Transcript Circuit electricity

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 structure
Because 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)
The Ampere is a
measure of how much
electrical current is
flowing and is measured
in units of amps
Q
I = ---t
I = amps Q = charge (in
coulombs)
and t = time ( in seconds)
Potential Difference or Voltage
Alessandro
Volta
Potential difference, or
voltage, is the electrical
potential energy per
coulomb of charge.
E
V = ---Q
V = voltage E = energy in
Joules Q = charge (in
coulombs)
Resistance
Georg Ohm
Resistance is a
measure of opposition
to the flow of charge
and is measured in
ohms ()
V
I = ---R
I = current V = voltage
R = resistance in ohms
Ohms Law (three versions)
V
I = ---R
V = IR
V
R = ---I
Electrical Power
Power is the rate of using
energy in joules per
second
P=E
t
or E = Pxt
Electrical Power
From previous slides we
know that
E
V = ---Q
and
Q
I = ---t
Electrical Power
Combine the two and
cancel the Q from each
E
V = ---Q
X
Q
I = ---t
Leaving E/t so electrical power is
P=VxI
Electrical Power Equation
variations
P=VxI
P = I2R
P = V2/R
These were obtained by using Ohm’s
law to substitute for V and I
Kirchoff’s Laws
Kirchoff’s first Law
The total current flowing into a
junction is the total current
flowing out of the circuit
I2
I1
I1 = I 2 + I 3
I3
Kirchoff’s Second Law
1. The sum of the potential
differences around an
electrical
circuit 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 series
The 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 parallel
The 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
Total resistance = R1 + R2 =150Ω
Total (supply) current V/I = 12/150 =0.08 amps.
Voltage across R1 = 50 x 0.08 = 4 volts
Voltage across R2 = 100 x 0.08 = 8 volts
Power loss in R1 = V x I = 4 x 0.08 = 0.32 Watts
R1 = 50Ω, R2 = 100Ω and supply voltage = 12V.
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.
Example
1/Total resistance = 1/100 +1/1000.= 10/1000
+ 1/1000 = 11/1000
Total resistance = 1000/11 =90.9Ω
Example
Total current = V/R = 12/90.9 = 0.132 amps
Current through R1, V/R1 = 12/100 = 0.12 amps
Current 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 R1
IxR
=0.048 x 200
= 9.6 volts
Example
Voltage across R1 & R2
V=IxR
0.048 x 50
= 2.4 volts
Current through R2
I = V/R
=2.4/100
=0.024amps
Example
Power dissipated
P=VxI
2.4 x 0.024
= 0.058 watts