Transcript Basic Relativity

IJSO Training Course
Phase II
Module: Batteries and Bulbs
Time allocation: 10 hours
1
Objectives:
Introduce a model of electrical conduction
in a metal, and the concepts of resistance
and internal resistance.
Define and apply the concepts of current,
and the use of ammeters and voltmeters.
Draw circuit diagrams with accepted circuit
symbols.
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1. Electrical Conduction in
Metals
A solid piece of metal, at room temperature,
consists of metal ions arranged in a regular
pattern called a crystal lattice and free
electrons moving in the spaces between the
ions.
The motion of the free electrons is random.
We say they have random thermal motion
with an average speed which increases with
temperature.
3
The figure below represents a piece of metal
which does not have current flowing
through it. The arrows represent the random
thermal motion of the electrons (their
average speed at room temperature is
hundreds of kms-1).
4
If a current is flowing in the piece of metal,
then another motion is added to the random
thermal motion (see the figure below). This
motion is more regular and results in a
general drift of electrons through the metal.
A typical drift velocity for electrons in
metals is less than 1mm/s . The magnitude
of the drift velocity depends on the current,
the type of metal and the dimensions of the
piece of metal.
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The resistance of a piece of metal is due to
collisions between the free electrons and the
metal ions.
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During a collision, some of the kinetic
energy possessed by the electron can be
transferred to the ion thus increasing the
amplitude of the lattice vibrations.
Therefore, resistance to the flow of current
causes the temperature to increase or in
other words, resistance causes electrical
energy to be converted into thermal energy
(internal energy).
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At higher temperature, the amplitude of the
lattice vibrations increases, the collisions
between the free electrons and the metal
ions are more often. This suggests that the
resistance of a piece of metal should
increase with temperature.
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Note: Not all materials behave in this way:
the resistance of semi-conductors (e.g.
silicon and germanium) decreases with
temperature.
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Conductors / Insulators
Electrical conductors readily conduct
electric charges, small resistance .
Electrical insulators conduct electric
charges poorly, large resistance .
Examples:
Good
conductors
Poor
conductors
Good insulators
Metals, carbon
moist air, water,
human body
Rubber, dry air
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2. Electric Circuits
When drawing diagrams to represent
electric circuits, the following symbols are
used.
Wires crossing but not
connected:
Wires crossing and
connected:
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Unless otherwise stated, we assume that
connecting wires are made of a perfect
conductor, i.e., no resistance.
Switch
Battery
A.C. supply
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Resistor
Variable resistor
Push button
Filament lamp
Voltmeter
Ammeter
Transformer
Rheostat (variable
resistor)
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Exercise
A:_________, B: _______, C: ________
D: _________, E: _________
push button, A.C. supply, rheostat,
voltmeter, bulb
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3. Electric Current
Generally speaking, an electric current is a
flow of charged particles. For examples, a
current in a metal is due to the movement of
electrons. In a conducting solution, the
current is due to the movement of ions.
Current is measured using an ammeter. An
ammeter measures the rate of flow of charge.
For simplicity, an ammeter gives a reading
which is proportional to the number of
electrons which pass through it per second.15
The unit of current is the Ampere, A.
An ammeter is always connected in series
with other components. The resistance of
an ammeter must be low compared with
other components in the circuit being
investigated.
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Current in Series Circuits
A current of 2A corresponds to a certain number of
electrons flowing in the circuit per second. So if I1 =
2A, I2 and I3 must also be 2A because in a series
circuit, the electrons have only one path to follow.
Conclusion: The current is the same at all points in a
series circuit.
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If the three current I1, I2 and I are measured
it is found that I1 + I2 = I
This result is called Kirchhoff’s current law, stated as
follows.
The total current flowing towards a junction in a
circuit is equal to the total current flowing away from
that junction.
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As an analogy, consider vehicles at a road junction.
The number of vehicles passing point 1, per minute,
must be equal to the number of vehicles passing
point 2 per minute plus the number of vehicles
passing point 3 per minute.
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Relation between Current,
Charge and Time
Another analogy is often found to be
helpful. Consider a pipe through which
water is flowing. If the rate of flow of water
through the pipe is, for example, 25l min-1,
then in 15 minutes, the total quantity of
water which has moved through the pipe is
25 x 15 = 375l . The quantity of liquid is
equal to the rate of flow multiplied by the
time.
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Similarly, when considering a flow of electric
charges, the quantity of charge which passes is
given by Q = I × t
quantity of charge = rate of flow of charge × time
the unit of charge is the Coulomb. The Coulomb
can be now defined as follows:
1 C is the quantity of charge which passes any
point in a circuit in which a current of 1A flows
for 1sec.
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It should be noted that the Coulomb is a
rather large quantity of charge. 1 C is the
quantity of charge carried by
(approximately) 6×1018 electrons!
Hence, each electron carries
–1.6022 × 10–19 C. This is the basic unit of
electric charge.
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Exercises
1. A current of 0.8A flows through a lamp.
Calculate the quantity of electric charge
passing through the lamp filament in 15
seconds.
12 C
2. A current of 2.5A passes through a
conductor for 3 minutes. Calculate the
quantity of charge passes through the
conductor.
450 C
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4. Voltage
When a body is falling through a gravitational
field, it is losing gravitational potential energy.
Similarly, when a charge is "falling" through
an electric field, it is losing electric potential
energy.
Water has more gravitational potential energy
at B than at A so it falls. The potential energy
lost by 1 kg of water in falling from level B to
level A is the gravitational potential difference
(J/kg) between A and B.
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The flow of water can be
maintained using a pump.
A flow of electrons can be maintained
using a battery. The battery maintains
an electrical potential difference
between points A and B.
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To measure voltage we use a voltmeter. The unit of
voltage is the volt.
A voltmeter gives us a reading which indicates the
amount of energy lost by each Coulomb of charge
moving between the two points to which the
voltmeter is connected. 1V means 1 JC-1.
A voltmeter is always connected in parallel with
other components. The resistance of a voltmeter
must be high compared with other components in
the circuit being investigated.
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What is an ideal voltmeter?
An ideal voltmeter can measure the
potential difference across two points in a
circuit without drawing any current.
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Voltages in Series Circuits
Consider the simple series circuit above.
Energy lost by each Coulomb of charge moving from
A to B is V1.
Energy lost by each Coulomb of charge moving from
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B to C is V2.
Energy lost by each Coulomb of charge
moving from C to D is V3.
Obviously the total amount of energy lost
by each Coulomb of charge moving from A
to D must be V1 + V2 + V3 (= V).
Conclusion: The total voltage across
components connected in series is the sum
of the voltages across each component.
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Voltage across Components in
Parallel
All points inside the dotted ellipse
on the right must be at the same
potential as they are connected by
conductors assumed to have
negligible resistance.
Similarly for all points inside the
dotted ellipse on the left.
So the three voltmeters are
measuring the same voltage.
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Conclusion: Components connected in
parallel with each other all have the same
voltage.
Again, this does not depend on what the
components are.
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5. Resistance
The resistance of a conductor is a measure
of the opposition it offers to the flow of
electric current. It causes electrical energy
to be converted into heat.
The resistance of a conducting wire is given
by
R = ρl /A
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The unit of resistant is Ohms (W). It depends
on the length of the piece of metal l and the
cross-sectional area of the piece of metal A,
and r is called the resistively, units Wm, which
depends on type of metal.
In a circuit, the resistance is defined by
Resistant = voltage / current
R=V/I
Where V is the voltage across the resistor and
I is the current flows through it.
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Ohm’s law
The Ohm’s law states:
For a metal conductor at constant temperature, the
current flowing through it is directly proportional to the
voltage across it.
As voltage divided by current is resistance,
this law tells us that the resistance of a piece
of metal (at constant temperature) is constant.
Note: the resistance of a piece of metal
increases as its temperature increases.
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Exercises
1. If the resistance of a wire, of length l and
uniform across sectional area A, is 10W.
What is the resistance of another wire made
of the same material but with dimensions of
twice the length and triple cross sectional
area?
unchanged
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2. A uniform wire of resistance 4 W is
stretched to twice its original length. If its
volume remains unchanged after stretching,
what is the resistance of the wire?
16 W
3. A current of 0.8A flows through a lamp.
If the resistance of the lamp filament is
1.4W, calculate the potential difference
across the lamp.
1.12 V
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Effective Resistances
If two or more resistors are connected to a
battery, the current which will flow through
the battery depends on the effective
resistance (or equivalent resistant), RE, of
all the resistors. We can consider RE to be
the single resistor which would take the
same amount of current from the same
battery.
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Resistors in Series
A
B
The effective resistance of circuit A is
RE = R1 + R2 + R3
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Resistors in Parallel
A
B
The effective resistance of circuit A is
1/RE = 1/R1 + 1/R2 + 1/R3
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Exercise
1. A hair dryer consisting of two identical
heating elements of resistance 70W each is
connected across the 200V mains supply.
The two elements can be connected in series
or in parallel, depending on its setting.
Calculate the current drawn from the mains
in each setting.
1.4A; 5.7A
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6. Potential Dividers
In the circuit, let v1 be the voltage across
R1 and v2 the voltage across R2.
It can be shown that
V1/V2 = R1/R2
Circuits of this type are often called
potential dividers.
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Exercises
1. Two resistors are connected in series,
show that
and
.
2. Two resistors are connected in parallel,
show that
and
.
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Variable Resistors
A variable potential divider can be made using
all three connections of a variable resistor (also
be called a rheostat) .
(i) Rotating variable resistor (internal view)
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(ii) Linear variable resistor
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Using Variable Resistors
In the circuit below, notice that only two of
the connections to the variable resistor are
used.
The maximum resistance of the variable
resistor is 100W.
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When the sliding contact, S, is moved to A
the voltmeter will read 6V (it is connected
directly to both sides of the supply). This is,
of course the maximum reading of voltage
in this circuit.
What is the reading of the voltmeter when
the sliding contact is moved to B?
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We have, in effect, the following situation.
Therefore, the voltmeter will read 3V.
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Variable Resistor used as a
Variable Potential Divider
What is the reading of the voltmeter when the
sliding contact is moved to B?
The voltmeter reading can be reduced to zero by
moving the sliding contact to B. The wire "x"
(assumed to have zero resistance) is in parallel
with the 100W resistor. This circuit is useful in
experiments in which we need a variable voltage
supply.
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Exercise 1
In the circuits above, a variable resistor of resistance 100 W is
connected to a 50 W resistor by means of a sliding which can be
moved along the variable resistor.
(a) Determine the maximum and minimum currents delivered by the
battery, which has an e.m.f. of 10 V and negligible internal resistance,
in the two circuits.
(b) Determine also the currents delivered by the battery when the
sliding contact is at the mid-point of the wire in both cases.
(a) 0.2A, 0.067A; 0.3A, 0.1A
(b) 0.1A; 0.13A
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7. Electrical Power and Energy
Any component which possesses resistance
will convert electrical energy into thermal
energy.
Consider the simple circuit shown below.
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The current, I, is a measure of the number
of Coulombs of charge which pass through
the resistor per second.
The voltage, V, is a measure of the number
of Joules of energy lost by each Coulomb of
charge passing through the resistor.
So, the energy per second (power) supply
by the battery is P = VI
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To calculate the power consumed by a
resistor:
In series
P = I2 R
In parallel P = V2 / R
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Thus electrical energy can be expressed in the
engineering unit kilowatt-hour (i.e., energy
dissipated in an appliance of 1 kW rating operated
for 1 hour)
Electricity is supplied to our house through the
mains. The voltage supplied is alternating and
correspondingly an alternating flow of charges
occurs in the wire. This kind of electricity is called
alternating current (a.c.) in contrast to the direct
current (d.c.) as supplied by a battery.
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Live wire: brown in colour. In Hong Kong, the
voltage at the live wire changes from +220 V to 220V continuously and alternately, so that the
current flows backwards and forwards round the
circuit. A switch and a fuse can be installed in live
wire to prevent the appliance to go 'live'.
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Exercises
1. An electric cooker with a fuse and a
switch in series with the heating element is
to be connected to the pins of a socket. The
correct connections should be
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2. Three lamps A, B and C of resistance 250 W, 350 W
and 600 W respectively are connected across a 200V
supply as shown.
(a) Calculate the potential difference across the lamps.
(b) Calculate the current passing through the lamps.
(c) Calculate the power dissipated in the lamps.
(d)List the lamps in ascending order of brightness.
a. 200V, 117V, 83V
b. 0.33A, 0.33A, 0.33A
c. 27.8W, 38.9W, 66.7W
d. A, B, C
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8. Battery and its Internal
Resistance
The metal contacts which are used to connect a
battery into a circuit are called its terminals.
For this reason, when the voltage of a battery is
measured, we often describe the result as the
terminal potential difference of the battery.
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A battery converts chemical energy into electrical
energy.
The electrical energy given to each Coulomb of
charge is called the e.m.f.1, denoted as , of the
battery. So the unit of e.m.f is also Volt.
The term "e.m.f." originally came from the phrase
"electro-motive force". This is now considered an
inappropriate term as emf is a quantity of energy
not a force. However, the abbreviation is still
used.
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In the following circuits, the voltmeter is
assumed to have infinite resistance (a
modern digital voltmeter has a resistance of
around 107 W). The voltmeter reading is
equal to the e.m.f. of the battery.
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However, the substances of which the
battery is made have some resistance to the
flow of electric current. This is called the
internal resistance of the battery. A more
complete symbol to represent a battery is
shown below.
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The resistor, r, represents the internal
resistance of the battery. The reading of the
voltmeter across A and B will be
V=
- Ir
The terminal potential difference is only
equal to the e.m.f. of the battery if the
current flowing through the battery or the
internal resistant is zero.
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Suppose there is an external resistance R in
the circuit, it can be considered as in series
to the internal resistance, so we have
= I (R + r)
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Exercises:
1. A battery of e.m.f. 3 V and internal resistance
1.5 W is connected to another battery of e.m.f. 3 V
and internal resistance 6 W, same polarities being
wired together as shown in the figure. A student
says the rate at which electrical energy is
converted into internal energy is zero. Do you
agree? Explain briefly.
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2. A cell of e.m.f. 2 V is connected in series with a
variable resistor of resistance R and an ammeter of
resistance 0.4 W. By varying R, a series of
ammeter readings, I, are taken. A graph of R
against 1/I is then plotted. The value of the yintercept is found to be -3 W. What is the internal
resistance of the cell?
2.5 W
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3. A student is given two identical batteries, each
of e.m.f. 2V and negligible internal resistance and
two identical resistors, each of resistance 4.5 W.
Determine the current through each resistor in the
circuits shown in the figure
(a) 0.444A, (b) 1.78A, (c) 0.222A, (d) 0.889A and
0.444A
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Note: Combination of batteries
Batteries in series
- effective e.m.f = E1 + E2+ E3
- effective internal resistance = r1 + r2 + r3
Identical batteries in parallel
- effective e.m.f. = E
- effective internal resistance = r/N, where N is
the number of batteries in parallel.
- It can supply a current N times larger than
that can be supplied by one battery alone.
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