Download PPT1 - Animated Science

Electricity - Basic ideas…
Electric current is when electrons start to flow around a
circuit. We use an _________ to measure it and it is
measured in ____.
Potential difference (also called _______) is how big the
push on the electrons is. We use a ________ to measure
it and it is measured in ______.
Resistance is anything that resists an electric current. It
is measured in _____.”
(Words: volts, amps, ohms, voltage, ammeter, voltmeter)
Charge (Q)
As we said, electricity is when electrons move around a
circuit and carry energy with them. Each electron has a
negative CHARGE. Charge is measured in Coulombs (C).
We can work out how much charge flows in a circuit using
the equation:
Charge = current x time
(in C)
(in A)
Q
(in s)
1.6x10-19C
Charge on electron =
One coulomb = I amp per second
I
T
Example questions
Charge (C)
Current (A)
Time (s)
5
2
0.4
1
20
0.5
50
250
3
60
1) A circuit is switched on for 30s with a current of 3A. How much
charge flowed?
2) During electrolysis 6A was passed through some copper chloride
and a charge of 1200C flowed. How long was the experiment on
for?
3) A bed lamp is switched on for 10 minutes. It works on a current of
0.5A. How much charge flowed?
More basic ideas…
If a battery is
added the current
will ________
because there is a
greater _____ on
the electrons
If a bulb is added
the current will
_______ because
there is greater
________ in the
circuit
Current in a series circuit
If the current
here is 2
amps…
The
current
here will
be…
The current
here will
be…
And the
current
here will
be…
In other words, the current in a series
circuit is THE SAME at any point
Current in a parallel circuit
A PARALLEL circuit is one where the current has a “choice
of routes”
Here comes the current…
Half of the current
will go down here
(assuming the bulbs
are the same)…
And the rest will
go down here…
Current in a parallel circuit
If the
current
here is 6
amps
And the
current here
will be…
The current
here will be…
The current
here will be…
The current
here will be…
Voltage in a series circuit
If the voltage
across the
battery is 6V…
V
…and these
bulbs are all
identical…
…what will the
voltage across
each bulb be?
V
V
2V
Voltage in a series circuit
If the voltage
across the
battery is 6V…
…what will the
voltage across
two bulbs be?
V
V
4V
Voltage in a parallel circuit
If the voltage across
the batteries is 4V…
What is the
voltage here?
4V
V
And here?
V
4V
Summary
In a SERIES circuit:
Current is THE SAME at any point
Voltage SPLITS UP over each component
In a PARALLEL circuit:
Current SPLITS UP down each “strand”
Voltage is THE SAME across each”strand”
An example question:
6V
A3
3A
A1
V1
A2
V2
V3
Resistance
Resistance is anything that will
RESIST a current. It is measured
in Ohms, a unit named after me.
Georg Simon Ohm
1789-1854
The resistance of a component can be
calculated using Ohm’s Law:
Resistance
(in )
=
V
Voltage (in V)
Current (in A)
I
R
An example question:
Ammeter
reads 2A
A
V
Voltmeter
reads 10V
1) What is the resistance across
this bulb?
2) Assuming all the bulbs are the
same what is the total resistance
in this circuit?
Current-voltage graphs
I
I
V
1. Resistor
2. Bulb
I
V
V
3. Diode
Explain the shape of each graph
Voltage - Current Graphs:
Diode
Only lets current flow in one direction. Forward biased –
forward bias: high resistance (initially gives small
current) at ≈ 0.6V, resistance decreases rapidly
(current increases)
reverse bias: high resistance (gives ~ zero or
slightly negative current. At breakdown, resistance ~
zero (and very large current)
I
V
Voltage - Current Graphs:
Filament Lamp
• As Voltage increases current
increases.
• Current heats the filament
• Therefore resistance increases
• Same applies in negative direction
I
V
Current/Voltage Graphs:
Ohmic conductor
• Straight line through origin –
constant gradient
• V and I increasing proportionally
I
V
Resistors in Series
• Suppose all the resistors could be replaced with one resistor of
equivalent resistance RT
VT = IRT and VT=V1+V2+V3 Eqn1
V1=IR1
V2=IR2
V3=IR3
• Therefore substituting into 1
• I RT = IR1 + IR2 + IR3
( I cancels )
• Therfore in series:
RT = R1 + R2 + R3
Resistors in Parallel
•
•
I = I1+I2+I3
Equation 1
Suppose all the resistors could be replaced with one resistor of
equivalent resistance RT. I=V/RT
I1 = V/R1
I2 = V/R2
I3 = V/R3
Substituting into Equation 1
•
V/RT = V/R1 + V/R2 + V/R3
Cancelling V
•
Therefore 1/RT = 1/R1 + 1/R2 + 1/R3
Questions
1.
Find the combined resistance of two resistors of 4Ω and 6Ω
connected in a) series
b) parallel.
2.
Find the combined resistance of three resistors of 4Ω and 5Ω
and 20Ω connected in a) series
b) parallel.
3.
What is the combined resistance of 5 resistors, each of value
10Ω, connected in a) series and b) parallel.
4.
What resistance must be connected in parallel with a resistor of
16Ω to give a combined resistance of 3.2Ω.
Resistivity
• The resistance of a piece of conductor depends on
RL
length, area, temp and material. Assuming
temp is
1
R
constant
A
L
1
R
A
L
R
A
L
R 
R
• R = Resistance (Ω)
• L = Length (m)
• A = Area (m2)
• Ρ is a constant of
Proportionality
Resistivity units are Ωm.
R
L
A
A
Table of Resistivities
Material
Resistivity(Ωm)
Silver
1.6x10-8
Copper
1.7x10-8
Aluminium
2.7x10-8
Silicon
2300
Glass
1010 - 1014
Polystyrene
1015
DC and AC
V
DC stands for “Direct
Current” – the current only
flows in one direction:
Time
1/50th s
AC stands for “Alternating
Current” – the current
changes direction 50 times
every second (frequency =
50Hz)
240V
T
V
Power and fuses
Power is “the rate of doing work”.
The amount of power being used in
an electrical circuit is given by:
Power = voltage x current
in W
in V
in A
P
V
I
Using this equation we can work out the fuse rating for any
appliance. For example, a 3kW (3000W) fire plugged into a
240V supply would need a current of _______ A, so a
_______ amp fuse would be used (fuse values are usually 3,
5 or 13A).
Power and fuses
Copy and complete the following table:
Appliance
Power rating
(W)
Voltage (V)
Toaster
720
240
Fire
2000
240
Hairdryer
300
240
Hoover
1000
240
Computer
100
240
Stereo
80
240
Current
needed (A)
Fuse needed
(3, 5 or 13A)
Energy and charge
The amount of energy that flows in a circuit will depend on
the amount of charge carried by the electrons and the
voltage pushing the charge around:
Energy transferred = charge x voltage
(in J)
(in C)
(in V)
W
V
Q
Electromotive force and Internal
Resistance
• Let an amount of energy E be supplied when a
charge Q passes through the source. We call the
energy supplied per unit charge the e.m.f. of the
source Є.
Є = E/Q
Where E is measures in Joules, J
Q is measured in coulombs, C
Electromotive Force and Internal
Resistance
•
•
For the dry cell battery, as the current is increased, the reading of
the voltmeter across the battery decreased.
Missing Voltage is proportional to Current
•
E.M.F. of battery
(
ε – V) proportional current
– (ε – V) = Ir ( r is the internal resistance of the cell.
Chemicals inside the battery offer some resistance to
current flow)
– ε = V + Ir
= IR + Ir
– therefore
ε = I(R+r)
Example questions
1) In a radio circuit a voltage of 6V is applied and a charge
of 100C flows. How much energy has been transferred?
2) In this circuit the radio drew a current of 0.5A. How
long was it on for?
3) A motor operates at 6V and draws a current of 3A. The
motor is used for 5 minutes. Calculate: a) The motor’s
resistance, b) the charge flowing through it, c) the
energy supplied to it
4) A lamp is attached to a 12V circuit and a charge of
1200C flows through it. If the lamp is on for 10 minutes
calculate a) the current, b) the resistance, c) the energy
supplied to the bulb.