capacitors capacitors store charge + ++ – – – e +q –q ++++++ +++++ ––––––––...
TRANSCRIPT
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CapacitorsCapacitors store charge
+ ++ ++
– – – ––
E
+Q
–Q
++++++ +++++
–––––––– ––––––––
d
dA
C
Q=CV
dV
E
11120 mAsV10854.8 dt
dQI
dtdV
CI
Parallel plate capacitor
plate area
Dielectric permittivity ε=εrε0
Units: farad [F]=[AsV-1] more usually μF, nF, pF
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Capacitor typesRange Maximum voltage
Ceramic 1pF–1μF 30kV
Mica 1pF–10nF 500V (very stable)
Plastic 100pF–10μF 1kV
Electrolytic 0.1μF–0.1F 500V
Parasitic ~pF Polarised
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CapacitorsCapacitors in parallel: C1 C2 C3
321n
nTotal CCCCC
Capacitors in series: C1 C2 C3
321n nT C
1C1
C1
C1
C1
21 V
2
2
V
1
1
T CQ
CQ
CQ
V V1 V2
+Q1 –Q1 +Q2 –Q2
Q1=Q2
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Capacitors – energy stored
2
21
V
0
t
0
t
0
t
0
CV'dV'VC
'dt'tV'dt'tdV
C'dt'tV'tI'dt'tPW
C2Q
QVCVW2
212
21
Energy stored as electric field
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RC circuits
+V0
R
C
1
2
Initially VR=0 VC=01
Switch in Position “1” for a long time.
Then instantly flips at time t = 0.
Capacitor has a derivative!How do we analyze this?
There are some tricks…
1.) What does the Circuit do up until the switch flips?(Switch has been at pos. 1 for a VERY long time. easy)
2). What does the circuit do a VERY long time AFTER the switch flips? (easy)
3). What can we say about the INSTANT after switch flips? (easy if you know trick)
I
+
+
ALWAYS start by asking these three questions:
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The Trick!!• Remember • Suppose the voltage on a 1 farad Capacitor changes
by 1 volt in 1 second.– What is the current?– What if the same change in V happens in 1 microsecond?– So…What if the same change in V happens instantly?
• Rule: It is impossible to change the voltage on a capacitor instantly!– Another way to say it:
The voltage at t = 0-e is the same as at t = 0+ .e
Note for Todd : Put problem on board and work through these points! 6
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RC circuits
+V0
R
C
1
2
Initially at t = 0- VR=0 VC=0 I=0
1
2
CI
dtdI
R0
C
QIR
VVV
VVV
CR
CR
0
0
0
differentiate wrt t
Then at t = 0+ Must be that VC=0
If VC=0 then KVL says VR=V0 and I=V0/R.Capacitor is acting like a short-circuit.
Finally at t = +∞VR=0 VC=V0 I=0
I
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CI
dtdI
R0
dtdI
RCI
dII1
dtRC
1
bIln
blnIln
aIlnRC
t
bIe RCt
RCt
0
RCt
eI
eb1
I
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t0 eRV
tI+V0
R
RC
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Integrator Capacitor as integrator
R
CVi Vint
t
int C 0
ti C
C 0
t
i0
1V V Idt '
C1 V V
V dt 'C R
1Vdt '
RC
If RC>>t
VC<<Vi
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Differentiation
RC
Vi Vdiff
dtdV
dtdV
RCRIVV RiRdiff
Small RC dt
dVdtdV Ri
dtdV
RCV idiff
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Inductors
I
Solenoid N turns
NI
B 0
Electromagnetic induction
B
Vind
BdA flux Magnetic
dtd
Vind
dtdI
NN
AV
cetanInduc
21
0ind
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Self Inductance
Mutual Inductance
I
dtdI
LV
2
0
NAL
I
dtdI
MV M
21
0
NNAM Units: henrys [H] = [VsA-1]
for solenoid
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InductorsWire wound coils - air core
- ferrite core
Wire loops
Straight wire
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Series / parallel circuitsL1 L2 L3
Inductors in series: LTotal=L1+L2+L3…
n
nT LL
L1 L2
Inductors in parallel
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n nT
L1
L1
L1
L1
dtdI
LdtdI
LdtdI
LV 321
I
I1 I2
2
2
1
1
21
LVdtdI
LVdtdI
LVdtdI
III
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Inductors – energy stored
t
0'dt'tPW
221
I
0
t
0
t
0
LI
'IdIL
'dtdt
'tdIL'tI
'dt'tV'tI
Energy stored as magnetic field
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The Next Trick!!• Remember • Suppose the current on a 1 henry Inductor changes
by 1 amp in 1 second.– What is the voltage?– What if the same change in I happens in 1 microsecond?– So…What if the same change in I happens instantly?
• Rule: It is impossible to change the current on an inductor instantly!– Another way to say it:
The current at t = 0-e is the same as at t = 0+ .e– Does all this seem familiar? It is the concept of
“duality”.17
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RL circuits
+V0
R
L
1
2
20R
Initially t=0-
VR=V0 VL=0 I=V0/R
And at t=∞ (Pos. “1”)VR=0 VL=0 I=0
At t=0+ Current in L MUST be the sameI=V0/R So…
VR=V0 VR20=20V0
and VL = -21V0 !!!
I
+
++
dt
dILIR
dt
dILIRIR
VVV LRR
210
200
0 20
For times t > 0
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texp0ItI
ttI
I
dt
I
dI
IL
R
dt
dI
00
021
t0ItI
ln
R
L
21
00tI R
VI 00
t
R
VtI exp0 0
R
VtI 00
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t
R
VtI exp0
+V0
R
L
Notice Difference in Scale! 20
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+V0
R2
R1L
I2
t<0 switch closed I2=?t=0 switch openedt>0 I2(t)=?
voltage across R1=?
Write this problem down and DO attempt it at home 21
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Physics 3 June 2001 Notes for Todd: Try to do it cold and consult the backupslides if you run into trouble.
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LCR circuit
+V0
R
C
1
2
L
V(t)
0t
0tfor
V
0tV
0
I(t)=0 for t<0
0ILC1
dtdI
LR
dtId
VCQ
dtdI
LIR
VVVV
0t
2
2
0
0CLR
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R, L, C circuits
• What if we have more than just one of each?
• What if we don’t have switches but have some other input instead?– Power from the wall socket comes as a
sinusoid, wouldn’t it be good to solve such problems?
• Yes! There is a much better way!– Stay Tuned!
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BACKUP SLIDES
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RL circuits
+V0
R
L
1
2Initially VR=0 VL=01
2
dtdi
LiR
VVV LR0
texp0ItI
t
0
tI
0I
dtidi
0iLR
dtdi
t0ItI
ln
RL
00tI
RVt
exp0ItI 0
RV
0I 0
texp1
RV
tI 0
LV
iLR
dtdi 0
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texp1
RV
tI 0+V0
R
L
RL
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This way up32
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V0
R1
R1
R0
L
R1
R0
L
R1
0
1
V
R
R0
L
0
1
V
R1R
2R0
L
0V
2
1R
2
Norton
0 10
V R dII R L
2 2 dt 33
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This way up
0 10
V R dII R L
2 2 dt
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This way up35
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VolH2
NIBH HA
IN
A2
LIW
20
0
202
1
20
20
221
Energy stored per unit volumeInductor Capacitor
VolE2
VdA
CVW
20
2021
221
dA
C 0
dV
E
36