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Electricity Flashcards

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1
Q

What is current

A

The rate of flow of charge
Carried through the wires by electrons
Q = It
Charge (coulombs)= current (amps) x time
Defined as the number of coulombs transferred per second

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2
Q

Coulomb definition

A

One coulomb is defined as the amount of charge that passes in one second when current is 1 ampere

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3
Q

Ammeter

A

You can measure the current flowing through a circuit using an ammeter
Always need to attach an ammeter in series, so that the current through the ammeter is the same as the current through the component

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4
Q

What is potential difference

A

To make electric charge flow through a circuit, you need to transfer energy to the charge.
This energy is supplied by the power source ( battery or cell )
When a charge flows through the source it’s raised through a potential and energy is transferred to the charge as electrical potential energy
Defined as the work done in moving a unit charge between points
Defined as energy transferred per coulomb

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5
Q

Equation for potential difference with work done and charge

A

V = W/Q
V = potential difference in volts(v)
W = work done in joules(j)
Q = charge in coulombs (C)

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6
Q

Voltmeter

A

Can measure the potential difference across a circuit using a voltmeter
You need a voltmeter to be attached in parallel

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7
Q

Resistance

A

If you put a potential difference across an electrical component, a current will flow and the amount of current you get for a particular potential difference depends on its resistance
How difficult it is to get current to flow through it
Measured in ohms

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8
Q

Resistance equation

A

R = V/I
R = resistance in ohms
V = potential difference in volts(V)
I = current in amperes(A)

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9
Q

Ohms law

A

Ohms law states that provided the physical conditions, such as temperature, remain constant, the current through an ohmic conductor directly proportional to the potential difference across it

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10
Q

I-V characteristics

A

Fancy way of saying a graph which shows how the current flowing through a component changes as the potential difference across it is increased
The shallower the gradient of a characteristic I-V graph, the greater the resistance of the component

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11
Q

Ideal voltmeters and ammeters

A

Voltmeters are assumed to have an infinite amount of resistance ( so no current flows through them )
Ammeters are assumed to have no resistance and so will have no potential difference across them

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12
Q

Filament lamps

A
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13
Q

Diodes

A

Diodes are made from semiconductors and are designed to let current flow in one direction only.
Most diodes require a voltage of about 0.6V in the forward direction before they will conduct - this is called the threshold voltage, in reverse bias the resistance of the diode is very high and the current that flow is very tiny

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14
Q

Resistivity

A

The resistivity of a material tells you how difficult it is for current to flow through it
Measured in ohmic-meters
p = RA/ L
p= resistivity
R = resistance in ohms
A = cross-sectional area in meters
L = length in meters
The lower the resistivity of a material the better at conducting electricity

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15
Q

Resistance depends on

A

-Length: the longer the wire, the more dufficult it is to make a current flow through it. The resistance is proportional to the length of the wire
- area: the wider the wire, the easier it’ll be for electrons to pass along
- resistivity: this is a measure of how much a particular materials resists the flow. Depends on structure of the material as well as environmental factors such as temperature and light intensity

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16
Q

Semiconductors

A

A group of materials that aren’t as good at conducting electricity as metals as they have far fewer charge carriers available
But if energy is supplied to semiconductors, e.g. by an increase in temperature, more carriers are are released and the resistivity of the material decreases.
Means they make excellent sensors for detecting changes in the environment
Three common semiconductors
-thermistors
- diodes
- light dependent resistors (LDRs)

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17
Q

Thermistors

A

Is a component with a resistance that depends on its temperature
NTC thermistors - ‘negative temperature coefficient’ thermistors
This means that the resistance decreases as the temperature goes up
Warming the thermistors gives more electrons enough energy to escape from their atoms
Means that there are more charge carriers available so resistance is lower
This sensitivity to temperature makes them really good temperature sensors

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18
Q

Superconductors

A

normally all materials have resitivity
That means that whenever electricity flows through them, they heat up and some electrical energy is wasted as heat
You can lower the resistivity of many materials like metals by cooling them down

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19
Q

Transitional temperature

A

You can lower the resistivity of many materials like metals by cooling them down
If you cool down some materials to below a critical temperature called the transitional temperature, their resistivity disappears and they entirely become a superconductor without resistance none of the electrical energy is turned into heat, so none of it’s wasted

20
Q

Uses of superconductors

A

Superconducting wires can be used to make:
- power cables that transmit electricity without any loss of power
- really strong electromagnets that have lots of applications, e.g. in medicine and maglev trains
- electronic circuits that work really fast with minimal energy loss, because there is no resistance to slow the current down

21
Q

Finding the resistivity of a metal

A

You’ll have to find the resitance of a test wire made from a material you want to find the resistivity of
You can then use the resistance and the dimensions of the wire to calculate resistivity of the material using the equation

22
Q

Calculating the cross-sectional area

A

Measure the diameter of the wire in at least three different directions places along the wire using a micrometer
Find the mean diameter and halve it to get the radius and use pie r squared

23
Q

Power

A

Defined as the rate of transfer of energy
Measured in watts
1 watt is equivelent to 1 joule per second

24
Q

Equations for power(3)

A

P = E/t
P = power in watts
E = energy in joules(j)
t = time in seconds

P = IV
I = current in amperes
V = potential difference in volts

P = IIR

25
Q

What id e.m.f

A

The amount of electrical energy the battery produces and transfers to each coulomb is called the electromotive force (e.m.f)
measured in volts
e = E / Q
E = lectrical energy in joules
Q = charge in coulombs (C)

26
Q

internal resistance

A

resistance comes from electrons colliding with atoms and losing energy
in a battery, chemical energy is used to make lectrons move. As they move, they collide with atoms inside the battery - so batterys must have resitance
This is called internal resistance. IR is what makes batteries and cells warm up when used

27
Q

load resistance

A

the total resistance of all the components in the external circuit, might see it as referred as external resistance

28
Q

terminal potencial difference

A

pd across the load resistance is the energy transferred when one coulomb of charge flows through the load resistance . this pd is known as terminal pd

29
Q

lost volts

A

terminal pd would mbe the same as e.m.f however, in real power supplies, there is always some energy lost overcoming internal resistance. The energy wasted per coulomb overcoming the internal resistance is called the lost volts

30
Q

Calculations using e.m.f and internal resistance
Using current, internal and load resitance

A

e = I(R+r)
e = e.m.f in volts
R = load resistance
r = internal resistance

31
Q

Calculating e.m.f
using terminal pd and lost volts

A

e = V + v
e = e.m.f in volts
V = terminal potential difference
v = lost volts

== V = e - v
V=Ir so V = e - Ir

32
Q

Measuring internal resistance and e.m.f

A
  1. Set up a variable resistor(load resistance) to its highest resistance close the switch and record current and pd. open and close the switch again to get two more sets of results for this load resistance. Calculate the mean of and current for this resistance from your results
  2. Decrease the resistance of the variable resistor by a small amount and repeat step 1
  3. Repeat steps 1 and 2 till you have a set of mean currents and pds for 10 different load resistances
  4. Plot a V-I graph for the mean data and draw a line of best fit ( straight line graph)
  5. Make sure all other variables are kept constant including external factors like temperature
33
Q

Kirchhoffs first law

A

The total current entering a junction = the total current leaving it

34
Q

Kirchhoffs second law

A

Total amount of e.m.f around a series circuit = the sum of the pds across each component
e = ∑IR

35
Q

Conservation of charge

A

As charge flows through a circuit, it doesn’t get used up or lost. Since current is the rate of flow of charge, it follows that whatever current flows into a junction is the same as the current flowing out of it

36
Q

Conservation of energy

A

Energy is conserved. In electrical circuits, energy is transferred around the circuit. Energy transferred to a charge is e.m.f and energy transferred from a charge is pd. in a closed loop these two quantities must be equal if energy is conserved

37
Q

Series circuit

A
  • there will be the same current at all points of the circuit
  • the e.m.f is split betweeen the components so( Kirchhoffs 2nd law):
    e = V1+ V2 + V3
    -voltage splits proportionally to the resistance as V = IR so if I is constant:
    IR1+IR2+IR3 = IRT( total resistance)
  • cancelling I give Rt = R1+R2+R3
38
Q

Parallel circuits

A

-current is split at each junction so:
I =I1+I2+I3
-there is the same pd across all components as the pd across each branches of the circuit is equal to the e.m.f I=V/R
V/Rt=V/R1 + V/R2 + V/R3
Cancelling Vs give:
1/Rt = 1/R1 + 1/R2 + 1/R3

39
Q

Cells in series

A

You can calculate total e.m.f of their combination by adding their individual e.m.fs
Makes sense as each charge goes through each of the cells and so gains e.m.f (electrical energy) from each one
et = e1+e2+e3

40
Q

Cells in parallel

A

The total e.m.f of the combination of cells is the same size as the e.m.f of each individual cells. This is because the amount of charge flowing in the circuit doesn’t increase by adding cells in parallel but the number of paths of the charge can take does.
The current will split equally between identical cells. The charge only gains e.m.f from the cells it travels through - so the overall e.m.f in the circuit doesn’t increase

41
Q

What is a potential divider

A

A potential divider is a circuit with a voltage source and a couple of resistors in series. The pd across the voltage source is split across the resistors in the ratio of the resistances

42
Q

Potential dividers to supply pd

A

You can use potential dividers to supply pd between zero and the pd across the power supply.

43
Q

Light sensors

A

A light-dependent resistor(LDR) has a very high resitance in the dark but lower resistance in the light

Can be used as one of the resistors in a potential divider giving an output voltage that varies with the light level

44
Q

Temperature sensor

A

An NTC thermistors has a high resistance at Low temperatures but a lower resistance at high temperatures
Can be used as one of the resistors in a potential divider giving an output voltage that varies with the temperature level

45
Q

Potentiometers

A

Has a variable resistor replacing R1 and R2 in the potential divider
You move a slider or turn on a knob to adjust the relative sizes of R1 or R2 which is useful if you want to change voltage continuously

Physics (12 decks)

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