

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 50 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor cont  Casting ODE in Standard form  Recognize Time Const  Next Using IC (t = 0 + )  Thus the ODE Solution  Also Note FC  Thus

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 51 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Solution Process Summary 1.ReWrite ODE in Standard Form Yields The Time-Constant,  2.Analyze The Steady-State Behavior Finds The Final Condition Constant, K 1 3.Use the Initial Condition Gives The Exponential PreFactor, K 2 4.Check: Is The Solution Consistent With the Extreme Cases t = 0+ t → 

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 52 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Solutions for f(t) ≠ Constant 1.Use KVL or KCL to Write the ODE 2.Perform Math Operations to Obtain a CoEfficient of “1” for the “Zero th ” order Term. This yields an Eqn of the form 3.Find a Particular Solution, x p (t) to the FULL ODE above This depends on f(t), and may require “Educated Guessing”

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 53 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Solutions for f(t) ≠ Constant 4.Find the total solution by adding the COMPLEMENTARY Solution, x c (t) to previously determined x p (t). x c (t) takes the form: The Total Solution at this Point → 5.Use the IC at t=0 + to find K; e.g.; IC = 7 Where M is just a NUMBER

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 54 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Example  At t=0+ Apply KCL  Assume a Solution of the Form for v c  For Ckt Below Find v o (t) for t>0 (note f(t) = const; 12V)

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 55 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Cap Exmp cont  Step1: By Inspection of the ReGrouped KCL Eqn Recognize   Now Examine the Reln Between v o and v C a V-Divider  Step-2: Consider The Steady-State In This Case After the Switch Opens The Energy Stored in the Cap Will be Dissipated as HEAT by the Resistors  With x p = K 1 = SSsoln

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 56 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Cap Exmp cont  Now The IC  If the Switch is Closed for a Long Time before t =0, a STEADY-STATE Condition Exists for NEGATIVE Times  v o (0 − ) by V-Divider  Recall Reln Between v o and v C for t ≥ 0 Recall: Cap is OPEN to DC

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 57 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Cap Exmp cont  Step-3: Apply The IC  Now have All the Parameters needed To Write The Solution  Note: For f(t)=Const, puttingthe ODE in Std Form yields  and K 1 by Inspection:  Recall v o = (1/3)v C

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 58 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Thevenin/Norton Techniques  Obtain The Thevenin Voltage Across The Capacitor, Or The Norton Current Through The Inductor  With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find Time Constant using R TH Steady-State Final Condition using v TH (if v TH a constant) Thévenin

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 59 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Thevenin Models for ODE  KVL for Single Loop  KCL at node a

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 60 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Find ODE Soln By Thevenin  Break Out the Energy Storage Device (C or L) as the “Load” for a Driving Circuit  Analyze the Driving Ckt to Arrive at it’s Thévenin (or Norton) Equivalent  ReAttach The C or L Load  Use KCL or KVL to arrive at ODE  Put The ODE in Standard Form

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 61 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Find ODE Soln By Thevenin  Recognize the Solution Parameters  For Capacitor  = R TH C K 1 (Final Condition) = v TH = v OC = x p  For Inductor  = L/R TH K 1 (Final Condition) = v TH /R TH = i SC = i N = x p  In Both Cases Use the v C (0 − ) or i L (0 − ) Initial Condition to Find Parameter K 2

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 62 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example  The Variable Of Interest Is The Inductor Current  The Thévenin model  Since this Ckt has a CONSTANT Forcing Function of 0V (the 24V source is switched OUT at t = 0), Then The Solution Is Of The Form  Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit”

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 63 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example cont  The f(t)=const ODE in Standard Form  The Solution Substituted into the ODE at t > 0  Combining the K 2 terms shows shows K 1 = 0, thus  From This Ckt Observe

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 64 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example cont.2  Now Find K 2  Assume Switch closed for a Long Time Before t = 0 Inductor is SHORT to DC  Analyzing Ckt with 3H Ind as Short Reveals The Reader Should Verify the Above  Then the Entire Solution

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 65 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Untangle to find i O (0 −) Untangle Analyze  Be Faithful to Nodes

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 67 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work This Problem WWell, Maybe NEXT time…

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 68 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 69 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Appendix DE Approach

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 70 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Differential Eqn Approach cont  Math Property When all Independent Sources Are CONSTANT, then for ANY variable y(t); i.e., v(t) or i(t), in The Circuit The Solution takes the Form  The Solution Strategy Use The DIFFERENTIAL EQUATION And The FINAL & INITIAL Conditions To Find The Parameters K 1 and K 2

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 71 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Differential Eqn Approach cont  If the ODE for y is Known to Take This Form  We Can Use This Structure to Find The Unknowns. If:  Then Sub Into ODE  Equating the TRANSIENT (exponential) and CONSTANT Terms Find

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 72 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Differential Eqn Approach cont  Up to Now  Next Use the Initial Condition  So Finally  If we Write the ODE in Proper form We can Determine By Inspection  and K 1

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 73 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example  The Model for t>0 → KVL on single-loop ckt  Rewrite In Std Form  Recognize Time Const  For The Ckt Shown Find i 1 (t) for t>0  Assume Solution of the Form

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 74 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example cont  Examine Std-Form Eqn to Find K 1  For Initial Conditions Need the Inductor Current for t<0  Again Consider DC (Steady-State) Condition for t<0  The SS Ckt Prior to Switching Recall An Inductor is a SHORT to DC  So

BMayer@ChabotCollege.edu ENGR-43_Lec-04a_1st_Order_Ckts.pptx 75 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Example cont.1  Now Use Step-3 To Find K 2 from IC Remember Current Thru an Inductor Must be Time-Continuous Recall that K 1 was zero  Construct From the Parameters The ODE Solution  The Answer 0

ENGR-43_Lec-04a_1st_Order_Ckts.pptx 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical.

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ENGR-43_Lec-04a_1st_Order_Ckts.pptx 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical.

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