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Electricity Flashcards

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AP® Physics flashcards Physics 101 flashcards
1
Q

DC

A

Direct current supply has one size and one direction.

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2
Q

AC

A

An alternating current supply is a current which changes direction and instantaneous value with time. This is shown as a sine wave on an oscilloscope screen. Time is measured using x axis and voltage on y axis.

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3
Q

Frequency of ac supply

A

Found from xaxis (timebase) using these steps:
1. find no boxes to complete one wave
2. find period by x the boxes by timebase (e.g. 4x(2x10^-3)ms
3. use equation f=1/T (Hz)

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4
Q

Peak voltage of AC supply

A

Found by counting no boxes which = amplutude (centre up)
Multiply by y-gain or voltage gain value. (e.g 2 x 3V=6V)

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5
Q

True value of ac voltage

A

Average voltagw for one whole wave. This is called r.m.s volatge (Vrms). Mains r.m.s is 230v while peak v is 325V.

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6
Q

rms and peak voltage equation

A

vrms = Vpeak/squareroot 2

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7
Q

rms current and peak current equation

A

Ipeak = squareroot2 x Irms

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8
Q

IMPORTANT info on calculating rms values

A

When doing a calculation ALWAYS use r.m.s. values. If peak values are used in a
calculation the answer will be wrong! For example;
If Ipeak = 2 A and Power = 8 W, calculate Vrms.
You must find Irms first then use the equation P = IV or the answer will be wrong.
(Answer is Vrms = 5.7 V)

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9
Q

Current

A

The rate of flow of charge or number of coulombs of charge passing a point each second.

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10
Q

relationship between charge and current

A

Q=It

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11
Q

Voltage

A

Energy given to each coulomb of charge

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12
Q

Potential difference

A

Same thing as voltage

Word done (energy) in moving one coulomb of charge between 2 points.

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13
Q

Relationship between p.d, current and resistance

A

V=IR
When describing p.d is ‘across’ component and current is ‘through’ component.

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14
Q

As resistance increases

A

Current decreases

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15
Q

Power related to ohm’s law

A

P=IV

P=I^2R

P=v^2/R

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16
Q

In a series circuit

A

Current is same at all points

Supply V is shared across components

Total resistance is found by adding resistors together

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17
Q

Parallel Circuit

A

Current split up through each branch

Vsupply is same as p.d across each branch.

Total resistance 1/Rt = 1/R1 +1/R2 +1/R3

When resistor in parallel is added, total resistance decreases.

If 2 resistors in parallel have same value, then total r will be half one of resistors.

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18
Q

Potential dividers

A

v1/v2 = r1/r2

v2 = (r2/(r1+r2)) x Vs

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19
Q

Bridge voltage

A

Voltage across V2 for both potential dividers then difference between 2 values

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20
Q

current and internal resistance

A

same as rest of circuit

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21
Q

e.m.f

A

electromotive force is the voltage across the battery when no current is in the supply

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22
Q

definition of e.m.f

A

Energy given to each coulomb of charge that passes through the supply. You could also say the voltage across the battery when there is an open circuit.

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23
Q

T.p.d

A

terminal potential difference is the voltage across the supply where there is a current in the supply (switch closed). The voltage across the battery is the same as what would be across the load resistor (resistors outside the battery) in the circuit.

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24
Q

‘lost volts’

A

Difference between e.m.f and t.p.d values. This occurs as the battery has a small internal resistance and when a current passes through battery a potential difference is generated across internal resistance.

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25
Q

internal resistance equation

A

E = V + IR
E= I (R+r)

if emf is used then total resitance is E = I (R+r)

if its t.p.d then load resistance must be used Vtpd = IR

If its lost volts the internal resistance must be used V Iv = IR

remember current will always be same as it is a series circuit, unless load resistance or internal resistance is changed

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26
Q

common exam q for load

A

A common way the load resistance can be altered
is to add another resistor in parallel. This would
reduce the reading on the voltmeter because;

 The load resistance decreases which decreases
the total resistance in the circuit.

 The current in the circuit will increase.

 The lost volts across the internal resistance
increases OR share of potential difference across the parallel branch
decreases.

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27
Q

Short circuit

A

Where a wire is placed across the terminals of the battery meaning load resistance is no longer forming part of the circuit (R=0).

A short circuit will increase current as there is less resistance. To find current in the circuit E = I (R+r) changes to E=Ir

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28
Q

Using graph to determine the emf and internal resistance

A

EMF is found by determining y-intercept (when current is 0)

Internal resistance determined by gradient (neg gradient but written pos ohms)

Internal resistance can also be calculated E= Ir where E is y intercept and I is the x intercept (short circuit)

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29
Q

Capacitor

A

Device used to store charge, measured in Farads.

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30
Q

Capacitance

A

Ratio of charge stored to potential difference. Capacitance is gradient of line in charge against V graph, an so is always a constant.

C=Q/V

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31
Q

If a capacitor of larger capacitance is used then charge stored will be…

A

greater

The v across capacitor can never be greater than supply v

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31
Q

Capacitor plates

A

Has 2 plates which store charge. Plate is connected to negative side of battery will be negatively charged as extra electrons are added to it. The plate connected to the positive side of battery will be positively charged as electrons are removed from it. Electrons do not move across gap between plates as insulator.

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31
Q

Capacitor Charge

A

Charge when connected to a battery. The capacitor stores energy during charging process. This happens because work done (energy) by the capacitor as negatively charged electrons experience a repelling force as they join the negatively charged plate. Work is done to overcoming this repelling force

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32
Q

Energy stored by a capacitor on graph

A

= area under charge against v graph

therefore, E=1/2 QV

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33
Q

Equations for energy stored by capacitor

A

E=1/2QV

E=1/2 CV^2

E=1/2Q^2C

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34
Q

If maximum energy wanted in energy stored by capacitor question then…

A

supply v and max charge should be used

35
Q

graphs for capacitor charging

A

voltage against time: starts from origin n curve up and left, at origin switch closed at straight top line capacitor fully charged (supply voltage)

current against time: starts top at current axis then goes down to 0 current at right time (start top current is switch closed and bottom end of time is capacitor fully charged.

36
Q

As a capacitor will be in a series circuit with a resistor, the potential difference
across the resistor and capacitor will…

A

always equal the supply voltage.

37
Q

This means
that the potential difference across the resistor will equal the supply voltage
when the potential difference across the capacitor is…

A

0

38
Q

The potential
difference across the resistor will be zero when the capacitor is

A

Fully charged

39
Q

The value of the resistance in the circuit will determine…

A

the maximum charging
current. A greater resistance will result in a smaller maximum charging current.

40
Q

capacitor charging description

A

When the switch is in position A the capacitor is charging. Electrons leave the
negative side of the cell or battery through point A to the capacitor plate (1). The
size of the maximum charging current is determined by the resistance in the
circuit. When the capacitor is charging this would
be due to resistor (x). The capacitor charges as it
is connected to a power supply. When the switch
is changed to position B the electrons leave the
plate (1) and go through point B to the positive
plate (2). This means the current is in the
opposite direction. The size of the maximum
discharging current can be different from the
charging current as it depends on the resistance
of resistor (y).

41
Q

capacitance of capacitor and charging/discharging time

A

A greater capacitance will require more charge to be fully charged and so will take greater time to charge.

Small capacitance will therefore take less time. The size of max charging/discharging current will not be affected and the size of p.d across capacitor will also be unaffected.

41
Q

Capacitor discharging graphs

A

The potential difference
across the capacitor will be equal to the supply voltage and then decrease to zero
when the capacitor is fully discharged.

As we saw from the previous diagram, the current will be in the opposite direction. This means the current is negative but the size of the maximum discharge current will be determined by the resistance
in the circuit. The current will then decrease to zero when the capacitor is fully
discharged.

41
Q

next furthest band from nucleus (above valence band)

A

conduction band

41
Q

band furthest from nucleus

A

has max number of electrons

called valence band

41
Q

Greater resistance and charging/discharging time

A

Results i lower current and greater charging/ discharging time as there is less charge in the circuit the current is lower Q=It

P.d across capacitor when it is fully charged will not change.

41
Q

In conductors what happens to the bands

A

The valence and conduction band overlap. This allows ‘free’ electrons to move from atom to atom in the conduction band when a potential difference is applied.

41
Q

space between bands

A

band gap, electrons are not allowed here

41
Q

Insulators and band gap

A

The conduction band has no ‘free’ electrons. The band gap between valence and conduction band is very large so electrons in valence band cannot jump to conduction band, even when a p.d is applied.

41
Q

conductors, semi-conductors and insulators all have what

A

full valence band of electrons, but inly conductors have ‘free’ electrons which are found in conduction band

41
Q

Pure intrinsic semiconductors

A

Conduction band also has no ‘free’ electrons at low temperatures. However, band gap from valence to conduction band is small and so at high temperatures (room temp) a small number of electrons are able to jump from valence to conduction band. This increases conductivity of semiconductors and therefore reduces resistance.

41
Q

How an conductivity of a semiconductor be increased

A

Process called doping

These are called extrinsic semiconductors

41
Q

conductors

A

have ‘free’ electrons in conduction band, so high conductivity (low resistance)

41
Q

level between valence band and conduction band

A

fermi level

42
Q

Doping

A

Doping involves the addition of an impurity atom which has either 5 outer electron (n-type) or 3 outer electrons (p-type)

42
Q

P-type

A

In a p-type semiconductor, because the impurity atom has 3 outer electrons there will be one electron missing, which is called a hole. Electrons will fill the hole which results in the movement of charge. this means the conductivity of the semiconductor has increased.

42
Q

Charge of n-tpe and p-typ

A

Both n-type and p-type semiconductors have a neutral charge. The reason they are given names n-type and p-type is to show the nature of charge that is free to move, negative electrons and positive holes.

42
Q

n-type

A

In a n-type semiconductor, because the impurity atom has 5 outer electrons there will be one extra electron which is ‘free’ to move. this again, results in the movement of charge meaning the conductivity od semiconductor is increased.

43
Q

Instrinsic semiconductor

A

Will have ‘free’ electrons in conduction band and ‘holes’ in valence band. In the n-type semiconductor the ‘free’ electrons are found in conduction band an p-type semiconductor ‘holes’ are found in valence band. The fermi level is located where charge carriers (free electrons or holes are likely to be). This is why the fermi-level changes for n-type an p-type.

44
Q

p-n junction

A

When p-type semiconductor and n-type semiconductor are joined together a p-n junction is produced. Another name is a diode. The n-type side of diode is right and p-type left.

45
Q

p-n jucntion basic work

A

‘Holes’ from p-type cross into n-type while ‘free’ electrons from n-type cross to p-type. the ‘free’ electrons fill the ‘holes’ which creates a layer called the depletion layer. Charge cannot flow through the depletion layer unless p-n junction is connected in the correct bias and potential difference is large enough.

46
Q

forward bias

A

A p-n junction is connected i forward bias is p-type is connected to positive terminal (bottom of triangle). The depletion layer is broken down if connected this way and charge can flow through the junction, if p.d is large enough

47
Q

reverse bias

A

A p-n junction is in reverse bias is n-type is connected to positive terminal (point of triangle). When connected this way the depletion layer is widened and no charge can flow, even when a large p.d is applied.

48
Q

LED

A

Triangle point facing right with arrows coming right out. It is a p-n junction connected in forward bias which emits light.

49
Q

Band theory in LED

A

Band theory is used to explain how they work. Electrons from n-type conduction band move to p-type conduction band. When the electrons in p-n junction they fall from conduction to valence band where they pair with a ‘hole’. A photon is then emitted. One electron that pair with one hole will emit one photon.

50
Q

colour of light emitted

A

Predicted by determining wavelength if photon. Visible light is 700 to 400 nanometers. red is 650nm green is 520nm and blue is 480nm.

51
Q

visible light nm

A

700-400

52
Q

red nm

A

650nm

53
Q

green nm

A

520nm

54
Q

blue nm

A

480nm

55
Q

wavelength equation

A

v= frequency x wavelength

56
Q

frequency of wavelength

A

E = h f

h is plancks constant which is 6.63x10^-43 Js

57
Q

Energy of photon =

A

gap between conduction and valence band

58
Q

Solar cell

A

P-n junction which produces a potential difference when photons when photons enter depletion layer. Known as photovoltaic effect.

59
Q

solar cell photon movement

A

When photon arrives at depletion layer it causes electron-hole pairs to separate. Electrons move from valence band to conduction band while hole remain in valence band. this means there will be. a greater number of charge carriers and p.d is produced.

The greater the number of photons that arrive at the depletion layer the greater the p.d produced by solar cell.

60
Q

semiconductor thermistor

A

a thermistor is a devise whose resistance decreases as temperature increases

input - heat
output - electrical

61
Q

semi conductor LDR

A

resistance decreases as light increases

input - light
output - electrical

62
Q

what is meant by a doped semiconductor

A

a semiconductor that has (specific) impurities added

63
Q

A voltage is applied across the LED so that it is forward
biased and emits light. Using band theory, explain how the
LED emits light.

A

(Voltage applied causes) electrons ( ) to move from the
conduction band of the n-type (semiconductor) towards the
conduction band of the p-type (semiconductor). (1)
To access this mark, the direction the electrons move must
be clear.
Electrons ( ) ‘fall’ from the conduction band into the
valence band (on either side of the junction) (1)

64
Q

valence ban is the

A

lower energy levels of electrons

65
Q

conduction band is the

A

higher energy levels of electrons

66
Q

for conduction to occur what 2 things must there be

A

electrons free to move into conduction band

spaces in energy bands for electrons to move into

67
Q

conductor

A

no band gaps and the valence n conduction overlap

The conduction band is only partially filled. This means there are spaces for electrons to move into. When electrons for the valence band move into the conduction band they are free to move. This allows conduction.

68
Q

insulators

A

has a large gap between valence band and conduction band

The valence band is full as no electrons can move up to the conduction band. As a result, the conduction band is empty.

Only the electrons in a conduction band can move easily, so because there aren’t any electrons in an insulator’s conduction band, the material can’t conduct.

69
Q

semi conductor

A

smaller gap between bands

At room temperature there is sufficient energy available to move some electrons from the valence band into the conduction band. This allows some conduction to take place.

An increase in temperature increases the conductivity of a semiconductor because more electrons will have enough energy to move into the conduction band.

70
Q

LED

A

The voltage applied causes electrons to move away from the n-type material towards the
junction.

Electrons ‘fall’ from the conduction band to the valence band.

A photon is emitted when an electron recombines with a hole in the junction.

71
Q
A

Higher Physics (1 decks)

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