capacitors.pdf

CAPACITORS AND CAPACITANCE.  INTRODUCTION ❑A capacitor is a passive element designed to store energy in its electric field. ❑Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communications, computers, a...

CAPACITORS AND CAPACITANCE.  INTRODUCTION ❑A capacitor is a passive element designed to store energy in its electric field. ❑Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communications, computers, and power systems. ❑For example, they are used in the tuning circuits of radio receivers and as dynamic memory elements in computer systems  INTRODUCTION ❑In many practical applications, the plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica. FIG 1 TYPES OF CAPACITORS. SYMBOLS OF CAPACITORS. ❑when a voltage source v is connected to the capacitor. ❑the source deposits a positive charge q on one plate and a negative charge −q on the other. ❑ The capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that q=Cv ❑where C, the constant of proportionality , is known as the capacitance of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867). From Eq. (1), we may derive the following definition. ❑Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F). ❑1 farad = 1 coulomb/volt. ❑Alternatively, capacitance is the amount of charge stored per plate for a unit voltage difference in a capacitor. CAPACITOR CONNECTED  FACTORS THAT CAPACITANCE DEPENDS ON ❑Although the capacitance C of a capacitor is the ratio of the charge q per plate to the applied voltage v, it does not depend on q or v. ❑It depends on the physical dimensions of the capacitor. For example, for the parallel plate capacitor shown in Fig.1 , the capacitance is given by. ϵ𝐴 C= 𝑑 ❑where ; A is the surface area of each plate, d is the distance between the plates, and ϵ is the permittivity of the dielectric material between the plates. applies to only parallel-plate capacitors, we may infer from it that, in general, three f actors determine the v alue of the capacitance:  FACTORS AFFECTING CAPACITANCE. three factors determine the value of the capacitance are: 1. The surface area of the plates—the larger the area, the greater the capacitance. 2. The spacing between the plates—the smaller the spacing, the greater the capacitance. 3.. The permittivity of the material—the higher the permittivity, the greater the capacitance  ELECTRO STATISTICS. ❑ electrostatics is that branch of science which deals with the phenomena associated with electricity at rest. ❑positive electrification of a body results from a deficiency of the electrons whereas negative electrification results from an excess of electrons. ❑The total deficiency or excess of electrons in a body is known as its charge.  ABSOLUTE AND RELATIVE PERMITTIVITY OF A MEDIUM ❑While discussing electrostatic phenomenon, a certain property of the medium called its permittivity plays an important role. ❑Every medium is supposed to possess two permittivities : 1. absolute permittivity (ε) and 2. relative permittivity (ε𝑟 ).  PERMITTIVITY ❑At any point in an electric field, the electric field strength E maintains the electric flux and produces a particular value of electric flux density D at that point. For a field established in vacuum (or for practical purposes in air), the ratio D/E is a constant ε0. D/E=ε0. Where D= electric flux density ,E= electric flux , ε0 = absolute flux density in a vacum  RELATIVE PERMITTIVITY ❑where εr, the relative permittivity of the insulating material, indicates its insulating power compared with that of vacuum. flux density in material relative permittivity,= lux density in vacuum ❑The insulating medium separating charged surfaces is called a dielectric. Compared with conductors, dielectric materials have very high resistivities. They are therefore used to separate conductors at different potentials, such as capacitor plates or electric power lines.  RELATIVE PERMITIVITY ❑When measuring relative permittivity, vacuum or free space is chosen as the reference medium. It has an absolute permittivity of 8.854 × 10−12 F/m. ❑Absolute permittivity ε0 = 8.854 × 10−12 F/m. ❑Relative permittivity, ε𝑟 = 1 ❑Now, take any other medium. If its relative permittivity, as compared to vacuum is ε𝑟 , then its absolute permittivity is ε = ε0 ε𝑟 F/m ❑for example, relative permittivity of mica is 5, then, its absolute permittivity is  SOLUTION ❑ε = ε0 ε𝑟 = 8.854 × 10−12 × 5 = 44.27 × 10−12 F/m  LAWS OF ELECTROSTATICS ❑First Law. Like charges of electricity repel each other, whereas unlike charges attract each other. ❑Second Law. According to this law, the force exerted between two point charges i. is directly proportional to the product of their strengths ii. is inversely proportional to the square of the distance between them. ❑The second law is also known as coulombs law.  COULOMBS LAW EQUATION 𝑄1 𝑄2 F ∝ 2 𝑑 𝑄1 𝑄2 F =K 2 𝑑  COULOMB'S LAW ❑states that the force between two point charges is proportional to each charge and inversely proportional to the square of the distance between them.  OTHER WAYS TO WRITE COULOMBS LAW 𝑄1 𝑄2 F = 4πε𝑑2 𝑄1 𝑄2 F = 4πε0 ε𝑟 𝑑2  OTHER WAYS TO WRITE COULOMBS LAW 9 𝑄1 𝑄2 F =9× 10 2 (in a medium) ε𝑟 𝑑 9 𝑄1 𝑄2 F =9× 10 2 (in air or vacuum). 𝑑 K= 9× 10 is called coulomb’s constant 9  DEFINITION OF 1 COULOMB. ❑one coulomb of charge may be defined as that charge (or quantity of electricity) which when placed in air (strictly vacuum) from an equal and similar charge repels it with a force of 9× 109 N.  SOME RELATIVE PERMITTIVITY. 1. air =1 2. water =81 3. paper =between 2 and 3, 4. glass between =5 and 10. 5. mica between 2.5 and 6  EXAMPLES. 1. Calculate the electrostatic force of repulsion between two α-particles when at a distance of 10−13 m from each other. Charge of an α-particles is 3.2 × 10−12 C. If mass of each particle is 6.68 × 10−7 N-𝑚2 /𝐾𝑔2. 2. Calculate the distance of separation between two electrons (in vacuum) for which the electric force between them is equal to the gravitation force on one of them at the earth surface. mass of electron = 9.1 × 11−31 kg, charge of electron = 1.6 × 10−19 C  EXAMPLES 3. The small identical conducting spheres have charges of 2.0 × 10−9 C and − 0.5 × 10−9 respectively. When they are placed 4 cm apart, what is the force between them ? If they are brought into contact and then separated by 4 cm, what is the force between them ?  ELECTRIC FIELD. ❑It is found that in the medium around a charge a force acts on a positive or negative charge when placed in that medium. ❑If the charge is sufficiently large, then it may create such a huge stress as to cause the electrical rupture of the medium, followed by the passage of an arc discharge. ❑Electric field can be further defined as a spherical region around a charged particle where test charges experiences attractive force or repulsive force. ELECTRIC FIELD  ELECTRIC FIELD ❑The region in which the stress exists or in which electric forces act, is called an electric field or electrostatic field. ❑The stress is represented by imaginary lines of forces. ❑The direction of the lines of force at any point is the direction along which a unit positive charge placed at that point would move if free to do so.  ELECTRIC FIELD ❑It was suggested by Faraday that the electric field should be imagined to be divided into tubes of force containing a fixed number of lines of force. ❑He assumed these tubes to the elastic and having the property of contracting longitudinally the repelling laterally.  ELECTRIC FIELD ❑With the help of these properties, it becomes easy to explain (i) why unlike charges attract each other and try to come nearer to each other and (ii) why like charges repel each other. ❑However, it is more common to use the term lines of force. These lines are supposed to emanate from a positive charge and end on a negative charge  ELECTRIC FIELD LINES ❑ the imaginary lines that indicate the direction of electric field.. ❑These lines always leave or enter a conducting surface normally.  ELECTRIC FLUX ❑ the number of electric field line passing through a particular area normally is called electric flux. Electric flux( ψ) =electric field * area.  ELECTRIC FLUX DENSITY ❑Unit flux is defined as emanating from a positive charge of 1 coulomb. Thus electric flux ψ is measured in coulombs, and for a charge of Q coulombs, the flux ψ = Q coulombs. Electric flux density D is the amount of flux passing through a defined area A that is perpendicular to the direction of the flux: D=Q/A Where D=electric flux density(C/𝑚2 )= charge(C),A= area in metres squared  ELECTRIC FIELD STRENGTH. The fig on the next slide shows two parallel conducting plates separated from each other by air. They are connected to opposite terminals of a battery of voltage V volts. There is therefore an electric field in the space between the plates. If the plates are close together, the electric lines of force will be straight and parallel and equally spaced, except near the edge where fringing will occur Over the area in which there is negligible fringing, E=V/d ELECTRIC FIELD STRENGTH.  DIELECTRIC STRENGTH ❑The maximum amount of field strength that a dielectric can withstand is called the dielectric strength of the material. Dielectric strength, E=V/d  EXAMPLE. ❑A capacitor is to be constructed so that its capacitance is 0.2µF and to take a p.d. of 1.25 kV across its terminals. The dielectric is to be mica which, after allowing a safety factor of 2, has a dielectric strength of 50MV/m. Find (a) the thickness of the mica needed, and (b) the area of a plate assuming a two-plate construction. (Assume εr for mica to be 6.)  ELECTRIC FLUX DENSITY ❑Electric flux density is also called charge density, σ  ENERGY STORED IN CAPACITORS ❑The energy, W, stored by a capacitor is given by 1 W= C𝑉 2 2 Where C= capacitance. V=voltage across the capacitor.  EXAMPLE Determine the energy stored in a 3µF capacitor when charged to 400 V. (b) Find also the average power developed if this energy is dissipated in a time of 10µs.  EXAMPLES 1. Two parallel rectangular plates measuring 20 cm by 40 cm carry an electric charge of 0.2µC. Calculate the electric flux density. If the plates are spaced 5mm apart and the voltage between them is 0.25 kV determine the electric field strength. A direct current of 4 A flows into a previously uncharged 20µF capacitor for 3ms. Determine the p.d. between the plates.  EXAMPLES. 3. Two parallel plates having a p.d. of 200 V between them are spaced 0.8mm apart. What is the electric field strength? Find also the electric flux density when the dielectric between the plates is (a) air, and (b) polythene of relative permittivity 2.3  SERIES AND PARALLEL CAPACITORS ❑We know from resistive circuits that the series-parallel combination is a powerful tool for reducing circuits. ❑This technique can be extended to series-parallel connections of capacitors, which are sometimes encountered. We desire to replace these capacitors by a single equivalent capacitor 𝐶𝑒𝑞.  PARALLEL CAPACITORS. ❑In order to obtain the equivalent capacitor 𝐶𝑒𝑞 of N capacitors in parallel, consider the circuit below  PARALLEL CIRCUITS ❑ apply KCL to the node. i= 𝑖1 + 𝑖2 + 𝑖3 + ⋯ … … … … … ….. +𝑖𝑁. 𝑑𝑣 𝑖𝑘 = 𝐶𝑘. 𝑑𝑡 Hence. 𝑑𝑣 𝑑𝑣 𝑑𝑣 𝑑𝑣 𝑑𝑣 i= 𝐶1 + 𝐶2 + 𝐶3 ……….. +𝐶𝑁 = 𝐶𝑒𝑞 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐶𝑒𝑞 = 𝐶1 +𝐶2 +𝐶3 + ⋯ … … … … … …. +𝐶𝑁. ❑The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.  SERIES CAPACITORS. ❑In order to obtain the equivalent capacitor 𝐶𝑒𝑞 of N capacitors in parallel, consider the circuit below.  SERIES CAPACITORS. ❑ applying KVL to the loop. V=𝑣1 + 𝑣2 + 𝑣3 + ⋯ … … … … … … 𝑣𝑁. 1 1 1 1 1 ❑ = + + + ⋯…………………….+ 𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3 𝐶𝑁 ❑The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.  SERIES CAPACITORS. ❑Note that capacitors in series combine in the same manner as resistors in parallel. For N = 2. 1 1 1 = + 𝐶𝑒𝑞 𝐶1 𝐶2 𝐶1 𝐶2 𝐶𝑒𝑞 = 𝐶1 +𝐶2  EXAMPLES. Find the equivalent capacitance seen between terminals a and b of the circuit in FIGURE below.  SOLUTION. ❑The 20- μF and 5- μF capacitors are in series; their equivalent capacitance is 𝐶1 𝐶2 ❑ 𝐶𝑒𝑞 = 𝐶1 +𝐶2 20×5 ❑𝐶𝑒𝑞 = =4µF. 20+5 ❑This 4-μF capacitor is in parallel with the 6- μF and 20- μF capacitors; their combined capacitance is 4 + 6 + 20 = 30 µF.  SOLUTION ❑This 30-μF capacitor is in series with the 60- μF capacitor. Hence, the equivalent capacitance for the entire circuit is 30×60 𝐶𝑒𝑞 = =20µF. 30+60  EXAMPLE 2 ❑find the voltage across each capacitor.