al capacitor p.34. a capacitor is an electrical device for storing electric charge and energy. -...
TRANSCRIPT
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AL Capacitor
P.34
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P.34
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P.34
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P.34
Capacitor
Resistor
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P.34
A capacitor is an electrical device for storing electric charge and energy.
- consists of two parallel metal plates with an insulator (dielectric) between the plates
Dielectric – air, paper, wax, ceramic, mica
Capacitor
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P.34
Air capacitor Disc ceramic capacitor Film capacitor
Dipped mica capacitor Metallized paper c
apacitor
Electrolytic capacitor
Types of capacitors
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P.34
fixed capacitor
variable capacitor
Types of capacitors
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P.34Charging
+++
---
Vx
Q σ E=σ/ε V=Ed
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P.35Discharging
+++
---
Vx
Q σ E=σ/ε V=Ed
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P.35
ε/R
-ε/R
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P.35Capacitance C
Q σ E=σ/ε V=Ed
VQ
V
QC Unit : Farad (F)
a
QV
4
1For a point charge,a
V
QC 4
The capacitance of a capacitor is one farad (1 F), if the capacitor stores one coulomb (1 C) of charge when there is a potential difference of one volt (1 V) applied across it.
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P.36 Experiment
+++
- - -
Rmax I= V/R
Set R to max I is min.
Reduce R Keep I constant Vc
VR
ε = VC + VR
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P.37 If Q is constant,
++++++++++
----------
V
QC
A +Q -Q B
A
Q
d
VE E is constant,
If d is reduced, V is reduced
C is increased
V
QC
If A is increased, E is reduced
C is increased
V is reduced
d
A
V
QC
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P.37
Increase C by:(1) increasing overlapping area (A)(2) placing two plates closer (d)(3) replacing dielectric with higher permittivity ()
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P.38 Dielectric material
External E field
E field by dipole
Resultant E field
For all insulators, r > 1.0dielectric increases capacitance (charge-storing ability)
- molecules of dielectric are polarized- one end of each molecule has excess +ve charge- other end has excess –ve charge- charges appear on surface- potential difference is induced- By C = Q/V, C increases
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P.38 Dielectric material
V+V-
VC = V+ - V-More and more charges are needed to increase VC to ε
Qd
A
V
QC
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P.38
Material Relative permittivity (r)
Vacuum 1.0000
Air 1.0005
Oil 2 – 5
Paper 2 – 6
Glass 8
Ceramic 80 – 1 200
Capacitance of capacitors with different materials between plates
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Conducting material
External E field
E field by induced charges
Zero resultant E field
P.38
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P.39
CVQ
)(1025.0
10100 46
F
dtdVdtdQ
C
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P.39
d
VE
)/(108101
800 53
mVE
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P.39
d
AC o
Cd
A
d
A
d
AC ooo 22
2/''
VV '
a.
b.
c. QVCVCQ 2))(2('''
d. EEoo
22'
'
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P.39
V
QC
+++
+++
A +? +? B
++++
++
+900μ +100μ
Q=400μ)(100
4
400F
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P.39
''
d
AC
C
d
A
d
A
6
5
6
5
56
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d
AC
C
QV
P.40
Q is constant Overlapping area is decreased
C is decreased
V is increased
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P.40
Q is constantd
AC
d is decreased C is increased
C
QV V is decreased
A
QE E remains unchanged
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P.40
VC is constant d
AC
d is increased
C is decreased CVQ Q is decreased
A
QE
E is decreased
VP is decreased
W is decreased
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P.41 Use of reed switch for measuring capacitance
Charging
+-
Discharging
Vibrating switch- vibrates between A and B at frequency f- at A, capacitor is charged Q = CV- at B, capacitor discharges
A B
fV
ICfCV
t
QI ,
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P.41 Use of reed switch for measuring capacitance
Charging
Discharging
VC
Protective resistor ->reduce I
Mean VC
Mean IC
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P.42
Gradient)(1
f
fV
I
V
QC
If C = 2 F , V = 6 V and the average current I = 0.5 mA, what should be the frequency f ?
CV
If
6102
105.06
3
)(7.41 Hz
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Capacitance exists in (1) an insulated conductor and metal framework(2) two conductors in an electric cable(3) the turns of a coil of wire
electric field lines leak to conductors- gives rise to stray capacitance
C=Co+Cs
Determination of capacitance by reed switch (Stray capacitance)
P.42
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P.43
d
AC
3
12
101
)1085.8)(5)(4.0(
CVfI )(10062.1)50)(120)(1077.1( 48 A
)(1077.1 8 F
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P.43
d
AC
)1085.8)(5.2)(1040(
)10105.1)(102(123
326
d
wl
w
Cdl )(33 m
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P.44
CRO measures VC + VR = E during charging
Reading of CRO is constant during charging
CRO measures VC + VR = 0 during discharging
Reading of CRO is zero during discharging
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P.44Parallel Connection
21 QQQ
V is the same
V +Q C
- Q
VCVCCV 21
21 CCC
If C1 < C2
then Q1 < Q2
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P.44Parallel Connection
321 QQQQ
V is the same
V
+Q - Q
C
VCVCVCCV 321
321 CCCC
QCCC
CQ
321
11
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P.44Series Connection
21 VVV
Q is the same
21 C
Q
C
Q
C
Q
V
-Q + Q
C
21
111
CCC
21
21
CC
CCC
V
CCCC
CV
21
21
11
1
V
CC
C
21
2
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P.44Series Connection
321 VVVV
Q is the same
321 C
Q
C
Q
C
Q
C
Q
V
-Q + Q
C321
1111
CCCC
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P.45
)102())105.0()105.0((
)102))(105.0()105.0((666
666
C
6103
2 C
)(1046103
2 66 CCVQ
)(104 62 CQQ
)(1022
1 65.0 CQQ
2
2
C
C
V
V )(2)6(
102
1032
6
6
22 VV
C
CV
5.0
5.0
C
C
V
V
)(4)6(101
1032
6
6
5.05.0 VV
C
CV
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P.45
)(103301010 46)(10 CCVQ i
What is the connection ?
V is the same Parallel connection
)(50)(10)(10 ffi QQQ
VCVCQ i 5010)(10
)(510501010
10366
4
5010
)(10 VCC
QV i
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P.45
)(108.1200109 36)(9 CCVQ i
V is the same
)(3)(9)(9 ffi QQQ
VCVCQ i 39)(9
)(1035.1103109
108.1)109( 366
36
)(939
9)(9 CQ
CC
CQ if
)(939
9)(9 i
n
nf QCC
CQ
n
12
95.0 41.2n 3
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P.45
CVQQ oldp )(
What is the type of connection ?Apoldp QQQQ )(
Parallel connection
A B
+ - + -
+ -
QA QB
QP
Bpoldp QQQQ )(
BA QQ 0
BA QQ
BA VV
BAp VVV
BAp QQQ
Ap QQ 2
QQp 3
2
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P.45
A. Yes. Charges cannot be destroyed or created
B. Yes. Voltage will be the same. By the equation. Q = C V, comparing the capacitance, C2 > C1, then Q2 > Q1
C. Yes. C1 shares charges with C2, so V1 decreases as Q1 decreases.
D. Yes. Both of them are fully charged. No p.d. between them, so V1 = V2
E. No. There is charge flowing through R. P=I2R , there is energy loss.
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P.46
)(42222 FQ
Q22 and Q8 are the same
822 QQ
outVCVC 82222
outVV 622
6 108104
outVV 222
outVV 2245
outout VV 245
)(15 VVout
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P.46
C
QQVCVE
22
2
1
2
1
2
1
Q=I t, x-axis relates to Q
2QE
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P.46
Connected to X
212
C
CC
CCC
1 and 2 will be charging up
1
2 3
4
Connected in series
Charge in C12 CC
Q 3)6(212
Connected to Y 3 and 4 will be charging up 1 will be discharging
234
C
CC
CCC
341 VV
34
341
C
Q
C
Q 341 2QQ
By conservation of charges 3413 QQC CQ 21
1Q1Q
)(21 VV
21 QQ
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P.46
212
C
CC
CCC
1 2
3
CCC
C2
3
2123
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P.47
Connected in series
3
10
105
10510,5
C 20
3
106105
QQ )(2
10
2010 VVVp
)(1651
11 V
kk
kVV kQ
Switch K is on )(1 VVp
1010 nQ 25555 nQ
5μF 10μF
25205 Q
102010 Q
+5μ +10μ
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P.47
VC/2(1) , V2C(2) and E are the same when steady state is obtained.
2
C
2
R
R2
C2
22
)1(2 4
1
22
1CEE
CEc
22
)2(2 4
142
2
1CEECE C
4
1
)2(2
)1(2 C
C
E
E
Whole circuit is independent of R when steady state is obtained.
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P.47
(1) Yes. By E = (1/2) Q V
(2) Yes. By W = (1/2) F e
(3) No. By W = P V
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P.48
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P.48
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P.49
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P.48
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P.48
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P.50Charging mechanism of capacitors
(1) electrons from –ve terminal of battery accumulate on one plate of capacitor(2) equal amount of +ve charges induced on opposite plate(3) until p.d. across the capacitor = e.m.f. of battery
Charge stored in capacitor
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P.50Charging at constant rate
+++
---
R set to maximum value
R I=V/R Q=I t = C V t = R C
Keep I constant
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P.50Charging at varying current
+++
---
R is not fixed, I will be changed
RCQ
V
R
VV
R
V
dt
dQI CR
00
QCVdt
dQRC 0
Qt
QCV
dQdt
RC 00
0
1
QQCVtRC 00ln1
0
0ln1
CV
QCVt
RC
QCVeCVt
RC
0
1
0
tRC
tRC eQeCVQ
1
0
1
0 11
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P.51 VR
tRCeVV1
0 1
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P.51Discharging at varying current
+++
---
RC
Q
R
V
R
V
dt
dQI CR
dt
dQRCQ
Q
Q
t
Q
dQdt
RC 00
1
QQQt
RC 0ln
1
tRCeQQ1
0
tRCeVV1
0
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P.51
tRCeQ
RCdt
dQI
1
0
1
tRC
tRC eIe
R
VI
1
0
10
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P.52Time constants τ
Charging
Discharging
tRCeQQ1
0 1
t
eQ 10
If t=τ, then
10 1 eQQ 0632.0 Q
tRCeQQ1
0
t
eQ
0
If t=τ, then1
0 eQQ 0368.0 Q
If Q=0.5Q0, then
21
005.0
t
eQQ
21
2lnt
2ln21 RCt
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P.52
e.m.f. of battery (E)= p.d. across R + p.d. across C= VR + V = IR + Q/C
1. Variation of I
At t = 0:00 RI
C
QIRE
R
EI 0
At t:CR
t
eII
0
Charging and Discharging
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P.52
2. Variation of Q
CR
t
eCECVQ 1
CR
t
eQQ 10
3. Variation of V
VReIVIRE CR
t
0
CR
t
eEV 1
Charging and Discharging
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P.52Energy of a charged capacitor
C
Qdq
C
qdWW
Q2
0 2
1
QVCVC
QW
2
1
2
1
2
1 22
- Charge on each plate (q) = CV- Suppose a charge of +dq is moved from –ve plate to +ve plate:
dqC
qVdqdW
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P.52Energy of a charged capacitor
Difference in energy
If capacitor is charged by a battery of e.m.f. (E)- energy stored in capacitor = ½ QE
QEQEQE2
1
2
1
energyin Difference
The energy escapes in the form of HEAT in the connecting wires and EM WAVEs are emitted
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P.53Energy changes (Constant V)
V is the same
V +Q C
- Q
V = constant = E d
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy returns to battery
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P.53Energy changes (isolated capacitor)
Q is the same
V +Q C
- Q
Q = constant
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy supplied by battery to overcome the attractive force between plates
= constant
V = E d
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P.53Energy changes (inserting dielectric material with constant V)
V is the same
V +Q C
- Q
V = constant = E d
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy supplied by battery to accumulate more charges inside the capacitor
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P.53Energy changes (inserting dielectric material into isolated capacitor)
Q is the same
V +Q C
- Q
Q = constant
A
QE
QVCVC
QW
2
1
2
1
2
1 22
Energy returns to battery because +ve work has been done by electric force to attract the dielectric material into the capacitor.
V = E d
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P.54
C
QE
2
2
1
C should be constant for an isolating conducting sphere
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P.54
QA and QB are the same
r
QV
04
1
rA < rB
VA > VB
QVE2
1 EA > EB
V
QC CA < CB
When they are connecting by wire, there is p.d.. The charges flow from high potential (A) to low potential (B)
QA < QB
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P.54Energy loss on joining capacitors
2
22
1
21
2
1
2
1
C
Q
C
QEi
21
221
2
2
1
2
1
CC
C
QE f
21
221
2
22
1
21
2
1
2
1
2
1
CC
C
Q
C
QE f
2121
22112
2
1
CCCC
QCQC
0
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P.55Energy loss on joining capacitors
0 fE
2121
22112
2
1
CCCC
QCQC
02112 QCQC
2
2
1
1
C
Q
C
Q
21 VV
No movement of charges, so no energy loss
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P.55Types of capacitors
Series or parallel connection ?
Parallel connection
Greater total capacitance
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P.56Electrolytic capacitors
Greater capacitance
Polarity of capacitor is fixed, for d.c. only
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P.56Variable capacitors
Varying capacitance by
changing the overlapping area
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P.57Spooning charge
Extra high internal resistance
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P.57As voltmeter
No other connection
If the meter is full scale deflection (f.s.d.), then V=1 (V)
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P.57As ammeter
R= 1010 F V= 1V
If the meter is full scale deflection (f.s.d.), then I = V/R = 10-10 A
If V=0.5 V, then I = V/R = 0.5 x 10-10 A
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P.57As meter to measure charge
C= 10-8 F V= 1V
If the meter is full scale deflection (f.s.d.), then Q = CV = 10-8 C
If V=0.5 V, then Q = CV = 0.5 x 10-8 C
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P.59Spooning charge
For each spoon, a certain amount of +ve charge will be transfer to electrometer, the corresponding reading will be shown on the voltmeter
Different size of spoon will have different amount of +ve charge carried.
Different size of p.d. of EHT will have different amount of +ve charge carried.
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P.59
P.d. across voltmeter equals to p.d. across capacitor
VCd
AQ V
Charges are isolated
22
11 VC
d
AVC
d
AQ VV
25102
)1.0(7.4100
102
)1.0(3
20
3
20
VV CC
VV CC 320
320 102)1.0(7.4108)1.0(4
20
3 )1.0(7.0106 VC
1110244.3 VC
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P.60
Before K is closed
Capacitors are fully charged
Capacitors are connected in series
Q1 : Q2 = 80μ : 80μ
C1 : C2 = 10μ : 20 μ
V1 : V2 = 2 : 1 = 8 : 4
VR1 : VR2 = 4 : 8
P T
After K is closed
VP = 8 , VT = 4
VP = 8 = VT
Q2f = (20 μ)(8) = 160 μ
Q1f = (10 μ)(4) = 40 μ
80 μ ->40 μ
80 μ ->160 μ
80μ
40μ
120μ
Charge flowing through K is 120 μC
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P.60
Initially, C is fully charged
When P breaks, C starts to discharge
When Q breaks, discharging stops
VC = 12 (V)
VC drops
VC = 3.2(V)
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P.60
When P breaks, C starts to discharge
)7.4)(21(122.3 kk
t
e
)(01864.0 st
t
dv
01864.0
1v
)/(64.53 smv
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P.60
QVCVC
QW
2
1
2
1
2
1 22
Half of W is dissipated
2
02
1
Q
Q
02
1QQ
)1)(2(0
M
t
eQQ
2
2
1 t
e
)(693.0 st
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P.60
2
1
22
11 CR
CR
1
2
2
1 R
R
2
1
1
2
2
1 C
C
4
1
2
1 C
C
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P.60
RC
t
e1
1106
))6.0ln()4.0(ln(221 RCtt
Charging to 6V by 10V battery
RC
t
e2
1104
Charging to 4V by 10V battery
RC
t
e1
4.0
RC
t14.0ln
4.0ln1 RCt
RC
t
e2
6.0
RC
t26.0ln
6.0ln2 RCt
)(9.4 sRC
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P.61
RC
t
eVV 10
)2)(1(
5.0
15 MeV
)(106.1 VV
Charging for 0.5s by 5V battery
Charging for t s to 5V by 10V battery
)2)(1(
2
1105 M
t
e
)(386.12 st Charging for t s to 1.106V by 10V battery
)2)(1(
1
110106.1 M
t
e
)(2344.01 st )(1516.12344.0386.112 sttt
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P.61
Slope = VAC / t = Q / (Ct) = I / C
Slope = const => I = const => VR = const
Charging with const. I
Fully charged, I=0
Discharging with const. I in opposite direction
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P.61
(1) C = Q / V C = (1 C) / (1 V) = 1(F)
Yes
(2) E = (1/2) Q2 /C = (1/2) (J)
No
(3) Q = I t = (1 A) (1 s) = 1 (C)
No
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P.62
RC
t
eVV
0
)47(
8.7
82 Re
)47(
8.7
25.0 Re
)47(
8.7)25.0ln(
R
)25.0ln()47(
8.7
R
kR 7.119
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P.62
(1) Q = C V = (100000μ) (20) = 2(C)
Yes
(2) Time const RC = (10) (100000 μ) = 1(s)
No
(3) E = (1/2) CV2 = (1/2) (0.1)(20)2 = 20 (J)
No
mean I= Q / t = 2 / 2000 = 1 (mA)
After 1s, only 63% of initial charge has discharged
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P.62
Time constant t = RC
2 x 10-3 = (1 x 103) C
C = (2 x 10-6) F
C = 2 μ F
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P.62
I1 = V /R
No
NoNo
I2 = V /(2R) = (1/2) I1
V1 = V2
E dissipated in R independent of R
E dissipated in R = E in Cap = (1/2) CV2
Total charge Q stored depends on V and C only
RC
t
eQQ
0
RC
t
eQQ1
01
CR
t
eQQ )2(02
2
2ln1 RCt 2ln22 RCt
Yes
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P.62
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P.63
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P.63
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P.63
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P.63
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P.63
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P.64
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P.64
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P.64
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P.65
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P.65
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P.65
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P.65
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P.65
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P.65
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P.65
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P.65