CHAPTER 7: ELECTRICITY

Transcription

Physics Module Form 5
Teacher’s Guide
Chapter 7: Electricity
CHAPTER 7: ELECTRICITY
7.1 CHARGE AND ELECTRIC CURRENT
Van de Graaf
1. What is a Van de Graaff generator? Fill in each of the boxes the name of the part shown.
A device that produces and store electric charges at high voltage on its dome
+
+
+
+
Metal dome
dome
+
+
+
+
+
+
roller
rubber belt
roller
motor
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Teacher’s Guide
2. How are electrical charges produced by a Van de Graaff generator? And what type of
charges is usually produced on the dome of the generator?
 When the motor of the Van de Graaff generator is switched on, it drives the
rubber belt.
 This causes the rubber belt to against the roller and hence becomes charged.
 The charge is then carried by the moving belt up to the metal dome where it is
collected.
 A large amount of charge is built up on the dome
 Positive charges are usually produced on the dome of the generator.
+
3. What will happen if the charged dome of
the Van de Graaff is connected to the earth
via a micrometer? Explain.
 There is a deflection of the pointer of
the meter.
 This indicates an electric current flow.
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+ + +
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Teacher’s Guide
4. Predict what will happen if a discharging
metal sphere to the charged dome.
+
+
 When the discharging metal sphere is
brought
near
the
charged
+ + + + +
dome,
sparkling occurs.
 An electric current flow.
5.
Predict what will happen if hair of a
student is brought near to the charged
dome. Give reasons for your answer.
 The metal dome attracts the hair and
the hair stand upright.
 This is because of each strand of hair
receives positive charges and repels
each other.
6. The flow of electrical charges produces electric current.
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Teacher’s Guide
Electric Current
1. Electric current consists of a flow of electrons
2. The more charges that flow through a cross
section within a given time, the larger is the
current.
3. Electric current is defined as the
rate of flow of electric charge
Each second, 15 coulombs of charge cross
the plane. The current is I = 15 amperes.
One ampere is one coulomb per second.
4. In symbols, it is given as:
I=Q
t
where I = electric current
Q = charge
t = time
(i) The SI unit of charge is (Ampere / Coulomb / Volt)
(ii) The SI unit of time is (minute / second / hour)
(iii) The SI unit of current is (Ampere / Coulomb / Volt) is equivalent to
(Cs // C-1s // Cs-1)
I
(iv)By rearranging the above formula, Q = ( It / t
t
/ I
)
4. If one coulomb of charge flows past in one second, then the current is one ampere.
5. 15 amperes means in each second, 15 coulomb of charge through a cross section of a
conductor.
6. In a metal wire, the charges are carried by electrons.
7. Each electron carries a charge of 1.6 x 10-19 C.
8. 1 C of charge is 6.25 x 1018 electrons.
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Teacher’s Guide
Electric Field
a) An electric field is a region in which an electric charge experiences a force.
b) An electric field can be represented by a number of lines indicate both the magnitude and
direction of the field
c) The principles involved in drawing electric field lines are :
(i) electric field lines always extend from a positively-charged object to a
negatively-charged object to infinity, or from infinity to a negatively-charged object,
(ii) electric field lines never cross each other,
(iii)electric field lines are closer in a stronger electric field.
Demo 1 : To study the electric field and the effects of an electric field.
Apparatus & materials
Extra high tension (E.H.T) power supply (0 – 5 kV), petri dish, electrodes with different
shapes (pointed electrode and plane electrode), two metal plates, talcum powder, cooking oil,
polystyrene ball coated with conducting paint, thread and candle.
Method
DEMO
A)
1. Set up the apparatus as shown in the above figure
2. Switch on the E.H.T. power supply and adjust the voltage to 4 kV
3. Observed the pattern formed by the talcum powder for different types of electrodes.
4. Draw the pattern of the electric field lines.
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Teacher’s Guide
ELECTRIC FIELD AROUND A POSITIVE CHARGE
ELECTRIC FIELD AROUND A NEGATIVE CHARGE
ELECTRIC FIELD AROUND A POSITIVE AND NEGATIVE CHARGE
ELECTRIC FIELD AROUND TWO NEGATIVE CHARGES
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Teacher’s Guide
ELECTRIC FIELD AROUND TWO POSITIVE CHARGES
ELECTRIC FIELD AROUND A NEGATIVE CHARGE AND A
POSITIVELY CHARGED PLATE
ELECTRIC FIELD AROUND A POSITIVE CHARGE AND A
NEGATIVELY CHARGED PLATE
ELECTRIC FIELD BETWEEN TWO CHARGED
PARALLEL PLATES
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Teacher’s Guide
EFFECT OF AN ELECTRIC FIELD ON A POLYSTYRENE BALL
Observation:
The polystyrene ball oscillated between the
two plates, touching one plate after
another.
Explanation:
 When the polystyrene ball touches the
negatively charged plate, the ball
receives negative charges from the plate
and experiences a repulsive force.
1. Place the polystyrene ball between the
two metal plates.
 The ball will then move to the positively
charged plate.
2. Switch on the E.H.T and displace the
polystyrene ball slightly so that it
 When the ball touches the plate, the ball
loses some of its negative charges to the
touches one of the metal plates
plate and becomes positively charged.

It then experiences a repulsive force.
This process continues.
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Teacher’s Guide
EFFECT OF AN ELECTRIC FIELD ON A CANDLE FLAME
C)
Observation:
The candle flame splits into two portions in
opposite direction. The portion that is
attracted to the negative plate is very much
larger than the portion of the flame that is
attracted to the positive plate.
1) Switch of the E.H.T and replace the
polystyrene ball with a lighted candle.
Explanation:
 The heat of the flame ionizes the air
molecules to become positive and
2) Sketch the flame observed when the
E.H.T. is switched on.
negative charges.
 The positive charges are attracted to the
negative plate while the negative
charges are attracted to the positive
plate.
 The flame is dispersed in two opposite
directions but more to the negative
plate.
 The positive charges are heavier than
the negative charges. This causes the
uneven dispersion of the flame.
Conclusion
1. Electric field is a region where an electric charge experiences a force.
2. Like charges repel each other but opposite charges attract each other.
3. Electric field lines are lines of force in an electric field. The direction of the field
lines is from positive to negative.
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Teacher’s Guide
Exercise 7.1
1. 5 C of charge flows through a wire in 10 s. What is the current in the wire?
Q
I
2.
=
It
=
Q/t
=
5 / 10
=
0.5 A
A charge of 300 C flow through a bulb in every 2 minutes. What is the electric
current in the bulb?
Q
=
It
I
=
Q/t
=
300 / 120
=
2.5 A
3.
The current in a lamp is 0.2 A. Calculate the amount of electric charge that passes
through the lamp in 1 hour.
Q
4.
=
=
=
It
0.2 (60 x 60)
720 C
If a current of 0.8 A flows in a wire, how many electrons pass through the wire in one
minute? (Given: The charge on an electron is 1.6 x 10-19 C)
Q
=
=
=
It
0.8 (60)
48 C
Convert: 1 minute = 60s
1.6 x 10 -19 C of charge

1 electron.
Hence, 2880 C of charges is brought by
48 C
1.6 x 10 -19 C
= 3 x 1020 electrons
5.An electric current of 200 mA flows through a resistor for 3 seconds, what is the
(a)
electric charge
(b)
the number of electrons which flow through the resistor?
a) Q
=
=
=
b)
1.6 x 10 -19 C of charge

1 electron.
Hence, 2880 C of charges is 0.6 C
= 3.75 x 1018 electrons
-19
1.6 x 10 C
It
200 x 10-3 (3)
0.6 C
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Teacher’s Guide
Ideas of Potential Difference
(a)
(b)
X
P
Y
Q

 Pressure at point P is greater than the pressure
 Gravitational potential energy at X is greater than
at point Q
the gravitational potential energy at Y.
 Water will flow from P to Q when the valve is
 The apple will fall from X to Y when the apple is
opened.
released.
 This due to the difference in the pressure of
 This due to the difference in the gravitational
water
potential energy.
(c) Similarly,
 Point A is connected to positive terminal
 Point B is connected to negative terminal
 Electric potential at A is greater than the electric potential at
B.
 Electric current flows from A to B, passing the bulb in the
A
Bulb
B
circuit and lights up the bulb.
 This is due to the electric potential difference between the two
terminals.
 As the charges flow from A to B, work is done when electrical
energy is transformed to light and heat energy.
 The potential difference, V between two points in a circuit is
defined as the amount of work done, W when one coulomb of
charge passes from one point to the other point in an electric
field.
 The potential difference,V between the two points will be
given by:
W
Work
V = Quantityofch arg e = Q
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where W is work or energy in Joule (J)
Q is charge in Coulomb (C)
Teacher’s Guide
Device and symbol
ammeter
A
voltmeter
V
connecting wire
Cells
Switch
Constantan wire //
eureka wire
bulb
resistance
rheostat
Measuring Current and Potential Difference/Voltage
Measurement of electricity
Measurement of potential difference/voltage
(a) Electrical circuit
(a) Electrical circuit
(b) Circuit diagram
(b) Circuit diagram
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Teacher’s Guide
1.Name the device used to measure electrical
1.Name the device used to measure
current.
potential difference.
An ammeter
A voltmeter
2.(a) What is the SI unit for current?
Amperes
2.(a) What is the SI unit for potential
difference?
Volts
(b) What is the symbol for the unit of
current?
A
(b) What is the symbol for the unit of
potential difference?
V
3.How is an ammeter connected in an
electrical circuit?
In series
3.How is an voltmeter connected in an
electrical circuit?
In parallel
4.The positive terminal of an ammeter is
connected to which terminal of the dry
4.The positive terminal of a voltmeter is
cell?
connected to which terminal of the dry
Positive
cell?
Positive
5.What will happen if the positive terminal of
the ammeter is connected to the negative
terminal of the dry cell?
The ammeter needle will deflect and show
reading below zero.
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Teacher’s Guide
Exp 1:
To investigate the relationship between current and potential difference
for an ohmic conductor.
(a)
(b)
Figure (a) and figure (b) show two electrical circuits. Why do the ammeters show different
readings? Why do the bulbs light up with different intensity?
Referring to the figure (a) and (b) complete the following table:
(a) Inference
The current flowing through the bulb is influenced by the potential difference across it.
(b) Hypothesis
The higher the current flows through a wire, the higher the potential difference across
(c) Aim
it.
To determine the relationship between current and potential difference for a
constantan wire.
(i)
(d) Variables
manipulated variable
(ii) responding variable
(iii) fixed variable
: current, I
: potential difference, V
: length of the wire // cross sectional area //
temperature
Apparatus /
materials
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Teacher’s Guide
Method
:
1.
Set up the apparatus as shown in the figure.
2.
Turn on the switch and adjust the rheostat so that the ammeter reads the
current, I= 0.2 A.
Tabulation of
3.
Read and record the potential difference, V across the wire.
4.
Repeat steps 2 and 3 for I = 0.3 A, 0.4 A, 0.5 A, 0.6 A and 0.7 A.
:
data
Analysis of data
:
Current,I/A
Volt, V/V
0.2
1.0
0.3
1.5
0.4
2.0
0.5
2.5
0.6
3.0
0.7
3.5
Draw a graph of V against I
Potential difference, V /V
4.0 3.0 2.0 1.0 -
0.2
0.4
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0.6
Current, I /A
0.8
Teacher’s Guide
Discussion
:
1.
From the graph plotted.
(a) What is the shape of the V-I graph?
The graph of V against I is a straight line that passes through origin
(b) What is the relationship between V and I?
This shows that the potential difference, V is directly proportional to the
current, I.
(c) Does the gradient change as the current increases?
The gradient ≡ the ratio of
2.
V
I
is a constant as current increases.
The resistance, R, of the constantan wire used in the experiment is equal to the
gradient of the V-I graph. Determine the value of R.
3.5
o.7
3.
=5
What is the function of the rheostat in the circuit?
It is to control the current flow in the circuit
Conclusion
:
The potential difference, V across a conductor increases when the current, I passing
through it increases as long as the conductor is kept at constant temperature.
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Teacher’s Guide
Ohm’s Law
(a)
Ohm’s law states
that the electric current, I flowing through a conductor is directly proportional to
the potential difference across the ends of the conductor,
if temperature and other physical conditions remain constant
(b) By Ohm’s law:
V

I
= constant  I
or
V
= constant
I
(c) The constant is known as resistance, R of the conductor.
(d) The resistance, R is a term that describes the opposition experienced by the electrons
as they flow in a conductor. It is also defined as the ratio of the potential difference
across the conductor to the current, I flowing through the conductor. That is
V
R= I
and
V=IR
(e) The unit of resistance is volt per ampere (V A-1) or ohm ()
(f) An ohmic conductor is one which obeys Ohm’s law, while a conductor which does not
obey Ohm’s law is known as a non-ohmic conductor
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Teacher’s Guide
Factors Affecting Resistance
1. The resistance of a conductor is a measure of the ability of the conductor to (resist /
allow) the flow of an electric current through it.
2. From the formula V = IR, the current I is (directly / inversely) proportional to the
resistance, R.
3. When the value of the resistance, R is large, the current, I flowing in the conductor is
(small / large)
4. What are the factors affecting the resistance of a conductor?
a) the length of the conductor
b) the cross-sectional area of the conductor
c) type of material of the conductor
d) the temperature of the conductor
5. Write down the relevant hypothesis for the factors affecting the resistance in the table
below.
The temperature of
the conductor
The type of the
material of the
conductor
The cross-sectional
area of the
conductor, A
Length of the
conductor, l
Factors
Diagram
Hypothesis
The longer the conductor, the
higher its resistance
Resistance is directly proportional
to the length of a conductor
The bigger the cross-sectional
area, the lower the its resistance
Resistance is inversely
proportional to the crosssectional area of a conductor
Different conductors with the
same physical conditions have
different resistance
The higher temperature of
conductor,
the
higher
the
resistance
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Graph
Teacher’s Guide
6. From, the following can be stated:
 Resistance of a conductor,
R

length
 Resistance of a conductor,
R

1
cross-sectional area
 Hence, resistance of a conductor, R

length
cross-sectional area
Or
R
l
or
R=  l
A
A
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where  =
resistivity of the
substance
Teacher’s Guide
Exercise 7.2
1.
Tick (√) the correct answers
True
(a) Unit of potential difference is J C-1
√
(b) J C-1 ≡ volt, V
√
False
The potential difference between two points is 1 volt if 1 joule
(c) of work is required to move a charge of 1 coulomb from one
√
point to another.
(d)
2 volt is two joules of work done to move 2 coulomb of charge
(e) Potential difference ≡ Voltage
2.
√
from one to another in an electric field.
√
I
t
/
)
t
I
i) Electric charge,
Q = ( It /
ii) Work done,
W = (QV /
V Q
/
)
Q V
iii) Base on your answer in 2(i) and (ii) derive the work done, W in terms of I, V and t.
W
=
QV
=
ItV
3. If a charge of 5.0 C flows through a wire and the amount of electrical energy converted
into heat is 2.5 J. Calculate the potential differences across the ends of the wire.
W
=
QV
2.5
=
5.0 (V)
V
=
0.5 V
4. A light bulb is switched on for a period of time. In that period of time, 5 C of charges
passed through it and 25 J of electrical energy is converted to light and heat energy. What
is the potential difference across the bulb?
W
=
QV
20
=
6 (V)
V
=
3.33 V
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Teacher’s Guide
5. The potential difference of 10 V is used to operate an electric motor. How much work is
done in moving 3 C of electric charge through the motor?
W
=
QV
=
3 (10)
=
30 J
6. When the potential difference across a bulb is
20 V, the current flow is 3 A. How much work
done to transform electrical energy to light and
heat energy in 50 s?
W
=
VIt
=
20 (3) (50)
=
3000 J
Bulb
3A
A
20 V
7. What is the potential difference across a light bulb
of resistance 5  when the current that passes
through it is 0.5 A?
V
=
IR
=
0.5 (5)
=
2.5 V
8. A potential difference of 3.0 V applied across a resistor of resistance R drives a current of
2.0 A through it. Calculate R.
V
=
IR
3.0
=
2.0 (R)
R
=
1.5 
9. What is the value of the resistor in the figure, if
the dry cells supply 2.0 V and the ammeter
reading is 0.5 A?
V
=
IR
2.0
=
0.5 (R)
R
=
4
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Teacher’s Guide
10. If the bulb in the figure has a resistance of 6 ,
what is the reading shown on the ammeter, if the
dry cells supply 3 V?
V
=
IR
3.0
=
6 (R)
R
=
0.5 
11. If a current of 0.5 A flows through the resistor of
3  in the figure, calculate the voltage supplied
by the dry cells?
V
R
=
IR
=
0.5 (3)
=
1.5 
12. The graph shows the result of an experiment to
determine the resistance of a wire. The resistance
of the wire is
From V-I graph, resistance
V/V
1.2
= gradient
1.2
= 5
= 2.4
0
13. An experiment was conducted to measure the
current, I flowing through a constantan wire when
the potential difference V across it was varied.
The graph shows the results of the experiment.
What is the resistance of the resistor?
= gradient
=
8 x10 3
4
= 2.0 x 10-3
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5
I/A
Teacher’s Guide
14.Referring to the diagram on the right, calculate
(a) The current flowing through the resistor.
V
=
IR
12
=
I (5)
I
=
2.4 A
I
5
12 V
(b) The amount of electric charge that passes
through the resistor in 30 s
Q
=
It
=
2.4 (30)
=
72 C
(c) The amount of work done to transform the
electric energy to the heat energy in 30 s.
W
=
QV
or
W
=
72 (12)
= 12(2.4)(30)
=
864 C
= 864 C
15. Figure shows a torchlight that uses two 1.5 V dry
cells. The two dry cells are able to provide a
current of 0.3 A when the bulb is at its normal
brightness. What is the resistance of the filament?
V
=
IR
3.0
=
0.3(R)
I
=
10
16. The diagram shows four metal rods of P, Q, R
and S made of the same substance.
a) Which of the rod has the most
resistance?
P
b) Which of the rod has the least
resistance?
S
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= VIt
+ 1.5 V -
+ 1.5 V -
Teacher’s Guide
17. The graph shows the relationship between the
V/V
potential difference, V and current, I flowing
X
through two conductors, X and Y.
8
Y
a) Calculate the resistance of conductor X.
= gradient
2
8
0
=2
0
I/A
2
= 4
b) Calculate the resistance of conductor Y.
= gradient
2
=2
= 1
c) If the cross sectional area of X is 5.0 x 10-6
m2, and the length of X is 1.2 m, calculate its
resistivity.
R
l
= A
ρ
= l
RA
4( 5.0 x10 6 )
=
1 .2
= 1.67 x 10-5m
18. The graph shows a graph of I against V for three
conductors, P, Q and R.
i) Compare the resistance of conductor P, Q and R.
Q
I/A
P
Q
Rr > RQ >Rp
R
ii) Explain your answer in (a)
From V-I graph, resistance = gradient
The greater the gradient, the greater the resistance
Gradient of R > Gradient of Q > Gradient of P
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V/V
Teacher’s Guide
19. Figure shows a wire P of length, l with a crosssectional area, A and a resistance, R. Another
wire, Q is a conductor of the same material with
a length of 3l and twice the cross-sectional area
of P. What is resistance of Q in terms of R?
Conductor P
R
l
= A
Conductor Q
R’
l'
= A' (notes: P and R have the same resistivity, ρ)
=
 ( 3l )
2A
=
3
= 2R
20. PQ, is a piece of uniform wire of length 1 m
with a resistance of 10. Q is connected to an
ammeter, a 2  resistor and a 3 V battery. What
is the reading on the ammeter when the jockey
is at X?
Resistance in the wire
R is directly proportional to l
= 10 
100 cm
20
Hence, 20 cm = 100 (10)
R
= 2
Total resistance
2 + 2 = 4
Current, I
V
= R
3
= 4
= 0.75 A
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Teacher’s Guide
21. Figure shows the circuit used to investigate the relationship between potential
difference, V and current, I for a piece of constantan wire. The graph of V against I
from the experiment is as shown in the figure below.
(a)
What quantities are kept constant in this experiment?
Length // cross-sectional area // type of material // temperature of the wire
(b)
State the changes in the gradient of the graph, if
i) the constantan wire is heated
R , gradient  // the resistance increases, hence the gradient increases
ii) a constantan wire of a smaller cross-sectional area is used
R , gradient  // the resistance increases, hence the gradient increases
iii)a shorter constantan wire is used
R , gradient  // the resistance decreases, hence the gradient decreases
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Teacher’s Guide
7.3 SERIES AND PARALLEL CIRCUITS
Current Flow and Potential Difference in Series and Parallel Circuit
SERIES CIRCUIT
1
the current flows through each bulb/resistor is
PARALLEL CIRCUIT
1
the same
bulb/resistor
I = I1 = I2 = I3
2
the potential difference is the same across each
V = V1 = V2 = V3
the potential difference across each bulb /
2
the current passing through each bulb / resistor is
resistor depends directly on its resistance. The
inversely proportional to the resistance of the
potential difference supplied by the dry cells is
resistor. The current in the circuit equals to the
shared by all the bulbs / resistors.
sum of the currents passing through the bulbs /
resistors in its parallel branches.
V = V1 + V2 + V3
where V is the potential
difference across the
I = I1 + I2 + I3
where I is the total current
battery
3
If Ohm’s law is applied separately to each bulb /
from the battery
3
If Ohm’s law is applied separately to each bulb /
resistor, we get :
resistor, we get :
V = V1 + V2 + V3
IR = IR1 + IR2 + IR3
I = I1 + I2 + I3
V
V
V
V
R = R1 + R2 + R3
If each term in the equation is divided by I, we
If each term in the equation is divided by V, we
get the effective resistance
get the effective resistance
1
R
R = R1 + R2 + R3
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1
1
1
= R + R + R3
1
2
Teacher’s Guide
Identify series circuit or parallel circuit
(a)
(b)
(c)
(d)
Series
Parallel
A, B - series
Q, S - parallel
Ammeter reading ≡ Current
3A
0.5
1
5
3
2
2
1.5
Voltmeter reading ≡ Potential difference ≡ Voltage
5
6
1
2
2
3
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Teacher’s Guide
Effective resistance, R
R = 20 + 10 + 5= 35 
(a)
(b)
1/R
= ½ +1/5 + 1/10 = 4/5
Effective R = 1.25 
1/R
(c)
E
(d)
=1/16 + 1/8 + 1/8
=5/16
Effective R = 3.2 
1/R = 1/8 + 1/8= 1/8
R=4
Effective R = 20 + 10 + 4 = 34 
(e)
1/R = 1/4 + 1/2=3/4
R = 1.33 
Effective R = 1.33 + 1 = 2.33 
Effective R = 2+5+3+10
= 20 
1/R = 1/5 + 1/10=3/10
R = 3.33 
(g)
(i)
(f)
1/R = 1/4 + 1/12=1/3
R=3
Effective R = 3 + 2 = 5 
(h)
1/R = 1/20 + 1/20=1/10
R = 10 
Effective R = 10 + 10 + 5 =25 
(j)
1/R = 1/10 + 1/20=3/20
R = 8 + 20/3 = 14.67 
- 29 -
Teacher’s Guide
Solve problems using V = IR
V = IR
9 =I(18)
= 0.5
V = IR
240 = 6(R)
R =40 
20 
1/R = 1/10 + 1/10 =2/10
R=5
1/R = 1/5 + 1/20=1/4
R=4
V = IR
= 2(5) = 10 V
V = IR
12 =I(5)
= 2.4 A
Exercise 7.3
1.
The two bulbs in the figure have a resistance of 2 and 3
respectively. If the voltage of the dry cell is 2.5 V, calculate
(a) the effective resistance, R of the circuit
(b) the main current, I in the circuit
(c) the potential difference across each bulb.
2: V = IR = (0.5)(2) = 1V
3: V = IR = (0.5)(3) = 1.5 V
V = IR
2.5 =I(5)
= 0.5 A
2.
There are two resistors in the circuit shown. Resistor R1 has a
resistance of 1. If a 3V voltage causes a current of 0.5A to flow
through the circuit, calculate the resistance of R2.
V = IR
3=0.5(1+R2)
R2 = 5 
- 30 -
Teacher’s Guide
The electrical current flowing through each branch, I 1 and I2, is 5
3.
A. Both bulbs have the same resistance, which is 2. Calculate
the voltage supplied.
Parallelcircuit;V =V1=V2 = IR1 or
= IR2
= 5(2)
= 10 V
4.
The voltage supplied to the parallel is 3 V. R1 and R2
have a resistance of 5 and 20. Calculate
(a) the potential difference across each resistor
3 V (parallel circuit)
(b) the effective resistance, R of the circuit
1/R = 1/5 + 1/20 =1/4
R=4
(c) the main current, I in the circuit
(d) the current passing through each resistor
5:
V = IR
3 =I(4)
= 0.75 A
5.
20 :
V = IR
3 =I(20)
I = 0.15 A
In the circuit shown, what is the reading on the ammeter
when switch, S
(a) is open?
(b) is closed?
Effective R = 6 
V = IR
12 =I(6)
I=2A
6.
V = IR
3 =I(5)
I = 0.6 A
Effective R = 4 
V = IR
12 =I(4)
I=3A
Determine the voltmeter reading.
(a)
Determine the ammeter reading.
R = 12 
I = 24/12
= 2A
(a)
V= IR
= (2)(8)
= 16 V
(b)
R = 12 
I = 6/12
= 0.5A
V at 9 : V= IR
= (0.5)(9)
= 4.5 V
V reading : 6 – 4.5 = 1.5 V
- 31 -
R =9 
I = 4.5/9
= 0.5A
A reading : 0.5/2= 0.25 A
Notes: Divide 2 because
the resistors have similar
resistance.
Teacher’s Guide
7.
Calculate
(d) (i) The potential difference across 8
(a) The effective resistance, R
resistor.
R = 12 
V = IR
= 2(8) = 16 V
(b) The main current, I
I=2A
(ii) The potential difference across 2.5
(c) The current passing through 8 and 2.5
resistor.
V = IR
resistors.
I=2A
= 2(2.5) = 5 V
(e) The current passing through 6  resistor.
V = V8 + V2.5 +Vparallel
24 = 16 + 5 + Vparallel
Vparallel = 3V
V = IR
3 = I(6)
I = 0.5 A
- 32 -
Teacher’s Guide
8.
The electrical components in our household appliances are connected in a combination of series and
parallel circuits. The above figure shows a hair dryer which has components connected in series and
parallel. Describe how the circuit works.
Suggested answer






The hair dryer has three switches A, B and C
When switch A is switched on, the dryer will only blow air at ordinary room temperature
When switches A and B are both switched on, the dryer will blow hot air.
As a safety feature to prevent overheating, the heating element will not be switched on if the fan is
not switched on
The hair dryer has an energy saving feature. Switch C will switch on the dryer only when it is
held by the hand of user
The body of the hair dryer must be safe to hold and does not get hot easily
- 33 -
Teacher’s Guide
7.4 ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE
Electromotive force
Figure (a)
Figure (b)
Voltmeter reading,
potential difference, V < e.m.f.,
E
Voltmeter reading,
e.m.f.
E,r
R
Current flowing
No current flow
1. An electrical circuit is set up as shown in figure (a). A high resistance voltmeter is
connected across a dry cell which labeled 1.5 V.
a) Figure (a) is (an open circuit / a closed circuit)
b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up /
lights up)
c) The voltmeter reading shows the (amount of current flow across the dry cell / potential
difference across the dry cell)
d) The voltmeter reading is (0 V / 1.5 V / Less than 1.5 V)
e) The potential difference across the cell in open circuit is (0 V / 1.5 V / Less than 1.5 V).
Hence, the electromotive force, e.m.f., E is (0 V / 1.5 V / Less than 1.5 V)
f) It means, (0 J / less than 1.5 J / 1.5 J / 3.0 J) of electrical energy is required to move 1 C
charge across the cell or around a complete circuit.
- 34 -
Teacher’s Guide
2. The switch is then closed as shown in figure (b).
a) Figure (b) is (an open circuit / a closed circuit)
b) There is (current flowing / no current flowing) in the circuit. The bulb (does not light up /
lights up)
c) The voltmeter reading is the (potential difference across the dry cell / potential difference
across the bulb / electromotive force).
d) The reading of the voltmeter when the switch is closed is (lower than/ the same as /
higher than) when the switch is open.
e) If the voltmeter reading in figure (b) is 1.3 V, it means, the electrical energy dissipated by
1C of charge after passing through the bulb is (0.2 J / 1.3 J / 1.5 J)
f) The potential difference drops by (0.2 V/ 1.3 V / 1.5 V). It means, the potential difference
lost across the internal resistance, r of the dry cell is (0.2 V/ 1.3 V / 1.5 V).
g) State the relationship between e.m.f , E , potential difference across the bulb, VR and drop
in potential difference due to internal resistance, Vr.
Electromotive force, e.m.f., E = Potential Difference + Drop in Potential Difference
across resistor, R
= VR + Vr
due to internal resistance,r
where VR = IR and Vr = Ir
= IR + Ir
= I (R + r)
- 35 -
Teacher’s Guide
3.
a) Why is the potential difference across the resistor not the same as the e.m.f. of the
battery?
The potential drops as much as 0.4
V across the internal resistance
b) Determine the value of the internal resistance.
Since
E
=
1.5 =
r
=
V
+
1.1 +
Ir
0.5 r
0.8 
Therefore, the value of the internal resistance is
0.8 
c) Determine the value of the external resistor.
Since
V
=
1.1 =
R
=
IR
0.5 R
2.2 
Therefore, the value of the external resistance is
- 36 -
2.2 
Teacher’s Guide
Activity :
To determine the values of the electromotive force (e.m.f.) and
the internal resistance, r of the cell
Voltmeter
V
Internal resistance
+
Dry cell
-
Ammeter
Switch
Rheostat
To determine the values of the electromotive force (e.m.f.) and
Aim
the internal resistance, r of the cell
Apparatus /
Dry cells holder, ammeter (0 – 1 A), voltmeter(0 – 5 V), rheostat (0 – 15 ), connecting
materials
wires, switch, and 2 pieces of 1.5 V dry cell.
Method
:
a)
Set up the circuit as shown in the figure.
b) Turn on the switch, and adjust the rheostat to give a small reading of the
ammeter, I, 0.2 A.
c)
Read and record the readings of ammeter and voltmeter respectively
d) Adjust the rheostat to produce four more sets of readings, I = 0.3 A, 0.4 A, 0.5
A and 0.6 A.
Tabulation of
data
:
Current,I/A
Volt, V/V
0.2
2.6
0.3
2.5
0.4
2.4
0.5
2.2
0.6
2.0
0.7
1.9
- 37 -
Teacher’s Guide
Analysis of data
:
Potential difference, V /V
Draw a graph of
V against I
3.0 2.0 1.0 -
0.2
Discussion
:
0.4
0.6
Current, I /A
0.8
1. From the graph plotted, state the relationship between the potential difference, V
across the cell and the current flow, I?
The potential difference, V across the cell decreases as the current flow increases.
2. A cell has an internal resistance, r. This is the resistance against the movement of
the charge due to the electrolyte in the cell. With the help of the figure, explain the
result obtained in this experiment.
When the current flowing through the circuit increases, the quantity of charge
flowing per unit time increased. Hence, more energy was lost in moving a larger
amount of charge across the electrolyte. Because of this, there was a bigger drop
in potential difference measured by the voltmeter.
3. By using the equation E = V + Ir
(a) write down V in terms of E, I and r.
V = -rI + E
(b) explain how can you determine the values of E and r from the graph plotted
in this experiment.
E = the vertical intercept of the V – I graph
R = the gradient of the V – I graph
(c) determine the values of E and r from the graph.
By extrapolating the graph until it cuts the vertical axis,
E = 2.9 V
r = - gradient
= 1.4 
- 38 -
Teacher’s Guide
Exercise 7.4
1
A voltmeter connected directly across a battery gives a reading of 1.5 V.
The voltmeter reading drops to 1.35 V when a bulb is connected to the
battery and the ammeter reading is 0.3 A. Find the internal resistance of
the battery.
E = 3.0 V, V = 1.35 V, I = 0.3 A
Substitute in :
E = V + Ir
1.5 = 1.35 + 0.3(r)
r = 0.5 
2. A circuit contains a cell of e.m.f 3.0 V and internal resistance, r. If the external resistor has a value of 10.0
 and the potential difference across it is 2.5 V, find the value of the current, I in the circuit and the internal
resistance, r.
E = 3.0 V, R = 10 , V = 2.5 V
Calculate current : V = IR
Calculate internal resistance : E = I(R + r)
r = 2.0 
3
A simple circuit consisting of a 2 V dry cell with an internal resistance of 0.5. When the switch is
closed, the ammeter reading is 0.4 A.
Calculate
(a) the voltmeter reading in open circuit
The voltmeter reading = e.m.f. = 2 V
(b) the resistance, R
(c) the voltmeter reading in closed circuit
E = I(R + r)
2
V = IR
= 0.4(R + 0.5)
= 0.4 (4.5)
R = 4.5 
4
= 1.8 V
Find the voltmeter reading and the resistance, R of the
resistor.
E = V + Ir
e.m.f.
12 = V + 0.5 (1.2)
V = 11.4 V
V = IR
11.4 = 0.5 (R)
R = 22.8 
- 39 -
Teacher’s Guide
5
/V
A cell of e.m.f., E and internal resistor, r is connected to
a rheostat. The ammeter reading, I and the voltmeter
6
reading, V are recorded for different resistance, R of the
rheostat. The graph of V against I is as shown.
2
From the graph, determine
a)
the electromotive force, e.m.f., E
E = V + Ir
Rearrange
:V = E - Ir
Equivalent
: y = mx + c
/A
2
b) the internal resistor, r of the cell
r = - gradient
= - (6 - 2)
2
=2
Hence, from V – I graph : E = c = intercept of V-axis
=6V
6
V/V
The graph V against I shown was obtained from an experiment.
1.5
a) Sketch a circuit diagram for the experiment
0.2
5
1/A
b) From the graph, determine
i) the internal resistance of the battery
ii) the e.m.f. of the battery
r = -gradient
E = c = intercept of V-axis
= 0.26
7
= 1.5 V
A graph of R against 1/I shown in figure was obtained
R/
from an experiment to determine the electromotive force,
1.3
e.m.f., E and internal resistance, r of a cell. From the
graph, determine
a)
the internal resistance of the cell
E = I(R + r)
-1
0.5
- 0.2
1 (A )
Rearrange
I
:R=
E
- r,
I
Hence, r = -gradient = -(-0.2) = 0.2
b) the e.m.f. of the cell
e.m.f. = gradient = 3 V
- 40 -
Teacher’s Guide
7.5 ELECTRICAL ENERGY AND POWER
Electrical Energy
1.
Energy Conversion
battery
(chemical energy)
(a)
current
(b)
current
battery
(chemical energy)
current
current
Light and heat
energy
Energy Conversion:
Electrical energy  Light energy
+ Heat energy
Energy Conversion:
Electrical energy  Kinetic
energy
2. When an electrical appliance is switched on, the current flows and the electrical energy
supplied by the source is transformed to other forms of energy.
3. Therefore, we can define electrical energy as : The energy carried by electrical charges
which can be transformed to other forms of energy by the operation of an electrical
appliance.
- 41 -
Teacher’s Guide
Electrical Energy and Electrical Power
1. Potential difference, V across two points is the energy,E dissipated or transferred by a
coulomb of charge, Q that moves across the two points.
2. Therefore,
Potential difference, V = Electrical energy dissipated, E
Charge, Q
3. Hence, E = VQ
4. Power is defined as the rate of energy dissipated or transferred.
5. Hence, Power, P = Energy dissipated, E
time, t
Electrical Energy, E
From the definition of potential
difference, V
Electrical Power, P
Power is the rate of transfer of electrical energy,
P= E
t
V= E
Q
Electrical energy converted, E
E = VQ
Hence,
E = VI t
2
Hence,
E=IR
Hence,
2
V
t
E=
R
SI unit : Joule (J)
P = VQ
t
; where Q = It
; where V = IR
P = VI
2
P= I R
t
; where I = V
R
P = I2 R
SI unit : Joule per
- 42 -
second // J s-1 // Watt(W)
Teacher’s Guide
Power Rating and Energy Consumption of Various Electrical Appliances
1. The amount of electrical energy consumed in a given period of time can be calculated by
Energy consumed
E
=
Power rating x
Time
=
Pt
energy, E is in Joules
where
power, P is in watts
time, t is in seconds
2. The unit of measurement used for electrical energy consumption is the
kilowatt-hour, kWh.
1 kWh
=
1000 x 3600 J
=
3.6 x 106 J
=
1 unit
3. One kilowatt-hour is the electrical energy dissipated or transferred by a 1 kW device in
one hour
4. Household electrical appliances that work on the heating effect of current are usually
marked with voltage, V and power rating, P.
5. The energy consumption of an electrical appliance depends on the power rating and the
usage time, E = Pt
6. Power dissipated in a resistor, three ways to calculate:
R= 100, I=0.5 A, P=?
P = I2R
= (0.5)2 100
= 25 watts
R= 100, V=50 W, P=?
P = (V/R)2 R
= V2/R
= (50)2 /100
= 2500/100
= 25 watts
- 43 -
V=50 V, I=0.5 A, P=?
P = I2(V/I)
= IV
= (0.50)50
= 25 watts
Teacher’s Guide
Cost of energy
Appliance
Quantity
Power / W
Power / kW
Time
Energy
Consumed
(kWh)
Bulb
5
60
0.06
8 hours
2.4
Refrigerator
1
400
0.4
24 hours
9.6
Kettle
1
1500
1.5
3 hours
4.5
Iron
1
1000
1.0
2 hours
2.0
Total energy consumed, E
= (2.4 + 9.6 + 4.5 + 2.0)
= 18.5 kWh
= 18.5 kWh x RM 0.28
Cost
= RM 5.18
Comparing Various Electrical Appliances in Terms of Efficient Use of Energy
1. A tungsten filament lamp changes electrical energy to
useful light energy and unwanted heat energy
2. A fluorescent lamp or an ‘energy saving lamp’
produces less heat than a filament lamp for the same
amount of light produced.
3. a) Efficiency of a filament lamp :
Efficiency
=
Output power x 100
Input power
=
3 x 100
60
=
5%
b) Efficiency of a fluorescent lamp and an ‘energy
saving lamp’
Efficiency
=
Output power x 100
Input power
=
3 x 100
12
=
25 %
- 44 -
Teacher’s Guide
Exercise 7.5
1. How much power dissipated in the bulb?
R = 10
(a)
P
=
=
5V
(b)
R = 10
P
=
=
5V
I
V= 15V
R1=2
V2
R
52 / 10
2.5 W
=
R = 10
2.
V2
R
52 / 10
2.5 W
=
R2=4
R3=4
Calculate
(a) the current, I in the circuit
Total resistance, R
V
I
=
=
=
=
(b) the energy released in R 1 in 10 s.
= (2 + 4 + 4) 
= 10 
E
IR
V/R
15 / 10
1.5 A
(b) the electrical energy supplied by the battery in 10 s.
E
=
=
=
I2Rt
(1.5)2 (10)(10)
225 J
- 45 -
=
=
=
I2Rt
(1.5)2 (2)(10)
45 J
Teacher’s Guide
2. A lamp is marked “12 V, 24 W”. How many joules of electrical energy does it consume
in an hour?
E
=
=
=
Pt
24 (1 x 60 x 60)
86 400 J
3. A current of 5A flows through an electric heater when it is connected to the 24 V mains
supply. How much heat is released after 2 minutes?
E
=
=
=
VI t
24 (5) (2 x 60)
144 000 J
4. An electric kettle is rated 240 V 2 kW. Calculate the resistance of its heating element and
the current at normal usage.
P
I
=
=
=
=
IV
P/V
2000 / 240
8.3 A
5. An electric kettle operates at 240 V and carries current of 1.5 A.
(a) How much charge will flow through the heating coil in 2 minutes.
Q
=
=
=
It
(1.5) (2 x 60)
180 C
(b) How much energy will be transferred to the water in the kettle in 2 minutes?
E
=
=
=
QV
180 (240)
4.32 kJ
(c) What is the power dissipated in the kettle?
P
=
=
=
IV
1.5 (240)
360 W
- 46 -
Teacher’s Guide
6. An electric kettle is labeled 3 kW, 240 V.
(a) What is meant by the label 3 kW, 240 V?
The electric kettle dissipates electrical power 3 kW if it operates at 240 V
(b) What is the current flow through the kettle?
P =
3000 =
I =
IV
I (240)
12.5 A
(c) Determine the suitable fuse to be used in the kettle.
12 A
(d) Determine the resistance of the heating elements in the kettle.
P =
3000 =
R =
I2 R
(12.5)2 R
19.2 
7. Table below shows the power rating and energy consumption of some electrical appliances
when connected to the 240 V mains supply.
Appliance
Quantity
Power rating / W
Time used per day
Kettle jug
1
2000
1 hour
Refrigerator
1
400
24 hours
Television
1
200
6 hours
Lamp
5
60
8 hours
Electricity cost: RM 0.218 per kWh
Calculate
(a) Energy consumed in 1 day
Energy consumed
Kettle jug,
= Quantity x Power rating (kW) x Time used
=1x2x1
= 2 kWh
Refrigerator
= 1 x 0.4 x 24 = 9.6 kWh
Television
= 1 x 0.2 x 6
= 1.2 kWh
Lamp
= 5 x 0.06 x 8
= 2.4 kWh
Total energy consumed
= 15.2 kWh
- 47 -
Teacher’s Guide
(b) How much would it cost to operate the appliances for 1 month?
= 16.58 kWh x 30 x RM 0.218
Cost
= RM 108.43
8. A vacuum cleaner consumes 1 kW of power but only delivers 400 J of useful work per
second. What is the efficiency of the vacuum cleaner?
Efficiency
=
Output power x 100 %
Input power
400 x 100 %
1000
40 %
=
=
9. An electric motor is used to lift a load of mass 2 kg to a height 5 m in 2.5 s. If the supply
voltage is 12 V and the flow of current in the motor is 5.0 A, calculate
(a) Energy input to the motor
E
=
=
=
VIt
12 (5.0) (2.5)
150 J
(b) Useful energy output of the motor
U
=
=
=
mgh
2 (9.8) (5)
98 J
(c) Efficiency of the motor
Efficiency
=
=
=
Output power x 100 %
Input power
98 x 100 %
150
65.3 %
- 48 -
Teacher’s Guide
Reinforcement Exercise Chapter 7
Part A: Objective Questions
1. What is the unit of electric charge?
3. Which of the following graphs shows
A. Ampere, A
the correct relationship between the
B. kelvin,K
potential difference, V and current, I
C. Coulomb, C
for an ohmic conductor?
D. Volt, V
A.
2. Which of the following diagrams
shows the correct electric field?
A.
B.
B.
C.
C.
D.
- 49 -
Teacher’s Guide
4. A small heater operates at 12 V, 2A.
How much energy will it use when it is
run for 5 minutes?
A. 90 J
C.
B. 120 J
C. 1800 J
D. 7200 J
D.
5. The electric current supplied by a
battery in a digital watch is 3.0 x 10-5
A. What is the quantity of charge that
flows in 2 hours?
A. 2.5 x 10-7 C
B. 1.5 x 10-5 C
7. Why is the filament made in the
C. 6.0 x 10-5 C
shape of a coil?
D. 3.6 x 10-3 C
A. To increase the length and produce
E. 2.2 x 10-1 C
a higher resistance.
B. To increase the current and produce
6. Which of the following circuits can be
used to determine the resistance of the
bulb?
more energy.
C. To decrease the resistance and
produce higher current
A.
D. To decrease the current and produce
a higher potential difference
8. Which of the following will not
affect the resistance of a conducting
wire.
B.
A. temperature
B. length
C. cross-sectional area
D. current flow through the wire
- 50 -
Teacher’s Guide
9. The potential difference between two
12. Which two resistor combinations have
points in a circuit is
the same resistance between X and Y?
A. the rate of flow of the charge from
one point to another
B. the rate of energy dissipation in
moving one coulomb of charge
from one point to another
C. the work done in moving one
coulomb of charge from one point
to another
D. the work done per unit current
flowing from one point to another
10.
A. P and Q
B. P and S
An electric kettle connected to the
C. Q and R
240 V main supply draws a current
D. R and S
of 10 A. What is the power of the
E.
kettle?
A. 200 W
B. 2000 W
C. 2400 W
D. 3600 W
E. 4800 W
13. In the circuit above, what is the
11. An e.m.f. of a battery is defined as
ammeter reading when the switch S
is turned on?
A. the force supplied to 1 C of charge
A. 1.0 A
B. the power supplied to 1 C of charge
B. 1.5 A
C. the energy supplied to 1 C of
C. 2.0 A
charge
D. 9.0 A
D. the pressure exerted on 1 C of
E. 10.0 A
charge
- 51 -
Teacher’s Guide
14. A 2 kW heater takes 20 minutes to
16. An electric motor lifts a load with a
heat a pail of water. How much
potential difference 12 V and fixed
energy is supplied by the heater to
current 2.5 A. If the efficiency of the
the water in this period of time?
motor is 80%, how long does it take
A. 1.2 x 106 J
to lift a load of 600 N through a
B. 1.8 x 106 J
vertical height of 4 m
C. 2.4 x 106 J
A. 20 s
D. 3.6 x 106 J
B. 40 s
E. 4.8 x 106 J
C. 60 s
D. 80 s
15. All bulbs in the circuits below are
E. 100 s
identical. Which circuit has the
smallest effective resistance?
17. The kilowatt-hour (kWh) is a unit of
A.
measurement of
A. Power
B. Electrical energy
C. Electromotive force
B.
C.
D.
18. The circuit above shows four
identical bulbs to a cell 6 V. Which
bulb labeled A, B, C and D is the
brightest?
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Teacher’s Guide
19. A 24  resistor is connected across
the terminals of a 12 V battery.
20. Which of the following quantities can
Calculate the power dissipated in the
be measured in units of JC-1
resistor.
A. Resistance
A. 0.5 W
B. Potential difference
B. 2.0 W
C. Electric current
C. 4.0 W
D. 6.0 W
E. 8.0 W
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Teacher’s Guide
Part B: Structured Questions
1.
Gradient
0 .6  0 .2
3 .6  0
-1
= 0.11 A V
=
The figure above shows a graph of electric current against potential difference for three
different conductors X, Y and Z.
(a) Among the three conductors, which conductor obeys Ohm’s law?
Conductor Y
(b) State Ohm’s law.
The potential difference across a conductor is directly proportional to the current that
flows through it, if the temperature and other physical quantities are kept constant.
(c) Resistance, R is given by the formula R = V/I. What is the resistance of X when the
current flowing through it is 0.4 A? Show clearly on the graph how is the answer
obtained.
From the graph I against V;
resistance, R
= reciprocal of gradient, 1/gradient
1
= 0.11
= 9.09 
(d) Among X, Y and Z, which is a bulb? Explain your answer.
X, because as I increases, the gradient decreases. Hence, the resistance X increases
as I increases which is a characteristic of a bulb.
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Teacher’s Guide
2.
The figure below shows an electric kettle connected to a 240 V power supply by a
flexible cable. The kettle is rated “240 V, 2500 W”.
The table below shows the maximum electric current that is able to flow through
wires of various diameters.
(a)
diameter of wire / mm
maximum current / A
0.80
8
1.00
10
1.20
13
1.40
15
What is the current flowing through the cable when the kettle is switched
on?
P = IV
I = P/V
(b)
= 2500 / 240 = 10.4 A
Referring to the table above,
i. What is the smallest diameter wire that can be safely used for this
kettle?
1.20 mm
ii. Explain why it is dangerous to use a wire thinner than the one selected
in b(i)
As resistance is inversely proportional to cross-sectional area,
a thinner wire will have a higher resistance thus the wire will
become very hot. This could probably cause a fire to break
out.
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Teacher’s Guide
(c)
State one precautionary measure that should be taken to ensure safe usage of
the kettle.
Do not operate kettle with wet hands.
(d)
Mention one fault that might happen in the cable that will cause the fuse in the
plug to melt.
Short circuit might occur if the insulating materials of the wires in the cable are
damaged.
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Teacher’s Guide
Part C: Essay Questions
1.
Figure 1 shows the reading of the voltmeter in a simple electric circuit
Figure 2 shows the reading of the same voltmeter
(a) What is meant by electromotive force (e.m.f.) of a battery?
(b) Referring to figure (a) and figure (b), compare the state of the switch, S, and
the readings of the voltmeter. State a reason for the observation on the
readings of the voltmeter.
(c) Draw a suitable simple electric circuit and a suitable graph, briefly explain
how the e.m.f. and the quantity in your reason in (b) can be obtained.
(d)
The figure above shows a dry cell operated torchlight with metal casing
(i)
What is the purpose of the spring in the torchlight?
(ii)
Why it is safe to use the torchlight although the casing is made of metal?
(iii)
What is the purpose of having a concave reflector in the torchlight?
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Teacher’s Guide
Suggested Answers
1. (a) The work done by a battery to move a unit charge around a complete circuit.
(b) - Switch in figure 1 is turned off
- Switch in figure 2 in turned on
- Reading of voltmeter in figure 1 is higher than in figure 2
- This is due to the presence of an internal resistance in the battery
(c)
Voltmeter
V
Internal resistance
+
Dry cell
Ammeter
Switch
Rheostat
Potential difference, V/V
emf
Currrent, I/A
0
(d)
e.m.f = intercept on the v-axis
internal resistance = -(gradient of the graph)
(i)
(ii)
(iii)
To improve the contact between the dry cells and the terminals of the
torchlight
Current flowing through the torchlight is very small, will not cause
electric shock
To converge the light rays to obtain increase the intensity of the light rays
projected by the torchlight.
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Teacher’s Guide
2. A group of engineers were entrusted to choose a suitable cable to be used as the
transmitting cable for a long distance electrical transmission through National Grid
Network.
Four different cables and their characteristic of the cables were given. The length and
diameter of all the cables are similar.
(a) Define the resistance of a conductor.
(b) The table below shows the characteristic of the four cables, A, B, C and D.
Resistivity /
m
Maximum load
before breaking/
N
Density /
kgm-3
Rate of
expansion
A
0.020
500
2800
Low
B
0.056
300
3200
Low
C
0.031
400
5600
Medium
D
0.085
200
3800
High
Base on the above table:
(i)
Explain the suitability of each characteristic of the table to be used for a long
distance electricity transmission
(ii)
Determine the most suitable wire and state the reason
(c) Suggest how three similar bulbs are arranged effectively in a domestic circuit.
Draw a diagram to explain your answer. Give two reasons for the arrangement.
(d) An electric kettle is rated 2.0 kW.
(i)
Calculate how long would it take to boil 1.5 kg of water from an initial
temperature of 280 C.
[specific heat capacity of water = 4200 J kg-1 0C-1]
(ii)
What is the assumption made in the calculations above?
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Teacher’s Guide
Suggested Answers
2.(a) Resistance is the ratio of potential difference to current flowing in an ohmic conductor.
(b)
Characteristics
Explanations
A low resistivity
Energy loss during transmission is reduced
Max load before
Mass or weight reduced. Can be supported by transmission
braking is high
tower
A low density
Cable will not slag when it heated during transmission
Cable A is chosen because it has low resistivity, high max load before breaking, low
density and low expansion rate.
(c) (i) If one bulb is burnt the others is still be lighted up
(ii) Each bulb can be switch on and off independently
(d) (i)
Pt = mcθ
(2000)(t)
t
(ii)
=
(1.5)(4200)(100-28)
=
226.8 s
No heat is lost to the surroundings and absorbed by the kettle
END OF MODULE CHAPTER 7
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CHAPTER 7: ELECTRICITY

Transcription

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