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A Level Physics OCR B > 3: Sensing > Flashcards

3: Sensing Flashcards

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1
Q

What is current?

ai)

A

The rate of flow of charged particles

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2
Q

What is Kirchhoff’s first law?

A

The total current entering a junction = the total current leaving it

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3
Q

What is potential difference?

aii)

A

energy (work done) per unit charge

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4
Q

potential difference equation for change in energy

A

V = ΔE / Q

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5
Q

Do you connect a voltmeter in parallel or series. Why?

A

Parallel because the p.d across components in parallel is the same

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6
Q

3.1what is joule heating, equation to calculate joule heating or wasted energy

A

potential difference that is lost by electrons in a wire getting obstructed by positive ions. this lost pd does work on the wire, heating it up, this wasted energy is known as dissipation

W = VIT

W = work done

also P = I2R

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7
Q

What is power? what’s it measured in and units

A

The rate of transfer of energy (the rate of work done)

measured in J s-1, which is the equivalent of the Watt (W)

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8
Q

equations for power

A

P = I V

P = I2 R (DISSIPATION)

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9
Q

How can you reduce the power dissipated during transmission of mains electricity?

A

P=IV Mains electricity is transmitted at a high voltage and low current to minimise the power dissipated

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10
Q
A

kirchoff’s first law means that current entering junction is the same as that leaving it, so in that parallel section there’s 0.1 A, so 0.4 - 0.1 = 0.3A going through I1,

as V1 is part of that parallel circuit, and pd is the same everywhere in a parallel circuit, V1 is 3.4V

so V2 is 6 - 3.4V = 2.6V

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11
Q

What is resistance?

equation

units?

A

A measure of how difficult it is to get a current to flow through a component

R = V / I

V A-1 (ohms Ω)

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12
Q

What is Ohm’s Law? describe the graph as well

A

If temperature is constant, the current through an ohmic conductor is directly proportional to the p.d across it (V=IR) The gradient of the IV graph is constant (so resistance is constant) and the graph goes through the origin

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13
Q

what is conductance

units, unit equivalence

A

how well electrons can flow through a conductor

conductance = 1 / R = I / V

the unit of conductance is A V-1 which is the Siemen (S)

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14
Q

how to work out resistance and conductance in series and parallel circuits

A
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15
Q

pd, current, conductance and resistances in parallel and series

A
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16
Q

How do you reduce the effect of random errors when investigating the I-V characteristic of a component?

A

Repeat your measurements and take averages

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17
Q

Describe and draw the I-V characteristic of a filament lamp, include pd against resistance and conductance

Why is it this shape?

A

current: A curve, which starts steep but then gets shallower as the p.d rises

Current flowing though the lamp increases its temperature, but current also dissipates energy through joule heating at a rate of (P = I²R) so its resistance increases. because of that, conductance decreases.

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18
Q

How can you investigate the I-V characteristic of a component using a test circuit?

A

1) Use a variable resistor to alter the p.d across the component and so the current flowing through it, record V and I
2) Plot a graph of current against p.d difference from your results. This graph is the I-V characteristic of the component

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19
Q

what is a conductor

A

Materials (such as metals) which have a high proportion of mobile charge carriers (free electrons), and can conduct electricity well.

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20
Q

semiconductors

A

Materials with a low proportion of mobile charge carriers, but the number of mobile charge carriers increases with a factor like light or temperature, enabling them to conduct electricity.

Under normal conditions, however, they do not conduct very well.

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21
Q

insulators

A

Materials with no (or very few) mobile charge carriers, which do not conduct electricity

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22
Q

What does the resistance of a wire depend on? Explain each one

A

1) Length. The longer the wire, the more difficult it is to make a current flow
2) Area. The wider the wire, the easier it will be for the electrons to pass along it
3) Resistivity. This depends on the material the wire’s made from, as the structure of the material may make it easy or difficult for charge to flow. Resistivity also depends on external factors like temperature

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23
Q

What affects how conductive a material is?

A

It’s number density of mobile charge carriers - the number of free electrons (or ions that are free to move) there are per cubic metre of the material.

The more mobile charge carriers a material has per unit volume, the better a conductor it will be

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24
Q

relationship between conductance and resistance with length of wire

A

resistance doubles if the length of the wire doubles so R ∝ L, and so conductance is G ∝ 1 / L

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25
Q

relationship between cross sectional area of a wire with resistance and conductance

A

conductance doubles if c.s.a doubles so G ∝ A, and resistance R ∝ 1 / A

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26
Q

relationship between conductivity and resistivity using their symbols

A

σ = 1 / p

σ = conductivity

p = resistivity

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27
Q

equation for conductivity using its constant

A

G = σA / L

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28
Q

3.3 equation for resistivity using its constant

A

R = pL / A

29
Q

Why are metals good conductors?

A

They have loads of free electrons (charge carriers). The number density of mobile charge carriers is high

30
Q

Explain how a metal’s conductivity is affected by temperature

A

As the electrons move, they scatter from the metallic lattice

As the temperature increases the lattice vibrates more, increasing the electron scattering, so the electrons are slightly less free to move

This means that as the temperature increases, the conductivity of a metal will slightly decrease

31
Q

what is mean drift velocity

A

distance travelled of the electrons along the wire per second as they collide constantly

32
Q

What is the equation linking current, the number of electrons, and the velocity of the electrons?

A

I = nAve

I = current (A)

n = n. density of electrons (m⁻¹)

A = cross sectional area (m²)

v = drift velocity (ms⁻¹)

e = charge of an electron (C)

33
Q

What are Thermistors and LDRs, use for each

thermistor symbol

A

Examples of semiconductors.

In thermistors, increasing temp liberates electrons, enabling them to conduct and increasing conductivitiy, so reducing resistance

in LDRs it is light which liberates them. light falling on the LDRs increases conductivity, reducing resistance.

This means they can be used for temperature or light sensors.

34
Q

What type of thermistor do we look at?

I-V graph for thermistor, temp and resistance graph

A

NTC negative temperature coefficient, as the temperature increases the resistance decreases

35
Q

Describe an LDR and its circuit symbol

A

Light dependent resistor, sensitive to light The more light falls on it, the lower the resistance

Symbol: resistor in a circle with 2 arrows outside the circle, pointing towards centre of circle

36
Q

Describe diodes

symbol for led and diodes

A

They are designed to let current flow in one direction only

37
Q

What does forward bias (direction) mean when talking about diodes?

A

The direction in which the current is allowed to flow

38
Q

Most diodes require a [] voltage of about 0.6 V in the [] direction before they will conduct

A

Threshold Forward

39
Q

What happens in reverse bias (direction) with diodes?

A

The resistance of the diode is very high and the current that flows is very tiny

40
Q

IV characteristic graph of a diode

symbol for diode and LED

A

Slightly negative current before threshold voltage where the current increases in a rough straight line

41
Q

How is the p.d split in a potential divider circuit?

A

The p.d of the voltage source is divided in the ratio of the resistances

42
Q

What can you use a potential divider circuit for?

A

Calibrating voltmeters,

you can choose the resistances to get the voltage you want

check whether the voltmeter reading is the same as the actual voltage across the component

43
Q

ratio of resistances and voltages between 2 resistors

A

V1 / V2 = R1 / R2

44
Q

potential divider equation for Vout and Vin

A

Vin = supply voltage

Vout = Vin * reistance of component at Vout/ total resistance

45
Q

Why do batteries have resistance? What is it called? what does this do to the battery

A

In a battery, chemical energy is used to make electrons move. As they move, they collide with atoms inside the battery.

Internal resistance

internal resistance is what heats batteries up

46
Q

What is e.m.f? what is it measured in

A

The amount of energy the battery (source) provides per unit charge (coulomb)

measured in volts

47
Q

What is Kirchhoff’s second law?

A

The total emf around a series circuit = the sum of the p.ds across each component

48
Q

What is load resistance?

A

The total resistance of all the components in the external circuit. Also called external Resistance

49
Q

What is the terminal p.d?

A

The potential difference across the load resistance. It is the energy transferred when one coulomb of charge flows through the load resistance

50
Q

What would happen if there was no internal resistance in the battery?

A

The terminal p.d would be the same as the e.m.f. However in real power supplies, there’s always some energy lost, as heat energy, overcoming the internal resistance

51
Q

What are lost volts?

A

The energy wasted per coulomb overcoming the internal resistance

52
Q

equation to find out Energy per coulomb supplied by the source

A

Energy per coulomb supplied by the source = energy per coulomb transferred in load resistance + energy per coulomb wasted in internal resistance

53
Q

equations for emf and internal resistance

A

ε = V + v ε = I(R + r)

V = ε - v V = ε - Ir

ε = emf

V = terminal pd v = lost volts

I = current

R = load resistance r = internal resistance

54
Q

an assumption that is made when calculating emf across a circuit

A

emf = Voltage at every component + internal resistance (lost voltage within the battery itself) we assume that there is no resistance from the connecting wires

55
Q

How can you calculate emf and internal resistance from a I against V graph?

A

Start with V = ε - Ir Since ε and r are constants, this is an equation of a straight line. So the intercept with the vertical axis is ε And the gradient is -r

y = mx + c -> V = (-r)I + ε

56
Q

how can a decrease in load resistance lead to a decrease in terminal pd

A

when load resistance R decreases, current increases because the total resistance R + r has decreased, but this also means the lost voltage, Ir, increases because I has so reducing the terminal PD

57
Q

how to work out total emf in series circuits

and what is the condition

A

for cells in series you can calculate the total emf by adding the individual emfs

this requires all cells to be connected in the same direction

εtotal = ε1 + ε2 + ε3 …

58
Q

How do you find the total emf of cells in series pointing in opposite directions?

A

Subtract emf of the ones facing in the opposite direction from the ones in the other direction

59
Q

how to find total emf for identical cells in parallel

A

for identicall cells in parallel, the total emf is the same size as the emf of each individual identical cell.

this is because the current will split equally between identical cells.

εtotal = ε1 = ε2 = ε3 …

60
Q

What is an easier way of measuring the emf of a power source?

A

Connect a voltmeter across its terminals. The current through the voltmeters 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

61
Q

How can you investigate internal resistance and emf?

A

1) Vary the current in the circuit by changing the value of the load resistance using the variable resistor 2) Measure the p.d for several different values of current 3) Record the data for V and I, and plot the results in a graph of V against I, then calculate emf and internal resistance

62
Q

What assumption do you make in any experiment using voltmeters and ammeters?

A

You can assume that the voltmeter has a very high resistance, and the ammeter has a very low resistance

63
Q

Why is it hard to choose the value for the load resistance, when investigating emf and internal resistance?

A

A low load resistance will give a large current, which will reduce the percentage uncertainty in the ammeter reading of the current. But large currents will cause significant heating in the wires, which will invalidate your results

64
Q

Why does including an ammeter in the circuit not affect the current trough the variable resistor? (Investigating emf and internal resistance)

A

The ammeter has a resistance that’s so low it’s negligible, and so the voltage across it is also negligible

65
Q

Why does including a voltmeter in the circuit not affect the current through the variable resistor? (Investigation of emf and internal resistance)

A

Voltmeters have a very high internal resistance, so the current through them is so low you can usually assume it is negligible

66
Q

When investigating internal resistance and emf, how can you reduce the effect of heating the wires?

A

Include a switch in your circuit to turn off the current whenever possible to reduce the effect of heating in the wires on the resistance of the circuit

67
Q
A

answer on q

68
Q

thermistor practical

how to improve

A

place thermistor in beaker of boiling water containing crushed ice and measure temperature with a thermometer at selected regular intervals (5 degrees), make sure to stir before taking a reading to get an accurate measurement

can improve using water bath with thermostat

cool slowly to reduce temp fluctuations

A Level Physics OCR B (57 decks)

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