Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

ALevel Physics > Electricity > Flashcards

Electricity Flashcards

You may prefer our related Brainscape-certified flashcards:
Physics 101 flashcards AP® Physics flashcards
1
Q

What is electric current?

A

The net flow of charged particles.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Metal wire?

A

A wire consists of millions of atoms. Most of the electrons are held tightly to their atoms, but each atom has one or two electrons which are loosely held.
The metal wire is made up of a lattice of positive ions, surrounded by ‘free electrons’.
The ions can only vibrate about their fixed positions, but the electrons are free to move randomly from one ion to another.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Current in a metal wire?

A

When a battery is attached to the wire, the free electrons are repelled by the negative terminal and attracted to the positive one. They still have random movement, but they now also move in the same direction through the wire with a steady drift velocity.
As the electrons carry a charge, we now have a flow of charge - an electric current.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Current in liquids?

A
  • Salt solution is an electrolyte. It can conduct electricity.
  • Electrolytes contain both positive and negative ions.
  • When a power supply is connected, the positive ions move towards the negative terminal and the negative ions move towards the positive terminal.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Equation for current?

A

-Current is the rate of flow of charge.

I (A) = ΔQ (C) / Δt (s)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

The coulomb?

A

Since ΔQ = IΔt, the coulomb is:

The quantity of electric charge that passes a given point in a circuit in 1s when a current of 1A is present.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What is current direction?

A

The flow of positive charge!

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What is potential difference?

A

The p.d between two points in a circuit is the amount of electrical energy changed to other forms of energy per coulomb of charge flowing between them.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Equation for potential difference?

A

V (V) = W (J) / Q (C)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Equation for work done?

A

W=QV
but Q=It
therfore W=ItV

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

The volt?

A

1 volt is the p.d. between two points in a circuit in which 1J of energy is converted to other forms when 1C of charge passes between them.
1V = 1J/1C

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

What is electrical resistance?

A

A measure of a conductor’s opposition to flow of charge.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Equation for resistance?

A

R (Ω) = V (V) / I (A)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

The ohm?

A

A conductor has a resistance of 1Ω if a current of 1A flows through it when a p.d. of 1V is applied across it.
1Ω = 1VA^-1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Ohm’s law?

A

The current through a metal wire is in directly proportional to the p.d. across it (providing the temperature remains constant).
Materials that obey Ohm’s law are called ohmic conductors.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

I-V characteristics (Ohmic Conductor)?

A

A straight line through the origin is obtained, showing that Ohm’s law is obeyed.
Since the gradient is constant, the resistance is constant for both directions of current flow.
m = I/V
V = I/R thus I/V = 1/R
(only for ohmic conductor)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

I-V characteristics (Filament Lamp)?

A

The curve shows that resistance increases as current increases. This is due to the rise in temperature of the filament, caused by the heating effect of the current.
The resistance of a pure metallic conductor increases with temperature.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

I-V characteristics (Semiconductor Diode)?

A

The curve shows that the semiconductor diode allows virtually no current when a negative p.d. is applied (reverse bias), and that a forward-biased p.d. of about 0.6V is needed before the diode will conduct in the forward direction.
When the p.d. is greater than this, resistance is very low, allowing a large increase in current for a small increase in p.d.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Resistivity?

A 
The resistance of uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.
R = pl/A
where:
R (resistance) in Ω
p (resistivity) in Ωm
l (length) in m
A (cross-sectional area) in m^2
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Effect of temperature on metals?

A

Electrical resistivity of metals increases as their temperature rises. At high temperatures, the resistance of a metal increases linearly with temperature.
Metals contain large numbers of free electrons. As these electrons move through the metal lattice, they collide with vibrating metal ions. These collisions oppose the flow of electrons and so the metal has resistance.
As the temperature rises, the ions vibrate faster, and with greater amplitude, and it is more difficult for the electrons to pass through the lattice. The resistance of the metal has increased.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Superconductor?

A

Below the critical temperature, a metal can lose all of its resistance - it becomes a superconductor.
This happens to mercury at 4.15K (-269°C).
Superconducting wires do not become hot, because electrons can flow through them without any transfer of energy.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Insulators?

A

In insulators at room temperature, there are a few free electrons available for conduction.
At high temperatures, some electrons have enough energy to escape from their atoms and the insulator is able to conduct.
So in insulators, resistance decreases as the temperature rises.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Semiconductors?

A

Silicon is one of the best known semiconductor materials.
At low temperatures, it is a poor conductor. As its temperature rises, more and more electrons break free from their atoms and so it becomes a better conductor.
However, at about 150°C, breakdown occurs and the silicon is permanently damaged.
Like silicon, the resistance of many semiconducting materials decreases as their temperature rises. These materials have a negative temperature coefficient of resistance (NTC).
Semiconductor materials are used to make thermistors.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

What is electrical power?

A

The rate at which electrical energy is converted into other forms of energy.
The unit of power is the Watt, W.
1W = 1J per second
This means a 60W light-bulb uses 60J of electrical energy per second.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Equation for electrical power?

A

P = W/t but W =ItV
so P = ItV/t = IV
P (W) = I (A) x V (V)

26
Q

1kWh?

A

3,600,000J

27
Q

Other power equations?

A 
When a current is flowing through a resistor, electrical energy is transferred to heat.
We can combine equations:
P=IV and V=IR
to give:
P=I^2R and P=V^2/R
28
Q

Conventional current and electron flow?

A

The direction of conventional current is taken to be the direction of flow of positive charge.
In liquids, gases and some semiconductors, the positive ions move in this direction while any negative charge carriers e.g. electrons, flow in the opposite way.
In metals, the only charge carrier is the electron, so only negative charge flows.
(Electrons flow in opposite direction to conventional current).

29
Q

Currents in circuits?

A

In series, the current is the same at all points.

In parallel, the current leaving and returning to the supply is the sum of the currents in the separate branches.

30
Q

Kirchoff’s first law?

A

The sum of the currents flowing into any junction in a circuit is equal to the sum of currents flowing out of that junction, i.e. charge is conserved.

31
Q

Kirchoff’s second law?

A

Around any closed loop in a circuit, the sum of the e.m.fs is equal to the sum of the p.ds.

32
Q

KSL explanation?

A
  • This is a statement of the conservation of energy in a circuit.
  • A coulomb gains electrical energy as it moves through each e.m.f, and loses electrical energy as it moves through each p.d.
  • After one loop of the circuit, the energy it has gained must be equal to the energy it has dissipated.
33
Q

EMF?

A

Electromotive Force

34
Q

Energy transfer in series?

A

Two identical lamps in series have equal brightness, but each lamp is less bright than if connected to the battery on its own.
The total p.d across both lamps is 6V. This is shared between the two lamps, so each lamp has a p.d. of 3V across it.
In series, the total p.d. across all components is the sum of the p.ds across the separate components.

35
Q

Series (Model of Energy Transfer)?

A

Each coulomb collects 6J of energy from the battery. A single lamp connected to the battery would receive all 6J of energy.
In this circuit, each coulomb transfers 3J at the first lamp and then 3J at the second lamp.
The energy is shared equally because the lamps are identical.

36
Q

Energy transfer in parallel?

A

Each lamp is as bright as if it were connected to the battery on its own.
The p.d. across each lamp is 6V.
In parallel, the p.d. across each branch is the same.

37
Q

Parallel (Model of Energy Transfer)?

A

Each coulomb of charge transports 6J of energy and transfers it all to one of the lamps. Each lamp recieves the same energy as if it were connected to the battery on its own.
Twice as many coulombs pass per second through the battery as when a single lamp is connected, and so the battery runs down more quickly.

38
Q

Equation for EMF?

A

ΣE = IR
Each emf needs to be in the same direction, otherwise you subtract them.
If there’s more than one branch, use simultaneous equations.

39
Q

What is EMF?

A

The emf of a source (battery, generator, thermocouple, etc.) is defined as the energy (chemical, mechanical, thermal, etc.) converted into electrical energy when unit charge (i.e. 1C) passes through it.
p.d. = conversion of electrical energy to other forms
Σemf = Σpd

40
Q

Calculating EMF?

A

Series:
For cells in series, the total emf of their combination is the sum of their individual emfs.
Each charge passes through each cell so gains energy from all three.
Parallel:
For identical cells in parallel, the total emf is the same size as each of the cells individually.
Each charge only passes through one cells so gains energy from the single cell.

41
Q

Resistors in series?

A

-The current through each resistor is the same.
-The total p.d. across the resistors is the sum of the p.d.s across the separate resistors i.e. V = V1+V2+V3.
-The combined resistance is calculated by:
R = R1+R2+R3

42
Q

Resistors in parallel?

A

-The p.d. across each resistor in parallel is the same.
-The current in the main circuit is the sum of the currents in each parallel branch i.e. I = I1+I2+I3
-The combined resistance is calculated by:
1/RT = 1/R1+1/R2+1/R3

43
Q

4 resistors in square?

A

A AND B:
3 resistors in series, in parallel with a single resistor.
A AND C:
Opposites in parallel and in series with other opposite pair.

44
Q

What resistance provides a perfect voltmeter?

A

∞ as then no current is able to flow so the voltmeter wouldn’t alter the p.d. it is measuring.

45
Q

What is a potential divider?

A

A linear circuit that uses resistors/thermistors/LDRs to supply a variable potential difference.
If there are two identical resistors, they share the applied voltage equally.
If one resistor is 2/3 of the total resistance, then the voltmeter across it will read 2/3 of the applied voltage.

46
Q

Equation for potential dividers?

A

Vout=Vin(R1/R1+R2)

R1 is the one you are measuring across.

47
Q

Thermistors and LDRs?

A

They can be used as sensors in electronic circuits.

48
Q

Thermistor?

A

In a potential divider, R1 or R2 can be replaced with a thermistor.
The circuit now has an output p.d. that changes with temperature.
If R1 is replaced, as the thermistor cools, its resistance rises. It takes a larger share of the input voltage, so Vout rises. The circuit could be used to switch on a heater.
If R2 is replaced, as the thermistor heats up, its resistance falls. It takes a smaller share of the input voltage. R1 takes a larger share of input voltage so Vout rises. The circuit could be used to switch on an air conditioning unit.

49
Q

LDR?

A

In a potential divider, R1 or R2 can be replaced with an LDR.
The circuit now has an output p.d. that changes with light intensity.
If R1 is replaced, as the LDR receives less light, its resistance rises. It takes a larger share of the input voltage, so Vout rises. The circuit could be used to switch on street lights.
If R2 is replaced, as the LDR receives more light, its resistance falls. It takes a smaller share of the input voltage. R1 takes a larger share of input voltage so Vout rises. The circuit could be used to open automatic blinds.

50
Q

Role of charge carriers?

A

When there’s more light or heat energy, more charge carriers are released thus there’s a lower resistance.

51
Q

E.m.f and terminal p.d.?

A

The e.m.f of a cell (or other device) which supplies electrical energy is equal to the terminal p.d. (the potential difference across the terminals of a cell) when there is no current in the cell (i.e. the p.d. across the terminals in the open circuit).
As soon as a current is drawn up from the cell, the terminal p.d. will drop.
The energy delivered by each coulomb is less than the energy supplied to each coulomb.
This is because the cell has internal resistance.
This means that each coulomb of charge gains energy as it travels through the cell, but some of this energy is wasted or ‘lost’ as the coulomb moves against the resistance of the cell itself.

52
Q

Equations for e.m.f?

A

If current=0, emf=terminal p.d.
If current≠0, emf=terminal p.d. + lost volts (p.d. across internal resistance).
emf=I(R+r)

53
Q

How to calculate R and r?

A

R=V/I where V=new reading on voltmeter

r=V/I where V=new reading - old reading

54
Q

Short circuit?

A

The 1.5V dry cell is short-circuited by the low resistance copper wire connected to its terminals. The only significant resistance in this circuit is 0.5Ω, the internal resistance of the cell itself.
Using e=I(R+r), where R=0, the maximum possible current that this dry cell can supply is 3A.
Some sources, such as mains supply, have very low internal resistance, and so they can supply high short-circuit currents. These currents are dangerous because of the heat they produce.

55
Q

Car battery?

A

A 12V car battery must have a very low internal resistance because a starter motor needs a current of over 100A.

56
Q

What happens if a driver starts a car with the head-lamps on?

A

The current through the battery is so large that the ‘lost volts’ are high - even though the battery’s internal resistance is low.
The terminal p.d. drops to around 8V and the head lamps go dim.

57
Q

EMF question?

A

Total resistance in external circuit: R
The current supplied by the battery: e=I(R+r)
The terminal p.d. of the battery: V=IR
The current through each parallel branch:
I=V/R1 and I=V/R2 where V is the same for both.

58
Q

What happens to terminal p.d. of supply if another lamp is connected in series?

A

The current is lower and so the drop in p.d. due to internal resistance would be less.

59
Q

What happens to energy transfer to a lamp if it has higher resistance?

A
  • The current will reduce.
  • The p.d. across the internal resistance drops.
  • A smaller proportion of energy is wasted by internal resistance.
60
Q

Power dissipation?

A

P=V^2/R

If both circuits have the same voltage, the circuit with the higher resistance dissipates less power.

ALevel Physics (8 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions