January 2020 ~ INFOTECHNO

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^-6x 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 X_c and its unit is ohm. The amount of capacitive reactance is inversely proportional to the capacity and the frequency

Where, X_C= 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

X_C = capacitive reactance, ohm

R = capacitor's resistance, ohm

Z_C= 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, Q_T = Q₁+ Q₂+ Q₃+ …

or C_T V = C₁ . V + C₂ . V + C₃ .V+ …

Or C_T = 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

C_AB = 30 + 30

= 60 pF

Total capacitance between B and C

C_BC= 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.