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Capacitors
Capacitors store charge
++++++
+Q
+++++
E
+
++ ++
–
Q=CV
dQ
I
dt
––––––––
d
––––––––
–– ––
Parallel plate
capacitor
E
V
d
dV
IC
dt
–Q
plate area
A
C
d
Dielectric permittivity ε=εrε0
0  8.854  10 1 2 AsV1m1
Units: farad [F]=[AsV-1] more usually μF, nF, pF
1
Capacitor types
Range
Maximum voltage
Ceramic
1pF–1μF
30kV
Mica
1pF–10nF
500V
Plastic
100pF–10μF
1kV
Electrolytic 0.1μF–0.1F
Parasitic
~pF
(very stable)
500V
Polarised
2
Capacitors
C1
Capacitors in parallel:
C T otal 
C
n
n
 C1  C2  C3 
C1
Capacitors in series:
1

CT
C2 C3
C2
1
1
1
1





C1 C2 C3
n Cn
Q
Q1 Q2
V



CT
C1 
C2

V1
C3
Q1=Q2
+Q1 –Q1 +Q2 –Q2
V1
V2
V2
3
Capacitors – energy stored
dV t'
W   Pt'dt'   It'V t'dt'   C
V t'dt'
0
0
0
dt'
t
t
V
 C  V' dV' 
0
t
1
2
CV 2
2
Q
W  12 CV 2  12 QV 
2C
Energy stored as electric field
4
1
RC circuits
2
V0
+
+
Initially VR=0 VC=0
R
Switch in Position “1” for a long time.
1
Then instantly flips at time t = 0.
I
+
C
Capacitor has a derivative!
How do we analyze this?
There are some tricks…
ALWAYS start by asking these three questions:
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)
5
3). What can we say about the INSTANT after switch flips? (easy if you know trick)
The Trick!!
• Remember  𝐼 =
𝑑𝑉
𝐶
𝑑𝑡
• 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- is the same as at t = 0+.
Note for Todd : Put problem on board and work through these points!
6
1
RC circuits
2
V0
+
R
Initially at t = 0VR=0 VC=0 I=0
1
I
C
2
 V0  VR  VC  0
V0  VR  VC
Q
 IR 
C
differentiate wrt t
dI I
0R

dt C
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
7
dI I
0R

dt C
I
dI

RC dt

1
dt 
RC

1
dI
I
t
 ln I  a
RC
 ln I  ln b
 lnbI
e
 t 


 RC 
 bI
1
I   e
b 
 I0e
 t 


RC


 t 


 RC 
8
V0
+
R
V0  t 
It  
e
R
  RC
9
Integrator
R
Vi
C
Vint
Capacitor as integrator
1 t
Vint  VC   Idt '
C 0
1 t Vi  VC
VC  
dt ' If RC>>t
0
C
R
VC<<Vi
1 t

Vdt
'
i

0
RC
10
Differentiation
Vi
Vdiff
C
R
Vdiff
 dVi dVR 
 VR  RI  RC


dt 
 dt
Small RC 
Vdiff
dVi
dVR

dt
dt
dVi
 RC
dt
11
Inductors
Solenoid
Electromagnetic
induction
N turns
0NI
B

I
B

Vind
Vind
N1
dI
 0 A
N2
 dt
I nduc tan c e
Vind
d

dt
Magnetic flux    BdA
12
Self Inductance
I
dI
V L
dt
N2
L  0 A

for solenoid
Mutual Inductance
I
M
Units: henrys [H] = [VsA-1]
dI
V M
dt
N1N2
M  0 A

13
Inductors
Wire wound coils - air core
- ferrite core
Wire loops
Straight wire
14
Series / parallel circuits
L1
L2
L3
LT 
L
n
n
dI
dI
dI
V  L1
 L2
 L3
dt
dt
dt
Inductors in series: LTotal=L1+L2+L3…
I
I1
I2
L1
L2
I  I1  I2
dI dI1 dI2


dt 
dt 
dt

V
V
V
L
L1
L2
Inductors in parallel
1

LT
1
n L
n
1
1



L1 L 2
15
Inductors – energy stored
t
W   Pt'dt'
0

 It'Vt'dt'
t
0
dIt'
  It'L
dt'
0
dt
t
I
 L  IdI'
0
2
 LI
1
2
Energy stored as magnetic field
16
The Next Trick!!
• Remember  V =
𝑑𝐼
𝐿
𝑑𝑡
• 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- is the same as at t = 0+.
– Does all this seem familiar? It is the concept of
“duality”.
17
Initially t=0VR=V0 VL=0 I=V0/R
RL circuits
2
V0
+
+
R
1
+
20R
+
For times t > 0
0  VR 20  VR  VL
dI
0  20 IR  IR  L
dt
dI
0  21IR  L
dt
I
L
At t=0+  Current in L
MUST be the same
I=V0/R So…
VR=V0 VR20=20V0
and VL = -21V0 !!!
And at t=∞ (Pos. “1”)
VR=0 VL=0 I=0
dI 21R

I 0
dt
L
I t  dI
t dt
I 0  I   0 
L

21R
 It  
t
  
ln

 I0 

It   I0 exp 

t


It  0  0
V0
 t
I t  0   exp  
R
 
V0
I 0  
R
V0
I t  0 
R 19
V0
+
R
V0 
 t
I t    exp 
R
 

 

L
Notice Difference in Scale!
20
I2
R2
V0
+
R1
L
t<0 switch closed I2=?
t=0 switch opened
t>0 I2(t)=?
voltage across R1=?
Write this problem down and DO attempt it at home
21
Physics 3 June 2001
Notes for Todd:
Try to do it cold and consult the backup
slides if you run into trouble.
22
LCR circuit
2
V0
+
t0
0
Vt    for
t0
V0
R
1
L
V(t)
I(t)=0 for t<0
C
t0
VR  VL  VC  V0
dI Q
IR  L

 V0
dt C
d2I R dI
1


I0
2
L dt LC
dt
23
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!
24
BACKUP SLIDES
25
26
27
28
29
1
RL circuits
2
V0
+
R
Initially VR=0 VL=0
2
1
V0  VR  VL
L
di R
 i0
dt L
It  di
t dt
I0 i   0 
 It  
t
  
ln

 I0 

It   I0 exp 


L
R
di
 iR  L
dt
V0
di R
 i
dt L
L

It   I0 exp 

It  0  0
t


V0
It  
R
t  V0

 R
 V0
I0 
R


1  exp 


t 
 
 
30
V0
+
R
V0
It  
R
L


1  exp 


t 
 
 
L

R
31
This way up
32
R1
V0
Norton
R0
R0
V0
R1
R1
R1
R1
L
L
R1
2
V0
2
R0
V0
R1
R1
2
R0
L
L
V0
dI
R

 I  1  R0   L
2
dt
 2

33
V0
dI
 R1

 I   R0   L
2
dt
 2

This way up
34
This way up
35
Energy stored per unit volume
Inductor
Capacitor
2
W
2

W  LI
1
2
0 N 2

A
I0
2


1
2
2
0 AH
B
NI
H

0

1
2
1
2
CV 2
0 A 2
V
d
0 2

E  Vol
2
0 A 

C  d 


V

E  d 


0 2

H  Vol
2
36