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OLD St Aloysius H Physics (Pre Feb 24) > Booklet 3 - Electricity (Factual) > Flashcards

Booklet 3 - Electricity (Factual) Flashcards

1
Q

What is meant by alternating current?

A

An alternating current regularly changes direction.

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2
Q

Describe how to measure frequency using an oscilloscope.

A

Adjust the timebase until a number of complete waveforms can be seen on the screen.

Measure the distance between two neighbouring crests.

Multiply this distance by the timebase setting to get the period of the wave.

Use f = 1/T to calculate the frequency.

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3
Q

What is the voltage and frequency of the UK mains supply?

A

230V, 50Hz

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4
Q

What is meant by the Root Mean Square Voltage? (Vrms)

A

It is the value of d.c. voltage that will deliver the same amount of power as an a.c. supply.

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5
Q

What can be said when comparing Vrms to Vpeak?

A

Vpeak is always greater than the Vrms

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6
Q

When calculating power should you use peak or rms voltage and current?

A

rms

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7
Q

What is the definition for current?

A

Current is the number of Coulombs of charge passing a point in one second.

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8
Q

What is the definition of 1 amp?

A

1 Coulomb of charge passing a point in 1 second.

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9
Q

What is potential difference?

A

The number of Joules of energy transferred per Coulomb of charge.

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10
Q

What is the definition of 1 volt?

A

A potential difference of 1 Volt exists when 1 Joule of energy is transferred per Coulomb of charge

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11
Q

What is the definition for emf?

A

The number of Joules of energy given to each Coulomb of charge by a supply.

OR

The voltage across a supply when no current is flowing.

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12
Q

Give two sources of emf

A
  • chemical cell
  • Thermocouple
  • Piezo-electric generator
  • Photovoltaic cell
  • Electromagnetic generator
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13
Q

What is meant by internal resistance?

A

This is the effective resistance within a power supply that can be used to model power losses within the supply itself.

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14
Q

What is a load resistor?

A

The is the total resistance in a circuit when current is flowing (or it is under load).

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15
Q

What is the terminal potential difference (tpd), V?

A

It is the potential difference across the load resistor when the circuit is complete and current flows.

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16
Q

What are the lost volts, Vlost?

A

This is the potential difference unavailable to the circuit because of the internal resistance of the supply.

It is difference between the emf and the tpd.

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17
Q

In the following circuit what quantity will the voltmeter measure and why?

A

The e.m.f as NO current is flowing as the circuit is incomplete. When no current flows no voltage is lost across the internal resistance and tpd = emf.

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18
Q

In the following circuit what quantity will the voltmeter measure and why?

A

The tpd (V) as there is a complete circuit and current is flowing so, some voltage will be lost.

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19
Q

What is the short circuit current?

A

The maximum current a supply can give - this is achieved when the terminals of the supply are joined with a short thick wire (effectively zero external resistance) and all of the emf is applied across the internal resistance.

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20
Q

Describe how you can measure the emf and internal resistance of a cell?

A

Set up the apparatus as shown.

Measure a range of voltages and currents, by changing the variable resistor.

Plot a graph of voltage versus current.

The internal resistance is -gradient.

The emf is the y-intercept.

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21
Q

From the graph, how can you calculate the internal resistance of the cell?

A

internal resistance = - gradient

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22
Q

From the graph, how do you find the emf ?

A

The y-intercept i.e. the voltage where current = 0A.

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23
Q

From the graph, how do you find the short circuit current?

A

The x-intercept i.e. the current when the tpd (V) = 0V.

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24
Q

Explain what happens to the reading on the voltmeter when the circuit is changed from figure 1 to figure 2.

A

Total resistance in the circuit decreases.

Total current in the circuit increases.

Lost volts will increase Vlost = Ir

Vtpd = E - Vlost

So the reading on the voltmeter will decrease as it measures the Vtpd.

25
Q

Explain the concept of load matching i.e. the conditions under which maximum power will be transferred from a power source.

A

As you change the load resistance R in a circuit, the current I and tpd V will both change also. Therefore, the power delivered to the load resistance changes. The maximum power is delivered when the external resistance equals the internal resistance. This is known as load matching.

26
Q

What is capacitance?

A

The charge stored per unit voltage.

27
Q

What is meant by 1 Farad?

A

1 Coulomb of charge stored per volt

28
Q

Explain why work must be done to charge a capacitor.

A

Work must be done to charge a capacitor because any charge already stored on the plates will repel any further charge, requiring work to be done to overcome this force.

29
Q

How is the energy stored in a capacitor calculated from a charge-voltage graph?

A

The energy stored in a capacitor is the area under a charge-voltage graph.

30
Q

What could be altered in the circuit below to increase the charging time of the capacitor?

A
  • Increase the resistance of the resistor
  • Increase the capacitance of the capacitor
31
Q

What could be altered in the circuit below to decrease the charging time of the capacitor?

A
  • Decrease the resistance of the resistor
  • Decrease the capacitance of the capacitor
32
Q

Describe the shape of a graph of voltage across capacitor against time for charging a capacitor.

A

Starts from 0V. Increases to the supply voltage.

33
Q

Describe the shape of a graph of voltage across capacitor against time for discharging a capacitor.

A

Starts from the supply voltage Decreases to 0V.

34
Q

Describe the shape of a graph of current in a capacitance circuit against time for charging a capacitor.

A

Starts from a maximum value (Vs/R) and decreases to zero.

35
Q

Describe the shape of a graph of current in a capacitance circuit against time for discharging a capacitor.

A

Starts from a maximum value (-Vs/R) and decreases to zero, current is in the opposite direction from the charging current.

36
Q

Explain the shape of an I-t graph for charging and discharging a capacitor.

A
  1. When charging, a large current flows initally (limted only by the resistance in the circuit, I = Vs/R), but as charge builds up on the plates this repels any further charge trying to flow on to the plates, reducing the current.
    The capacitor is fully charged when the repulsion equals the forward ‘push’ of the supply (i.e. Vc = -Vs) and the current reduces to zero.
  2. When discharging, a large current flows initially in the opposite direction (negative values) because of the repulsion of all the charge on the plates. As some charge leaves, there is less charge left to repel further charge so the current reduces.
    The capacitor is fully discharged when all the charge has left (i.e. Vc = 0) and no current flows.
37
Q

Explain the shape of an Vc-t graph for charging and discharging a capacitor.

A
  1. For a capacitor voltage and charge are directly proportional.
  2. When charging, a large current flows initally giving a large increase in charge and voltage (Vc), but as charge builds up on the plates this repels any further charge trying to flow on to the plates, reducing the increase in charge and voltage.
    The capacitor is fully charged when the repulsion equals the forward ‘push’ of the supply (i.e. Vc = -Vs).
  3. When discharging, the initial voltage Vc = Vs but as charge leaves the voltage reduces meaning the current decreases and charge leaves more slowly.
    The capacitor is fully discharged when all the charge has left (i.e. Vc = 0) and no current flows.
38
Q

Describe the effect of changing the resistance in a capacitor charging circuit on a graph of voltage versus time.

A

A smaller resistance will make the capacitor charge up more quickly, larger resistance more slowly.

39
Q

Describe the effect of changing the capacitance on a graph of voltage versus time when charging a capacitor.

A

A capacitor with smaller capacitance will charge up more quickly, larger capcitance more slowly.

40
Q

Describe the effect of changing the capacitance on a graph of current versus time when charging a capacitor.

A

A smaller capacitance will charge up more quickly, larger capacitance more slowly.

41
Q

Describe the effect of changing the resistance in a capacitor charging circuit on a graph of current versus time.

A

A smaller resistance will make the initial current greater and result in the capacitor charging up more quickly, larger resistance smaller initial current and slower charging.

NB Since resistance doesn’t affect the total charge stored in the capacitor, both graphs will have the same area under graph since this is the total charge.

42
Q

Describe and explain some of the applications of capacitors.

A
  • Flashing Neon Light - Charging and discharging of a capacitor can be used to make a gas discharge lamp repeatedly flash. The frequency of flashing can be controlled by the values of resistance and capacitance in the circuit.
  • Smoothing - A capacitor and large resistance can be used with a diode to “flatten out” any varying voltage as the capacitor slowly discharges through the resistor. This is an important part of AC to DC convertors.
  • Timing circuits - Capacitors are used to provide timing in electronic circuits since the charging time is controlled by the resistance and capacitance. When the voltage across the capacitor reaches a certain value it can be used to trigger some other function in the circuit e.g. switch on a transistor.
  • Interference suppression - When sparks occur, e.g., in the contacts of switches connected to the spark plugs in cars, high frequency electromagnetic waves are radiated. By fitting a small capacitor across the contacts much of the high frequency pulse can be safely by-passed, so reducing interference caused to radio and television reception.
  • Camera flash units - The fact that capacitors store charge is used in electronic flash units. In these a capacitor is charged and then discharged quickly through a discharge tube which provides a short pulse of intense light.
  • Capacitive Touch Screen - Most touch screens are made up of a grid of capacitors underneath a glass screen. When the screen is touched with a finger it reduces the capacitance at that point and the corresponding change in voltage is measured by sensors at the end of each row and column in the grid, giving an accurate coordinate for where the touch took place. This can be applied to multiple fingers at the same time.
43
Q

Explain the concept of energy bands in solids.

A
  • When many atoms are combined into a crystalline solid the energy levels of the individual atoms become ‘blurred’ so that electrons can have a range of energies known as a band.
  • There will also be groups of energies that are not allowed, in what is known as a band gap.
  • As with energy levels, electrons will always try to fill the lowest band first.
  • Partially filled bands can conduct electricity.
44
Q

From this energy level diagram, what type of material is being represented?

A

Insulator

45
Q

Using this energy level diagram, explain why an insulator is a poor conductor.

A

The valence band is full and the conduction band is empty. The large band gap makes it unlikely that electrons will have sufficient energy to jump across to the valence band and contribute to conduction. This can happen in extreme conditions such as lightning.

46
Q

From this energy level diagram, what type of material is being represented?

A

Semiconductor

47
Q

Explain how a pure semiconductor conducts.

A

The conduction band is empty and the valence band is full at absolute zero. There is a small band gap. If the temperature is increased the electrons in the valence band can gain enough energy to jump the gap into the conduction band. These electrons are free to move and so conduction increases.

48
Q

From this energy level diagram, what type of material is being represented?

A

Conductor

49
Q

Using this energy level diagram, explain conduction in conductors?

A

In a conductor the highest occupied band (the conduction band) is partially filled which allows the electrons to move in and out from neighbouring atoms and therefore conduct easily.

50
Q

Explain how an n-type semiconductor can be created.

A

The semiconductor has impurities added that have five electrons in its outer shell. Four of these are used to fill the valence band. The fifth electron is in the conduction band. This is free to move and so conduction increases.

51
Q

Explain how a p-type semiconductor can be created.

A

The semiconductor has impurities added that have three electrons in its outer shell. These three do not completely fill the valence band. So there are electrons free to move in this band so conduction increases.

52
Q

How is the internal electric field created in a pn junction?

A
  • Electrons from the conduction band of the n-type move into the conduction band of the p-type and drop into the valence band of the p-type material.
  • This leaves the area around the junction with no free charge carriers, known as the depletion layer.
  • It also leaves the n-type material slightly positively charged and the p-type material slightly negatively charged around the junction.
  • This creates a potential difference which gives an electric field.
  • The electrons in the conduction band of the n-type, do not have enough energy to overcome the potential difference of the electric field, to pass into the conduction band of the p-type.
53
Q

How is a p-n junction connected in forward bias?

A

Connect n-type to negative of supply connect p-type to positive of supply

54
Q

How is a p-n junction connected in reverse bias?

A

Connect n-type to positive of supply Connect p-type to negative of supply

55
Q

Explain why a forward biased p-n junction conducts.

A
  • The applied voltage (Va) raises the energy levels of the n-type semiconductor
  • This reduces the effect of the internal electric field (Vi).
  • The electrons in the conduction band of the n-type now gain enough energy to overcome the internal electric field and pass into into the conduction band of the p-type.
  • The p-n junction will conduct.
56
Q

Explain why a reverse biased p-n junction will not conduct.

A
  • The applied voltage (Va) raises the energy levels of the p-type semiconductor
  • This increases the effect of the internal electric field (Vi).
  • The electrons in the conduction band of the n-type do not have enough energy to overcome the larger potential difference of the electric field.
  • Electrons cannot pass into the conduction band of the p-type so the p-n junction will not conduct.
57
Q

Explain how an LED produces light.

A
  • The electrons from the n-type move towards the conduction band of the p-type across the junction.
  • Electrons drop from the conduction band to the valence band.
  • The electron gives out energy as a photon of light.
58
Q

Explain how a photovoltaic cell produces an electric current.

A
  • A photon is absorbed by an electron in the valence band of the junction.
  • If this energy is high enough the electron can jump into the conduction band, leaving a hole in the valence band.
  • The separation of the electron and hole generates a voltage.

OLD St Aloysius H Physics (Pre Feb 24) (17 decks)

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