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CAPACITORS
©
John Parkinson
1
BASIC CONSTRUCTION
The Parallel Plate Capacitor
CONDUCTOR
+
-
INSULATOR
©
John Parkinson
CONDUCTOR
TWO
OPPOSITELY
CHARGED
CONDUCTORS
SEPARATED
BY AN
INSULATOR WHICH MAY
BE AIR
2
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
©
John Parkinson
3
Charge stored [Q] depends on p.d. [Volts] applied [V]
Q
Gradient = C =
V
Q
V
Hence Q = C V
C is measured in FARADS, [F], - more often : F, nF or pF
The capacitance, C, is defined as
the charge required to raise the potential by one volt.
©
John Parkinson
4
Charging a Capacitor
Q
Q
+ + + -
C
Q
Q = CV
+Q -Q
©
John Parkinson
5
As VOLTS = Joules per Coulomb
WORK DONE BY THE CELL
W=QV
The pd across the capacitor builds up as more charge is added
Voltage
V
Charge
0
Q
0 V
1
W [
] Q  QV
2
2
©
John Parkinson
Work done in charging =
average volts x charge
6
The work done in
adding such an
infinitesimally small
amount of charge, q,
that V
remains constant is
Charge Q given by:
Volts V
v
q
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
1
E  QV
2
Or if we combine with Q = CV,
or
©
John Parkinson
1
E  CV 2
2
1 Q2
E
2C
7
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??
©
John Parkinson
8
Charging a Capacitor
Initially Vc = 0,
so V = Vr and so
initial current = ?
V
VR
VC
R
C
At all times after the switch is closed V = VR + VC
V
I0 
R
Finally I =0 when capacitor is
charged and VC = V
Volts
I0
VC
time
©
John Parkinson
0
time
9
CAPACITOR DISCHARGE
Initially V = V0
+ -
V
I0 
R
Hence initial
current, I0 = ?
V
R
I
Final Current = ?
Volts, V
V0
V  V0e

1
t
RC
i.e. the decay
is exponential
What if R and / or C is larger?
0
©
John Parkinson
Time, t
10
CAPACITOR DISCHARGE
V  V0e
1

t
RC
Q  Q0 e
The current also falls
exponentially and is given by :
1

t
RC
I  I 0e

1
t
RC
Current I
I0
Note the area
under this graph
is the initial
charge stored.
Time, t
©
John Parkinson
11
CAPACITOR DISCHARGE
R
times
C
V  V0e
has units of

1
t
RC
seconds
time constant
and is called the
of the circuit
If we put t=RC into the equation above, it becomes V=V0 e-1,
which works out to:
V=0.37V0
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
©
John Parkinson
12
CAPACITOR DISCHARGE
V  V0e

1
t
RC
Taking logs to the base “e”
1
ln V  ln V0 
t
RC
In the form
y = c + m x
Hence Plot
ln V
Gradient is negative and equal to
1

RC
t
©
John Parkinson
13
The Parallel Plate Capacitor
Area of
Plate overlap
= A
Medium relative
permittivity = r
d
A 0 r
C
d
©
John Parkinson
d = plate separation
0 = the permittivity of free
space = 8.86. X 10-12 F m-1
For air or a vacuum, r = 1
14
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?
(a)
(b)
V 100
I 
 0.050 A
R 2000
V  V0e

t
RC
 100e

4
20001000106
 100e2  13.5V
V 13.5
I 
 0.00675 A  6.75 mA
R 2000
©
John Parkinson
15
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?
(a)
(b)
Initially the p.d. is 100 V, so
V
100
R 
 500 
I 0.200
Q = I t = 0.200 x 10 = 2.0 C
The capacitor was initially charged to 200 V with 2.0 C of
charge
Q = C V, so 2.0 = C x 100 Hence C = 2 x 10-2 F
or 20 000 F
©
John Parkinson
16