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Summary Electricity (Year 1, 2015 AS Level)

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Subido en
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Escrito en
2022/2023

It is a brief summary of the AS-Level of Electricity for Edexcel. All of the topics are covered in short paragraphs with diagrams and equations to help explain the key topics.

Institución
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Escuela, estudio y materia

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Subido en
13 de abril de 2023
Número de páginas
1
Escrito en
2022/2023
Tipo
Resumen

Temas

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Charge Drift Velocity Resistivity
A charge comes in discrete
quantities of q.
ρL
R=
Protons = 1.6 x 10-19 A
Electrons = -1.6 x 10-19 Resistance is affected by
temperature, cross sectional area
and length of wire.
Drift velocity is the velocity of the
charge carriers which make up
current in a wire. I-V graph of a bulb
I = nAvq

The typical number of chare carriers
In this diagram the battery is doing
in:
the work, pushing the electrons
around. Conductors … 1 x 10^29

Semi- conductors … 1 x 10 ^19

The electrons transfer energy to the Insulators … 1 x 10^9
bulb, causing it to light up. The graph looks like this because as
Ohms Law and resistance the PD is increasing constantly the
current only increases up to a
Current = charge/time certain point, the equation V/I = R
It is the rate of change of charge can be used to describe this, so as a
with time. higher PD is used, a larger resistance
is going through the bulb and so
Voltage = work done/charge current begins to stay constant as
the bulb gets hotter.
Voltage is the work done –per
charge. A material is ohmic when potential
EMF
difference is proportional to current
at a constant temperature. Any component that supplies energy
is a source of EMF.
You will need to prove the
Power
resistance equations EMF of a supply is the work done by
P=VI the supply on both the external and
(Current constant in series)
2 internal resistances per coulomb of
P=I R (PD is constant in parallel) charge which flows.
2
V PD is the opposite of EMF as it is the
P=
R energy lost between two points.



ϵ=I ( R+r )=Ir +V
Where:
If lamp (Left) is shining brighter in
parallel, which one shines brighter in If you are to connect an ammeter I is current
series? and a voltmeter in a circuit, then:
R is load resistance
In parallel, voltage is constant, and The ammeter must be in series to try
This r is internal resistance
due to p=V^2/R, lamp (Left) has a and close to 0 resistance.
lower resistance than lamp (Right) V is terminal voltage
The voltmeter must be in parallel
and when its changed to series and have up to infinite resistance. Potential dividers
current is constant and P=I^2R, so
since Lamp ® has more R it will be R measured
V out =V ¿ [ ]
Total R
$4.10
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