Lecture 11 Inductance and Capacitance

Lecture 11Inductance and Capacitance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

GoalsGoals1 Find the current (voltage) for a capacitance1. Find the current (voltage) for a capacitance or inductance given the voltage (current) as a function of timefunction of time.

2. Compute the capacitances of parallel-plate capacitors.


Goals

3 Compute the stored energies in3. Compute the stored energies in capacitances or inductances.

4. Describe typical physical construction of capacitors and inductors and identify parasitic effects.

5 Find the voltages across mutually coupled5. Find the voltages across mutually coupled inductances in terms of the currents.


Capacitor

Energy is stored in gythe electric field that exists between the plates whenthe plates when the capacitor is charged.

qvqC =


Capacitance

CqvqC ==

t

∫

Cv

( ) ( ) ( )0tqdttitqt

+= ∫0t

( ) ( ) ( )1 t

∫( ) ( ) ( )0

0

1 tvdttiC

tvt

+= ∫


0

Capacitance

Does DC current flow through a capacitor? No!


Only AC current flows through a capacitor!

Determining Current for a Capacitance Given Voltage

Find i(t) given v(t):

ddtvtCvtq

)()()(10)()( 6−==

dttdv

dttdvCti )(10)()( 6−==


Determining Current for a Capacitance Given Voltage

td

sVx

sxV

dttdv

)(

10510210)( 6

6 ==−From 0 to 2 μs:

tidtdv

Axxdt

tdvCti

0)(0

510510)()( 66

=→=

=== −

from 2 to 4 μs:

( )tdv

sV

sxV

dttdv

dt

)(

1010110)( 7

6 −=−

=−

μ

from 4 to 5 μs:


( ) Adt

tdvCti 101010)()( 76 −=−== −

Determining Voltage for a Capacitance Given Current

0)0()10sin(50)( 4 === tqtti 0)0()10sin(5.0)( === tqtti



)0()()( =+= ∫ tqdttitqt

)10sin(50 4

0

= ∫

∫

dttt

)10(1050

)10sin(5.0

44

0

=

−

∫ dtt

t

[ ]1)10cos(105.0

)10cos(105.0

440

44

−−=

−=

− tx

tx


[ ])(


[ ])10cos(1500)()()( 47 ttqtqtv −=== [ ])(

10)( 7C −


Stored Energygy

dtdvvCtitvtp )()()( ==

CtqtqtvtCvvdvCdt

dtdvvCdttptw

tvt

t

t

t2

)()()(21)(

21)()(

22

)(

0

====== ∫∫∫tt 000

Ctqtv

tvtqC )()()()(

==Ctv )(


Current, Power and Energy for a Capacitance

311000101000)(

<<=<<=

ttforttv

)()(

53)5(500

=

<<−=tdvCti

tt

C = 10μF

101010)(

)(

3 <<= − tforxtidt

CtiC = 10μF

53105

3103 <<−=

<<=− tx

t



)()()( i

3101010

)()()(<<=

=tfort

titvtp

53)5(5.2310

<<−=<<=

ttt



)(1)( 2= tCvtw

105

)(2

)(

2 <<= tfort

tCvtw

53)5(25.1

3152 <<−=

<<=

tt

t


)(

Exercise 3.2The current through a 0.1 μF capacitor is shown below. Find the charge voltage power and stored energy as functions ofthe charge, voltage, power and stored energy as functions of time and plot them to scale versus time.


Charge, Voltage, Power and Stored Energy

∫ =+=t

tqdttitq )0()()(

∫

∫−

<<==

+

−−x

msttxdtx

tqdttitq

310233

0

20101101

)0()()(

∫ ∫

∫−

<<+−=+−=

<<

−−−−t x

mstmsxtxdtxdtx

msttxdtx

3

63102

33

0

42104101101101

20101101

∫ ∫−

<<++x

mstmsxtxdtxdtx3102 0

42104101101101



Ftq

Ctqtv

1.0)()()( ==μ

tt

mstt

421040

20104

4 <<=

μ


mstmst 421040 4 <<−=


mstttvtitp

2010)()()(

<<=

mstmstx

mstt

42101040

20103 <<+−=

<<=−



t

dttptwt

)()(0

= ∫

mstmstx

mstt

42)1040(1050

205247

20

<<

<<=−


mstmstx 42)1040(105.0 <<−=

Capacitances in Parallel

11 dtdVCi =

22

dVdtdVCi =

( )321321321

33

ddVC

ddVCCC

ddVC

ddVC

ddVCiiii

dtdVCi

eq=++=++=++=

=


( )

321

321321321

CCCCdtdtdtdtdt

eq

eq

++=

Capacitances in Series

ttt

∫∫∫0

30

20

1321

1111111

)(1)(1)(1

Q

dttiC

dttiC

dttiC

vvvv

ttttt⎤⎡

++=++=

∫∫∫∫∫

∫∫∫

0321

03

02

01

01111

)(111)(1)(1)(1)(1 dttiCCC

dttiC

dttiC

dttiC

dttiCC

Qveqeq

++=

⎥⎦

⎤⎢⎣

⎡++=++=== ∫∫∫∫∫


321 CCCCeq++=

Capacitance of the ParallelCapacitance of the Parallel-Plate Capacitorp

WLAAC ==ε WLAd

C ==

mF1085.8 120

−×≅ε

0εεε r=


Capacitance of the Parallel-Plate Capacitor

mxmmd

mmxmxcmcmWLA

10110

02.0)1020)(1010()20)(10(4

222

==

====−

−−

( ) nFFxmx

mm

FxdAC

mxmmd

77.110177010102.010854.8

1011.0

124

212

0 ==⎥⎥⎦

⎤

⎢⎢⎣

⎡==

==

−−

−ε


mxd 101 ⎥⎦⎢⎣

Design a 1 μF Capacitorg μ p

dWL

dAFxFC rr1011 00

6 ==== − εεεεμ

mxmd

mcmW

101515

02.026==

==−μ

mxFxCdL

polyesterr

9324)1015)(101(

)(4.366

=−−

ε

mmmFxW

Lr

93.24)02.0)(/10854.8)(4.3(

))((12

0===

−εε


Impractical!


Parasitic Elements

Rs = Resistivity of the capacitor plates

LS = inductance due to current flow into and out of the capacitor

RP = Leakage resistance of the dielectric


Parasitic Elements

At t = 0 C1 is charged up to 100V, C2 is discharged

The total energy wtotal:

mJVFvCw 5)100)(10(11 262 === −

Jw

mJVFvCw

50

5)100)(10(22

2

111

=

===


mJwwwtotal 521 =+=

Parasitic Elements

At t=0 the switch is closed, and the charge q on the capacitor plates redistributes:the capacitor plates redistributes:

CVFxvCq μ0

100)100)(101( 6111 === −

Cqqqq

total μ1000

21

2

=+==


Parasitic Elements

Cq

FCCCeq

100

221 =+=

μ

μ

mJVFvCw

VFC

Cq

veq

teq

251)50)(10(11

502

100

26211 ===

===

−

μμ

mJVFvCw

mJVFvCw

eq

eq

25.1)50)(10(21

21

25.1)50)(10(22

26222

11

===

===

−

Wh did th i i


mJwwwtotal 5.221 =+= Where did the missing energy go?

Inductance


Joseph Henry (1797 – 1878)y

Writes Henry: "I may however mention one fact which I have not seen noticed in any work ...when a small battery is moderately excited... if a wire thirty or forty feet long be used instead of the short wire, though no spark will be perceptible when the connection is made yet when it is broken bywhen the connection is made, yet when it is broken by drawing one end of the wire from its cup of mercury, a vivid spark is produced."


Inductance

The polarity of the voltage is such as to oppose the change in current (Lenz’s law).current (Lenz s law).


Inductance

( )dtdiLtv =dt

( ) ( ) ( )01 tidttvti

t

+= ∫( ) ( ) ( )0

0

tidttvL

tit

+∫didtditLititvtp )()()()( ==

( ) ( )tLiLididtdtdiLidttptw

tit t2

)(

21)( ∫∫ ∫ ====


( ) ( )dt

00 02∫∫ ∫

Inductor Current with Constant Applied Voltage

Addtt

)()10(1)()(1)( ∫∫ tAdtVH

tidttvL

tit

)5()10(21)()(1)(

00

0

==+= ∫∫

What happens if we open the switch at t = 1 sec?diLtv )(


dtLtv =)(

Water Hammer


Find the Current given the Voltageg g

t

V

tidttvL

tit

t

)1057(

)()(1)(

22 6

0

0+= ∫

AtHxs

Vxtdtx

Ltistfor

1

1.0)10150(2

)105.7()105.7(1)(20

40

22

06

66 ===<< ∫ −

μ


AtitidtL

tistfor 1.0)0()2(01)(422

====+=<< ∫μ

Series Inductances

( ))( diLLLdiLdiLdiLvvvdiLtv ++=++=++== ( )321321221)(

LLLL

dtLLL

dtL

dtL

dtLvvv

dtLtv eq ++=++=++==

321 LLLLeq ++=


Parallel Inductances

)0()(1)( iiitidtttit

∫ 320

1

)(1)(1)(1)(1

)0()()(

dddd

iiitidttvL

ti

tttt

eq

⎟⎞

⎜⎛

++==+=

∫∫∫∫

∫

0321030201

11

)()()()( dttvLLL

dttvL

dttvL

dttvL

=

⎟⎟⎠

⎜⎜⎝ ++

=++= ∫∫∫∫


321 LLLLeq ++=

Equivalent Inductance

HHHLeq 532 =+=HHHHLeq 5.225)5)(5( 2=== q

HHHeq 1055 +

HHHLeq 5.35.21 =+=


RS = Resistance of connecting wires

CP = Capacitance R R i f

connecting wires

CP Capacitance between the coil windings

RP = Resistance of the coil windings


Mutual InductanceFields are aiding Fields are opposing

Magnetic flux produced by one coil links the other coil


Documents

Lecture 11 Inductance and Capacitance