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## Capacitors

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**Capacitors**• A capacitor is a device which is used to store electrical charge ( a surprisingly useful thing to do in circuits!). • Effectively, any capacitor consists of a pair of conducting plates separated by an insulator. The insulator is called a dielectric and is often air, paper or oil.**Illustrating the action of a capacitor**Set up the circuit. Connect the flying lead of the capacitor to the battery. Connect it to the lamp. What do you observe. Try putting a 100Ω resistor in series with the lamp. What effect does it have? 6V 6V, 0.003A lamp 10 000μF**What is happening**• When the capacitor is connected to the battery, a momentary current flows. • Electrons gather on the plate attached to the negative terminal of the battery. At the same time electrons are drawn from the positive plate of the capacitor +++++ -------**What is happening**• When the capacitor is connected to the battery, a momentary current flows. • Electrons gather on the plate attached to the negative terminal of the battery. At the same time electrons are drawn from the positive plate of the capacitor +++++ -------**What is happening**• When the capacitor is connected to the lamp, the charge has the opportunity to rebalance and a current flows lighting the lamp. • This continues until the capacitor is completely discharged. +++++ -------**While the capacitor is charging**• Although the current falls as the capacitor is charging the current at any instant in both of the meters is the same, showing that the charge stored on the negative plate is equal in quantity with the charge stored on the positive plate. mA +++++ ------- mA**When the capacitor is fully charged**• When the capacitor is fully charged the pd measured across the capacitor is equal and opposite to the p.d. across the battery, so there can be no furthur current flow. V V +++++ -------**Capacitance**• The measure of the extent to which a capacitor can store charge is called its capacitance. It is defined by C= capacitance (unit farad (F)) Q = the magnitude of the charge on one plate (unit coulombs (C)) V = the p.d. between the plates ( unit volts (V)) Notice that in reality the total charge stored by the capacitor is actually zero because as much positive as negative charge is stored. When we talk about the charge stored (Q in this formula) it refers to the excess positive charge on on the positive plate of the capacitor. +Q ++++++++ -Q --------------**Putting a resistor in series with the capacitor increases**the charging time and increases the discharging time The effec of a resistance on the charging and discharging 6V 2 200μF**Kirchoff’s second la tells us that the e.mf. Must equal**the sum of the pd’s Vbattery = V resistor+Vcapacitor 6V Initially the capacitor is uncharged. At this time Vcapacitor =0 And V battery= Vresistor**Kirchoff’s second la tells us that the e.mf. Must equal**the sum of the pd’s Vbattery = V resistor+Vcapacitor As the capacitor charges Vcapacitor rises and so Vresistor falls. From 6V The current through the resistor (and therefore the whole circuit) falls**I max**Current A Small R For a large resistor the maximum current, (which is the initial current) is lower. The time taken to charge the capacitor is correspondingly larger. I max I max Large R Time/s**Finding the charge stored**Remember that the charge stored on each plate is the same. Finding the stored charge is another way of saying finding the charge stored on the positive plate. +++++++++ Charge stored (Q = It) --------------- Current/A mA The area under the curve is the charge stored Time/s**Discharging a capacitor**Here the 1 000μF capacitor is charged from a battery and discharged through a 100KΩ resistor. Try timing the discharge with a charging potential of 3V, 4.5V and 6V. Draw a current against time graph in each case and measure the area under the graph. This area will give you the charge on the capacitor. Calculate the capacitance of the capacitor in each case using V mA**Current/A**Q = I x t Time/s Discharging with a constant current If the series resistance is decreased continuously as the capacitor is discharged it is possible to keep the current constant while discharging the capacitor. The advantage of this is that the charge on the capacitor is easier to calculate.**Discharging with a constant current**100kΩ 1 000μF V 6V mA**Exponential decay**Whether charging or discharging the capacitor, the current time graph has this particular form. It is exponential in form. (The “mathematical” form of a curve like this never actually falls to zero though in practice it does). Current μA Time s**Exponential decay**The equation of the curve can be shown to be Io Current μA Where C is the capacitance of the capacitor and R is the resistance of theFIXED series resistor Note that the only variable on the right is t. When t=CR Time s So C x R is an important value and is known as the time constant e = 2.718 so 1/e = 0.368**Exponential decay**Current μA Io I = 0.368Io 0.368Io (0.368)2Io (0.368)3Io RC 2RC 3RC Time s The time it takes the current to fall by a factor of 1/e is a constant. That time interval is RC the time constant**C1**Q1 C2 Q2 C3 Q3 V Capacitors in parallel The capacitors are in parallel and therefore there is the same p.d. across each from A single capacitor which stores as much charge (Q =Q1+Q2+Q3) is represented by: So C= C1+C2+C3 It follows that capacitors in parallel have a total capacitance which is equal to the sum of their individual capacitances.**C3**C2 C1 Q1 Q2 Q3 V3 V1 V2 Capacitors in series adding i.e. V A single capacitor which has the same effect is: So:**Energy and Capacitors**Work is done in moving the electrons During charging the addition of electrons to the negative plate involves work in overcoming the repulsion of electrons already there. In the same way removal of electrons from the positive plate involves overcoming the attractive electrostatic force of the positive charge on the plate C +++++++ ------------**Energy and Capacitors**Remember that the voltage V is the work done per unit charge: Imagine the capacitor is partially charged so that the charge on the plates is Q It then acquires a little more charge δQ. This involves the work of moving charge δQ from one plate to the other. If δQ is very small V can be considered unchanged in which case Q Q+ δQ + + + + + V - - - - - C**Energy and Capacitors**And as Q+ δQ + + + + + V - - - - - We can substitute for V C So the total work done in giving the capacitor full charge from 0 to Qfull And in the limit as δQ→0**Writing Q fpr Qfull and making use of Q=VC**W =the energy stored by the charged capacitor (J) Q= the charge on the plates (C) V= the pd across the plates (V) C = the capacitance of the capacitor (F)