1. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Capacitors & Inductors

2. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitance & Inductance  Introduce Two Energy STORING Devices Capacitors Inductors  Outline Capacitors –Store energy in their ELECTRIC field (electroStatic energy) –Model as circuit element Inductors –Store energy in their MAGNETIC field (electroMagnetic energy) –Model as circuit element Capacitor And Inductor Combinations –Series/Parallel Combinations Of Elements

3. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Capacitor  First of the Energy- Storage Devices  Basic Physical Model  Circuit Representation Note use of the PASSIVE SIGN Convention  Details of Physical Operation Described in PHYS4B & ENGR45

4. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitance Defined  Consider the Basic Physical Model  Where A  The Horizontal Plate-Area, m 2 d  The Vertical Plate Separation Distance, m  0  “Permittivity” of Free Space; i.e., a vacuum –A Physical CONSTANT – Value = 8.85x10 -12 Farad/m  The Capacitance, C, of the Parallel-Plate Structure w/o Dielectric  Then What are the UNITS of Capacitance, C  Typical Cap Values → “micro” or “nano”

5. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Circuit Operation  Recall the Circuit Representation  LINEAR Caps Follow the Capacitance Law; in DC  The Basic Circuit- Capacitance Equation  Where Q  The CHARGE STORED in the Cap, Coulombs C  Capacitance, Farad V c  DC-Voltage Across the Capacitor  Discern the Base Units for Capacitance

6. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis “Feel” for Capacitance  Pick a Cap, Say 12 µF  Recall Capacitor Law  Now Assume That The Cap is Charged to hold 15 mC Find V c  Solving for V c  Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage Caps can Be DANGEROUS!

7. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Forms of the Capacitor Law  The time-Invariant Cap Law  If v C at −  = 0, then the traditional statement of the Integral Law  Leads to DIFFERENTIAL Cap Law  The Differential Suggests SEPARATING Variables  Leads to The INTEGRAL form of the Capacitance Law  If at t 0, v C = v C (t 0 ) (a KNOWN value), then the Integral Law becomes

8. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Integral Law  Express the VOLTAGE Across the Cap Using the INTEGRAL Law  Thus a Major Mathematical Implication of the Integral law  If i(t) has NO Gaps in its i(t) curve then  Even if i(y) has VERTICAL Jumps:  The Voltage Across a Capacitor MUST be Continuous  An Alternative View The Differential Reln  If v C is NOT Continous then dv C /dt → , and So i C → . This is NOT PHYSICALLY possible

9. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Differential Law  Express the CURRENT “Thru” the Cap Using the Differential Law  Thus a Major Mathematical Implication of the Differential Law  If v C = Constant Then  This is the DC Steady-State Behavior for a Capacitor  A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit  Cap Current Charges do NOT flow THRU a Cap –Charge ENTER or EXITS The Cap in Response to Voltage CHANGES

10. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Current  Charges do NOT flow THRU a Cap  Charge ENTER or EXITS The Capacitor in Response to the Voltage Across it That is, the Voltage-Change DISPLACES the Charge Stored in The Cap –This displaced Charge is, to the REST OF THE CKT, Indistinguishable from conduction (Resistor) Current  Thus The Capacitor Current is Called the “Displacement” Current

11. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Summary  The Circuit Symbol  From Calculus, Recall an Integral Property  Compare Ohm’s Law and Capactitance Law CapOhm  Now Recall the Long Form of the Integral Relation  Note The Passive Sign Convention  The DEFINITE Integral is just a no.; call it v C (t 0 ) so

12. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Summary cont  Consider Finally the Differential Reln  Some Implications For small Displacement Current dv C /dt is small; i.e, v C changes only a little Obtaining Large i C requires HUGE Voltage Gradients if C is small  Conclusion: A Cap RESISTS CHANGES in VOLTAGE ACROSS It

13. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis i C Defined by Differential  The fact that the Cap Current is defined through a DIFFERENTIAL has important implications...  Consider the Example at Left Shows v C (t) –C = 5 µF Find i C (t)  Using the 1 st Derivative (slopes) to find i(t)

14. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Energy Storage  UNlike an I-src or V-src a Cap Does NOT Produce Energy  A Cap is a PASSIVE Device that Can STORE Energy  Recall from Chp.1 The Relation for POWER  For a Cap  Then the INSTANTANEOUS Power  Recall also  Subbing into Pwr Reln  By the Derivative CHAIN RULE

15. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Energy Storage cont  Again From Chp.1 Recall that Energy (or Work) is the time integral of Power Mathematically  Integrating the “Chain Rule” Relation  Comment on the Bounds If the Lower Bound is −  we talk about “energy stored at time t 2 ” If the Bounds are −  to +  then we talk about the “total energy stored”  Recall also  Subbing into Pwr Reln  Again by Chain Rule

16. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Energy Storage cont.2  Then Energy in Terms of Capacitor Stored-Charge  The Total Energy Stored during t = 0-6 ms  Short Example  w C Units?  Charge Stored at 3 mS V C (t) C = 5 µF

17. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Some Questions About Example  For t > 8 mS, What is the Total Stored CHARGED?  For t > 8 mS, What is the Total Stored ENERGY? v C (t) C = 5 µF CHARGING Current DIScharging Current

18. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Summary: Q, V, I, P, W  Consider A Cap Driven by A SINUSOIDAL V-Src  Charge stored at a Given Time  Find All typical Quantities Note –120  = 60∙(2  ) → 60 Hz  Current “thru” the Cap  Energy stored at a given time i(t)

19. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitor Summary cont.  Consider A Cap Driven by A SINUSOIDAL V-Src  At 135° = (3/4)   Electric power absorbed by Cap at a given time i(t)  The Cap is SUPPLYING Power at At 135° = (3/4)  = 6.25 mS That is, The Cap is RELEASING (previously) STORED Energy at Rate of 6.371 J/s

20. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work See ENGR-43_Lec-06- 1_Capacitors_WhtBd.ppt  Let’s Work this Problem The VOLTAGE across a 0.1-F capacitor is given by the waveform in the Figure Below. Find the WaveForm Eqn for the Capacitor CURRENT + v C (t) - ANOTHER PROB 0.5 μF

21. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Inductor  Second of the Energy-Storage Devices  Basic Physical Model:  Details of Physical Operation Described in PHYS 4B or ENGR45  Note the Use of the PASSIVE Sign Convention Ckt Symbol

22. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Physical Inductor  Inductors are Typically Fabricated by Winding Around a Magnetic (e.g., Iron) Core a LOW Resistance Wire Applying to the Terminals a TIME VARYING Current Results in a “Back EMF” voltage at the connection terminals  Some Real Inductors

23. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductance Defined  From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law  Where the Constant of Proportionality, L, is called the INDUCTANCE  L is Measured in Units of “Henrys”, H 1H = 1 Vs/Amp  Inductors STORE electromagnetic energy  They May Supply Stored Energy Back To The Circuit, But They CANNOT CREATE Energy  For a Linear Inductor The Flux Is Proportional To The Current Thru it

24. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductance Sign Convention  Inductors Cannot Create Energy; They are PASSIVE Devices  All Passive Devices Must Obey the Passive Sign Convention

25. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Circuit Operation  Recall the Circuit Representation  Separating the Variables and Integrating Yields the INTEGRAL form  Previously Defined the Differential Form of the Induction Law  In a development Similar to that used with caps, Integrate −  to t 0 for an Alternative integral Law

26. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Model Implications  From the Differential Law Observe That if i L is not Continuous, di L /dt → , and v L must also →   This is NOT physically Possible  Thus i L must be continuous  Consider Now the Alternative Integral law  If i L is constant, say i L (t 0 ), then The Integral MUST be ZERO, and hence v L MUST be ZERO This is DC Steady-State Inductor Behavior –v L = 0 at DC –i.e; the Inductor looks like a SHORT CIRCUIT to DC Potentials

27. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor: Power and Energy  From the Definition of Instantaneous Power  Time Integrate Power to Find the Energy (Work)  Subbing for the Voltage by the Differential Law  Units Analysis J = H x A 2  Again By the Chain Rule for Math Differentiation  Energy Stored on Time Interval Energy Stored on an Interval Can be POSITIVE or NEGATIVE

28. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 28 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor: P & W cont.  In the Interval Energy Eqn Let at time t 1  Then To Arrive At The Stored Energy at a later given time, t  Thus Observe That the Stored Energy is ALWAYS Positive In ABSOLUTE Terms as i L is SQUARED ABSOLUTE-POSITIVE-ONLY Energy-Storage is Characteristic of a PASSIVE ELEMENT 0)( 2 1 1 2  tLi L )( 2 1 )( 2 t tw L 

29. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 29 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Given The i L Current WaveForm, Find v L for L = 10 mH  The Derivative of a Line is its SLOPE, m  The Differential Reln  Then the Slopes  And the v L Voltage

30. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 30 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Power & Energy  The Energy Stored between 2 & 4 mS  The Value Is Negative Because The Inductor SUPPLIES Energy PREVIOUSLY STORED with a Magnitude of 2 μJ  The Energy Eqn  Running the No.s

31. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 31 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Given The Voltage Wave Form Across L, Find i L if L = 0.1 H i(0) = 2A  The PieceWise Function  The Integral Reln  A Line Followed by A Constant; Plotting

32. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 32 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example - Energy  The Current Characteristic  The Initial Stored Energy  The Energy Eqn  The “Total Stored Energy” Energy Stored on Interval Can be POS or NEG  Energy Stored between 0-2  → Consistent with Previous Calculation

33. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 33 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Notice That the ABSOLUTE Energy Stored At Any Given Time Is Non Negative (by sin 2 ) i.e., The Inductor Is A PASSIVE Element  Find The Voltage Across And The Energy Stored (As Function Of Time)  For The Energy Stored

34. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 34 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis L = 5 mH; Find the Voltage

35. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 35 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Total Energy Stored  The Ckt Below is in the DC-SteadyState  Find the Total Energy Stored by the 2-Caps & 2-Inds  Recall that at DC Cap → Short-Ckt Ind → Open-Ckt  Shorting-Caps; Opening-Inds  KCL at node-A  Solving for V A

36. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 36 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Total Energy Stored  Continue Analysis of Shorted/Opened ckt  Using Ohm and V A = 16.2V  By KCL at Node-A  V C2 by Ohm  V C1 by Ohm & KVL

37. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 37 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Total Energy Stored  Have all I’s & V’s:  Next using the E-Storage Eqns  Then the E- Storage Calculations 16.2 V10.8 V −1.2 A1.8 A

38. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 38 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Caps & Inds  Ideal vs. Real  A Real CAP has Parasitic Resistances & Inductance:  A Real IND has Parasitic Resistances & Capacitance: Generally SMALL Effect

39. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 39 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Ideal vs Real  Ideal C & L  Practical Elements “Leak” Thru Unwanted Resistance

40. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 40 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitors in Series  If the v i (t 0 ) = 0, Then Discern the Equivalent Series Capacitance, C S  By KVL for 1-LOOP ckt  CAPS in SERIES Combine as Resistors in PARALLEL

41. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 41 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Find Equivalent C Initial Voltage  Spot Caps in Series  Or Can Reduce Two at a Time  Use KVL for Initial Voltage  This is the Algebraic Sum of the Initial Voltages Polarity is Set by the Reference Directions noted in the Diagram

42. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 42 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Two charged Capacitors Are Connected As Shown.  Find the Unknown Capacitance  Recognize SINGLE Loop Ckt → Same Current in Both Caps Thus Both Caps Accumulate the SAME Charge  Finally Find C by Charge Eqn  And Find V C by KVL V C = 12V-8V = 4V

43. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 43 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Capacitors in Parallel  Thus The Equivalent Parallel Capacitance  By KCL for 1-NodePair ckt  CAPS in Parallel Combine as Resistors in SERIES

44. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 44 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Complex Example → Find C eq

45. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 45 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductors in Series  Thus  By KVL For 1-LOOP ckt  INDUCTORS in Series add as Resistors in SERIES  Use The Inductance Law

46. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 46 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductors in Parallel  And  By KCL for 1-NodePair ckt  INDUCTORS in Parallel combine as Resistors in PARALLEL  Thus

47. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 47 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example – Find: L eq, i 0  Series↔Parallel Summary INDUCTORS Combine as do RESISTORS CAPACITORS Combine as do CONDUCTORS

48. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 48 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Ladder Network  Find L eq for L i = 4 mH  Place Nodes In Chosen Locations  Connect Between Nodes  When in Doubt, ReDraw Select Nodes

49. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 49 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Find L eq for L i = 6mH  ReDraw The Ckt for Enhanced Clarity  Nodes Can have Complex Shapes The Electrical Diagram Does NOT have to Follow the Physical Layout  It’s Simple Now

50. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 50 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis C&L Summary

51. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 51 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work This Problem  Find: v(t), t max for i max, t min for v min

52. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 52 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis By MATLAB % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid

53. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 53 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis By MATLAB % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid

54. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 54 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

55. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 55 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

56. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 56 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

57. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 57 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis L = 10 mH; Find the Voltage

58. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 58 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Appendix Complex Cap Example

59. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 59 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Given i C, Find v C  The Piecewise Fcn for i C  Integrating & Graphing C= 4µF v C (0) = 0 Parabolic Linear >

60. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 60 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Example  For The Previous Conditions, Find The POWER Characteristic C = 4 µF i C by Piecewise curve  From Before the v C  Using the Pwr Reln

61. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 61 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Power Example cont  Finally the Power Characteristic  Absorbing or Supplying Power?  During the CHARGING Period of 0-2 mS, the Cap ABSORBS Power  During DIScharge the Cap SUPPLIES power But only until the stored charge is fully depleted

62. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 62 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Energy Example  For The Previous Conditions, Find The ENERGY Characteristic C = 4 µF p C by Piecewise curve  Now The Work (or Energy) is the Time Integral of Power  For 0  t  2 mS

63. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 63 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Energy Example cont  For 2 < t  4 mS  Taking The Time Integral and adding w(2 mS)  Then the Energy Characteristic

64. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 64 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Current through capacitor Voltage at a given time t Voltage at a given time t when voltage at time to<t is also known V Charge at a given time C Voltage as a function of time V Electric power supplied to capacitor Energy stored in capacitor at a given time W J “Total” energy stored in the capacitor J SAMPLE PROBLEM If the current is known...

65. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 65 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis SAMPLE PROBLEM Compute voltage as a function of time At minus infinity everything is zero. Since current is zero for t<0 we have In particular Charge stored at 5ms Total energy stored Total means at infinity. Hence Before looking into a formal way to describe the current we will look at additional questions that can be answered. Now, for a formal way to represent piecewise functions.... Given current and capacitance

66. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 66 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Formal description of a piecewise analytical signal

67. BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 67 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Find C eq for C i = 4 µF