lecture 11 inductance and capacitance
TRANSCRIPT
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.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Capacitor
Energy is stored in gythe electric field that exists between the plates whenthe plates when the capacitor is charged.
qvqC =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Capacitance
CqvqC ==
t
∫
Cv
( ) ( ) ( )0tqdttitqt
+= ∫0t
( ) ( ) ( )1 t
∫( ) ( ) ( )0
0
1 tvdttiC
tvt
+= ∫
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0
Capacitance
Does DC current flow through a capacitor? No!
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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−==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) Adt
tdvCti 101010)()( 76 −=−== −
Determining Voltage for a Capacitance Given Current
0)0()10sin(50)( 4 === tqtti 0)0()10sin(5.0)( === tqtti
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Determining Voltage for a Capacitance Given Current
)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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
[ ])(
Determining Voltage for a Capacitance Given Current
[ ])10cos(1500)()()( 47 ttqtqtv −=== [ ])(
10)( 7C −
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Stored Energygy
dtdvvCtitvtp )()()( ==
CtqtqtvtCvvdvCdt
dtdvvCdttptw
tvt
t
t
t2
)()()(21)(
21)()(
22
)(
0
====== ∫∫∫tt 000
Ctqtv
tvtqC )()()()(
==Ctv )(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Current, Power and Energy for a Capacitance
)()()( i
3101010
)()()(<<=
=tfort
titvtp
53)5(5.2310
<<−=<<=
ttt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Current, Power and Energy for a Capacitance
)(1)( 2= tCvtw
105
)(2
)(
2 <<= tfort
tCvtw
53)5(25.1
3152 <<−=
<<=
tt
t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)(
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.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charge, Voltage, Power and Stored Energy
Ftq
Ctqtv
1.0)()()( ==μ
tt
mstt
421040
20104
4 <<=
μ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
mstmst 421040 4 <<−=
Charge, Voltage, Power and Stored Energy
mstttvtitp
2010)()()(
<<=
mstmstx
mstt
42101040
20103 <<+−=
<<=−
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charge, Voltage, Power and Stored Energy
t
dttptwt
)()(0
= ∫
mstmstx
mstt
42)1040(1050
205247
20
<<
<<=−
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
mstmstx 42)1040(105.0 <<−=
Capacitances in Parallel
11 dtdVCi =
22
dVdtdVCi =
( )321321321
33
ddVC
ddVCCC
ddVC
ddVC
ddVCiiii
dtdVCi
eq=++=++=++=
=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( )
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
++=
⎥⎦
⎤⎢⎣
⎡++=++=== ∫∫∫∫∫
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
321 CCCCeq++=
Capacitance of the ParallelCapacitance of the Parallel-Plate Capacitorp
WLAAC ==ε WLAd
C ==
mF1085.8 120
−×≅ε
0εεε r=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Impractical!
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
=
===
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
=+==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
mJwwwtotal 5.221 =+= Where did the missing energy go?
Inductance
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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."
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Inductance
The polarity of the voltage is such as to oppose the change in current (Lenz’s law).current (Lenz s law).
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Inductance
( )dtdiLtv =dt
( ) ( ) ( )01 tidttvti
t
+= ∫( ) ( ) ( )0
0
tidttvL
tit
+∫didtditLititvtp )()()()( ==
( ) ( )tLiLididtdtdiLidttptw
tit t2
)(
21)( ∫∫ ∫ ====
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) ( )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 )(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
dtLtv =)(
Water Hammer
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
AtitidtL
tistfor 1.0)0()2(01)(422
====+=<< ∫μ
Series Inductances
( ))( diLLLdiLdiLdiLvvvdiLtv ++=++=++== ( )321321221)(
LLLL
dtLLL
dtL
dtL
dtLvvv
dtLtv eq ++=++=++==
321 LLLLeq ++=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Parallel Inductances
)0()(1)( iiitidtttit
∫ 320
1
)(1)(1)(1)(1
)0()()(
dddd
iiitidttvL
ti
tttt
eq
⎟⎞
⎜⎛
++==+=
∫∫∫∫
∫
0321030201
11
)()()()( dttvLLL
dttvL
dttvL
dttvL
=
⎟⎟⎠
⎜⎜⎝ ++
=++= ∫∫∫∫
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
321 LLLLeq ++=
Equivalent Inductance
HHHLeq 532 =+=HHHHLeq 5.225)5)(5( 2=== q
HHHeq 1055 +
HHHLeq 5.35.21 =+=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
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
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Mutual InductanceFields are aiding Fields are opposing
Magnetic flux produced by one coil links the other coil
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.