1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43
RC & RL 1st Order Ckts
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. C&L Summary

3. Introduction  Transient Circuits
In Circuits Which Contain Inductors And Capacitors, Currents & Voltages CanNOT Change Instantaneously
Even The Application, Or Removal, Of Constant Sources Creates Transient (Time- Dependent) Behavior

4. 1st & 2nd Order Circuits FIRST ORDER CIRCUITS SECOND 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

5. 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
Analysis
The DC Math Model
The Ckt

6. 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

7. 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

8. Basic Concept Inductors And Capacitors Can Store Energy
Under Certain Conditions This Energy Can Be Released
RATE OF ENERGY RELEASE Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element

9. Example: Flash Circuit
The Battery, VS, 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”

10. 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

11. 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
xp(t)  ANY Solution to the General ODE
Called the “Particular” Solution
xc(t)  The Solution to the General Eqn with f(t) =0
Called the “Complementary Solution” or the “Natural” (unforced) Response
i.e., xc is the Soln to the “Homogenous” Eqn
This is the General Eqn
Now By Linear Differential Eqn Theorem (SuperPosition) Let

12. 1st Order Response Eqns Given xp and xc 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:

13. 1st Order Response Eqns cont
Sub Into the General (Particular) Eqn xp and dxp/dt
Next Separate the Variables & Integrate
Recognize LHS as a Natural Log; so
Next, Divide the Homogeneous (RHS=0) Eqn by xc(t) to yield
Next Take “e” to The Power of the LHS & RHS

14. 1st Order Response Eqns cont
Then
For This Solution Examine Extreme Cases
t =0
t → ∞
Note that Units of TIME CONSTANT, , are Sec
Thus the Solution for a Constant Forcing Fcn
The Latter Case is Called the Steady-State Response
All Time-Dependent Behavior has dissipated

15. Effect of the Time Constant
Tangent reaches x-axis in one time constant
Response drop 63.2% from previous over one time constant
Decreases 63.2% after One Time Constant time
Drops to 1.8% after 4 Time Constants

16. Large vs Small Time Constants
Larger Time Constants Result in Longer Decay Times
The Circuit has a Sluggish Response
Slow to Steady-State
Quick to Steady-State

17. Time Constant Example Charging a Cap Now let
vC(t = 0 sec) = 0 V
vS(t)= VS (a const)
Rearrange the KCL Eqn For the Homogenous Case where Vs = 0
Use KCL at node-a
Thus the Time Constant

18. Time Constant Example cont
Charging a Cap
“Fully” Charged Criteria
vC >0.99VS OR
The Solution Can be shown to be

19. 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

20. Example Given the RC Ckt At Right with Find v(t) for t>0
Initial Condition (IC):
v(0−) = VS/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

21. 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
In This Case
Next Check Steady-State (SS) Condition
In SS the Time Derivative goes to ZERO
Now Use IC to Find K2

22. Example cont.2   At t = 0+ The Total/General Soln
The Model Solution
The Total/General Soln
Recall The IC: v(0−) = VS/2 = v(0+) for a Cap
Then
Check  

23. Inductor Exmpl Find i(t) Given
KVL
Find i(t) Given
i(0−) = 0
Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn
Assume Solution of Form
To Find the ODE Use KVL for Single-Loop ckt
Now Consider IC
By Physics, The Current Thru an Inductor Can NOT Change instantaneously
In This Case x→i

24. Inductor cont Casting ODE in Standard form Next Using IC (t = 0+)
Recognize Time Const
Also Note FC
Thus the ODE Solution
Thus

25. Solution Process Summary
ReWrite ODE in Standard Form
Yields The Time-Constant, 
Analyze The Steady-State Behavior
Finds The Final Condition Constant, K1
Use the Initial Condition
Gives The Exponential PreFactor, K2
Check: Is The Solution Consistent With the Extreme Cases
t = 0+
t → 

26. Solutions for f(t) ≠ Constant
Use KVL or KCL to Write the ODE
Perform Math Operations to Obtain a CoEfficient of “1” for the “Zeroth” order Term. This yields an Eqn of the form
Find a Particular Solution, xp(t) to the FULL ODE above
This depends on f(t), and may require “Educated Guessing”

27. Solutions for f(t) ≠ Constant
Find the total solution by adding the COMPLEMENTARY Solution, xc(t) to previously determined xp(t). xc(t) takes the form:
The Total Solution at this Point →
Use the IC at t=0+ to find K; e.g.; IC = 7
Where M is just a NUMBER

28. Capacitor Example
For Ckt Below Find vo(t) for t>0 (note f(t) = const; 12V)
Assume a Solution of the Form for vc
At t=0+ Apply KCL

29. Cap Exmp cont
Step1: By Inspection of the ReGrouped KCL Eqn Recognize 
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
Now Examine the Reln Between vo and vC
a V-Divider
So xp = K1

30. Recall: Cap is OPEN to DC
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
Recall: Cap is OPEN to DC
vo(0−) by V-Divider
Recall Reln Between vo and vC for t ≥ 0
Vc is the same a Vsw => V-divider 12V*(6/9) = 8V

31. Cap Exmp cont Step-3: Apply The IC Recall vo = (1/3)vC
Note: For f(t)=Const, puttingthe ODE in Std Form yields  and K1 by Inspection:
Now have All the Parameters needed To Write The Solution

32. Thevenin/Norton Techniques
Obtain The Thevenin Voltage Across The Capacitor, Or The Norton Current Through The Inductor
Thévenin
With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find
Time Constant using RTH
Steady-State Final Condition using vTH (if vTH a constant)

33. Thevenin Models for ODE
KCL at node a
KVL for Single Loop

34. 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

35. Find ODE Soln By Thevenin
Recognize the Solution Parameters
For Capacitor
 = RTHC
K1 (Final Condition) = vTH = vOC = xp
For Inductor
 = L/RTH
K1 (Final Condition) = vTH/RTH = iSC = iN = xp
In Both Cases Use the vC(0−) or iL(0−) Initial Condition to Find Parameter K2
K2 is the complementary solution exponential PreFactor

36. Inductor Example
Since this Ckt has a CONSTANT Forcing Function (24V), Then The Solution Is Of The Form
Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit”
The Variable Of Interest Is The Inductor Current
The Thévenin model

37. Inductor Example cont The f(t)=const ODE in Standard Form
The Solution Substituted into the ODE at t = 0+
From This Ckt Observe
From Std Form K1 = 0 So

38. Inductor Example cont.2 Analyzing Ckt with 3H as Short Reveals
The Reader Should Verify the Above
Now Find K2
Assume Switch closed for a Long Time Before t = 0
Inductor is SHORT to DC
Redraw ckt if needed
Then the Entire Solution

39. Untangle to find iO(0−)
Untangle
Analyze
Be Faithful to Nodes

41. WhiteBoard Work Well, Maybe NEXT time… Let’s Work This Problem
7e prob 6.16

42. WhiteBoard Work
None
Today

43. 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 K1 and K2

44. Differential Eqn Approach cont
If the ODE for y is Known to Take This Form
Then Sub Into ODE
Equating the TRANSIENT (exponential) and CONSTANT Terms Find
We Can Use This Structure to Find The Unknowns. If:

45. Differential Eqn Approach cont
Up to Now
So Finally
If we Write the ODE in Proper form We can Determine By Inspection  and K1
Next Use the Initial Condition

46. Inductor Example For The Ckt Shown Find i1(t) for t>0
Assume Solution of the Form
The Model for t>0 → KVL on single-loop ckt
Recognize Time Const
Rewrite In Std Form

47. Inductor Example cont Examine Std-Form Eqn to Find K1
The SS Ckt Prior to Switching
Recall An Inductor is a SHORT to DC
For Initial Conditions Need the Inductor Current for t<0
Again Consider DC (Steady-State) Condition for t<0 So

48. Inductor Example cont.1 The Answer Now Use Step-3 To Find K2 from IC
Remember Current Thru an Inductor Must be Time-Continuous
Recall that K1 was zero
Construct From the Parameters The ODE Solution