Electric Filed Strength And Electric Flux Density

Electric Filed Strength And Electric Flux DensityAll bodies argon made up of atoms, which consist of a nucleus containing protons (+ve) and neutrons (neutral) and surrounding the nucleus are orbiting electrons (-ve).When a body is unaerated it is galvanizingally neutral, it has the same controert bespeak as appointed eruption.If a conductor had a deficit of electrons it would exhibit a sack up positive peak and if it was to present a surplus of electrons it would exhibit a net negative pick (remember the previous study of the atom reference +ve/-ve ions). An imbalance in charge corporation be produced by friction (removing or depositing electrons victimization materials such as silk and fur, respectively) or induction (by attr moulding or repelling electrons using a second body which is, respectively, positively or negatively aerated).Coulombs Law soils that if charged bodies exist at two points, the force of attraction (if the charges are of opposite polarity) or repulsio n (if the charges have the same polarity) allow be proportional to the growth of the magnitude of the charges divided by the square of their distance away. Thus+++Direct Inverse Proportionality MathsQ1 and Q2 are the charges present at the two points (in Coulombs), d is the distance separating the two points (in metres), F is the force (in Newtons), and k is a mathematical continuous depending upon the medium in which the charges exist.In a vacuum or surrender space,0 is the permittivity of free space (8.854 x 10-12 F/m Farad per meter).The force exerted on a charged particle is a manifestation of the world of an galvanic plain stitch. The voltaic vault of heaven defines the direction and magnitude of a force on a charged object. The work itself is invisible to the human eye barely can be raddled by constructing lines which indicate the motion of a free positive charge within the field the number of field lines in a particular region being used to indicate the sex act strength of the field at the point in question.The figure above shows the galvanic fields in the midst of charges of the same and opposite polarity.The figure below shows the field which exists surrounded by two charged correspond rest homes.BAAs illustrated above, coats A and B are doped and charged to different say-sos. If an electron that has a negative charge is placed in the midst of the places, a force will act on the electron tending to push it away from the negative plate B and towards the positive plate A. Similarly, a positive charge would be acted on by a force tending to move it toward the negative plate.The region amidst the plates in which an electricalal charge experiences a force, is called an electrostatic field. The direction of the field is defined by the force acting on a positive charge placed in the field, i.e. the direction of the force is from the positive plate to the negative plate.Such a field may be represented in magnitude and direction b y lines of electric force drawn between the charged surfaces. The closeness of the lines is an indication of the field strength. Whenever a p.d. is established between two points, an electric field will always exist.The figure above shows two duplicate conducting plates disjointed from distributively other by air, and are connected to opposite terminals of a battery of potentiality 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 latitude and equally position, except near the limit where fringing will occur (see previous figure). Over the area in which there is negligible fringing,E is the electric field strength (V/m), V is the applied potential difference crosswise the agree plates (V) and d is the distance (m).**Note Electric Field Strength is as well called Potential/Voltage Gradient.A unit electric flux is defined as emanating from a positive charge of 1 c oulomb. Thus electric flux is measured in coulombs, and for a charge of Q coulombs, the electric flux is equal to Q coulombs. Electric flux density D is the amount of flux passing through a defined area A that is perpendicular style to the direction of the flux is the electric flux measured in coulombs, Q is the electric charge also measured in coulombs, and A is the area in m2 over which the flux is distributed. line 1Two parallel rectangular plates measuring 20cm by 40cm carry an electric charge of 0.2 C. (a) Calculate the electric ux density.(b) If the plates are spaced 5mm apart and the potentiality between them is 0.25 kV determine the electric field strength.Solution 1PERMITTIVITYAt 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 0, i.e.0 is called the permittivit y of free space or the free space constant.The value of 0 is 8.854 x 10-12 F/m Farad per meter.When a insulator (i.e. insulating medium separating charged surfaces), such as mica, study, plastic or ceramic is introduced into the region of an electric field, the ratio of D/E is modified.r is called the relative permittivity of the insulating material and indicates its insulating power compared with that of vacuum.r has no units and typical properties of round ballpark insulating dielectric materials are shown below.The product of 0 r is called the absolute permittivity, , i.e.As discussed earlier, the dielectric is an insulating medium separating charged surfaces and has the property of very full(prenominal) resistivity. They are therefore used to separate conductors at different potentials, such as content plates or electric power lines.The dielectric strength of an insulating dielectric is the uttermost electric field strength that can safely be applied to it before breakd own (conduction) occurs.The amount of charge produced for a given applied voltage on the two parallel plates shown earlier will depend not only on the physical dimensions but also on the insulating dielectric material that appears between the plates. Such materials need to have a very high value of resistivity (i.e. they must not conduct charge) coupled with an ability to withstand high voltages without breaking down.A more practical arrangement of parallel plates with an insulating dielectric material is shown.In this arrangement the ratio of charge, Q, to the potential difference, V, is given by the quest relationship.A = area of one on the plates, in m2D = thickness of the dielectric in m = absolute permittivity of the dielectric material*Later learning, i.e. the parallel plate condenser/ condenser and physical dimensions. single pair of plates arrangement of n plates hassle 1The ux density between two plates marooned by mica of relative permittivity 5 is 2C/m2. bump the vol tage gradient between the plates.Solution 1Problem 2Two parallel plates having a p.d. of 200V between them are spaced 0.8mm apart.What is the electric eld strength?Find also the electric ux density when the dielectric between the plates is (a) air, and (b) polythene of relative permittivity 2.3Solution 2self-importance ASSESSMENT (1-2)NOTE Where appropriate pay off 0 as 8.85 x 10-12 F/mA capacitance uses a dielectric 0.04mm thick and operates at 30V. What is the electric field strength across the dielectric at this voltage? Answer 750kV/mA two-plate electrical condenser has a charge of 25C. If the efficient area of individually plate is 5cm2 determine the electric ux density of the electric field. Answer 50 kC/m2A charge of 1.5C is carried on two parallel rectangular plates distributively measuring 60mm by 80mm. (a) Calculate the electric ux density. (b) If the plates are spaced 10mm apart and the voltage between them is 0.5kV determine the electric eld strength.Answer (a) 3 12.5C/m2, (b) 50kV/mTwo parallel plates are separated by a dielectric and charged with 10C. Given that the area of each plate is 50cm2, calculate the electric ux density in the dielectric separating the plates. Answer 2mC/m2The electric ux density between two plates separated by polystyrene of relative permittivity 2.5 is 5C/m2. Find the voltage gradient between the plates.Answer 226kV/mTwo parallel plates having a p.d. of 250V between them are spaced 1mm apart.(a) go under the electric eld strength.(b) Find also the electric ux density when the dielectric between the plates is(i) air and (ii) mica of relative permittivity 5.Answer (a) 250kV/m (bi) 2.213C/m2 (bii) 11.063C/m2 condenserS CAPACITANCEA capacitor is a device for storing electric charge. In effect, it is a reservoir into which charge can be deposited and then later extracted. In its simplest form a capacitor consists of two parallel metal plates which are separated by an insulating material known as a dielectric.CDocume nts and SettingsHarveyMy DocumentsMy PicturesPicturePicture 028.jpgBecause of the dielectric, oc flowing cannot flow from one plate to the other. When the capacitor is connected to a dc source, electrons heap up on the plate connected to the negative come out terminal. The negative charge repels electrons from the atoms of the other plate. These electrons flow away to the positive terminal of the dc source this leaves the plate positively charged.CDocuments and SettingsHarveyMy DocumentsMy PicturesPicturePicture 032.jpgIf the capacitor is disconnected from the supply, the charges remain. The capacitor investment companys the electric charge indefinitely.The symbols for a fixed capacitor and a variable capacitor used in electrical travel diagrams are shown below.Typical applications include reservoir and smoothing capacitors for use in power supplies, coupling a.c. signals between the stages of amplifiers, and decoupling supply rails (i.e. effectively cornerstone the supply rail s as far as a.c. signals are concerned).The following figures illustrate what happens to a capacitor when it is charging and discharging.If the switch is left open (position A), no charge will appear on the plates and in this condition there will be no electric field in the space between the plates nor will there be any charge stored in the capacitor.When the switch is moved to position B, electrons will be attracted from the positive plate to the positive terminal of the battery. At the same age, a similar number of electrons will move from the negative terminal of the battery to the negative plate. This sudden movement of electrons will manifest itself in a momentary surge of current (conventional current will flow from the positive terminal of the battery towards the positive terminal of the capacitor).Eventually, enough electrons will have moved to make the e.m.f. between the plates the same as that of the battery. In this state, the capacitor is say to be fully charged and an electric field will be present in the space between the two plates.If, at some later eon the switch is moved back to position A, the positive plate will be left with a deficiency of electrons whilst the negative plate will be left with a surplus of electrons. Furthermore, since there is no means for current to flow between the two plates the capacitor will remain charged and a potential difference will be maintained between the plates.Now assume that the switch is moved to position C. The excess electrons on the negative plate will flow through the resistor to the positive plate until a neutral state once again exists (i.e. until there is no excess charge on either plate). In this state the capacitor is said to be fully discharged and the electric field between the plates will rapidly collapse. The movement of electrons during the discharging of the capacitor will again result in a momentary surge of current (current will flow from the positive terminal of the capacitor and into the resistor).The figure below shows the direction of current flow during charging (i.e. the switch in position B) and discharging (i.e. the switch in position C). It should be noted that current flows momentarily in both circuits even though you may think that the circuit is broken by the gap between the capacitor platesThe charge Q (in coulombs) stored in a capacitor is given byI is the current in amperes and t is the measure in seconds.Charge Q on a capacitor is proportional to the applied voltage V, i.e. Q V.Direct Inverse Proportionality MathsQ = CVThe constant of proportionality C is the capacitance.The unit of capacitance C is the farad F (or more usually F =10-6F or pF =10-12F), and is defined as the capacitance when a p.d. of one volt appears across the plates when charged with one coulomb.Capacitance is the ability of a circuit or object (i.e. in this case a capacitor) to store electric charge.Problem 1(a) Determine the p.d. across a 4 F capacitor when charged with 5 mC (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.Solution 1Problem 2A direct current of 4A flows into a previously uncharged 20 F capacitor for 3 ms. Determine the p.d. between the plates.Solution 2Problem 3A 5F capacitor is charged so that the p.d. between its plates is 800V. Calculate how long the capacitor can provide an average discharge current of 2 mA.Solution 3SELF ASSESSMENT (3)Find the charge on a 10 F capacitor when the applied voltage is 250 V. (Answer 2.5 mC)Determine the voltage across a megabyteF capacitor to charge it with 2 C. (Answer 2 kV)The charge on the plates of a capacitor is 6 mC when the potential between them is 2.4 kV. Determine the capacitance of the capacitor. (Answer 2.5 F)For how long must a charging current of 2 A be fed to a 5 F capacitor to raise the p.d. between its plates by 500V. (Answer 1.25 ms)A direct current of 10 A flows into a previously uncharged 5 F capacitor for 1 ms. Determine the p.d. between the plates. (Answer 2 kV)A 16 F capacitor is charged at a constant current of 4 A for 2 minutes. Determine the final p.d. across the capacitor and the corresponding charge in coulombs. (Answer 30V, 480 C)A steady current of 10 A flows into a previously uncharged capacitor for 1.5 ms when the p.d. between the plates is 2 kV. Find the capacitance of the capacitor. (Answer 7.5F)CAPACITANCE AND PHYSICAL DIMENSIONS (Conventional Parallel Plate Capacitor)The capacitance of a capacitor depends upon the physical dimensions of the capacitor (i.e. the size of the plates and the separation between them) and the dielectric material between the plates. The capacitance of a conventional parallel plate capacitor is given byWhere, C = Capacitance, unit of measure farads (F)0 = Permittivity of free space or the free space constant (8.85 x 10-12 F/m)r = Relative permittivity of the dielectric medium between the plates(r has no units as it is a ratio of density material/vacuum)A = Area of one of the plates (m2)d = Thickness of the dielectric or separation between the plates (m)In order to increase the capacitance of a capacitor, many practical components employ multiple plates as shown.Ten plates are shown, forming cardinal capacitors with a capacitance nine times that of one pair of plates.Such an arrangement has n plates then capacitance C (n -1). Thus capacitance is then given byProblem 1A ceramic capacitor has an effective plate area of 4cm2 and separated by 0.1 mm of ceramic of relative permittivity 100. Calculate the capacitance of the capacitor in picofarads (F).If the capacitor in part (a) is given a charge of 1.2C what will be the p.d. between the plates?Solution 1Problem 2A waxed typography capacitor has two parallel plates, each of effective area 800 cm2. If the capacitance of the capacitor is 4425 pF determine the effective thickness of the paper if its relative permittivity is 2.5.Solution 2Problem 3A parallel plate capacitor has nineteen interleaved plates each 75 mm by 75 mm and separated by mica sheets 0.2 mm thick. Assuming that the relative permittivity of the mica is 5, calculate the capacitance of the capacitor.Solution 3n = 19, thus (n 1) = 18A = 75 x 75 = 5625mm2r = 5, 0 = 8.85 x 10-12 F/md = 0.2mm = 0.2 x 10-3mSELF ASSESSMENT (4)** Where appropriate take 0 as 8.85 x 10-12 F/m.A capacitor consists of two parallel plates each of area 0.01 m2, spaced 0.1 mm in air. Calculate the capacitance in picofarads (pF). Answer 885 pFA waxed paper capacitor has two parallel plates, each of effective area 0.2m2. If the capacitance is 4000 pF determine the effective thickness of the paper if its relative permittivity is 2. Answer 0.885 mmCalculate the capacitance of a parallel plate capacitor having 5 plates, each 30 mm by 20 mm and separated by a dielectric 0.75 mm thick having a relative permittivity of 2.3. Answer 65.14 pFHow many plates does a parallel plate capacitor have if its capacitance is 5nF, each plate is 40mm by 40mm and each dielectric is 0.1 02mm thick with a relative permittivity of 6? Answer 7A parallel plate capacitor is made from 25 plates, each 70mm by 120mm interleaved with mica of relative permittivity 5. If the capacitance of the capacitor is 3000pF determine the thickness of the mica sheet. Answer 2.97mmThe capacitance of a parallel plate capacitor is 1000pF. It has 19 plates, each 50mm by 30mm separated by a dielectric of thickness 0.40mm. Determine the relative permittivity of the dielectric. Answer 1.67CAPACITORS CONNECTED IN twin AND SERIESCAPACITORS CONNECTED IN PARALLELThe figure above shows threesome capacitors, C1, C2 and C3 connected in parallel with a supply voltage V applied across the arrangement. (Note just like resistors in parallel, the supply voltage V is the same across each parallel capacitor)V = V1 = V2 = V3When the charging current I reaches point A it divides, some flowing into C1, some flowing into C2 and some into C3. Hence the total charge QT (i.e. QT= I x t) is divided between the thr ee capacitors. The capacitors each store a charge and these are shown as Q1, Q2 and Q3 respectively. Hence,But, QT=CV (where C is the total similar circuit capacitance)And, Q1=C1VQ2=C2VQ3=C3VTherefore, CV = C1V + C2V + C3V (where C is the total equivalent circuit capacitance)Dividing throughout by the common V giving,C = C1 + C2 + C3 . + CnThe equivalent capacitance of a group of parallel connected capacitors is the sum of the capacitances of the individual capacitors.CAPACITORS CONNECTED IN SERIESThe figure above shows three capacitors, C1, C2 and C3 connected in serial across a supply voltage V. Let the p.d. across the individual capacitors be V1, V2 and V3 respectively as shown.Let the charge on the plate a of the capacitor C1 be +Q coulombs. This induces and equal but opposite charge of -Q coulombs on plate b. The conductor between plates b and c is electrically isolated from the rest of the circuit so that an equal but opposite charge of +Q coulombs must appear on plate c, wh ich, in turn, induces an equal and opposite charge of -Q coulombs on plate d, and so on.Hence when capacitors are connected in series the charge on each is the same.QT = Q1 = Q2 = Q3In a series circuit V = V1 + V2 + V3 (Similar to resistors in series)Since, then (where C is the total equivalent circuit capacitance)Dividing throughout by the common Q giving,(Where C is the total equivalent circuit capacitance)For series connected capacitors, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitance.For special case of two capacitors in series,Hence,i.e.Problem 1Calculate the equivalent capacitance of two capacitors of 6F and 4F connected in(a) Parallel(b) Series.Solution 1Problem 2What capacitance must be connected in series with a 30F capacitor for the equivalent capacitance to be 12F?Solution 2Problem 3Capacitances of 1F, 3F, 5F and 6F are connected in parallel to a direct voltage supply of 100V. Determine (a) the equivalent c ircuit capacitance, (b) the total charge and (c) the charge on each capacitor.Solution 3Problem 4Capacitances of 3F, 6F and 12F are connected in series across a 350V supply. Calculate (a) the equivalent circuit capacitance, (b) the charge on each capacitor, and (c) the p.d. across each capacitor.Solution 4Problem 5For the arrangement shown, name (a) the equivalent capacitance of the circuit, (b) the voltage across QR and (c) The charge on each capacitor.Solution 5SELF ASSESSMENT (5)Capacitors of 2F and 6F are connected (a) in parallel and (b) in series. Determine the equivalent capacitance in each case. Answers (a) 8F (b) 1.5FFind the capacitance to be connected in series with a 10F capacitor for the equivalent capacitance to be 6F. Answer 15FWhat value of capacitance would be obtained if capacitors of 0.15F and 0.10F are connected in (a) series and (b) parallel? Answers (a) 0.06F (b) 0.25FTwo 6F capacitors are connected in series with one having a capacitance of 12F. Find the tota l equivalent circuit capacitance. What capacitance must be added in series to obtain a capacitance of 1.2F? Answers (a) 2.4F (b) 2.4FFor the arrangement shown below, find (a) the equivalent circuit capacitance and (b) the voltage across a 4.5F capacitor. Answers (a) 1.2F (b) 100VThree 12F capacitors are connected in series across a 750V supply. Calculate (a) the equivalent capacitance, (b) the charge on each capacitor and (c) the p.d. across each capacitor. Answers (a) 4F (b) 3mC (c) 250VIf two capacitors having capacitances of 3F and 5F respectively are connected in series across a 240V supply, determine (a) the p.d. across each capacitor and (b) the charge on each capacitor. Answers (a) 150V, 90V (b) 0.45 mC on eachCapacitances of 4F, 8F and 16F are connected in parallel across a 200V supply. Determine (a) the equivalent capacitance, (b) the total charge and (c) the charge on each capacitor. Answers (a) 28 F (b) 5.6mC (c) 0.8mC, 1.6mC, 3.2mCDIELECTRIC STRENGTHThe maximum safe work ing voltage is the maximum voltage that can be applied to the terminals of a capacitor without causing damage to the capacitor.The manufacturer specifies this voltage. The limit is necessary so that the field strength in the dielectric does not exceed a value that would cause the dielectric to breakdown and loose its insulating properties. The figure quoted by the manufacturer for a capacitor is also known as the dielectric strength and will be in volts per metre.E is the dielectric strength (V/m), V is the applied potential difference across the parallel plates (V) and d is the distance (m).**Note Equation identical to Electric Field Strength (Potential/Voltage Gradient).Problem1A capacitor is to be constructed so that its capacitance is 0.2F and to take a p.d. of 1.25kV across its terminals. The dielectric is to be mica and 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 r for mi ca to be 6).Solution 1ENERGY STORED IN CAPACITORSThe get-up-and-go, W, stored by a capacitor is given by,Where,W is the energy (in Joules),C is the capacitance (in Farads), andV is the potential difference (in Volts).Problem 1(a) Determine the energy stored in a 3F capacitor when charged to 400V.(b) Find also the average power developed if this energy is dissipated in a time of 10s.Solution 1Problem 2A 12F capacitor is required to store 4J of energy. Find the p.d. to which the capacitor must be charged.Solution 2Problem 3A capacitor is charged with 10mC. If the energy stored is 1.2J, determine (a) the voltage and (b) the capacitance.Solution 3SELF ASSESSMENT (6)** Where appropriate take 0 as 8.85 x 10-12 F/m.When a capacitor is connected across a 200V supply the charge is 4C. Find (a) the capacitance and (b) the energy stored. Answer (a) 0.02F (b) 0.4mJFind the energy stored in a 10F capacitor when charged to 2kV. Answer 20 JA 3300pF capacitor is required to store 0.5mJ of energy. Find the p.d. to which the capacitor must be charged. Answer 550 VA capacitor is charged with 8mC. If the energy stored is 0.4J, determine (a) the voltage and (b) the capacitance. Answer (a) 100V (b) 80 FA capacitor, consisting of two metal plates each of area 50 cm2 and spaced 0.2mm apart in air, is connected across a 120V supply. Calculate (a) the energy stored (b) the electric ux density and (c) the potential gradient (i.e. electric field strength). Answer (a) 1.593J (b) 5.31C/m2 (c) 600kV/mD.C TRANSIENTSNetworks of capacitors and resistors (known as C-R circuits) form the basis of many timing and pulse shaping circuits and are thus often found in practical electronic circuits.When a d.c. voltage is applied to a capacitor C and resistor R connected in series, there is a short period of time immediately afterward when the voltage is connected that the current flowing in the circuit and voltages across C and R are changing.These changing values are called transients.CHARGING A CAP ACITORThe figure above shows a series connected C-R circuit.When the switch S is closed, then by Kirchhoffs valotage law V = Vc + VRThe battery voltage V is constant.The capacitor voltage Vc is given by,The voltage drop across R (i.e. VR) is given by,Hence at all timesAt the instant of closing S (i.e. initial circuit condition), assuming there is no initial charge on the capacitor, Q is zero (i.e. Q0), hence Vc is zero (i.e. VC0). (Note From equation Vc = Q / C).Thus from equation V = Vc + VR,V = 0 + VR (i.e. V = VR = IR)A short time later at time T1 seconds after closing S, the capacitor is partly charged to, say, Q1 coulombs because current has been flowing. The voltage VC1 is now,If the current flowing is I1 amperes, then the voltage drop across R has fallen toVR1 = I1R volts. Thus from equation V = Vc + VRA short time later still, say at time T2 seconds after closing S, the charge has increased to Q2 coulombs and VC has increased to,Since V = VC + VR and V is a constant, then VR decreases to I2R. Thus VC is increase and I and VR are decreasing as time increases.Ultimately, a few seconds after closing S (i.e. at the final or steady state condition), the capacitor is fully charged to, say Q coulombs, current no longer flows, i.e. I = 0, and hence VR = IR = 0. It follows from equation V = Vc + VR that V = VC.Curves demo the changes in VC, VR and I with time are shown below.The submit showing the variation of VC with time is called an exponential growth curve and the graph is called the capacitor voltage / time characteristic.The curves showing variations of VR and I with time are called exponential decay curves, and the graphs are called resistor voltage / time and current / time characteristics respectively.The name exponential shows that the shape can be expressed mathematically by an exponential mathematical equation, as shown below.Growth of capacitor voltage,Decay of resistor voltage,Decay of resistor current,TIME constant ( TAU) FOR A C-R CIRCUITAs shown earlier, if a constant d.c. voltage is applied to a series connected C-R circuit, a exponential transient growth curve of capacitor voltage VC results as shown below.With reference to the figure below, the constant voltage supply is replaced by a variable voltage supply at time t1 seconds. The voltage is varied so that the current flowing in the circuit is constant.Since the current flowing is a constant, the curve will follow a tangent, AB, drawn to the curve at point A.Let the capacitor voltage VC reach its final value of V at time t2 seconds.The time corresponding to (t2-t1) seconds is called the time constant of the circuit, denoted by the Greek letter tau, . The value of the time constant is CR seconds, i.e. for a series connected C-R circuit,(seconds)Where C is capacitance (F), R is the resistance () and is the time constant (s)DISCHARGING A CAPACITORWhen a capacitor is charged (i.e. with the switch in position A), and the switch is then moved to position B, the electr ons stored in the capacitor keep the current flowing for a short time.Initially, at the instant of moving from A to B, the current flow is such that the capacitor voltage VC is balanced by equal and opposite voltage (Kirchhoffs 2nd law), i.e. VC = VR = IR.Finally the transients decay exponentially as current is reduced to zero, i.e. VC = VR = 0.The transient curve representing the voltages and current are shown below.The equations representing the transient curves during discharge period of a series connected C-R circuit areDecay of voltage,Decay of current,When a capacitor has been disconnected from the supply it may still be charged and it may retain this charge for some considerable time. Thus precautions must be taken to ensure that the capacitor is automatically discharged after the supply is switched off. This is done by connecting a high value resistor across the capacitor terminals.Problem 1A capacitor is charg