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Capacitors1 THE NATURE OF CAPACITANCE All passive components have three electrical properties Resistance, capacitance and inductance Capacitance is a measure.

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Presentation on theme: "Capacitors1 THE NATURE OF CAPACITANCE All passive components have three electrical properties Resistance, capacitance and inductance Capacitance is a measure."— Presentation transcript:

1 Capacitors1 THE NATURE OF CAPACITANCE All passive components have three electrical properties Resistance, capacitance and inductance Capacitance is a measure of the ability of a component to store charge Capacitor – device manufactured to store charge Capacitance is defined as the ratio of the charge stored by a capacitor to the voltage across it: C = Q/V Unit of capacitance is the Farad (  F, nF, pF) A 6v source is required to store 24  C of charge on a certain capacitor. What is the capacitance of the capacitor? How much charge is stored when a 9v battery is used as a source? What voltage source is used when 16  C is stored on it? How much charge is stored by a 220pF capacitor when a 50v source is connected across it?

2 Capacitors2 WHAT IS A CAPACITOR? A capacitor is similar to a battery Both store electrical energy, but function differently Unlike a battery, a capacitor cannot produce new electrons – it only stores them The two terminals of a capacitor are connected to two metal plates separated by an insulator called a dielectric Dielectric can be air, paper, plastic – any type of insulator

3 Capacitors3 WHAT A CAPACITOR DOES Consider the following connection Capacitor plate attached to negative battery terminal accepts electrons produced by battery Capacitor plate connected to positive battery terminal loses electrons to battery Once charged, capacitor has same voltage as battery Lightning in the sky is essentially charge being released between two plates – one being the cloud, the other the ground

4 Capacitors4 EXAMPLE OF DISCHARGING Consider the following connection Assuming a large capacitor is being used, bulb would light up as current flows from battery to capacitor to charge it up Bulb would get progressively dimmer and finally go out once capacitor is fully charged On replacing the battery with a wire, current will flow from one plate of capacitor to the other Bulb will subsequently light, and get dimmer and dimmer, finally going out once capacitor has completely discharged

5 Capacitors5 CAPACITORS IN SERIES Every capacitor in a series circuit has the same charge, regardless of the individual capacitance values The total charge delivered to series-connected capacitors equals the charge on each capacitor Applying KVL: Yet Q T = Q 1 = Q 2 Cancelling the Q’s gives

6 Capacitors6 EXAMPLE 1 Find the total equivalent capacitance and the total charge delivered by the voltage source to the circuit Find the voltage across each capacitor

7 Capacitors7 CAPACITORS IN PARALLEL The total charge delivered to parallel connected capacitors is the sum of the charges on the capacitors Q T = Q 1 + Q 2 → C T V = C 1 V 1 + C 2 V 2 Since the voltage is the same in each line: C T = C 1 + C 2 The total equivalent capacitance of parallel-connected capacitors is the sum of the capacitances

8 Capacitors8 EXAMPLE 2 What is the total equivalent capacitance in the circuit on the left? Does the total charge delivered by the source equal the sum of the charges on the capacitors? Find the voltage across and charge on each capacitor

9 Capacitors9 PHYSICAL EXPLANATION OF HOW A CAPACITOR FUNCTIONS (1) A capacitor consists of two conducting surfaces separated by a non-conducting (dielectric) material When a source is connected to a capacitor, electrons will flow from the negative terminal of the battery and collect on one of the plates A negative charge develops on the plate collecting electrons from the negative terminal of the battery Electrons on this plate repel an equal amount of electrons from the other plate, forming a positive charge

10 Capacitors10 PHYSICAL EXPLANATION OF HOW A CAPACITOR FUNCTIONS (2) An electric field is established in the dielectric between the plates This electric field drives electrons off the positively charged plate These electrons are attracted to and received by the positive terminal of the battery The electric field stores energy A voltage exists between any two regions containing opposite types of charge A voltage is developed across the capacitor as a result of the charge stored on the plates As the charge builds up on the plates, the voltage across the capacitor increases A capacitor is fully charged when no more electrons can flow from the battery

11 Capacitors11 EFFECT OF DIELECTRIC MATERIAL ON CAPACITANCE (1) The permittivity  of dielectric material between capacitor plates strongly influences the value of its capacitance Since the dielectric is an insulator, its electrons are strongly bound to their parent atoms and are not free to travel under the influence of an electric field Electric field is established in dielectric by charging capacitor plates – causes dielectric electrons to only shift position which leaves one side of each atom +ve, and the other –ve (see next slide) Each atom in dielectric has a +ve side neutralised by -ve side of adjacent atom Atoms at boundaries of capacitor plates atoms have no adjacent atoms to neutralise their sides Consequently this creates an electric field E d that opposes field set up by the charged plates So net field intensity E – E d is smaller than it would be if there were no dielectric present Smaller value of E means smaller value of V required to deposit fixed amount of charge Q on the plates From C = Q/V, C is increased

12 Capacitors12 EFFECT OF DIELECTRIC MATERIAL ON CAPACITANCE (2)

13 Capacitors13 EFFECT OF DIELECTRIC MATERIAL ON CAPACITANCE (3) A dielectric is effective in increasing capacitance because it sets up an electric field in opposition to that created by the charged plates  is called the permittivity of the dielectric, and is a measure of how well the material permits the establishment of an electric field The greater  of a material, the greater the capacitance of a capacitor that uses the material as a dielectric Units of  are Fards/meter (Fm -1 ) Permittivity in a vacuum  0 = 8.84  10 -12 Fm -1 Other insulators have larger values of  as there are no atoms in a vacuum Relative permittivity  r of a material is defined as the ratio of its actual permittivity to that of a vacuum  r =  /  0 = dielectric constant of material If a material has an  r of 5, then its  is 5 times that of a vacuum, and a capacitor that has this material as a dielectric will have a capacitance 5 times greater than it would if the plates were separated by a vacuum Example When a 24V source is connected across a capacitor having an air dielectric (  r = 1.0006), 1800µC of charge is deposited on each plate. With the voltage source fixed, a porcelain dielectric (  r = 6) is used in the space between the plates. How much charge is then deposited on the plates?

14 Capacitors14 EFFECT OF CAPACITOR DIMENSIONS ON CAPACITANCE The physical dimensions that effect the capacitance of a parallel plate capacitor are the area A of its plates, and the distance d between them The greater the area of the plates, the greater the surface on which charge can be stored Capacitance is directly proportional to the area A of each parallel plate From E = V/d and C = Q/V : C = Q/ E d From above equation, capacitance is inversely proportional to the distance d between the parallel plates

15 Capacitors15 THE CAPACITANCE EQUATION It is possible to express the dependency of capacitance on the permittivity  of the dielectric, the area A between each plate and the distance d between the plates The capacitance equation can be expressed as C =  A/d =  r  0 A/d Example Each plate of a parallel plate capacitor measures 1cm by 2cm. The plates are separated by a 0.2mm thickness of mica (  = 44.2  10 -12 ). Find the capacitance of the capacitor. Find the capacitance if each dimension of each plate is doubled. Under these conditions, find the capacitance if the thickness of the mica separation is also doubled.

16 Capacitors16 CURRENT THROUGH AND VOLTAGE ACROSS A CAPACITOR (1) Current through a capacitor depends on the rate of change of the voltage across it Rate of change of voltage deduced from a graph of voltage v time = slope of different parts of the curve Slope at a = Δv/Δt = (v 2 -v 1 )/(t 2 -t 1 ) b = (v 2 -v 2 )/(t 3 -t 2 )=0; c = (v 1 -v 2 )/(t 4 -t 3 ) A new rate of change must be calculated over each line segment that has a different slope If the voltage is constant over a time interval, the rate of change is zero V t v2v2 v1v1 t1t1 t2t2 t3t3 t4t4 a b c

17 Capacitors17 CURRENT THROUGH AND VOLTAGE ACROSS A CAPACITOR (2) Since current is equal to the rate of change or charge, i.e. I = dQ/dt and Q = CV The above is in differential form, in integral form it is The differential form shows that the faster the voltage across the capacitor changes (i.e. for small Δt), the greater the current through it If there is no voltage change across the capacitor, i.e. dv/dt = 0, then there is no current flow, so the capacitor acts as an open circuit. The energy W stored by capacitance C when it is storing charge Q is

18 Capacitors18 CURRENT THROUGH & VOLTAGE ACROSS CAPACITOR EXAMPLE Above is a graph of the voltage across a 2.5µF capacitor. Plot and draw the current through the capacitor.


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