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

[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Capacitors &

Inductors



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

http://www.powerlabs.org/images/inductors.jpg



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



Capacitance Defined Consider the Basic

Physical Model Where

• A The Horizontal Plate-Area, m2

• 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 d

AC 0

Then What are the UNITS of Capacitance, C

Typical Cap Values →“micro” or “nano”



Capacitor Circuit Operation Recall the Circuit

Representation LINEAR Caps Follow the

Capacitance Law; in DC

The Basic Circuit-Capacitance Equation

tCvtq c

Where• Q The CHARGE

STORED in the Cap, Coulombs

• C Capacitance, Farad• Vc DC-Voltage Across

the Capacitor Discern the Base Units

for Capacitance

cCVQ

VoltCoulombFarad



“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 Vc

cCVQ

!!V! 1250Coul/Volt12x10

Coul15x106

3

c

c

VCQV

Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage• Caps can Be

DANGEROUS!



Forms of the Capacitor Law The time-Invariant

Cap Law If vC at − = 0, then the

traditional statement of the Integral Law

Leads to DIFFERENTIAL Cap Law

dttdvC

dttdqti C

cCVQ

The Differential Suggests SEPARATING Variables

tCdvdtti C Leads to The

INTEGRAL form of the Capacitance Law

tv

v

t C

C

dzCdyyi

t

C dyyiC

tv 1

If at t0, vC = vC(t0) (a KNOWN value), then the Integral Law becomes

t

tCC

t

t

t

C

dyyiC

tvtv

dyyiC

dyyiC

tv

0

0

0

1

11

0



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

t

CC dyyiCC

tqtv 1

tt

tCtdyyi

Cttv 1limlim

00

tvttv CCt

0lim

If vC is NOT Continous then dvC/dt → , and So iC → . This is NOT PHYSICALLY possible

dttdvCti C

C



Capacitor Differential Law Express the CURRENT

“Thru” the Cap Using the Differential Law

Thus a Major Mathematical Implication of the Differential Law

If vC = Constant Then

This is the DC Steady-State Behavior for a Capacitor

A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit

dttdvC

dttdqti C

C

0Ci Cap Current• Charges do NOT flow

THRU a Cap– Charge ENTER or EXITS

The Cap in Response to Voltage CHANGES



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



Capacitor Summary The Circuit Symbol From Calculus, Recall an

Integral Property

Compare Ohm’s Law and Capactitance LawCap Ohm

Now Recall the Long Form of the Integral Relation

Cv

Ci Note The Passive Sign Convention

)()( tdtdvCti c

C

t

CC dxxiC

tv )(1)( RR

RR

Riv

vR

i

1

t

t

tt

dxxfdxxfdxxf0

0

0

0

)(1)(1)(t t

tCCC dxxi

Cdxxi

Ctv

The DEFINITE Integral is just a no.; call it vC(t0) so

t

tCCC dxxi

Ctvtv

0

)(1)()( 0



Capacitor Summary cont Consider Finally the Differential Reln

dttdvCti C

C Some Implications

• For small Displacement Current dvC/dt is small; i.e, vC changes only a little

• Obtaining Large iC requires HUGE Voltage Gradients if C is small

Conclusion: A Cap RESISTS CHANGES in VOLTAGE ACROSS It



iC Defined by Differential The fact that the Cap

Current is defined through a DIFFERENTIAL has important implications...

Consider the Example at Left• Shows vC(t)

– C = 5 µF • Find iC(t)

FC 5

)()( tdtdvCti

mAsVFi 20

10624][105 3

6

mA60

elsewhereti 0)(

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



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

vip

titvtp

Cv

Ci

titvtp CCC

)()( tdtdvCti C

C

Subbing into Pwr Reln

dtdvtCvtp C

CC )()(

By the Derivative CHAIN RULE

)(

21)( 2 tv

dtdCtp CC

dtdv

dvd

dtd C

C



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 t2”

• If the Bounds are − to + then we talk about the “total energy stored”

Recall also

Subbing into Pwr Reln

2

1

)(),( 21

t

tCC dxxpttw

)(21)(

21),( 1

22

221 tCvtCvttw CCC

Ctqdxxi

Ctv C

t

CC)()(1)(

)()(1)( tdtdqtq

Ctp C

CC

Again by Chain Rule

)(

211)( 2 tq

dtd

Ctp cC

tvC tiC

dtdq

dqd

dtd C

C



Capacitor Energy Storage cont.2 Then Energy in

Terms of Capacitor Stored-Charge

The Total Energy Stored during t = 0-6 ms

Short Example wC Units?

Charge Stored at 3 mS

)0(21)6(

21)6,0( 22

CCC CvCvw

JVCoulVV

CoulVF 22

)(1)(1),( 12

22

21 tqC

tqC

ttw CCC

VC(t)C = 5 µF

][)24(*][10*521)6,0( 226 VFwC

)3()3( CC Cvq

CVFqC 60][12*][10*5)3( 6



Some Questions About Example For t > 8 mS, What is

the Total Stored CHARGED?

For t > 8 mS, What is the Total Stored ENERGY?

0)8( mStqC

vC(t)C = 5 µF

0)8( mStwC

CHARGING Current

DIScharging Current

222 05

219

21),( FtCvtw CC

22

22 0

51111),(F

tqC

tw CC



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

)()( tCvtq CC 0])[sin(*][10*233.81201 6 VCmSqC

)(tdtdvCi C

C

mA9833.81201

)cos(120*130*10*233.81201 6

mSimSi

C

C

FC 2

)120(sin130)( ttv

)(tv i(t)

)(21)( 2 tCvtw CC

mJ 16.92

sin130*][10*221

2401 226

C

C

w

Fw



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

FC 2

)120(sin130)( ttv

)(tv i(t)

)()()( titvtp CCC

mWmAVp

VvmAi

i

C

C

C

C

63713.699.9116019.91)75.0sin(1301601

3.691601)75.0cos(120*130*10*21601 6

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



WhiteBoard Work

v c( t) (V)

t(s )0 1 2 3 4 5

-12

F igure P 5 .14

-12

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

tAtiC 50000cos2+ vC(t) -

ANOTHER PROB0.5 μF



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



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



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 V•s/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

dtdvL

dtdiLvLi L

LL



Inductance Sign Convention Inductors Cannot

Create Energy; They are PASSIVE Devices

All Passive Devices Must Obey the Passive Sign Convention



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 t0 for an Alternative integral Law

dtdiLv L

L

t

LL dxxvL

ti )(1)(

00 ;)(1)()(0

ttdxxvL

titit

tLLL



Inductor Model Implications From the Differential

Law Observe That if iL is not Continuous, diL/dt → , and vL must also →

This is NOT physically Possible

Thus iL must be continuous

Consider Now the Alternative Integral law

If iL is constant, say iL(t0), then The Integral MUST be ZERO, and hence vL MUST be ZERO• This is DC Steady-State

Inductor Behavior– vL = 0 at DC– i.e; the Inductor looks like

a SHORT CIRCUITto DC Potentials

00 ;)(1)()(0

ttdxxvL

titit

tLLL

dtdiv L

L



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 A2

)()()( titvtp LLL

)()()( titdtdiLtp L

LL

)(

21 2 tLi

dtdtp LL

Again By the Chain Rule for Math Differentiation

2

1

)(21),( 2

21

t

tLL dxxLi

dxdttw

Energy Stored on Time Interval

)(21)(

21),( 1

22

221 tLitLittw LL

• Energy Stored on an Interval Can be POSITIVE or NEGATIVE

dtdi

did

dtd



Inductor: P & W cont. In the Interval Energy

Eqn Let at time t1

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 iL is SQUARED• ABSOLUTE-POSITIVE-ONLY Energy-Storage is

Characteristic of a PASSIVE ELEMENT

0)(21

12 tLiL

)(21)( 2 tLitw L



Example Given The iL Current

WaveForm, Find vL for L = 10 mH

The Derivative of a Line is its SLOPE, m

The Differential Reln

Then the Slopes

And the vL Voltage

dtdiLv L

L

sA

sAm 10

1021020

3

3

sAm 10

elsewhere

mstsAmstsA

dtdi

042)/(10

20)/(10

mVVtvHL

sAtdtdi

L 10010100)(1010

)/(10)( 3

3



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

)2(21)4(

21)4,2( 22

LL LiLiw

Running the No.s

µJ0.2)4,2(

)10*20(10*10*5.00)4,2( 233

ww



Numerical Example Given The Voltage Wave

Form Across L , Find iL if • L = 0.1 H• i(0) = 2A

The PieceWise Function

The Integral Reln

A Line Followed by A Constant; Plotting

2

)(Vv

2 )(st

t

dxxvL

iti0

)(1)0()(

20;2)(2)(0

ttdxxvtvt

stttiHL 20;202)(1.0

stsitittv 2);2()(2;0)(

22 )(st

)(Ai42



Numerical Example - Energy The Current

Characteristic The Initial Stored Energy

The Energy Eqn

The “Total Stored Energy”

22 )(st

)(Ai42

)(21)(

21),( 1

22

221 tLitLittw LL

• Energy Stored on Interval Can be POS or NEG

JAHw 2.0)2]([1.0*5.0)0( 2

JAHw 2.88)42(*][1.0*5.0)( 2

Energy Stored between 0-2

)0(21)2(

21)0,2( 22

LL LiLiw

J88)2,0()2(*1.0*5.0)42(*1.0*5.0)2,0( 22

ww

→ Consistent with Previous Calculation



Example

Notice That the ABSOLUTE Energy Stored At Any Given Time Is Non Negative (by sin2)• i.e., The Inductor Is A

PASSIVE Element

Find The Voltage Across And The Energy Stored (As Function Of Time)

)(tv

For The Energy Stored

)(twL



L = 5 mH; Find the Voltage)()( t

dtdiLtv

msmAm

120

)/(122010 sAm

Vvm

00

)/(34

100 sAm

mVv 50

mVv 50

mVmstsAHv 10010);/(20)(105 3

mVv 100



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 VA

66

6330

AAout

VVAI

V2.16V581

AV



Example Total Energy Stored Continue Analysis of

Shorted/Opened ckt

Using Ohm and VA = 16.2V

By KCL at Node-A

VC2 by Ohm

VC1 by Ohm & KVL

A 2.1 A 3A 8.1 A 3

1

21

L

LL

III

A 8.1V63V2.16

2

LI

V 8.10

6A 8.1 6

2

22

C

LC

VIV

V 2.7V 961.2AV 9

6V9

1

1

11

C

C

LC

VV

IV



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 A 1.8 A

2

2

21

21

LL

CC

LIw

CVw

mJ 46.13ktot ww



Caps & Inds Ideal vs. Real A Real CAP has

Parasitic Resistances & Inductance:

A Real IND has Parasitic Resistances & Capacitance:

GenerallySMALLEffect



Ideal vs Real Ideal C & L

Practical Elements “Leak” Thru Unwanted Resistance

)(tv

)(tv

)(ti )(ti

)()( tdtdvCti )()( t

dtdiLtv

)(ti

)(tv

)()()( tdtdvC

Rtvtileak

)(ti

)(tv

)()()( tdtdiLtiRtv leak



Capacitors in Series

If the vi(t0) = 0, Then Discern the Equivalent Series Capacitance, CS

By KVL for 1-LOOP ckt

CAPS in SERIES Combine as Resistorsin PARALLEL



Example Find

• Equivalent C• Initial Voltage

Spot Caps in Series

16

12361

31

211

SS

CC

F2

F1 Or Can Reduce Two at a Time

VVVVtv 31420

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



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

F30

C

+-

V8

V12

CVFQ 240)8)(30(

And Find VC by KVL• VC = 12V-8V = 4V

V4

μC 60V 4μC 240

C

C

VQC



Capacitors in Parallel

Thus The Equivalent Parallel Capacitance

By KCL for 1-NodePair ckt

CAPS in Parallel Combine as Resistors in SERIES



Complex Example → Find Ceq

F6

F2 F4

F3

F4

F12

eqC

F3FCeq 23



Inductors in Series

Thus By KVL For 1-LOOP ckt

INDUCTORS in Series add as Resistors in SERIES

Use The Inductance Law)()( t

dtdiLtv kk )()( t

dtdiLtv S



Inductors in Parallel

And By KCL for 1-NodePair ckt

INDUCTORS in Parallel combine as Resistorsin PARALLEL

Thus

N

jj titi

100 )()(



Example – Find: Leq, i0

mH4 mH2

AAAAti 1263)( 0

Series↔Parallel Summary• INDUCTORS Combine as do RESISTORS• CAPACITORS Combine as do CONDUCTORS



Inductor Ladder Network Find Leq for Li = 4 mH Place Nodes In Chosen

Locations Connect Between Nodes

mH2

mH2

a

b

c

d

When in Doubt, ReDraw• Select Nodes

a

b

cd

mH4

mH2mH2

mH4

eqL

mH6

mHmHmHmHLeq 4.42)4||6(



Find Leq for Li = 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

6||6||6

a

b

c

a

b

c

mH2

mH6mH6

mH6

eqL

mHLeq 7246

144866||)26(6

mHLeq 766



C&L Summary



WhiteBoard Work

HL 50

Let’s Work This Problem

02

004

tteti

ttit

Find: v(t), tmax for imax, tmin for vmin



ByMATLAB

% 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), griddisp('Showing Plot - Hit ANY KEY to Continue')pause%i = 2*t.*exp(-4*t);plot(t, iLn, t, iexp, t, i),griddisp('Showing Plot - Hit ANY KEY to Continue')pauseplot(t, i), griddisp('Showing Plot - Hit ANY KEY to Continue')pausevL_uV =100*exp(-4*t).*(1-4*t);plot(t, vL_uV)i_mA = i*1000;plot(t, vL_uV, t, i_mA), griddisp('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 minimumi_mA_TO = -i*1000;plot(t, i_mA_TO), griddisp('Showing Plot - Hit ANY KEY to Continue')pause[t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1);t_iLminiLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid



ByMATLAB

% 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), griddisp('Showing Plot - Hit ANY KEY to Continue')pause%i = 2*t.*exp(-4*t);plot(t, iLn, t, iexp, t, i),griddisp('Showing Plot - Hit ANY KEY to Continue')pauseplot(t, i), griddisp('Showing Plot - Hit ANY KEY to Continue')pausevL_uV =100*exp(-4*t).*(1-4*t);plot(t, vL_uV)i_mA = i*1000;plot(t, vL_uV, t, i_mA), griddisp('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 minimumi_mA_TO = -i*1000;plot(t, i_mA_TO), griddisp('Showing Plot - Hit ANY KEY to Continue')pause[t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1);t_iLminiLmin = -iLmin_TOplot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-50

0

50

100

150

200



Irwin Prob 5.26: I(t) & v(t)

0

25

50

75

100

125

150

175

200

-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0time (s)

Cur

rent

(mA

)

i(t) (mA)

file = Engr44_prob_5-26_Fall03..xls



Irwin Prob 5.26: I(t) & v(t)

0

25

50

75

100

125

150

175

200

-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0time (s)

Cur

rent

(mA

)

-40

-20

0

20

40

60

80

100

120

Elec

trica

l Pot

entia

l (µV

)

i(t) (mA)v(t) (µV)

file = Engr44_prob_5-26_Fall03.xls



L = 10 mH; Find the Voltage

)()( tdtdiLtv mVv 100

sAHv 3

33

1021020][1010



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Appendix Complex Cap

Example



Numerical Example Given iC, Find vC The Piecewise Fcn for iC

Integrating & Graphing

C= 4µFvC(0) = 0

20 t

0;)(1)0()(0

tdxxiC

vtvt

2;)(1)2()(2

tdxxiC

vtvt

Parabolic

Linear

mst 42 ][1082)( 3 Vttv

>



Power Example For The Previous

Conditions, Find The POWER Characteristic• C = 4 µF• iC by Piecewise curve

From Before the vC

Using the Pwr Relntti 3108)(

mstttp 20,8)( 3

mst 42 elsewheretp ,0)(



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



Energy Example For The Previous

Conditions, Find The ENERGY Characteristic• C = 4 µF• pC by Piecewise curve

Now The Work (or Energy) is the Time Integral of Power

For 0 t 2 mS38)( ttp



Energy Example cont For 2 < t 4 mS

Taking The Time Integral and adding w(2 mS)

Then the Energy Characteristic

42tptnt 128648 2



Cv

Ci

C

FC 2

][0;0

0;)(

5.0

mAtte

tit

C

Current through capacitor

Voltage at a given time t dxxiC

tvt

CC )(1)(

)0(Cv ][0 V

Voltage at a given time t when voltage at time to<t is also known t

tCCC dxxi

Ctvtv

0

)(1)()( 0

)2(Cv 2

0

5.01)0( dxeC

v xC

2

0

5.06 5.0

110*21

xe 61

6 10*6321.015.0

110*21

e V

Charge at a given time )()( tCvtq CC )2(Cq 6321.0*2 C

Voltage as a function of time dxxiC

tvt

CC )(1)(

0;0)( ttvC t

xCC dxe

Cvtv

0

5.01)0()(

0;00);1(10

)(5.06

tte

tvt

C VElectric power supplied to capacitor )()()( titvtp CCC

Energy stored in capacitor at a given time )(21)( 2 tCvtw C

W

J

“Total” energy stored in the capacitor )(21 2 CT Cvw 6266 10)10(*10*2

21

Tw J

SAMPLE PROBLEMIf the current is known ...



SAMPLE PROBLEM

sec)(mt5 10

Compute voltage as a function of timeAt minus infinity everything is zero. Sincecurrent is zero for t<0 we have

sec50 mttsAt

sAt

msAtiC ]/[10*3

10103

515)( 3

3

6

][10*410*3)(0)0(

06

3

VxdxtVVt

CC

][10*50];[810*3 32

3stVt

In particular ][875][

8)10*5(*10*3)5(

233

mVVmsVC

][10)(105 Atimst C

t

CC dxsAtVmVmsV310*5

66

3

]/)[10*10(10*41

810*75)(][

875)5(

][10*1010*5;][10*54

10810*75)( 333

3stVttVC

Charge stored at 5ms)()( tCVtq CC

][810*75*][10*4)5(

36 VFmsq

][)2/75()5( nCmsq

Total energy stored2

21

CCVE

Total means at infinity. Hence

][810*2510*4*5.0

236 JET

Before looking into a formal way to describe the currentwe will look at additional questions that can be answered.

Now, for a formal way to represent piecewise functions....

Given current and capacitance

0;0)( ttVC



][10;825

][105;54

10875

50;83

0;0

)(

2

mst

mstt

mstt

t

tVc][mV

Formal description of a piecewise analytical signal



Find Ceq for Ci = 4 µF

F8

F8

eqC F8

F8

F4F

123284

3328

38

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

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