10.1. CAPACITANCE
The
capability of storing electrical charge by the two conductor plates duely
separated by an insulator is known as capacity. In other words, the property of
a.c. circuits which opposes any change in the amount of voltage is known as
capacity or capacitance. Its symbol is C and its unit is Farad (F). The device
so formed is called a capacitor or a condenser.
When
a capacitor is connected to a battery, a momentary current flows in its circuit
which charges its one plate positive and the other negative. The quantity of
electrical charge stored in the capacitor is directly proportional to the
voltage applied across the capacitor. Thus,
where,
Q =
charge, coulombs
C
= capacity, farads
V
= voltage, volts.
1 farad. If a voltage
of one volt causes one coulomb of charge to he accumulated in a capacitor, its
capacity will be one farad. Thus –
Note. e.s.u. means
electro static unit.
Farad
is a large unit, therefore, smaller practical units used for denoting
capacitances are —
Micro farad, µF
= 10-6 F
Micro-micro farad or pico farad pF = 10-12
F
10.2. ELECTROSTATIC ACTION
A
continuous flow of current cannot be established through a capacitor because
there exists an insulator between the two conductor plates. But, a momentary
flow of current definitely takes place in the circuit. Consequently, the
capacitor is charged and an electrical energy is stored in it, in the form of
electrostatic field.
where,
W = energy stored,
joules
C = capacity, farads
V =
potential difference, volts
The
charge stored in a capacitor starts to reduce by itself and after some time it
gets lost. If a capacitor is large enough (greater than 1 µF), then it will take more time to get
discharged. But a capacitor can immediately be discharged by short-circuiting
its leads. Capacitors having capacity of the order of 1µF (350 V) and above produce an
electric spark on short-circuiting their leads.
Note. Before
starting a repairing job, a mechanic must ensure that no capacitor having a
large capacity is left charged so as to protect himself' against any possible
electric shock.
Example 10.1.
Calculate the charge stored by a capacitor of 40 µF when
connected across 220 V D.C.
Solution. Given : Capacity, C = 40 µF = 40 x 10-6 F
Voltage, V = 720 V
Therefore,
charge stored, Q = C. V
= 40 x 10-6 x
220
= 8800 x 10-6
= 8.8 x 10-3 coulombs Answer.
Example
10.2. Calculate the charge and electrostatic energy stored by a capacitor of
1000 p.F, 12 volts.
Solution. Given : Capacity, C = 1000 µF = 1000 x 10-6 F
Voltage,
V = 12 V
Therefore,
charge stored, Q = C . V
= 1000 x 10-6 x 12
= 12 x 10-3 coulombs Answer (i).
And
electrostatic energy stored in the capacitor,
10.3. CAPACITIVE REACTANCE
The
opposition offered by a capacitor to the flow of alternating current through it
is called capacitive reactance. Its symbol is Xc and its unit is
ohm. The amount of capacitive reactance is inversely proportional to the
capacity and the frequency
Where, XC =
capacitive reactance, ohms
f
= frequency, hertz.
C = capacitance, farads.
Example 10.3.
What will be the capacitive reactance of a 50 pF capacitor working at 1(10) MHz
frequency ?
Solution. Given : Capacitnce, C = 50 pF = 50 x 10-12F
Frequency, f = 100 MHz = 100 x 106 Hz
10.4. LAGGING OF VOLTAGE IN A
CAPACITIVE CIRCUIT
Initially
a capacitor has no charge, therefore the flow of current starts at its maximum
value. As the capacitor starts to store charge, the magnitude of circuit currents
starts to decrease. When the capacitor attains full charge, the magnitude of
circuit current reaches at zero and the potential difference across the
capacitor plates reaches at its maximum value. Consequently, the voltage lags
behind the current by 90'. If the resistance of the circuit is also taken into
consideration, the angle of lag is found to be less than 90C. The following
formula is used for its calculation:
 |
| Fig. 10.1 Lagging of voltage in a capacitive circuit |
where,
θ = angle of lag
XC =
capacitive reactance, ohm
R = capacitor's resistance, ohm
ZC = capacitor's
impedance, ohm
 |
Fig. 10.2 The angle of
lag of voltage in a
capacitor |
Example 10.4.
Determine the angle of lag of a capacitor having 75 ohms reactance and 5 ohm resistance.
Solution. Given: Capacitive reactance. = 75 ohms
Capacitor's
resistance. R = 5 ohms
[Note. The values of tan θ and cos θ can be noted from a trigonometric
table]
10.5. TIME CONSTANT
The
time taken by a capacitor in attaining 63.3% of its lid! charge is called time
constant. Its symbol is t and its unit is seconds.
where.
t = time
constant. second
C
= capacitance. farad
R
= resistance, ohms.
Example 10.5.
A capacitor of 10 microfarad is connected in series with a resistor of 1 mega
ohm, Calculate its time constant.
Solution. Given : Capacitance. C = 10 µF = 10 x 10-6 F
Resistance, R = 1 MΩ
= 1.0 x 106 Ω
Therefore,
time
constant. I = C • R
= Id x 10 x I x 10"
= 10 seconds
10.6. CAPACITORS IN SERIES
Capacitors
are connected in series as and when required. In a series circuit. the voltage
applied gets distributed over capacitors in accordance to their capacities.
Therefore,
In
this way, the total capacitance becomes lesser than that of the lowest
capacitance value. If the capacitance (C) of all the capacitors is the same and
their number is a, then
Example 10.6.
Calculate the total capacitance of three capacitors of 10 µF, 20 µF and 30 µF connected in
series.
Solution.
By using the formula
7.9. CAPACITORS IN PARALLEL
Capacitors
are connected in parallel for obtaining a higher capacitance value than that of
a single capacitor. Capacitors connected in parallel get equal voltage and
collect charges in accordance to their capacities.
Thus,
QT
= Q1+ Q2+ Q3+ …
or
CT V = C1
. V + C2 . V + C3 .V+ …
Or CT = C1 + C2 +
C3+ ...
In
this way, the total capacitance is equal to the sum of individual capacitance.
If the capacitance (C) of all the capacitors is the same and their number is n,
then
Example 10.7.
Three capacitors of 0.001 0.002 pf and 0.005 pf are connected in parallel,
calculate their total capacitance.
Solution.
By
using the formula CT = C1 + C2 +
C3
= 0.001 + 0.002 + 0.005
= 0.008µF
Example 10.8.
Determine the total capacitance combination shown in Fig. 10.5. All capacities
are in pico farads.
Solution.
Total
capacitance between A and B
 |
| Fig 10.4. Capacitors in parallel |
CAB = 30 + 30
= 60 pF
Total
capacitance between B and C
CBC=
20 + 20 + 20
=
60 pF
Now,
total capacitance between A and C
10.8. CAPACITY OF A PARALLEL PLATE
CAPACITOR
A
capacitor consisting of two or more plates placed parallel to each other is
called a parallel plate capacitor.
The
capacitance of a capacitor depends on the following factors:
(i)
Area of a plate C ∝ A
(ii)
Distance between plates,
(or thickness of the
dielectric used)
(iii) No. of plates C ∝ (N – 1)
(iv) Dielectric constant, C ∝ K
where.
C =
capacitance. microfarads
K = dielectric constant
A = area of a plate, cm2.
N = No. of plates
t = thickness of the
dielectric used, cm.
Table 10.1. Dielectric constant of common
insulators
Sl. No.
|
Name of
Material
|
Dielectric
Constant K
|
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
|
Vacuum
Air
Ordinary
paper
Paraffin
wax
Transformer
oil
Resin
Asbestos
Rubber
Polythene
Shellac
Dry
Oak wood
Quartz
fused
Glass
Bakelite
Varnish
Mica
Porcelain
Fibre
Distilled
Water
Ceramic
Titanium
dioxide
Barium
stromtium titanate
|
1.0
1.001
2.0 – 3.0
2.0 – 2.3
2.0 – 2.4
2.5
2.7
2.0 – 3.6
3.3
2.95 -3.75
3.6
4.4
4.1 – 5.0
4.5 – 6.0
4.5 – 5.5
5.0 – 9.0
5.5
6.5
81.0
90 – 170
7500
|
Example 10.9.
Calculate the capacitance of a 21 plate capacitor whose plates are
semi-circular in shape with 5 cm radius and distance between the plates being 1
mm. (Given, dielectric constant is 5)
10.9. STRAY CAPACITANCE
An
undesired capacitance is found between any two turns of a coil and between any
two conductors of an electronic circuit which is called stray capacitance or
distributed capacitance.
The
above stated effect can cause
(i)
the production of resonance condition in a coil,
(ii)
undesired transfer of r.f. energy from one circuit to another,
(iii)
the by-pass of r.f. currents to ground.
In
order to reduce the stray capacitance, a coil should be made to have good
length, small diameter and thin wire coated with an insulator having a lower
value of dielectric constant.
