2Demo Capacitors in every day use – Keynote Super Capacitor - Keynote [ lab power supply, ac/dc voltage converter/radio/tv/computer/VCR/ throwaway flash camera]A capacitor may be used with a resistor to produce a timer. Sometimes capacitors are used to smooth a current in a circuit as they can prevent false triggering of other components such as relays.Super Capacitor - Keynote
3CapacitorsIn its simplest form, a capacitor consists of two metal plates, separated from each other by an insulator (air, mica, paper etc.)A charged capacitor is shown below.The capacitor plates have an equal but opposite charge of magnitude Q.The net charge on the two plates is zero.+ Q- Q
4The conductor-insulator-conductor sandwich can be rolled into a cylinder or left flat
5ExperimentDetermining the relationship between charge, Q, on the capacitor and the pd, V, across the capacitor(Philip Allan – 172)
6CapacitanceThe charge, Q, on a capacitor is directly proportional to the potential difference, V, across the capacitor. That is,Q α VIntroducing a constant, C, known as the capacitance of the capacitor, we haveQ = CVCapacitance of a capacitor is defined as the ratio of charge on one of the capacitor plates to the potential difference between the plates.Charge Q is measured in coulombs, C.Potential difference, V, is measured in volts, V.Capacitance, C, is measured in farads, F.1 farad is 1 coulomb per volt: 1 F = 1 C V-11 farad is a very large unit. It is much more common to use the following:mF = 10-3 FμF = 10-6 FnF = 10-9 FpF = F
7Worked ExampleA capacitor of capacitance 250 μF is connected to a battery of emf 6.0 V. Calculate:a) the charge on one plate of the capacitorQ = CV = 6 x 250 x 10-6 = 1.5 x 10-3 Cb) the number of excess electrons on the negative plate of the capacitor.q = 1.6 x Cno. of electrons = Q / e = 1.5 x 10-3 / 1.6 x = 9.4 x 1015 electrons
8Fully charged – no current Charge StorageVVcapacitor = Vsupply+ q q+ Q QFully charged – no currentElectron flowThe current flows for a short time.Initially the current is large, then it decreases.
9Parallel Combination The p.d. across each capacitor is the same. QT= Q1 + Q2Apply Q = CV to each capacitor to find CT.QT= C1V + C2VQT= V (C1+ C2)QT / V = C1+ C2CT = C1 + C2
10Series Combination VT = V1 + V2 The charge, Q, on each capacitor is the same.Apply Q = CV to each capacitor.VT = Q/C1 + Q/C2VT / Q = 1/C1 + 1/C21/CT = 1/C1 + 1/C2
11Worked Example C A B Calculate: a) the capacitance between points B and CCT = C1 + C2 = = 350 μFb) the capacitance between points A and C1/CT = 1/C1 + 1/C2 = 1/ /350CT = 206 μFc) the charge on the 500 μF capacitorQ = CV = 6 x 206 x 10-6 = 1.24 x 10-3 Cd) the p.d across A and BV = Q/C = 1.24 x 10-3 / 500 x 10-6 = 2.48 V100 μF500 μFC250 μFAB6 V
12Energy stored in a Capacitor When the capacitor is connected across a battery, the pd across it increases from zero to a pd equal to the emf of the battery. When a small amount of charge ΔQ, is transferred from one plate of the capacitor and removed from the other, the battery does work, equal to the area of the shaded strip in the diagram.ΔE = VoΔQ = the area of the shaded strip in the diagram.The total energy transferred to a capacitor, E = total area under graph.E = ½ QVAs Q = CV, we also haveE = ½ CV2 and E = ½ Q2 / CPotential difference
13Worked Example A 100 000μF capacitor is connected to a 6 V battery. a) Calculate the energy stored by the capacitor.E = ½ CV2 = ½ x 0.1 x 6.02 = 1.8 Jb) The capacitor is discharged through a filament lamp. The lamp produces a flash of light lasting 12 ms. Calculate the average power dissipated by the lamp.Power = energy / time = 1.80 / 12 x 10-3 = 150 W
14Worked ExampleA μF capacitor is described as having a maximum working voltage of 25 V.a) Calculate the energy stored by the capacitor.E = ½ CV2 = ½ x 10,000 x 10-6 x 252 = Jb) If this capacitor were connected to a motor so that it could lift a mass of 100 g. What is the maximum height to which this could be raised?3.125 J = mg Δh = 0.1 x 9.8 x ΔhΔh = 3.2 m
15To do Capacitors 1 – Exam Questions Capacitors Worksheet (Philip Allan – 184)
16ExperimentDischarging a capacitor(Philip Allan – 186)
17Discharge of a Capacitor This circuit opposite can be used to show the discharge of a capacitor.It is charged by connecting the switch to A. It discharges through the resistor when the switch is at B.Io is approx 80 μA.Discharge time is approx 4 mins.Plot graphs of I, V and Q Against t.μAVt / sI / μAV / VQ / C
18Discharge of a Capacitor The graphs below show the variation with time, t.All graphs have a similar shape and all decrease with time.Vmax
19∆Q = I ∆t= area of shaded sectionQo = Area under graph
20Are these graphs of the type x α 1/t ? No!If I α 1/t, then when t = 0, the p.d across the capacitor would be infinite!The p.d decays exponentially with respect to time.
21A Quick Test for Exponential Decay If a particular quantity decays exponentially with respect to time, then, for equal time intervals, the ratio of the quantity will be the same.E.g. A charge-time graph has a constant-ratio property. For any constant interval of time, Δt:Q1/Qo = Q2/Q1 = Q3/Q2 = Qn/Qn-1 = constant0 t 2t 3tQ0Q1Q2Q3
22Testing Exponential Decay t / sQ / nCRatio Qn/Qn-110028046465184110331226
23Testing Exponential Decay t / sQ / nCRatio Qn/Qn-1100/2800.846465184110331226
24Does the constant-ratio rule apply for other intervals of time? Yes!Try for time intervals of 4 secs.
25Exponential DecayThe charge Q remaining on the capacitor after a time t is given byQ = Qo e-t/CRWhere e is the base of natural logarithms (~2.718), Qo is the initial charge on the capacitor, C is the capacitance of the capacitor and R is the resistance of the resistor in the discharge circuit.By subs: Q = CVV = Vo e-t/CR V0 = initial pd across the capacitorBy subs: V = IRI = Io e-t/CR Io = initial current in the resistor
26Time Constant, τ τ Q = Qo e-t/CR The product CR is known as the time constant of the capacitor-resistor circuit. The time constant is measured in seconds (s).The charge, Q, on the capacitor is given byQ = Qo e-t/CRAfter a time equal to CR, the charge left on the capacitor isQ = Qo e-1Q = Qo/e ~ 0.37 QoThe time constant for a capacitor-resistor circuit is defined as the time for the charge (or current or pd) to decrease to 1/e (~0.37) of its initial value.QoQo/eQo/e2Qo/e3τ2τ3τ4τQo/e4
27Worked ExampleA 5.0 μF capacitor is charged to a pd of 10 V. It is discharged through a resistor of resistance 1.0 MΩ. Calculate:a) the initial charge on the capacitorQ = CV = 10 x 5.0 x 10-6 = 5.0 x 10-5 Cb) the time constant of the circuittime constant = CR = 5.0 x 10-6 x 1.0 x 106 = 5.0 sc) the charge left on the capacitor after 28 sQ = Qo e-t/CR = 5.0 x 10-5 x e-(28/5.0) = 1.8 x 10-7 C
28Using logarithms to show that the pd across a capacitor decays exponentially Logarithm rulesLog AB = log A + log BLog A/B = log A – log BLn (ex) = xLog Ax = x log AV = Vo e-t/CRln V = ln V t/CR= - t/CR ln V0= - (1/CR) t + ln V0y = mx cA graph of ln (V) against t should be a straight line with a gradient equal to –(1/CR).
29To do Capacitors 2 – Exam Questions Discharge Curves Worksheet (Philip Allan – 184)