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Nov 15, 2013 (4 years and 7 months ago)

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Notes (Module: Batteries and Bulbs
)

IJSO Training: Batteries and Bulbs

Notes

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 ion
s.

The motion of the free electrons is

random
. We say they have random
thermal
motion
with an average speed which increases with temperature.

The figure below represents a piece of metal which does
not

have current flowing through
it. The arrows represen
t the random thermal motion of the electrons (their average speed
at room temperature is hundreds of kms
-
1
).

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 m
ore regular and results in a
general
drift

of electrons

through the metal. A typical

drift velocity for electrons in
metals is
less than
1 mm/s
. The magnitude of the drift velocity depends on
the
current
,
the
type of metal

and the
dime
nsions of the piece of metal
.

Notes (Module: Batteries and Bulbs
)

The
resistance

of a piece of metal is due to collisions between the free electrons and the
metal ions
.

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)
.

At higher temperature,
th
e 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
.

Note: Not all
materials

behave in this
way:
The resistance of semi
-
conductors (e.g.
silicon and germanium)
decreases

with temperature
.

This happens because the
density of charge carriers increases rapidly with temperature.

The resistance of superconductors falls gradually with decreasing temp
erature; however,
at a certain critical temperature
c
T
, the resistively drops suddenly to zero
.
Examples of superconductors: Aluminum with
1.19
c
T

7.22
c
T

K.
Some high
temperature,

138
c
T

K,

superconductors are made of ceramic materials
.

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

Notes (Module: Batteries and Bulbs
)

2.

Electric Circuits

When drawing diagrams to represent electric circuits, the following symbols are used.

Wires crossing but
not

connected
:

Wires crossing and connected
:

Unless otherwise stated, we assume that connecting wires are made of a perfect conductor, i.e.,
no resistance.

Switc
h

Battery

A.C. supply

Resistor

Variable resistor

Rheostat (variable resistor)

Filament
lamp

Voltmeter

Ammeter

Transformer

Push button

Notes (Module: Batteries and Bulbs
)

Exercise

2.1

A:__________________, B: __________________, C: __________________

D: __________________, E: __________________

3.

Electric Current

Generally speaking, a
n el
ectric 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.

The unit of current is the
Ampere, A
.

An ammeter is always connected in
series

with other components. The
resi
stance

of
an ammeter must be
low

compared with other components in the circuit being
investigated.

Current in Series Circuits

A current of 2A corresponds to a certain

number of electrons
flowing in the circuit p
er
second.

So if I
1

= 2A, I
2

and I
3

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
.

C

D

E

A

Notes (Module: Batteries and Bulbs
)

Currents in Parallel Circuits

If the three current I
1
, I
2

and I

are measured it is found that

I
1

+ I
2

= I

This result is called
Kirchhoff’s current law
Ⱐ獴慴e搠d猠景汬潷猺
1

The total current flowing
towards a junction

in a circuit is equal to the total current
flowing
away from that junction
.

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.

1

See Wikipedia, “
Kirchhoff's circuit laws” (4 May 2010, 21:47).

Notes (Module: Batteries and Bulbs
)

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, 25
min
-
1
, then in
15

minutes, the total quantity of water which has moved through the pipe is

25 15 375
 
.
The quan
tity of liquid is equal to the rate of flow multiplied by the time.

Similarly, when considering a flow of electric charges, the quantity of charge which
passes is given by

quantity of charge = rate of flow of charge × time

Q = I × t

the unit of charg
e 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 1

A flows for 1s
ec.

It should be noted that the Coulomb is a rather large quantity of charge. 1

C is the
q
uantity of charge carried by (approximately)

6×10
18

electrons!

Hence, each electron carries

1.6022

×

10

19

C
. This is the basic unit of electric charge.

Exercises

3.1

A current of 0.8A fl
ows through a lamp. Calculate the quantity of electric charge
passing through the lamp filament in 15 seconds.

12 C

3.2

A current of 2.5A passes through a conductor for 3 minutes. Calculate the quantity of
charge passes through the conductor.

450 C

Notes (Module: Batteries and Bulbs
)

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.

W
ater 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.

The flow of water can be maintained using a
pump.

A flow of elect
rons can be maintained using a
battery.

T h e b a t t e r y ma i n t a i n s a n e l e c t r i c a l
p o t e n t i a l d i f f e r e n c e b e t w e e n p o i n t s A a n d B.

T o me a s u r e v o l t a g e we u s e a
v o l t me t e r
.
T h e u n i t o f v o l t a g e i s t h e
v o l t
.

A v o l t me t e r g i v e s u s a r e a d i n g wh i c h i n d i c a t e s t h e a mo u n t o
f e n e r g y l o s t b y e a c h
Co u l o mb o f c h a r g e mo v i n g b e t we e n t h e t wo p o i n t s t o wh i c h t h e v o l t me t e r i s c o n n e c t e d.
1 V me a n s
1

J C
-
1
.

A v o l t me t e r i s a l wa y s c o n n e c t e d i n
p a r a l l e l

wi t h o t h e r c o mp o n e n t s. Th e
r e s i s t a n c e

o f a v o l t me t e r mu s t b e
h i g h

c o mp a r e d wi t h o t h e r c o
mp o n e n t s i n t h e c i r c u i t b e i n g
i n v e s t i g a t e d.

Wh a t i s a n i d e a l v o l t me t e r?

An i d e a l v o l t me t e r c a n me a s u r e t h e p o t e n t i a l d i f f e r e n c e a c r o s s t wo p o i n t s i n a c i r c u i t wi t h o u t
d r a wi n g a n y c u r r e n t.

A

B

W
a t e r f l o w s

i n
a p i p e

E l e c t r o n s

f l o w

i n a c i r c u i t

Notes (Module: Batteries and Bulbs
)

Voltages in Series Circuits

Consider the simple series circu
it above.

Energy lost by each Coulomb of charge moving from A to B is
V
1
.

Energy lost by each Coulomb of charge moving from B to C is
V
2
.

Energy lost by each Coulomb of charge moving from C to D is
V
3
.

Obviously the total amount of energy lost by each Coul
omb of charge moving from A to D
must be
V
1

+ V
2

+ V
3

(= V)
.

Conclusion
:
The total voltage across components connected in series is the
sum of the
voltages across each component
.

Voltage across Components in Parallel

All points inside the dotted ellips
e 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
.

Conclusion
:
Co
mponents connected in parallel with each other
all have the same voltage
.

Again, this does
not

depend on what the components are.

Notes (Module: Batteries and Bulbs
)

5.

Resistance

The resistance of a conductor is a measure of the opposition it offers to the flow of
electric current.

It
c
auses
electrical energy to be converted into heat
.

The r
esistance

of a conducting wire

is given by

The unit of resistan
ce

is
Ohm

(
Ω
)
.
It depends on
the length of the piece of metal

and
the cross
-
sectional area of the piece of metal

A
,
and

ρ

is called
the

resistivity
,

units

Ω
m, which depends on type of metal
.

In a circuit, the r
esistance is defined
by

Where V is the voltage

across the resistor and I is the current flows through it.

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.

Notes (Module: Batteries and Bulbs
)

Exercises

5.1

If the resistance of a wire, of length
l
and un
iform across sectional area A, is 10
Ω
. What
is the resistance of another wire made of the same material but with dimensions of
twice the length and twice the cross sectional area?

5.2

Ω

length. If its
volume remains unchanged after stretching, what is the resistance of the wire?

Ω

5.3

A current of 0.8A flows through a lamp. If the resistance of the lamp filament is
Ω
, calculate the potential difference across the lamp
.

1.12 V

E
ffective
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 resistance)
,

R
E
, of all the
resistors. We can consider R
E

to be the
single

resisto
r which would take the

same
amount of current from the same battery
.

Resistors in Series

A

B

The
effective resistance

of circuit A is

R
E

= R
1

+ R
2

+ R
3

Notes (Module: Batteries and Bulbs
)

Resistors in Parallel

A

B

The
effective resistance

of circuit A is

Exercise

5.4

Ω

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 i
n
each setting.

1.4 A; 5.7A

#1,2

Notes (Module: Batteries and Bulbs
)

6.

Potential Dividers

In the circuit, let v
1
be the voltage across R
1

and v
2

the
voltage across R
2
.

It can be shown that

C
ircuits of this type are often called
potential dividers
.

Exercises

6.1

Two resistors are c
onnected in series, show that

1
1
1 2
R
V V
R R

a n d
2
2
1 2
R
V V
R R

.

6.2

T w o r e s i s t o r s a r e c o n n e c t e d i n p a r a l l e l, s h o w t h a t

2
1
1 2
R
I I
R R

a n d
1
2
1 2
R
I I
R R

.

V a r i a b l e R e s i s t o r s

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)

(
ii) Linear va
riable resistor

Notes (Module: Batteries and Bulbs
)

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 100
Ω
.

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 contac
t is moved to B?

We have, in effect, the following situation.

Therefore, the voltmeter will read 3V.

Variable Resistor used as a Variable Potential Divider

What is the reading of the voltmeter when the sliding contact is moved to B?

The

ing 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 100
Ω

resistor.

This circuit is useful in experiments in which we
need a variable voltage supply.

Notes (Module: Batteries and Bulbs
)

Exercise

6
.1

In the two circuits below, a variable resistor of resistance 100
Ω

is connected to a 50
Ω

e variable resistor.

(a)

Determine the maximum and minimum currents delivered by the battery, which
has an e.m.f. of 10V and negligible internal resistance, in the two circuits.

(b)

Determine also the currents delivered by the battery when the sliding contact is a
t
the mid
-
point of the wire in both cases.

( a )

( b )

( a ) 0.2 A, 0.0 6 7 A; 0.3 A, 0.1 A

( b ) 0.1 A; 0.1 3 A

#3,4

7.

E l e c t r i c a l P o w e r a n d E n e r g y

A n y c o mp o n e n t w h i c h p o s s e s s
e s r e s i s t a n c e w i l l
c o n v e r t e l e c t r i c a l e n e r g y i n t o
t h e r ma l
e n e r g y
.

C o n s i d e r t h e s i mp l e c i r c u i t s h o w n b e l o w.

T h e c u r r e n t, I, i s a me a s u r e o f t h e n u mb e r o f
C o u l o mb s

o f c h a r g e w h i c h p a s s t h r o u g h
t h e r e s i s t o r
p e r s e c o n d
.

T h e v o l t a g e, V, i s a me a s u r e o f t h e n
u mb e r o f
J o u l e s

o f e n e r g y l o s t b y e a c h
C o u l o mb

o f
c h a r g e p a s s i n g t h r o u g h t h e r e s i s t o r.

S o,

t h e e n e r g y p e r s e c o n d (
p o w e r
) s u p p l y b y t h e b a t t e r y i s

P = VI

Notes (Module: Batteries and Bulbs
)

To calculate the
power consumed by a resistor:

In series

P

= I
2

R

In parallel

P

=
V
2

/
R

Thus el
ectrical 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 correspondin
gly 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.

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'.

Exercises

7
.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

7.2

Ω

Ω

Ω

respectively are
Notes (Module: Batteries and Bulbs
)

connected across a 200 V supply a
s 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) 83.3 V, 116.7 V, 200 V, (b) 0.67 A, 0.67 A,

0.67 A,

(c) 111 W, 156 W, 67 W, (d) C, A, B

#5

8.

Battery and its I
nternal
R
esistance

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 de
scribe
the result as the
terminal potential difference

of the battery.

A battery converts
chemical energy

into
electrical energy
.

The electrical
energy given to each Coulomb

of charge

is called the
emf
,
2

denoted as
, of the battery.

So the unit of emf
is also Volt.

2

The term "emf" originally came from the phrase "electro
-
motive force". This is now considered an
inappropriate term as emf is a quanti
ty of energy not a force. However, the
abbreviation

is still used.

Notes (Module: Batteries and Bulbs
)

In the following circuits, the voltmeter is assumed to have
infinite resistance

(a modern
digital voltmeter has a resistance of around 10
7
Ω
).
is equal to the
emf of the battery.

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 re
present a battery is shown below.

The resistor, r, represents the internal resistance of the battery.

The
voltmete
r

across A and B
will be

The terminal potential difference is only equal to the emf of the battery
if the current
flowi
ng through the battery
or the internal resistant
is zero
.

Suppose there is an external resistance R in the circuit, it can be considered as in series to
the internal resistance, so
we have

A

B

Notes (Module: Batteries and Bulbs
)

Exercises:

8.1

A battery of e.m.f. 3 V and internal resista
nce 1.5
Ω

is connected to another battery of
Ω
, 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? Exp
lain briefly.

8.2

A cell of e.m.f. 1.5 V is connected in series with a variable resistor of resistance
R

and
Ω
⸠Ky⁶ ry楮g
R
, a series of ammeter readings,
I
, are
taken. A graph of
R

against
1/
I

is t
hen plotted. The value of the
y
-
intercept is found
to be
-
Ω
⸠K桡琠t猠瑨e⁩湴e牮r氠le獩s瑡湣e⁯映瑨e⁣e汬?

2
Ω

Notes (Module: Batteries and Bulbs
)

8.3

A student is given two identical batteries, each of e.m.f. 1.5V and negligible internal
resistance and two identical resistors, each of

resistance 4
Ω
⸠Ke瑥t浩湥⁴ e⁣畲ue湴n

(a) 0.375A, (b) 0.75A, (c) 0.188A,

(d) 0.75A and 0.375A

Note: Combination of batteries

Batteries in series

effective e.m.f
1 2 3
E E E
  

eff
ective internal resistance
1 2 3
r r r
  

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.

#6,7