Lecture 11 Inductance and Capacitance - University of ... 11 Inductance and Capacitance ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008

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

Lecture 11Inductance and Capacitance


Goals1. Find the current (voltage) for a capacitance or inductance given the voltage (current) as a function of time.

2. Compute the capacitances of parallel-plate capacitors.


3. 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 coupled inductances in terms of the currents.

Goals


Capacitor

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

vqC =


Capacitance

( ) ( ) ( )0

0

tqdttitqt

t

+= ∫

( ) ( ) ( )0

0

1 tvdttiC

tvt

t

+= ∫

Cqv

vqC ==


Does DC current flow through a capacitor? No! Only AC current flows through a capacitor!

Capacitance


Determining Current for a Capacitance Given Voltage

Find i(t) given v(t):

dttdv

dttdvCti

tvtCvtq)(10)()(

)(10)()(

6

6

−

−

==

==


Determining Current for a Capacitance Given Voltage

( ) Adt

tdvCti

sV

sxV

dttdv

tidtdv

Axxdt

tdvCti

sVx

sxV

dttdv

101010)()(

1010110)(

0)(0

510510)()(

10510210)(

76

76

66

66

−=−==

−=−

=

=→=

===

==

−

−

−

−From 0 to 2 μs:

from 2 to 4 μs:

from 4 to 5 μs:


Determining Voltage for a Capacitance Given Current

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



[ ]1)10cos(105.0

)10cos(105.0

)10sin(5.0

)0()()(

440

44

0

4

0

−−=

−=

=

=+=

−

−

∫

∫

tx

tx

dtt

tqdttitq

t

t

t



[ ])10cos(150010

)()()( 47 ttq

Ctqtv −===

−


Stored Energy

CtqtqtvtCvvdvCdt

dtdvvCdttptw

dtdvvCtitvtp

tvt

t

t

t2

)()()(21)(

21)()(

)()()(

22

)(

000

======

==

∫∫∫

Ctqtv

tvtqC )()()()(

==


Current, Power and Energy for a Capacitance

53105

310101010)(

)()(

53)5(500311000101000)(

3

3

<<−=

<<=<<=

=

<<−=<<=<<=

−

−

tx

ttforxti

dttdvCti

ttttforttv

C = 10μF


53)5(5.23101010

)()()(

<<−=<<=<<=

=

ttttfort

titvtp




53)5(25.1

315105

)(21)(

2

2

2

<<−=

<<=<<=

=

tt

ttfort

tCvtw


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


∫ ∫

∫

∫

−

−

−

<<+−=+−=

<<==

=+=

−−−−

−−

t

x

x

x

t

mstmsxtxdtxdtx

msttxdtx

tqdttitq

3

3

3

102

63102

0

33

102

0

33

0

42104101101101

20101101

)0()()(

Charge, Voltage, Power and Stored Energy



mstmst

mstt

Ftq

Ctqtv

421040

2010

1.0)()()(

4

4

<<−=

<<=

==μ



mstmstx

mstttvtitp

42101040

2010)()()(

3 <<+−=

<<==

−



mstmstx

mstt

dttptwt

42)1040(105.0

205

)()(

247

20

<<−=

<<=

=

−

∫


Capacitances in Parallel

( )

321

321321321

33

22

11

CCCCdtdVC

dtdVCCC

dtdVC

dtdVC

dtdVCiiii

dtdVCi

dtdVCi

dtdVCi

eq

eq

++=

=++=++=++=

=

=

=


Capacitances in Series

321

03210302010

030201221

1111

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

)(1)(1)(1

CCCC

dttiCCC

dttiC

dttiC

dttiC

dttiCC

Qv

dttiC

dttiC

dttiC

vvvv

eq

ttttt

eqeq

ttt

++=

⎥⎦

⎤⎢⎣

⎡++=++===

++=++=

∫∫∫∫∫

∫∫∫


Capacitance of the Parallel-Plate Capacitor

WLAdAC == ε

mF 1085.8 120

−×≅ε

0εεε r=


Capacitance of the Parallel-Plate Capacitor

( ) nFFxmx

mm

FxdAC

mxmmd

mmxmxcmcmWLA

77.110177010102.010854.8

1011.0

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

124

212

0

4

222

==⎥⎥⎦

⎤

⎢⎢⎣

⎡==

==

====

−−

−

−

−−

ε


Design a 1 μF Capacitor

mmmFx

mxFxW

CdL

polyestermxmd

mcmWd

WLdAFxFC

r

r

rr

93.24)02.0)(/10854.8)(4.3(

)1015)(101(

)(4.3101515

02.02

1011

12

66

0

6

006

===

===

==

====

−

−−

−

−

εε

εμ

εεεεμ

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:

mJwwww

mJVFvCw

total 50

5)100)(10(21

21

21

2

262111

=+==

=== −


Parasitic Elements

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

Cqqqq

CVFxvCq

total μ

μ

1000

100)100)(101(

21

2

6111

=+==

=== −


Parasitic Elements

mJwww

mJVFvCw

mJVFvCw

VFC

Cq

v

FCCC

total

eq

eq

eq

teq

eq

5.2

25.1)50)(10(21

21

25.1)50)(10(21

21

502

100

2

21

26222

26211

21

=+=

===

===

===

=+=

−

−

μμ

μ

Where did the missing energy go?


Inductance


Joseph Henry (1797 – 1878)

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 by drawing one end of the wire from its cup of mercury, a vivid spark is produced."


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

Inductance


( )dtdiLtv =

( ) ( ) ( )0

0

1 tidttvL

tit

t

+= ∫

( ) ( )tLiLididtdtdiLidttptw

tit t2

)(

00 021)( ∫∫ ∫ ====

Inductance

dtditLititvtp )()()()( ==


Inductor Current with Constant Applied Voltage

tAdtVH

tidttvL

titt

t

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

00

0

==+= ∫∫

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

dtdiLtv =)(


Water Hammer


Find the Current given the Voltage

AtitidtL

tistfor

AtHxs

Vxtdtx

Ltistfor

tidttvL

tit

t

1.0)0()2(01)(42

1.0)10150(2

)105.7()105.7(1)(20

)()(1)(

4

2

2

0

22

06

66

0

0

====+=<<

===<<

+=

∫

∫

∫

−

μ

μ


Series Inductances

( )

321

321321221)(

LLLL

dtdiLLL

dtdiL

dtdiL

dtdiLvvv

dtdiLtv

eq

eq

++=

++=++=++==


Parallel Inductances

321

0321030201

320

1

11

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

)0()(1)(

LLLL

dttvLLL

dttvL

dttvL

dttvL

iiitidttvL

ti

eq

tttt

t

eq

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=++=

++==+=

∫∫∫∫

∫


Equivalent Inductance

HHHLeq 532 =+=HHH

HHHHLeq 5.2

1025

55)5)(5( 2

==+

=

HHHLeq 5.35.21 =+=


CP = Capacitance between the coil windings

RP = Resistance of the coil windings

RS = Resistance of connecting wires


Mutual InductanceFields are aiding Fields are opposing

Magnetic flux produced by one coil links the other coil

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Lecture 11 Inductance and Capacitance - University of ... 11 Inductance and Capacitance ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008