1 capacitors 2 basic construction insulator conductor + - two oppositely charged conductors...
TRANSCRIPT
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CAPACITORSCAPACITORS
2
BASIC CONSTRUCTION
INSULATOR
CONDUCTOR
CONDUCTOR
+-
TWO OPPOSITELY
CHARGED CONDUCTORS
SEPARATED BY AN
INSULATOR - WHICH MAY
BE AIR
The Parallel Plate Capacitor
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CAPACITORS STORE
• CHARGE&
• ENERGY
USES
1. Storing energy as in flash photography
2. Time delays in electronic circuits
3. As filters in electronic circuits
4. In tuning circuits
© 4
Charge stored [Q] depends on p.d. [Volts] applied [V]
Q
V
Gradient = C = V
Q
The capacitance, C, is defined as
the charge required to raise the potential by one volt.the charge required to raise the potential by one volt.
Hence Q = C V
C is measured in FARADS, [F], - more often : F, nF or pF
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Q
Q
Q
C
---
+++
+Q -Q
Q = CV
Charging a Capacitor
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WORK DONE BY THE CELL W = Q V
As VOLTS = Joules per Coulomb
The pd across the capacitor builds up as more charge is added
Voltage
Charge0
Q
V
Work done in charging = average volts x charge
QVQV
W2
1]
2
0[
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Volts V
Charge Qq
v
The work done in adding such an infinitesimally small amount of charge, q, that Vremains constant is given by:
w = V.q
This is the area of the strip. Hence the total work done = the energy stored in the capacitor is the area under
the graph
QVE21 Or if we combine with Q =
CV,
2
2
1CVE
CQ
E2
21or
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What is the charge and the energy stored when a 50 F capacitor is charged to (a) 100 V, (b) 200 V?
(a) Q = CV = 50 x 10-6 x 100 = 0.005 C
E = 0.5CV2 = 0.5 x 50 x 10-6 x 1002 = 0.25 J
(b) Q = CV = 50 x 10-6 x 200 = 0.01 C
E = 0.5CV2 = 0.5 x 50 x 10-6 x 2002 = 1.00 J
Why does doubling the p.d. in (b) quadruple the energy stored??
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At all times after the switch is closed V = VR + VC
Initially Vc = 0, so V = Vr and so initial current = ?
Finally I =0 when capacitor is charged and VC = V
time
Volts
VC
RV
I 0
I0
time0
VC
CR
V
VR
Charging a Capacitor
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CAPACITOR DISCHARGE
R
+ -
I
V
Initially V = V0
Hence initial current, I0 = ? R
VI 0
Final Current = ?
Volts, V
Time, t0
V0t
RCeVV1
0
i.e. the decay is exponential
What if R and / or C is larger?
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CAPACITOR DISCHARGE
The current also falls exponentially and is given by :
tRCeII1
0
tRCeVV1
0
Current I
Time, t
I0Note the area under this graph is the initial charge stored.
tRCeQQ1
0
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CAPACITOR DISCHARGE tRCeVV1
0
R times C has units of seconds and is called the
time constant of the circuit
If we put t=RC into the equation above, it becomes V=V0 e-1, which works out to:
i.e. when time is equal to R x C the p.d. across
the capacitor has dropped to 37% of its original p.d.
The capacitor is almost discharged in 5 time constants.
Try putting t = 5 x R x C
V=0.37V0
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CAPACITOR DISCHARGE tRCeVV1
0
Taking logs to the base “e”
tRC
VV1
lnln 0
y = c + m x
In the form
Hence Plot
ln V
t
Gradient is negative and equal to
RC
1
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The Parallel Plate Capacitor
Area of Plate
overlap = A
d d = plate separation
Medium relative permittivity = r
d
AC r0
0 = the permittivity of free space = 8.86. X 10-12 F m-1
For air or a vacuum, r = 1
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Question 1 A 1000 F capacitor is charged to 100 V and then discharged through a 2000 resistor.
(a) What will be the initial current?
(b) What will be the current after 4 s?
AR
VI 050.0
2000
100 (a)
(b)VeeeVV RC
t
5.13100100 261010002000
4
0
mAARV
I 75.600675.020005.13
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Question 2 A capacitor is charged to 100 V and then discharged through a resistor whose value
is gradually reduced in order to maintain a constant current of 200 mA though it. The capacitor becomes discharged after 10 s.
(a) What is the initial value of the resistor?
(b) What is the capacitance of the capacitor?
500200.0100
IV
R(a) Initially the p.d. is 100 V, so
(b) Q = I t = 0.200 x 10 = 2.0 C
Q = C V, so 2.0 = C x 100 Hence C = 2 x 10-2 F or 20 000 F
The capacitor was initially charged to 200 V with 2.0 C of charge