sensing Flashcards

Question

what is the equation for total energy transferred (aka total work done)? SENSING

Answer 1

total energy transferred (aka total work done) = power x time W = Pt so W = VIt

Answer 2

if you put pd across an electrical component, current will flow

Answer 3

how much current flows through a component for a particular pd depends on the resistance of the component

Answer 4

resistance is the measure of how difficult is it is for a current to flow through a component the symbol for resistance: R the unit for resistance is: Ohms, Ω

Answer 5

• a component has a resistance of 1 Ω if a pd of 1 V across the component makes a current of 1 A flow through the component • think of R = V/I

Answer 6

R = V/I (the equation DEFINES resistance also) or V = IR where: R = resistance V = voltage I = current

Answer 7

the inverse of resistance is conductance

Answer 8

conductance is the measure of how easy it is for a current to flow through a component the symbol for conductance is: G the unit for conductance is: Siemens, S or Ω^-1

Answer 9

G = 1/R so G = I/V

Answer 10

P = (I^2)R (this equation works out power dissipation) P = (V^2)/R

Answer 11

power dissipation is the rate at which a component converts electrical energy into other types of [unwanted] energy (eg, heat) the equation used to work out power dissipation is: P = (I^2)R

Answer 12

to minimise power dissipation when transmitting mains electricity, mains electricity is transmitted at a high voltage (so current would lower, provided resistance stays constant) (+ low I) to minimise power dissipated during transmission (bc I has lowered due to high voltage, P lost as power dissipation would decrease also)

Answer 13

I-V characteristic graphs show how the resistance of a component varies

Answer 14

⋅ 'I-V characteristic' refers to a graph which shows how the current (I) flowing through a component changes as the pd (V) across the component is increased

Answer 15

you can investigate the I-V characteristic of a component using a test circuit like this one: INPUT DIAGRAM

Answer 16

to investigate the I-V characteristic of a component using a test circuit: 1) set up equipment as shown below: 2) use a variable resistor to alter the pd across the component and thus the current flowing through the component 3) record varying pds and their corresponding currents in a table 4) take multiple measurements for a particular pd and take an average to reduce the effect of random errors on your results 5) plot a graph of current against pd from your table 6) this graph is the I-V characteristic of the component and you can use it to see how the resistance of a component changes INSERT DIAGRAM

Answer 17

for an ohmic conductor, R is constant

Answer 18

ohmic conductors are conductors that obey ohm's law ohmic conductors are mostly metals and examples include: metal wires and resistors

Answer 19

Ohm's law states that: "Provided external factors such as temperature are constant, the current through an ohmic conductor is directly proportional to the potential difference across it."

Answer 20

V ∝ I so V = kI which is just V = IR

Answer 21

description of I-V characteristic of an ohmic conductor: ⋅ as you can see from graph, doubling the pd doubles the current (bc pd ∝ I according to ohm’s law) ⋅ the gradient is constant, which means that resistance is constant notes: ⋅ remember, Ohm's law is only true for ohmic conductors where external factors like temperature are constant ⋅ non-ohmic conductors don't have this correlation (ohm's law) between current and pd

Answer 22

description of the I-V characteristic graph of a filament lamp: ⋅ the I-V characteristic graph for a filament lamp is a curve, which starts steep but gets shallower as pd rises ⋅ as the current flowing through the lamp increases, the temperature of the lamp will increase, so the lamps resistance will increase (which is shown by curve becoming more shallow as current increases) note: ⋅ a filament lamp does NOT have the same I-V characteristic graph as a metal conductor even though the filament in a filament lamp is just coiled up metal wire

Answer 23

a thermistor is a semiconductor with a resistance that depends on the temperature a thermistor's circuit symbol is: [you only need to know about NTC (Negative Temperature Coefficient) thermistors]

Answer 24

thermistors can be used as temperature sensors bc their resistance depends on the temperature

Answer 25

the resistance of an NTC thermistor decreases as temperature increase (hence the NEGATIVE temperature COEFFICIENT, (-)coefficient acts as gradient)

Answer 26

⋅ if you increase the current through a thermistor (or any component), it increases the temperature of the component ⋅ if you look at the graph, the increasing gradient of the I-V characteristic tells you that resistance is decreasing as current increases linking stuff together to help understanding: [⋅ so if temp increases as current increases and resistance decreases (bc R is high when grad is low) as current increases, this means the resistance of the thermistor decrease as the thermistor heats up]

Answer 27

a LDR is a semiconductor with a resistance that depends on the amount of light falling on it ⋅ the more light that falls on it, the lower the LDR's resistance an LDR's circuit symbol is:

Answer 28

diodes are components that let current flow in only one direction (their forward bias) an example of a diode is a light emitting diode (LED)

Answer 29

⋅ the forward bias of a diode is the direction that the diode allows current to flow ⋅ a diode's forward bias is the direction that the triangle of its circuit symbol points in

Answer 30

most diodes require a threshold voltage of about 0.6 V in the forward direction (direction of the forward bias) before they will conduct

Answer 31

description of the I-V characteristic of a diode: ⋅ in reverse bias (the opposite direction to which a diode lets current flow) the resistance of the diode is very high and any current that flows in reverse bias is very tiny (it's negligible) ⋅ the graph starts increasing once the threshold voltage is reached (once x-axis passes certain value)

Answer 32

the resistance of a material depends on: ⋅ length, L: the longer the component, the more difficult it is to make a current flow through the component ⋅ cross-sectional area, A: the wider the component, the easier it will be for electrons to pass along it (bc less of component for electrons to collide with) ⋅ resistivity, ρ: the resistivity of the component depends on the material the component is made from, as the structure of the material may make it easier or more difficult for a charge to flow

Answer 33

the resistivity of a material is the resistance of a 1 m length of the material with a 1 m^2 cross-sectional area the symbol for resistivity is: ρ the unit for resistivity is: Ohm metre, Ω⋅m (NOT OHMIC METRE)

Answer 34

resistivity = (resistance x cross-sectional area)/length ρ = RA/L or RA = Lρ (remember "ralp")

Answer 35

typical values of the resistivity for conductors are really small eg) copper (at 25 °C) resistivity = 1.72 x10^-8 Ohm metres

Answer 36

• yes • the resistivity of the material depends on its temperature, so you can only find the resistivity of the material for/at that certain temperature

Answer 37

the inverse of resistivity is conductivity σ = 1/ρ

Answer 38

the conductivity of a material is the conductance of a 1 m length of the material with a 1 m^2 cross-sectional area the symbol for conductivity is: σ the unit for conductivity is: Siemens per metre, S m^-1

Answer 39

conductivity = (conductance x length)/cross-sectional area σ = GL/A or GL = Aσ (remember "glass") (bc σ = sigma, makes "ss" noise)

Answer 40

to find the resistivity of a wire you need to find its resistance first

Answer 41

1) set out circuit as shown below 2) using MICROMETER, measure DIAMETER of wire in at least 3 different points along its length + take an AVERAGE. using ave diameter, find cross-sectional area of wire using A = πr^2 3) CLAMP TEST WIRE to ruler, with circuit attached to wire where ruler reads zero 4) attach FLYING LEAD to test wire (flying lead is just wire with crocodile clip on end to allow connection to any point along test wire) 5) record LENGTH of test wire CONNECTED in circuit, VOLTMETER READING + AMMETER READING 6) use your readings to calculate RESISTANCE of length of wire, using R = V/I 7) repeat this measurement at least 3 times + calculate average resistance for length 8) repeat for several DIFFERENT lengths, eg) between 0.10 and 1.00 m 9) plot results on graph of R against L, + draw LINE OF BEST FIT 10) GRADIENT of line of best fit R/L is equal to ρ/A, so MULTIPLY GRADIENT of line of best fit by CROSS-SECTIONAL AREA of wire to find resistivity of wire material note: ⋅ other components of circuit also have resistance, but gradient of line of best fit isn't affected by resistance within rest of circuit (bc it would be zero error?, + zero errors don't affect line of best fit gradients bc effects all data points by same amount) ⋅ current flowing in test wire can cause temperature to increase, so you need to try to keep temperature of test wire CONSTANT to calculate correct resistivity (eg. by only having small currents flow through wire) bc resistivity is affected by temperature

Answer 42

do the same experiment as when you are finding the resistivity of a wire, but then do 1/(final calculated resistivity) bc σ = 1/ρ

Answer 43

you could find the ρ for a single length of wire, using the resistivity formula, but this will give you an overestimate as it'll include the resistance of the ammeter, all the wires, and their connections

Answer 44

different materials have different numbers of charge carriers

Answer 45

⋅ how conductive a material is depends on its number density of mobile charge carriers (aka charge carrier density) ⋅ the more mobile charge carriers a material has per unit volume, the better a conductor it will be

Answer 46

charge carrier density is the number of mobile charge carriers (usually free electrons or ions that are free to move) there are per cubic metre of material

Answer 47

⋅ in metals, the charge carriers are free (aka delocalised) electrons ⋅ metals are good conductors bc they have a high number density of mobile charge carriers

Answer 48

if you increase the temperature of a metal, the number of mobile charge carriers stays mostly constant

Answer 49

as the temperature increases, the resistivity (thus resistance bc resistance depends on resistivity) of a metal slightly increases and the conductivity (thus conductance - same logic as before) slightly decreases

Answer 50

⋅ in semiconductors, the mobile charge carriers are free electrons ⋅ semiconductors have a much lower charge carrier number density (so as they have fewer free electrons than metals, they have a lower conductivity)

Answer 51

⋅ as you increase the temperature of a semiconductor, more electrons are freed to conduct

Answer 52

⋅ bc as you increase the temperature of a semiconductor, more electrons are freed to conduct ⋅ this means that as temperature increases, the conductivity of the semiconductor rapidly increase

Answer 53

perfect insulators wouldn't have any mobile charge carriers, so the insulator would not be able to conduct at all

Answer 54

thermistors are made up of semiconductors whose conductivity change with temperature, as shown:

Answer 55

⋅ LDRs are made up of semiconductors whose conductivity is mostly controlled by light (rather than heat like thermistors) ⋅ the semiconductors' conductivity increases with increasing light levels

Answer 56

resistance in general comes from electrons colliding with other atoms and losing energy to other forms

Answer 57

⋅ in a battery, chemical energy is used to make electrons move ⋅ as the electrons move, they collide with atoms inside the battery ⋅ so batteries must have resistance

Answer 58

internal resistance - when electrons collide with other atoms in a battery, they lose energy as heat

Answer 59

remember to draw the box around the battery and the resistor next to it, to signify the resistor is the internal resistance (box can also be drawn as a box with a border of short intermittent lines that act as dots)

Answer 60

EMF (electromotive force) is: ⋅ the amount of electrical energy the battery produces for each coulomb of charge ⋅ the total power/energy input by the power supply ⋅ the total energy gained by the electrons [when passing through the power supply] ⋅ the electromotive force is the terminal potential difference when the cell is in an open circuit (no current flows) the symbol for the emf is: ε the unit for the emf is: Volts, V

Answer 61

⋅ the pd across the load resistance is called the terminal pd ⋅ the potential difference across the load resistance is the energy transferred when 1 coulomb of charge flows through the load resistance

Answer 62

⋅ load resistance (R) is the total resistance of all the components in the external circuit ⋅ (so basically R of circuit sans internal resistance of power supply)

Answer 63

if there was no internal resistance (so an ideal power supply), terminal pd would be the same as emf ⋅ however, in real power supplies, there's always some energy lost (as heat energy) in overcoming the internal resistance

Answer 64

lost volts (v) is the energy wasted per coulomb in overcoming the internal resistance [of power supply]

Answer 65

the conservation of energy tells us that: the energy per coulomb [voltage] supplied by the source = the energy per coulomb [voltage] transferred in the load resistance + energy per coulomb [voltage] wasted in the internal resistance emf = terminal pd + lost volts ε = V + v

Answer 66

ε = V + v or V = ε - v ε = I(R + r) V = ε - Ir ⋅ all of these equations can be used to work out each other, knowing that V = IR and v = Ir

Answer 67

⋅ to work out the total ε in a series circuit of emfs: total ε = ε1 + ε2 + ε3 + ... (bc each charge goes through each of the cells [power supplies] + so gains ε [electrical energy] from each one ⋅ this requires all the cells to be connected in the same direction ⋅ if one is connected in the opposite direction, you subtract its emf from the total ε rather than adding it

Answer 68

⋅to work out the total ε for identical cell in parallel: total ε = ε1 = ε2 = ε3 = ... (bc current will split equally between identical cells [remember rules of how I and V slow in parallel circuits], a charge only gains ε from cells it travels through - so the overall ε in the circuit doesn't increase) ⋅ the total ε of the combination of the parallel cells is the same size as the ε of each of the individual cells

Answer 69

1) set up circuit as shown 2) VARY CURRENT in circuit by changing value of LOAD RESISTANCE (R) using variable resistor + MEASURE PD (V) for several different values of CURRENT (I) ⋅ include SWITCH in your circuit to TURN OFF current whenever possible to REDUCE effect of HEATING in wires on RESISTANCE of circuit 3) record your data for V and I in table + PLOT RESULTS in graph of V against I 4) to find EMF and INTERNAL RESISTANCE: y-intercept of graph = ε and m = -r (so to work out actual internal resistance, multiply gradient by -1) ⋅ analysing the graph: 1) to find EMF and internal resistance of cell, start with equation: V = ε - Ir 2) rearrange to give V = -rI + ε (this is in form y = mx + c, where y = V, m = -r, x = I and c + ε) (we can draw this equation as equation of straight line bc ε and r are constants) 3) y-intercept of graph = ε and m = -r (so to work out actual internal resistance, multiply gradient by -1)

Answer 70

an example of a difficulty you may encounter when investigating the internal resistance of a battery is that: ⋅ choosing values for the load resistance is difficult 1) a low load resistance will give you a large current, which will then reduce the percentage uncertainty in an ammeter reading of the current 2) but large currents will also cause significant heating in wires, which will invalidate your results (bc heating will change R of wires, thus changing R of load resistance which will invalidate your results) 3) so a middle ground must be reached

Answer 71

some of the assumptions that you make when investigating the internal resistance of a battery are: 1) the voltmeter has a very high internal resistance, so the current through the voltmeter is so low it's negligible (assume it's zero) ⋅ this makes sure that the voltmeter in the circuit does not affect the current through the variable resistor ⋅(bc current won't want to flow through smth with really high R (V = IR -> I = V/R so high R will result in small current), so almost all current flows through variable resistor + so voltmeter doesn't affect current) 2) the resistance of the ammeter is very low (negligible), so the pd across it is negligible ⋅ this makes sure that including the ammeter in the circuit doesn't affect the pf across the variable resistor)

Answer 72

⋅ an easier way to measure the emf of a power source is by connecting a voltmeter across the power source's terminals ⋅ as in the other experiment, the current through the voltmeter is assumed to be negligible and so any difference between your measurements and the emf will be so small that the difference isn't usually significant

Answer 73

⋅ yes, charge is conserved - it does not 'leak away'

Answer 74

⋅ as charge flows through a closed circuit, it cannot be used up or lost ⋅ this means that whatever charge flows into a junction will flow out ⋅ since the current is the rate of flow of charge, it makes sense that whatever current flows into a junction is the same as the current flowing out of it ⋅ this logic makes up the basis for Kirchhoff's first law

Answer 75

kirchhoff's first law states: "the sum of the current entering a junction is equal to the sum of the current leaving the junction"

Answer 76

kirchhoff's first law is about the law of the conservation of charge

Answer 77

⋅ in closed electrical circuits, energy is transferred around the circuit ⋅ the energy transferred to the charge is the emf... ⋅ ...and the energy transferred from the charge is the pd ⋅ so in a closed circuit energy is conserved ⋅ so in a closed loop: energy transferred to a charge and the energy transferred from the charge must both be equal to each other if energy is to be conserved

Answer 78

kirchhoff's second law states that: "the total emf around a series circuit is equal to the sum of the potential differences across each component" or the law can be written as ε = ΣIR

Answer 79

kirchhoff's second law is about the law of the conservation of energy

Answer 80

in a series circuit: 1) there is the same current at all points of the series circuit (since there are no junctions), so : total I = I1 = I2 = I3 = ... 2) the emf is split between components in series circuit (by kirchhoff's second law), so: ε = V1 + V2 + V3 + ... or total V = V1 + V2 +V3 + ... 3) bc V = IR if I is constant, results in IR = IR1 + IR2 + IR3 + ... (basically subbed V = IR into total V equation), you can then cancel out the Is bc I is constant in a series circuit, so: total R = R1 + R2 + R3 + ... 4) 1/total G = 1/G1 + 1/G2 + 1/G3 + ... (remember to do reciprocal of 1/total G to get total G)

Answer 81

you use a potential divider to get a fraction of the input voltage in some areas of the circuit

Answer 82

at its simplest, a potential divider is a circuit with a voltage source and a couple of resistors in series

Answer 83

⋅ the pd of the voltage source is divided into the ratio of resistances in a circuit ⋅ this can be written as V1/R1 = V2/R2 ⋅ or as V1:R1 = V2:R2 ⋅ this means that you can choose resistances to get the voltage output you want across one of the components

Answer 84

a potential divider circuit can be used for calibrating voltmeters, which have a very high resistance

Answer 85

⋅ DO NOT connect something with a RELATIVELY LOW RESISTANCE across R2 ⋅ if you do, the component and R2 will effectively act as 2 resistors in parallel ⋅ two resistors in parallel will always have a total resistance that is LESS than R2 ⋅ so the actual V out will be less than what you've calculated and will depend on what is connected across R2

Answer 86

⋅ potential dividers can be made into sensors by including components whose resistances change with external factors (like LDRs and thermistors) ⋅ this would mean that V out varies with light or heat, so you can make a potential divider that works as a light or heat sensor ⋅ if you add an LDR to the potential divider, you can make a light sensor ⋅ if you add a thermistor to the potential divider, you can make a heat sensor

Answer 87

⋅ before you conduct an experiment that uses a potential divider, you must first calibrate the circuit so you know how the voltage across the component and the V out varies as external factors change ⋅ eg) knowing the voltage across the thermistor at a given temperature

Answer 88

1) set up equipment as shown, then measure TEMPERATURE of water using THERMOMETER, + record VOLTAGE across resistors ⋅ pick fixed resistor carefully - if its resistance is too high, V out won't vary enough with temperature, + if it's too low, V out might vary over bigger range than your voltmeter can handle 2) HEAT beaker GEANTLY using bunsen burner (make sure water is well-stirred), + record temperature + voltage at REGULAR INTERVALS over SUITABLE RANGE (eg. at 5 °C intervals over range of 0-100 °C) 3) plot GRAPH of temperature against voltage from your results - this graph is thermistor's CALIBRATION CURVE + you can use it to find temperature of thermistor from voltage across it, without needing thermometer

Answer 89

⋅ a potentiometer uses a variable resistor to give a variable voltage ⋅ a potentiometer has a variable resistor instead of the R1 and R2 of a potential divider, but it uses the same idea ⋅ sometimes people call a potentiometer a potential divider too (even though it isn't archetype one) just to confuse things, so be wary

Answer 90

⋅ they are all non-ohmic conductors ⋅ you can tell by looking at their I-V characteristic graphs: their graphs are not linear and/or they do not go through the origin so the I and V are not directly proportional ⋅ for things to be directly proportional, the graph’s line must be linear, go through the origin, and have the variables change in the same way (eg. increase in x, increase in y OR decrease in x, decrease in y)

Answer 91

concepts of them: ⋅ think of current in wire like flow rate of water in pipe ⋅ just as flow rate is measure of how much water goes through pipe in given time interval, current is measure of number of charged particles that move past point in wire in given time ⋅ for pd, pd is like pressure that's forcing water along pipe relationships: ⋅ so knowing V = IR and after looking at picture, you can understand that: ⋅ I ∝ 1/R ⋅ and V ∝ I ⋅ but it does NOT mean V ∝ R

Answer 92

⋅ the shallower the gradient of the I-V characteristic, the greater the resistance of the component (bc less current is flowing through the current for a particular voltage, so current increases only a small amount as voltage increases, meaning it is MORE DIFFICULT for current to flow through component = high resistance) ⋅ a curved I-V characteristic line shows that the resistance of a component changes as the pd across it changes

Answer 93

• pick a point on the line of the graph • read the current and the voltage at this point from the graph • divide the voltage by the current • this gives you the resistance of the component AT THIS POINT • the resistance of a component is not the gradient of the whole graph (bc R doesn’t stay constant sometimes, which is reflected in I-V characteristic graphs curving)

Answer 94

1) as electrons move, they scatter from the metallic lattice 2) as the temperature increases, the lattice vibrates more, increasing electron scattering, so electrons are slightly less free to move (bc now they keep bumping into each other) 3) this means that as the temperature increases, the conductivity of the metal will slightly decrease (thus resistivity will slightly increase)

Answer 95

1) just as in metals, the semiconductor's atomic lattice will vibrate more, scattering free electrons as it moves 2) but its 'increase in resistivity' effect is much smaller than the huge increase in mobile charge carriers allowing more current to flow (also due to the increase in temperature)