ELECTRICITY AS Flashcards

1
Q

Define a potential divider circuit.

A

a set of resistors connected in series with a source of p.d across them.
- divides an input voltage into a set of smaller output voltages

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2
Q

Define the e.m.f of a cell.

A

The work done by the cell converting chemical potential energy into electrical energy per 1 C of charge flowing through it

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3
Q

Define electric current.

A

The rate of flow of charge.

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4
Q

Define 1 Volt.

A

1 volt is the difference in electrical potential between 2 points when 1 Joule of work is done to move a charge of 1 coulomb between those points.

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5
Q

Definition of 1 ohm.

A

The resistance of a component when 1 A of current flows through it when 1 V is applied across it.

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6
Q

Define the resistivity of a material.

A

The resistance of a uniform wire made out of the material with a cross-sectional area of 1 m^2, and a length of 1m.

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7
Q

Explain, in reference to electrons and metallic structure, why resistance increases for a metal when the temperature is increased.

A

Increasing the temperature means:
-positive ions have more energy
-so they vibrate with greater amplitude
-and so there are more frequent collisions between positive ions and electons
-causing greater resistance to the flow of electrons

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8
Q

Explain, in reference to electrons, why the resistance of a thermistor decreases as the temperature increases.

A

As the temperature increases,
- energy of the semiconductor increases
- so more electrons are released into the lattice
- more charge carriers available to carry a greater current for the same voltage
- resistance decreases, as R = V/I.

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9
Q

Draw and explain the I-V graph for a diode.

A

NEGATIVE P.D
- reverse bias, resistance is very high
- no/little current flows
POSITIVE P.D
- below threshold voltage, no/little current flows
- above threshold voltage, current increases as p.d increases

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10
Q

State 2 applications of superconductors and explain why they are superior to traditional conductors in these applications.

A
  • superconducting transmission wires - superconductors have 0 resistance below or at the critical temperature, so no energy losses to Joule heating.
  • maglev trains - superconductors can carry a huge amount of current without overheating, so they can act as really strong electromagnets.
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11
Q

What are the current limitations of using superconductors?

A
  • they must be cooled below the critical temperature, which is expensive as liquid helium or liquid nitrogen is required.
  • superconduncting elements are made out of rare elements which are hard to obtain
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12
Q

Define what is meant by a superconducting material.

A
  • A material that has zero resistivity at or below a critical temperature.
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13
Q

Define power.

A

The rate of energy transfer per unit time

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14
Q

A series circuit consists of a 4V cell and 2 resistors of value 2 and 6 ohms, connected in series.

Explain why the total power dissipated across both resistors must equal the power of the cell.

A
  • Power is the rate of energy transfer per unit time, and energy must be conserved
  • so each second, the total energy dissipated by the resistors must = total energy transferred by the cell
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15
Q

State Ohm’s Law.

A
  • For a conductor at constant temperature, the potential difference across it is d.p to the current flowing through it.
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16
Q

Describe how you would make a direct measurement of the emf of a cell, stating the type of meter you would use.
Explain why in practice, this method only gives an approximate value for the e.m.f.

A

Use a voltmeter, connected across the ends of the cell, without any additional connections/load.

In practice though, a small current will flow through the voltmeter, so the measured p.d is smaller than the actual e.m.f.

17
Q

Explain why an ideal voltmeter has infinite resistance.

A
  • so that virtually no current flows through the voltmeter
  • so the current flowing between the 2 points measured is not affected, and so the p.d reading is not affected
18
Q

A wire of uniform cross-section has a resistance of R. If it is drawn to three times the length, but the volume remains constant, what is it’s resistance?

A

9R, since R is d.p to L/A is d.p to L^2

19
Q

Communications satellites use solar cells to generate electrical power. Discuss why solar cells are appropriate for this task.

Your answer should refer to:
- any additional features that would be needed to ensure that the satellite’s electrical systems operate continuously
-whether solar cell arrays are appropriate for space probes that travel to the edge of the solar system.

A

SATELLITE SPECIFIC POINTS
SOLAR VS FUEL VS NUCLEAR
- solars cells provide a virtually non-ending source of energy from the Sun, since the satellite is always in it’s proximity
- extra rocket fuel is not required for power generation, which saves precious fuel that can be used for orbital manouvres, and also, any type of fuel (chemical or nuclear) would eventually run out over time
AREA/SPACE REQUIREMENTS
- the area of solar cells present on a satellite can be varied significantly to satellite requirements: solar cells can be folded, then unfolded in space, so as to not take up much space.
MASS
- solar cells are relatively light
RELIABILITY
- solar cells have no moving parts so malfunction is rare
SHIELDING
- however, solar cells require shielding from micrometeorites and cosmic rays
CONCLUSION
- for communication satellites in LEO or GEO, solar cells are a dependable and reliable source of continuous energy compared to other sources.

CONTINUOUS OPERATION SPECIFIC POINTS
BATTERIES
- batteries (which are heavy) required to store energy for when the satellite is in shadow behind the Earth
TRACKING/PIVOTING
- solar cells should be made to track the sun, but also, can be purposefully aligned to reduce or increase orbital drag (e.g night-slicer mode on the ISS)
LIMITING DEMAND
- must ensure to limit energy-intensive activities to ensure supply can always meet demand

USE ON A SPACE PROBE POINTS
- inverse-square laws means the intensity of solar radiation drops off with the square of the distance from the sun
- thus the light intensity will be too low for the probe, requiring solar cells which are infeasibly large
DEGREDATION
- solar cells degrade with extreme temperature variations and prolonged duration exposure to cosmic rays, so the solar cells will have degraded significantly.
CONCLUSION
- too heavy, unreliable in the context.

20
Q

The resistance of a length of copper wire is 6 ohms. A second piece of copper wire has twice the length and half the cross-sectional area. The resistance of the second piece of wire is?

21
Q

The heating element for an electric fire is made from a wire of resistance R. It is replaced with a wire of the same material which has the same length but is twice the diameter. The resistance of this second wire is:

22
Q

Explain why the efficiency of transmission lines would increase if conventional wires were replaced with superconducting wires, assuming they were cooled to the critical temperature.

A
  • at or below Tc, the resistance of the superconducting wires would be 0
  • so no energy lost due to joule heating
23
Q

Define 1 Amp.

A

When one coulomb of charge is passing a point per second

24
Q

The I-V characteristic of a filament lamp is shaped like an S.
Explain the I-V characteristic of a filament lamp, including reference to movement of electrons.

A
  • as the current increases, the temperature of the filament increases
  • metal ions have more energy and vibrate with greater amplitude
  • more frequent collisions between positive metal ions and delocalised electrons
  • greater resistance to flow of electrons
  • since I = V/R, the current increases at a decreasing rate, as R increases with V.
25
Q

A circuit consists of a cell, a light dependent resistor and a fixed resistor, all in series. A voltmeter is connected across the LDR.

If the lights are dimmed, explain what happens to the reading on the voltmeter.

A
  • as light intensity decreases, resistance of LDR increases
  • so proportion of total resistance across LDR increases
  • so greater voltage across LDR
26
Q

One use of a potential divider is in an automatic night light.
One such circuit consists of a cell, an LDR, and a fixed resistor connected in series, along with a bulb connected in parallel with the LDR.

Explain how the bulb automatically turns on when it gets dark.

A
  • as light intensity decreases, the resistance of the LDR increases
  • so there is a larger voltage across the LDR
  • since the bulb is in parallel with the LDR, larger voltage across bulb
  • since P = V^2/R, the power dissipated in the bulb increases
27
Q

Define internal resistance.

A

The resistance of a cell caused by electrons colliding with metal ions and losing energy.

28
Q

Define terminal p.d.

A

The voltage across the terminals of a cell, when there is no additional load/connections

29
Q

A 6V battery is connected to a 10 ohm resistor, along with a switch, all in series.

State what is meant by an e.m.f of 6V.

When the switch is open, the voltmeter reads 6V and when it is closed it reads 5.8V.
Explain why the readings are different.

A
  • when the switch is closed, current flows through the cell, and so some volts are lost due to internal resistance
  • when switch is open, no current flows, and so no volts are lost
30
Q

A battery with an e.m.f of 9.0V has an internal resistance of 12 ohms.

a) Calculate the potential difference across it’s terminals if it is supplying a current of 50mA.

b) Calculate the maximum current this battery could supply.

A

a) lost volts = 0.05 * 12 = 0.6V
so terminal p.d = 8.4V

b) maximum current when there is no load/ no additional resistance in circuit
so I = V/R = 9 / 12 = 0.75A

32
Q

How can resistivity be measured experimentally?
How can internal resistance be measured experimentally?

A

For resistivity:
- measure the resistance of different lengths of wire, using R = V/I
- plot a graph of resistance against length of wire
- R = gradient * A
For internal resistance:
- Using E = V + Ir, and a circuit with a cell, ammeter, variable resistor with a voltmeter across it,
- vary resistance and measure values of V and I for different values of R
- plot a graph of V against I
- gradient = -r
- y-intercept = E