### Document

```Nobel Prize in Physics, 2011
The Nobel Prize in Physics 2011 has been awarded to
Saul Perlmutter, Brian P Schmidt and Adam G Riess
for discovering the accelerating expansion of the
universe
Nobel Prize in Physics, 2010
"for groundbreaking experiments regarding the
two-dimensional material graphene"
Since it is practically transparent and a good
conductor, graphene is suitable for producing
transparent touch screens, light panels, and maybe
even solar cells. Graphene transistors are predicted
to be substantially faster than today’s silicon
transistors and result in more efficient computers.
Andre Geim
Konstantin
Novoselov
Geim and Novoselov extracted the
graphene from a piece of graphit such as
is found in ordinary pencils. Using
regular adhesive tape they managed to
obtain a flake of carbon with a thickness
of just one atom.
Lectures 10-11 (Ch. 26)
R-C circuits
(R and C in series, R and C in parallel)
II. Transient regimes
1.Charging the capacitor
a)
b)
initial conditions
temporal dynamics
2.Discharging the capacitor
a)
b)
initial conditions
temporal dynamics
III. Applications
NB: In steady state regime current does
not go through the capacitor: I=i(t→∞)=0!
=0!
=0!
1.C and R are in series
Switch has been closed for a long time.
I=i(t→∞)=0 →Vab=IR=0,
Vbc =ε, q(t →∞) = Q=εC (capacitor is fully charged)
R1
2. C and R are in parallel
I1
a
b I1
I=0
R2
Switch has been
closed for a long time.
c
I=i(t→∞)=0 but I1≠0 !
I1= ε/(R1+R2)
Vab= εR1 /(R1+R2)
Vbc =ε R2 /(R1+R2)
q(t →∞) = Q=CVbc =CεR2 /(R1+R2) (capacitor is fully charged)
Example. (Problem 26.74)
1. Switch is open. In the steady state regime find:
a)voltage at the points a and b; b) charge on each capacitor.
Current does no go through the capacitor in the steady
state regime. Hence there is no current in this circuit.
a) Vb=0x3Ω=0,
Va=18V-0x3Ω=18V
b) Q1=3μFx18V=54 μC
Q2=6μFx18V=108 μC
V=0
2. Switch is closed. In the steady state regime find:
a)voltage at the points a and b; b) charge on each capacitor.
Current does not go through the capacitors but it does go through 6Ω and 3Ω resistors.
I=18V/9Ω=2A , Va=Vb=2Ax3Ω=6V
Q1=3μFx6V=18 μC
Q2=6μFx12V=72 μC
Example.
1. Switch is open. In the steady state regime find:
a) voltage at the points a and b; b) charge on each capacitor.
Current does not go through the capacitors but it does
go through 14Ω and 10Ω resistors.
V=24V
14Ω
a
10Ω
Q1
S
2 μF
b
Q2
2 μF
I=24V/24Ω=1A
a) Va=1Ax10Ω=10V
Vb=24V/2=12V
b) Q1=Q2=Q=24VxCeq=24Vx1μF=24μC
V=0
2. Switch is closed. In the steady state regime find:
a) voltage at the points a and b; b) charge on each capacitor.
Current is the same as in the previous part:
I=24V/24Ω=1A
Va=1Ax10Ω=10V (also the same)
Vb=Va=10V
Q2=10Vx2μF=20μC; Q1=14Vx2μF=28μC
Example. Switch is closed at t=0. Find the final charge on the capacitor at t→∞.
i1+i2
S
12V
i2
10 Ω
i1
1Ω
a
1.5μF
9Ω
b
5Ω
o
Q  CVab , Vab  Vao  Vbo  i1 9  i2 5
12V
12V
i1 
 1.2 A, i2 
 0.8 A
10
15
Vab  10.8V  4V  6.8V
Q  10.2C
Charging the capacitor
I.C and R in series
1. Initial conditions
Switch is closed at t=0.
Capacitor is initially uncharged:
q( t=0)= 0→
Vbc(t=0)=q( t=0)/C=0 →
Vab (t= 0)=ε → i(t=0)=ε /R
Current appears at t=0.
at t=0
Example. All capacitors are initially uncharged. Switch is closed at t=0. a)Find Vab(t=0); b) Find
Vab(t→∞).
a
S
12V
a)t  0 : Vac  0 (becauseq  0),
3μF
c
5Ω
b
1Ω
4μF
Vab  Vac  Vcb
12V
Vcb 
 5  10V  Vab  10V
6
Q
b)t   : Vcb  0 (becausei  0),Vac 
3F
2μF
Q  12VC eq ,
1
1
1


,
Ceq 3F 6F
24C
Ceq  2F , Vac 
 8V
3F
Example. Problem 26.72 . The capacitor is initially uncharged. The switch is
closed at t=0. a) Immediately after the switch is closed, what is a current
through each resistor? b) what are the final currents and final charge ?
8Ω
i1+i2
S
i2
3Ω
6Ω
42V
i1
4μF
42V
a)t  0 : Vcb  0  Req  10  i1  i2 
 4.2 A
10
i1 6  i2 3  i2  2i1  3i1  4.2 A  i1  1.4 A, i2  2.8 A
42V
b)t   : i2  0, i1  i2  i1 
 3A
14
Q  18V  4F  72C
2. Transient regime
  iR  q / C  0, i 
dq
dt
dq C  q
dq
dt


 

dt
RC
q  C
CR
0
0
q
t

q  C
dt
q  C
ln
 

 e CR
 C
CR
 C
0
t

t

t
q  Q (1  e ); Q  C ,   CR
QV CV
[ ]  [
]
 1s
VI
VA
t
t
t

dq Q  C 


i

e 
e  i  Ie ; I 
dt CR
CR
R
Example. Verify an energy balance in the process of charging the capacitor.


U bat   Pout dt     idt  I
0

0


0

0
0
t
 e dt  I  
U R   PR dt  R  i 2 dt  RI 2  e
0


2t


 RC
R
  2C
RI 2 R 2 RC  2C
dt 


2
2
2R
2
Q 2  2C 2  2C
UC 


2C
2C
2
One half of the energy provided by the battery is stored in capacitor,
another half is dissipated in the resistor.
• Example. Which fraction of the maximum possible energy was
stored in capacitor by the moment when current droped 3
times?

t*
t*

I
1


i (t*)  Ie   e 
3
3
t*

2

q(t*)  Q(1  e )  Q
3
4U C max
q 2 (t*) 4Q 2
U C (t*) 


2C
9  2C
9
Example. Switch is closed at t=0. a)Find Vab(t*) if q(t*)=Q/3, where Q is the charge of the
capacitor when t→∞; b) Find characteristic time constant of this circuit.
a
S
12V
3μF
c
5Ω
b
1Ω
4μF
2μF
Vab  Vac  Vcb

t

t
Q
2

q (t*)  Q (1  e )   e 
3
3
t

2

i (t*)  Ie  I
3
2
Q
Vab  I  5 
3
3  3F
Q  12VC eq , Ceq  2 F , Q  24C
12V
I
 2A
6

Charging the capacitor.
2. C and R in parallel
1. Initial conditions.
Capacitor is initially uncharged. Switch is
closed at t=0. Current i1 appears at t=0.
q( t=0)= 0→
R1
i1+i2
R2
b
Vbc(t=0)=q( t=0)/C=0 →i2 (t=0)=0
Vab(t=0)= ε → i1+i2 (t=0)=i1=ε /R1
a
i1
c
i2
Example. Capacitor is initially uncharged. At t=0 the switch is closed. Find a)the initial and the
final current; b)the final charge.
S
10V
20Ω
c
0.5μF
40Ω
b
10V
a )t  0 : Vcb  0, i 
 0.5 A
20
10V
t i
 0.17 A
60
b)Q  Vcb (t  )  0.5F  0.17 A  40  0.5F  3.3C
2. Transient regime.
Charge increases, current, i1-i2 decreases till steady state regime considered above is
established at t →∞ .
q
1
q
  i1R1   0  i1  (  )
C
R1
C
q
  i1R1  i2 R2  0  i2 
CR2
CReq  qR1
dq

q
 i1  i2  

dt
R1 CReq
CReq
R1
i1
R2
b
i1-i2
1
1
1
R1  R2
 
 Req 
Req R1 R2
R1 R2
t
qR1  CReq
dq
dt
t


,
ln(
)


0 qR1  CReq 0 CReq
 CReq
CReq
q

t
q  Q(1  e  ), Q 
CR2
,   CReq
R1  R2
a
c
i2
Discharging the capacitor.
I.C and R in series
1.Initial conditions.
Capacitor is initially charged and
disconnected from the battery. Switch is
closed at t=0. Current appears at t=0.
q( t=0)= Q→
Vbc(t=0)=Q/C=0 →
Vab (t= 0)=-Q/C →
i(t=0)=-Q/CR
at t=0
2.Transient regime
 iR  q / C  0, i 
dq
dt
dq
q
dq
dt


 

dt
RC
q
CR
Q
0
q
t
t

q
t
ln
   q  Qe  ,  CR
Q

t
dq
Q 
i
 e
dt

Example. Verify an energy balance in the process of discharging the capacitor.
Q2
UC 
2C



U R   PR dt  R  i 2 dt  R 
0
0
0
Q2
2
e

2t

Q2R Q2
dt 

2
2C
Example. The switch was closed for a long time so that the capacitor was fully charged (Q=72 μ C, see
solution of this part of the problem above).Then the switch is opened at t=0. a) Immediately
after the switch is opened, what is a current through each resistor? b) what are the currents
through each resistor when charge decreased twice?
8Ω
S
42V
6Ω
3Ω
4μF
a )t  0 : through8 , c urrent  0;
72C
through6 and 3 : I 
 2A
4 F  9
b)q (t*)  Qe

i (t*)  Ie

t

Q/2e
t

 I / 2  1A

t

 1/ 2
Discharging the capacitor.
2. C and R in parallel
1. Initial conditions.
Capacitor is initially charged and disconnected from the battery. Switch is closed at t=0.
Current appears at t=0.
q( t=0)= Q→
Vbc(t=0)=Q/C→i2 (t=0)=Q/CR2
Vba(t=0)= Q/C → i1 (t=0)=Q /CR1
2. Transient regime.
Charge decreases , currents also decrease to zero.
q
dq
q
 i1 R1  i2 R2 ;
 (i1  i2 )  
C
dt
ReqC
t

1
1
1


; q  Qe  ,   ReqC
Req R1 R2

t

t

t
Qe 
Qe 
Qe 
i1 
, i2 
, i1  i2 
CR1
CR2
CReq
R1
i1
a
R2
b
i1+i2
c
i2
Example. Switch was closed for a long time. Then it was opened at t=0. At which
moment of time charge decreases up to 5% of its value at t=0.
S
12V
10 Ω
1Ω
1.5μF
9Ω
5Ω

t*
q (t*)  Qe   0.05Q  e
t*   ln 20,  CReq

t*

 0.05 
1
1
1


 Req  6.16
Req 11 14
Applications of R-C circuits
1. Sawtooth voltage (pacemakers, turn signals, windshield wipes, etc.)
V
Gas filled tube
2. Frequency filtration
HF output,LF filtr
(ώ<1/RC do not go
through)
ε(t)
~
LF output,HF
filtr (ώ>1/RC) do
not go through)
T
t
1. Example. You need to built a pacemaker with 72beats/min.
You have capacitor with C=2.5μF and it is known that the
discharge starts at v=0.45 ε. Which resistance you should
select?
V
ε
Gas filled tube
T
t
T
1 60s

 0.83s
f
72

T

T
V (t )   (1  e  )  0.45  e   0.55 
T
1
T
0.83s
 ln
 RC 
R
 0.2M
6

0.55
ln 1.82
2.5 10 F ln 1.82
```