Post on 22-Feb-2016
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Circuit electricity
Atomic structure
Atoms are composed of protons (+),
electrons (-) and neutrons. The nucleus
contains the protons and neutrons and the electrons surround the
nucleus.
Atomic structure
The outer layer of electrons in a metal is
incomplete which allows them to pass from atom to atom
Atomic structureBecause electrons can
pass from atom to atom. charge can pass through a conducting
material such as a metal.
Metals are conductors
Atomic structure
Some materials such as rubber and plastic have complete outer layers of electrons so they cannot pass from atom to atom. Charge cannot pass through
these materials.
These are called Insulators
Current
Because it is the electrons which move from atom to atom in reality negative charge flows
from negative to positive.This has the same effect as positive charge
moving from positive to negative
Conventional current flows from positive to negative
The Ampere (Named after Andre Marie Ampere)
QI = ---- t
The Ampere is a measure of how much
electrical current is flowing and is measured
in units of amps
I = amps Q = charge (in coulombs)and t = time ( in seconds)
Potential Difference or Voltage
Potential difference, or voltage, is the electrical
potential energy per coulomb of charge.
EV = ---- Q
Alessandro Volta
V = voltage E = energy in Joules Q = charge (in
coulombs)
Resistance
Resistance is a measure of opposition to the flow of charge and is measured in
ohms () VI = ---- R
Georg Ohm
I = current V = voltage R = resistance in ohms
Ohms Law (three versions)
VI = ---- R
V = IR VR = ---- I
Electrical Power
Power is the rate of using energy in joules per
second
P = E tor E = Pxt
Electrical Power
From previous slides we know that
EV = ---- Q
QI = ---- t
and
Electrical PowerCombine the two and
cancel the Q from each EV = ---- Q
QI = ---- t
X
Leaving E/t so electrical power is P = V x I
Electrical Power Equation variations
P = V x I
P = I2R
P = V2/R
These were obtained by using Ohm’s law to substitute for V and I
Kirchoff’s Laws
I1
Kirchoff’s first LawThe total current flowing into a junction is the total current flowing out of the circuit
I2
I3I1 = I2 + I3
Kirchoff’s Second Law
1. The sum of the potentialdifferences around an
electricalcircuit equals the supply
voltage.
Resistors in series The total resistance is
found by simply
adding the resistance
of each R1 + R2 +R3
etc
Resistors in series The supply voltage (pd) is shared across the resistors. The voltage across each
depends on the resistance of
each
Resistors in seriesThe current in a series
circuit is the same all the way round the circuit
(as per Kirchoff’s first Law). Current flowing into the
resistor is the same as the
current flowing out of the resistor)
Resistors in parallel
The total resistance is calculated as
below
Resistors in parallelThe current in a parallel
circuit is shared
between each resistor. (The amount in each depends
on the resistance)
Resistors in parallel The supply voltage (pd) across each resistor is
the same as the supply
voltage
Combined resistors
To calculate the total resistance of the circuit calculate the parallel set first and treat it as a single resistor in series with the other resistor
Example From the following diagram determine:a) Total resistance.b) Total (supply) current.c) Voltage across each resistor.d) Power loss in resistor R1.
R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
Example
R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
Total resistance = R1 + R2 =150ΩTotal (supply) current V/I = 12/150 =0.08 amps.Voltage across R1 = 50 x 0.08 = 4 voltsVoltage across R2 = 100 x 0.08 = 8 voltsPower loss in R1 = V x I = 4 x 0.08 = 0.32 Watts
Example From the following diagram determine:a) Total resistance.b) Total (supply) current.c) Current through each resistor.
R1 100Ω, R2 = 1kΩand supply voltage = 12V.
Example1/Total resistance = 1/100 +1/1000.= 10/1000 + 1/1000 = 11/1000 Total resistance = 1000/11 =90.9Ω
ExampleTotal current = V/R = 12/90.9 = 0.132 ampsCurrent through R1, V/R1 = 12/100 = 0.12 ampsCurrent through R2, V/R2 = 12/1000 = 0.012 amps
Example From the diagram below, determine:
a) The total resistance, and the supply current.
b) The voltage across the R1 resistor.
c) The current through R2 , and the power dissipated in
it.
R1 = 200Ω R2 and R3 are both 100Ω and the
supply voltage is 12 volts
Example Resistance of the parallel resistors
1/total = 1/100 +1/100=2/100
Total resistance = 100/2= 50Ω
Total resistance in circuit = 200+50 =
250Ω Current = V/R
=12/250=0.048 amps
Example
Voltage across R1I x R
=0.048 x 200= 9.6 volts
Example Voltage across R1 & R2V = I x R
0.048 x 50= 2.4 volts
Current through R2I = V/R
=2.4/100=0.024amps
Example
Power dissipatedP = V x I
2.4 x 0.024= 0.058 watts