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CHS Higher Physics > Electricity > Flashcards

Electricity Flashcards

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1
Q

Key area: Monitoring and measuring a.c.

What does a.c. stand for?

A

Alternating current

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2
Q

Key area: Monitoring and measuring a.c.

In terms of current, explain what a.c. means

A

a.c. refers to a current which changes in direction and value over time

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3
Q

Key area: Monitoring and measuring a.c.

What must be done to convert a peak voltage (Vp) into an r.m.s. voltage?

A

The peak voltage must be divided by √2.

√2 is approximately 1.41 which means this approach will give a value lower than Vpeak.

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4
Q

Key area: Monitoring and measuring a.c.

What must be done to convert an r.m.s. (Vr.m.s.) voltage into a peak voltage (Vp)?

A

The r.m.s. value must be multiplied by √2.

The peak voltage will always be bigger than the r.m.s., multiplying by root 2 does this.

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5
Q

Key area: Monitoring and measuring a.c.

State the relationship used to calculate Vr.m.s.

A

Vr.m.s. = Vpeak/√2

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6
Q

Key area: Monitoring and measuring a.c.

State the relationship used to calculate Vpeak

A

Vpeak = Vr.m.s. x √2

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7
Q

Key area: Monitoring and measuring a.c.

State the relationship used to calculate Ir.m.s.

A

Ir.m.s. = Ipeak/√2

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8
Q

Key area: Monitoring and measuring a.c.

State the relationship used to calculate Ipeak

A

Ipeak = Ir.m.s. x √2

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9
Q

Key area: Monitoring and measuring a.c.

Using an oscilloscope, how can Vpeak be found?

A

Check the setting of the Y-gain or V/div dial, each square on the screen represents the value on this dial.

Multiply the V/div setting by the number of squares in the amplitude of the wave. (remember a wave 8 squares high has an amplitude of 4 squares high)

If each square = 5V, Vpeak = 4 x 5 = 20V

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10
Q

Key area: Monitoring and measuring a.c.

Using an oscilloscope, how can the frequency of a wave be found?

A

Check the timebase setting on the oscilloscope.

Count the number of horizontal divisions for one complete wave and multiply this by the timebase setting to give the period of the wave (T).

Take this answer and use it in the equation f = 1/T

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11
Q

Key area: Current, potential difference, power and resistance

What is the relationship between the voltages in a series circuit?

A

The voltage across each component adds up to the supply voltage.

e.g: VS = V1 + V2 + V3 etc.

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12
Q

Key area: Current, potential difference, power and resistance

What can be said about the current in a series circuit?

A

The current is the same in all places.

e.g: IS = I1 = I2 = I3 etc.

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13
Q

Key area: Current, potential difference, power and resistance

How is the total resistance of the components in a series circuit calculated?

A

The resistances add to find RT.

e.g: RT = R1 + R2 + R3 etc.

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14
Q

Key area: Current, potential difference, power and resistance

What is the relationship between the voltages in a parallel circuit?

A

The voltages across each branch of a parallel circuit are equal to the supply voltage.

e.g: VS = V1 = V2 = V3 etc.

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15
Q

Key area: Current, potential difference, power and resistance

What is the relationship between the currents in a parallel circuit in relation to the supply current?

A

The currents in each branch of a parallel circuit add up to the supply current.

e.g: IS = I1 + I2 + I3 etc.

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16
Q

Key area: Current, potential difference, power and resistance

State the relationship used to calculate the total resistance of a parallel circuit.

A

1/RT = 1/R1 + 1/R2 + 1/R3 etc.

(The reciprocal button (x-1) on your calculator is good for this but don’t forget to solve for RT instead of leaving your answer as 1/RT)

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17
Q

Key area: Current, potential difference, power and resistance

State the three power equations used in this unit

A

P = IV

P = I2R

P = V2/R

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18
Q

Key area: Current, potential difference, power and resistance

What is the purpose of a potential divider circuit?

A

A potential divider circuit will split a voltage according to the ratio of the resistances in the circuit.

e.g: R1 = 100Ω, R2 = 200Ω. A 12V supply will be split 4V and 8V.

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19
Q

Key area: Current, potential difference, power and resistance

How does a wheatstone bridge work?

A

A wheatstone bridge is made of two potential dividers. The p.d. across e.g. R2 in both potential dividers will give a p.d. between the centres of potential divider 1 and potential divider 2. This p.d. can be used in a variety of applications.

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20
Q

Key area: Current, potential difference, power and resistance

If the resistances in a wheatstone bridge are of certain values that R1/R2 = R3/R4, what can now be said about the wheatstone bridge?

A

The wheatstone bridge is referred to as ‘balanced’, there is no potential difference between one side and the other.

21
Q

Key area: Electrical sources and internal resistance

Define the term ‘internal resistance’.

A

Internal resistance is the resistance present in a power supply which becomes evident as a current flows in the supply.

22
Q

Key area: Electrical sources and internal resistance

What happens to the internal resistance of a cell or battery as it goes ‘flat’?

A

The internal resistance increases.

23
Q

Key area: Electrical sources and internal resistance

If a voltmeter is placed across a lamp in a series circuit connected to a 6V battery and a reading of 5.6V is obtained, what is this reading known as?

A

The terminal potential difference or t.p.d.

24
Q

Key area: Electrical sources and internal resistance

What is the name given to the voltage drop caused by the internal resistance of a power supply.

A

Lost volts

25
Q

Key area: Electrical sources and internal resistance

What value can be obtained by calculating the gradient of a current-voltage graph?

A

Internal resistance

26
Q

Key area: Electrical sources and internal resistance

If a current-voltage graph is extended back to the y-axis, what is the value the line crosses the y-axis known as?

A

The e.m.f.

27
Q

Key area: Electrical sources and internal resistance

Define e.m.f.

A

The e.m.f. (electromotive force) is the maximum amount of energy supplied to each coulomb of charge by a power supply, measured when there is no current flowing in the circuit.

28
Q

Key area: Electrical sources and internal resistance

If a second component is added in parallel to a series circuit, what happens to the t.p.d. and why?

A

The t.p.d. reduces.

Adding a component in parallel will cause RT to decrease.

Current ‘I’ will then increase causing the lost volts ‘Ir’ to increase, t.p.d. therefore reduces as both the lost volts and the t.p.d. still need to add to equal the e.m.f.

29
Q

Key area: Electrical sources and internal resistance

How is it possible to calculate the short circuit current of a battery?

A

Place a thick piece of copper across the battery terminals, this will allow maximum current to flow. The resistance encountered will be the internal resistance of the battery.

Use I = E/r where I is the short circuit current, E is the e.m.f. and r is the internal resistance.

30
Q

Key area: Capacitors

Describe the voltage-time graph of a charging capacitor.

A

Initially the capacitor will charge quickly, the rate of charge will then reduce until the p.d. across the capacitor equals that of the supply.

31
Q

Key area: Capacitors

What happens to the current in a capacitor charging circuit as the capacitor fully charges?

A

The current starts off high but then reduces to zero when the capacitor is fully charged.

32
Q

Key area: Capacitors

As a charged capacitor discharges, what can be said about the direction of the discharge current?

A

The discharge current flows in the opposite direction to the charging current.

33
Q

Key area: Capacitors

Describe the value of the current as a capacitor discharges.

A

The discharge current starts high and then reduces to zero as the capacitor loses its charge.

34
Q

Key area: Capacitors

What two steps can be taken to increase the charging time of a capacitor?

A

Increase the resistance in the circuit or increase the capacitance of the capacitor.

35
Q

Key area: Capacitors

What does the area under a graph of charge against p.d. represent?

A

Total energy stored e.g:

E = ½QV

36
Q

Key area: Capacitors

Using the relationships Q = CV and E = ½QV, how can the equation E = ½CV2 be derived?

A

E = ½QV

substituting Q = CV into the equation above,

E = ½(CV)V

Therefore E = ½CV2

37
Q

Key area: Capacitors

Using the relationships Q = CV and E = ½QV, how can the equation E = ½Q2/C be derived?

A

E = ½QV

substituting V = Q/C into the equation above,

E = ½Q(Q/C)

Therefore E = ½Q2/C

38
Q

Key area: Conductors, semiconductors and insulators

How does the conduction band allow a metal to conduct?

A

The highest occupied band is not completely full and this allows the electrons to move and therefore the metal conducts.

39
Q

Key area: Conductors, semiconductors and insulators

Why does an insulator not conduct?

A

The highest occupied band (the valence band) is full. The gap between the valance band and the conduction band is too large at room temperature, therefore electrons will not move from the valence band to the conduction band.

40
Q

Key area: Conductors, semiconductors and insulators

What allows a semiconductor to conduct?

A

The gap between the valence band and the conduction band is small enough to allow transfer of some electrons from the valence band to the conduction band.

41
Q

Key area: p-n junctions

What property gives an n-type semiconductor its name?

A

An n-type semiconductor contains impurities in the form of a group 5 element such as Arsenic. The fifth electron in the outer shell of these atoms allows conduction to take place in an otherwise fully insulating surroundings of silicon.

This fifth electron gives rise to the ‘n-type’ term.

Note these atoms are not negatively charged, they remain neutral.

42
Q

Key area: p-n junctions

What property gives a p-type semiconductor its name?

A

A p-type semiconductor contains impurities in the form of a group 3 element such as Indium. The three outer electrons of these atoms allow a space for electrons to occupy in an otherwise fully insulating surrounding silicon material.

This ‘hole’ gives rise to the ‘p-type’ term.

Note these atoms are not positively charged, they remain neutral.

43
Q

Key area: p-n junctions

Is this semiconductor forward or reverse biased?

A

Forward biased

44
Q

Key area: p-n junctions

Is this semiconductor forward or reverse biased?

A

Reverse biased

45
Q

Key area: p-n junctions

Explain the purpose of the cell in this diagram:

A

The cell provides enough energy for electrons to overcome the depletion layer.

46
Q

Key area: p-n junctions

State how the cell affects the depletion layer when it is connected in this way.

A

The depletion layer becomes wider.

47
Q

Key area: p-n junctions

Explain how an LED works

A

LEDs are forward biased p-n junction diodes that emit photons when electrons ‘fall’ from the conduction band into the valence band of the p-type semiconductor.

48
Q

Key area: p-n junctions

Explain the purpose of the doping process when creating a semiconductor

A

In the doping process, a small amount of an impurity is mixed into a silicon material to change its conductive properties.

CHS Higher Physics (3 decks)

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