Capacitors

Demo 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

Capacitors In 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 - - - - - - -

The conductor-insulator-conductor sandwich can be rolled into a cylinder or left flat

Experiment Determining the relationship between charge, Q, on the capacitor and the pd, V, across the capacitor (Philip Allan – 172)

Capacitance The charge, Q, on a capacitor is directly proportional to the potential difference, V, across the capacitor. That is, Q α V Introducing a constant, C, known as the capacitance of the capacitor, we have Q = CV Capacitance 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-1 1 farad is a very large unit. It is much more common to use the following: mF = 10-3 F μF = 10-6 F nF = 10-9 F pF = 10-12 F

Worked Example A capacitor of capacitance 250 μF is connected to a battery of emf 6.0 V. Calculate: a) the charge on one plate of the capacitor Q = CV = 6 x 250 x 10-6 = 1.5 x 10-3 C b) the number of excess electrons on the negative plate of the capacitor. q = 1.6 x 10-19 C no. of electrons = Q / e = 1.5 x 10-3 / 1.6 x 10-19 = 9.4 x 1015 electrons

Fully charged – no current Charge Storage V Vcapacitor = Vsupply + q - q + Q - Q Fully charged – no current Electron flow The current flows for a short time. Initially the current is large, then it decreases.

Parallel Combination The p.d. across each capacitor is the same. QT= Q1 + Q2 Apply Q = CV to each capacitor to find CT. QT= C1V + C2V QT= V (C1+ C2) QT / V = C1+ C2 CT = C1 + C2

Series Combination VT = V1 + V2 The charge, Q, on each capacitor is the same. Apply Q = CV to each capacitor. VT = Q/C1 + Q/C2 VT / Q = 1/C1 + 1/C2 1/CT = 1/C1 + 1/C2

Worked Example C A B Calculate: a) the capacitance between points B and C CT = C1 + C2 = 100 + 250 = 350 μF b) the capacitance between points A and C 1/CT = 1/C1 + 1/C2 = 1/500 + 1/350 CT = 206 μF c) the charge on the 500 μF capacitor Q = CV = 6 x 206 x 10-6 = 1.24 x 10-3 C d) the p.d across A and B V = Q/C = 1.24 x 10-3 / 500 x 10-6 = 2.48 V 100 μF 500 μF C 250 μF A B 6 V

Energy 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 QV As Q = CV, we also have E = ½ CV2 and E = ½ Q2 / C Potential difference

Worked 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 J b) 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

Worked Example A 10 000μF 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 = 3.125 J b) 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 Δh Δh = 3.2 m

To do Capacitors 1 – Exam Questions Capacitors Worksheet (Philip Allan – 184)

Experiment Discharging a capacitor (Philip Allan – 186)

Discharge 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. μA V t / s I / μA V / V Q / C

Discharge of a Capacitor The graphs below show the variation with time, t. All graphs have a similar shape and all decrease with time. Vmax

∆Q = I ∆t = area of shaded section Qo = Area under graph

Are 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.

A 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 = constant 0 t 2t 3t Q0 Q1 Q2 Q3

Testing Exponential Decay t / s Q / nC Ratio Qn/Qn-1 100 2 80 4 64 6 51 8 41 10 33 12 26

Testing Exponential Decay t / s Q / nC Ratio Qn/Qn-1 100 / 2 80 0.8 4 64 6 51 8 41 10 33 12 26

Does the constant-ratio rule apply for other intervals of time? Yes! Try for time intervals of 4 secs.

Exponential Decay The charge Q remaining on the capacitor after a time t is given by Q = Qo e-t/CR Where 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 = CV V = Vo e-t/CR V0 = initial pd across the capacitor By subs: V = IR I = Io e-t/CR Io = initial current in the resistor

Time 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 by Q = Qo e-t/CR After a time equal to CR, the charge left on the capacitor is Q = Qo e-1 Q = Qo/e ~ 0.37 Qo The 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. Qo Qo/e Qo/e2 Qo/e3 τ 2τ 3τ 4τ Qo/e4

Worked Example A 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 capacitor Q = CV = 10 x 5.0 x 10-6 = 5.0 x 10-5 C b) the time constant of the circuit time constant = CR = 5.0 x 10-6 x 1.0 x 106 = 5.0 s c) the charge left on the capacitor after 28 s Q = Qo e-t/CR = 5.0 x 10-5 x e-(28/5.0) = 1.8 x 10-7 C

Using logarithms to show that the pd across a capacitor decays exponentially Logarithm rules Log AB = log A + log B Log A/B = log A – log B Ln (ex) = x Log Ax = x log A V = Vo e-t/CR ln V = ln V0 - t/CR = - t/CR + ln V0 = - (1/CR) t + ln V0 y = mx + c A graph of ln (V) against t should be a straight line with a gradient equal to –(1/CR).

To do Capacitors 2 – Exam Questions Discharge Curves Worksheet (Philip Allan – 184)