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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

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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

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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

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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

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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

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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 F 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!

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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

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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

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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

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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

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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 Ohms 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

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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

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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 F Find i C (t) Using the 1 st Derivative (slopes) to find i(t)

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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

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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

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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 F

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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 F CHARGING Current DIScharging Current

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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)

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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

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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 Lets 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 F

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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 J The Energy Eqn Running the No.s

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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

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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

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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

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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

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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

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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

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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 Is & Vs: Next using the E-Storage Eqns Then the E- Storage Calculations 16.2 V10.8 V 1.2 A1.8 A

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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

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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

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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

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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

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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

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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

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

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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

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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

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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 SeriesParallel Summary INDUCTORS Combine as do RESISTORS CAPACITORS Combine as do CONDUCTORS

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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

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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 Its Simple Now

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

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BMayer@ChabotCollege.edu ENGR-43_Lec-06-1_Capacitors.ppt 51 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work Lets Work This Problem Find: v(t), t max for i max, t min for v min

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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

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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

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

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

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

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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

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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

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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= 4F v C (0) = 0 Parabolic Linear >

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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 F i C by Piecewise curve From Before the v C Using the Pwr Reln

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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

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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 F p C by Piecewise curve Now The Work (or Energy) is the Time Integral of Power For 0 t 2 mS

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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

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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