chapter 7 - electricity(students copy)
TRANSCRIPT
JPN Pahang Physics Module Form 5Student’s Copy 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 ……………….. and ………………….. at high voltage on its dome
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dome+ +
+
+
+
+
++
+
+
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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
………………………..
This causes the rubber belt to against the …….……… and hence becomes …..………
The charge is then carried by the moving belt up to the …………… ……….
where it is collected.
A large amount of ……………. is built up on the dome
……………………. 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 …………………….. of the
pointer of the microammeter.
This indicates an electric current
…………………….
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+++
++
++
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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,
…………………………… occurs.
An electric current ……………
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 …………. the hair
and the hair stand ………………..
This is because of each strand of hair
receives ……………….. charges and
…………………….. each other.
6. The flow of electrical charges produces ………………….
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++
+ + + ++
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Electric Current
1. Electric current consists of a flow of …………......
2. The more charges that flow through a cross
section within a given time, the ………………
is the current.
3. Electric current is defined as the
rate of flow of ………………………….
4. In symbols, it is given as:
where I = …………………….…
Q = …………………….…
t = ………….....................
(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)
(iv) By rearranging the above formula, Q = ( It / / )
4. If one coulomb of charge flows past in one second, then the current is ………………….
ampere.
5. 15 amperes means in ………………second, …………….. coulomb of charge through a
cross section of a conductor.
6. In a metal wire, the charges are carried by………………….
7. Each electron carries a charge of ………………………..
8. 1 C of charge is……………………………..
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I =
Each second, 15 coulombs of charge cross the plane. The current is I = 15 amperes. One ampere is one coulomb per second.
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Electric Field
a) An electric field is a ………………. in which an……………… experiences
a…………..
b) An electric field can be represented by a number of lines indicate both
the……………. and ……………….. of the field
c) The principles involved in drawing electric field lines are :
(i) electric field lines always extend from a ……………… - charged
object to a
………………..-charged object
(ii) electric field lines never ………………….. each other,
(ii) electric field lines are ……………….. in a …………………..
electric field.
Demo: 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)
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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.
Draw the pattern of the electric field lines.
ELECTRIC FIELD AROUND A POSITIVE CHARGE
ELECTRIC FIELD AROUND A NEGATIVE CHARGE
ELECTRIC FIELD AROUND A POSITIVE AND NEGATIVE CHARGE
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
ELECTRIC FIELD AROUND TWO NEGATIVE CHARGES
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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
ELECTRIC FIELD BETWEEN TWO CHARGED
PARALLEL PLATES
EFFECT OF AN ELECTRIC FIELD ON A POLYSTYRENE BALL
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
1. Place the polystyrene ball between the
two metal plates.
2. Switch on the E.H.T and displace the
polystyrene ball slightly so that it
touches one of the metal plates
Observation:
The polystyrene ball oscillated between the
two plates, touching one plate after
another.
Explanation:
negatively charged plate, the ball
receives negative charges from the plate
and experiences a repulsive force.
charged plate.
loses some of its negative charges to the
plate and becomes positively charged.
It then experiences a repulsive force. This
process continues.
EFFECT OF AN ELECTRIC FIELD ON A CANDLE FLAME
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
C)
1) Switch of the E.H.T and replace the
polystyrene ball with a lighted candle.
2) Sketch the flame observed when the
E.H.T. is switched on.
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.
Explanation:
molecules to become positive and
negative charges.
negative plate while the negative
charges are attracted to the positive
plate.
directions but more to the negative
plate.
negative charges. This causes the
uneven dispersion of the flame.
Conclusion
1. Electric field is a ………………………………………………………………………..
2. Like charges ………………. each other but opposite charges …………… each other.
3. Electric field lines are ……………………in an electric field. The direction of the
field lines is from …………………….. to …………………………
Exercise 7.1
1. 5 C of charge flows through a wire in 10 s. What is the current in the wire?
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
2. A charge of 300 C flow through a bulb in every 2 minutes. What is the electric
current in the bulb?
3. The current in a lamp is 0.2 A. Calculate the amount of electric charge that passes
through the lamp in 1 hour.
4. 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)
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?
Ideas of Potential Difference
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(a)
Pressure at point P is ………………
than the pressure at point Q
Water will flow from ……to ……
when the valve is opened.
This due to the ……………….. in the
pressure of water
(b)
Gravitational potential energy at X is ……………
than the gravitational potential energy at Y.
The apple will fall from …… to …… when the
apple is released.
This due to the …………………….. in the
gravitational potential energy.
(c) Similarly,
Point A is connected to…………………. terminal
Point B is connected to …………………. terminal
Electric potential at A is ……………. than the electric
potential at B.
Electric current flows from A to B, passing the bulb in the
circuit and ……………….. the bulb.
This is due to the electric ………………. between the two
terminals.
As the charges flow from A to B, work is done when electrical
energy is transformed to ………….and …………… energy.
The potential difference, V between two points in a circuit is
defined as ……………………………………………………...
…………………………………………………………………
…………………………………………………………………
The potential difference,V between the two points will be
given by:
where W is ………………………….
Q is ………………………….
Device and symbol
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A BBulb
V = =
P Q
X
Y
water
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
ammeterCells
voltmeter Switch
connecting wireConstantan wire // eureka wire
resistancebulb
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
Turn to next page1. Name the device used to measure 1. Name the device used to measure
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
electrical current.
2. (a) What is the SI unit for current?
(b) What is the symbol for the unit of
current?
3. How is an ammeter connected in an
electrical circuit?
4. The positive terminal of an ammeter
is connected to which terminal of the dry
cell?
5. What will happen if the positive
terminal of the ammeter is connected to the
negative terminal of the dry cell?
potential difference.
2. (a) What is the SI unit for potential
difference?
(b) What is the symbol for the unit of
potential difference?
3. How is an voltmeter connected in an
electrical circuit?
4. The positive terminal of a voltmeter is
connected to which terminal of the dry
cell?
Experiment: To investigate the relationship between current and potential difference
for an ohmic conductor.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(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),
(i) Make one suitable inference.
(ii) State one appropriate hypothesis that could be investigated.
(iii) Design an experiment to investigate the hypothesis.
(a) Inference The current flowing through the bulb is influenced by the potential difference across it.
(b) Hypothesis
(c) Aim To determine the relationship between current and potential difference for a
constantan wire.
(d) Variables
(i) manipul
ated variable
(ii) respondi
ng variable
(iii) fixed
variable
:
:
:
Apparatus /
materials
Method :
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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.
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.
Tabulation 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
Analysis of data : Draw a graph of V against I .
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 is a constant as current increases.
2. The resistance, R, of the wire used in the experiment is equal to the gradient of
the V-I graph. Determine the value of R.
3. 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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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Ohm’s Law
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(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 = constant
(c) The constant is known as ………………………………. of the conductor.
(d) The resistance, R is a term that describes ……………………………………………..
…………………………………………………………………………………………..
It is also defined as the ratio…………………………………………………………….
…………………………………………………………………………………………..
That is
R = and V =
(e) The unit of resistance is …………………………………
(f) An ……………………….. is one which obeys Ohm’s law, while a conductor which
does not obey Ohm’s law is known as a ……………………….conductor
Factors Affecting Resistance
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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) …………………………………………………………….
b) …………………………………………………………….
c) …………………………………………………………….
d) …………………………………………………………….
5. Write down the relevant hypothesis for the factors affecting the resistance in the table
below.
Factors Diagram Hypothesis Graph
Len
gth
of th
e co
nduc
tor,
l
The …………… the conductor, the …………….. its resistance
Resistance is …………………. proportional to the length of a conductor
The
cro
ss-s
ecti
onal
ar
ea o
f th
e co
nduc
tor,
A
The ……………….….. the cross - sectional area, the …….………… the its resistance
Resistance is ……………...…….. proportional to the cross-sectional area of a conductor
The
type
of
the
mat
eria
l of
the
cond
ucto
r
Different conductors with the same physical conditions have ……………………. resistance
The
tem
pera
ture
of
the
cond
ucto
r
The …………………. The temperature of a conductor, the …………………... the resistance
6. From, the following can be stated:
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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 where = resistivity of the
A A substance
Exercise 7.2
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
1. Tick (√) the correct answers
True False
(a) Unit of potential difference is J C-1
(b) J C-1 ≡ volt, V
(c)
The potential difference between two points is 1 volt if 1 joule
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
from one to another in an electric field.
(e) Potential difference ≡ Voltage
2. i) Electric charge, Q = ( It / / )
ii) Work done, W = (QV / / )
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 V5. 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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
= 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
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
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
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3 A A
20 V
Bulb
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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 = IR
= 0.5 (3)
R = 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 = gradient== 2.4
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?
From V-I graph, resistance = gradient== 2.0 x 10-3
14.Referring to the diagram on the right, calculate
(a) The current flowing through the resistor.
V = IR
12 = I (5)
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V/V
I/A0 5
1.2
12 V
5 I
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
I = 2.4 A
(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 = VIt
= 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
17. The graph shows the relationship between the
potential difference, V and current, I flowing
through two conductors, X and Y.
a) Calculate the resistance of conductor X.
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1.5 V+ - 1.5 V+ -
I/A
V/V
00
X
Y
2
8
2
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
From V-I graph, resistance = gradient=
= 4
b) Calculate the resistance of conductor Y.
From V-I graph, resistance = gradient=
= 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.
18. The graph shows a graph of I against V for three
conductors, P, Q and R.
i) Q Compare the resistance of
conductor P, Q and R.
ii) Explain your answer in (a)
From V-I graph, resistance = gradientThe greater the gradient, the greater the resistanceGradient of P > Gradient of Q > Gradient of R
19. Figure shows a wire P of length, l with a cross-
sectional area, A and a resistance, R. Another
wire, Q is a conductor of the same material with
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V/V
I/AP
Q
R
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
a length of 3l and twice the cross-sectional area
of P. What is resistance of Q in terms of R?
Conductor P R =
Conductor Q R’ = (notes: P and R have the same resistivity, ρ)
=
=
R
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
100 cm = 10
Hence, 20 cm = (10)
R = 2
Total resistance
2 + 2 = 4
Current, I =
=
= 0.75 A
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.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(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
7.3 SERIES AND PARALLEL CIRCUITS
Current Flow and Potential Difference in Series and Parallel Circuit
SERIES CIRCUIT PARALLEL CIRCUIT
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
1 the current flows through each bulb/resistor is
the same.
I = I1 = I2 = I3
2 the potential difference across each bulb /
resistor depends directly on its ………………….
The potential difference supplied by the dry cells
is shared by all the bulbs / resistors.
V = V1 + V2 + V3 where V is the potential
difference across the
battery
3 If Ohm’s law is applied separately to each bulb /
resistor, we get :
V = V1 + V2 + V3
IR = IR1 + IR2 + IR3
If each term in the equation is divided by I, we
get the effective resistance
R = R1 + R2 + + R3
1 the potential difference is the same across each
bulb/resistor
V = V1 = V2 = V3
2 the current passing through each bulb / resistor is
inversely proportional to the resistance of the
resistor. The current in the circuit equals to the
sum of the currents passing through the bulbs /
resistors in its parallel branches.
I = I1 + I2 + I3 where I is the total current
from the battery
3 If Ohm’s law is applied separately to each bulb /
resistor, we get :
I = I1 + I2 + I3
If each term in the equation is divided by V, we
get the effective resistance
Identify series circuit or parallel circuit
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VR
VR1
VR2
+=VR3
+
1R
1R1
1R2
+= + 1R3
I
V
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(a)
(b)
(c)
(d)
Ammeter reading ≡ Current
Voltmeter reading ≡ Potential difference ≡ Voltage
Effective resistance, R
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(a)(b)
(c)
(d)
(e) (f)
(g) (h)
(i) (j)
Solve problems using V = IR
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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
Effective R = 2 + 3 = 5
(b) the main current, I in the circuit (c) the potential difference across each bulb.
V = IR 2: V = IR = (0.5)(2) = 1V 2.5 =I(5) 3: V = IR = (0.5)(3) = 1.5 V = 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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
3. The electrical current flowing through each branch, I1 and I2, is 5
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
V = IR 5: V = IR 20 : V = IR 3 =I(4) 3 =I(5) 3 =I(20) = 0.75 A I = 0.6 A I = 0.15 A
5. In the circuit shown, what is the reading on the ammeter when switch, S
(a) is open? (b) is closed?
Effective R = 6 Effective R = 4 V = IR V = IR 12 =I(6) 12 =I(4) I = 2 A I = 3 A
6. Determine the voltmeter reading.
(a)
(b)
Determine the ammeter reading.
(a)
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
7.
Calculate
(a) The effective resistance, R
R = 12
(b) The main current, I
I = 2 A
(c) The current passing through 8 and 2.5
resistors.
I = 2 A
(d) (i) The potential difference across 8
resistor.
V = IR
= 2(8) = 16 V
(ii) The potential difference across 2.5
resistor.
V = IR
= 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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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.
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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
7.4 ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE
Electromotive force
Figure (a) Figure (b)
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)
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No current flow
R
Voltmeter reading,e.m.f.
Voltmeter reading,potential difference, V < e.m.f., E
E , r
Current flowing
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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.
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 due to internal resistance,r
= VR + Vr where VR = IR and Vr = Ir
= IR + Ir
= I (R + r)
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
3.
i. 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
ii. Determine the value of the internal resistance.
Since E = V + Ir
1.5 = 1.1 + 0.5 r
r = 0.8
Therefore, the value of the internal resistance is 0.8
iii. Determine the value of the external resistor.
Since V = IR
1.1 = 0.5 R
R = 2.2
Therefore, the value of the external resistance is 2.2
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Activity : To determine the values of the electromotive force (e.m.f.) and
the internal resistance, r of the cell
AimTo determine the values of the electromotive force (e.m.f.) and
the internal resistance, r of the cell
Apparatus /
materials
Dry cells holder, ammeter (0 – 1 A), voltmeter(0 – 5 V), rheostat (0 – 15 ), connecting
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
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V
Dry cell
Internal resistance
+ -
Switch
Rheostat
Ammeter
Voltmeter
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Analysis of data
Based on the
above data,
draw a graph of
V against I
:
Discussion : 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,
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
E = 2.9 V
r = - gradient
= 1.4
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) V = IR
2 = 0.4(R + 0.5) = 0.4 (4.5)
R = 4.5 = 1.8 V
4 Find the voltmeter reading and the resistance, R of the
resistor.
E = V + Ir
12 = V + 0.5 (1.2)
V = 11.4 V
V = IR
11.4 = 0.5 (R)
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e.m.f.
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
R = 22.8
5A cell of e.m.f., E and internal resistor, r is connected to
a rheostat. The ammeter reading, I and the voltmeter
reading, V are recorded for different resistance, R of the
rheostat. The graph of V against I is as shown.
From the graph, determine
a) the electromotive force, e.m.f., E b) the internal resistor, r of the cell
E = V + Ir r = - gradient
Rearrange : V = E - I r = - (6 - 2)
Equivalent : y = mx + c 2
Hence, from V – I graph : E = c = intercept of V-axis = 2
= 6 V
6The graph V against I shown was obtained from an experiment.
a) Sketch a circuit diagram for the experiment
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 = 1.5 V
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6
2
2/A
/ V
1/A
V / V
1.5
0.2
5
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
7 A graph of R against 1/I shown in figure was obtained
from an experiment to determine the electromotive force,
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)
Rearrange : R =
- r,
Hence, r = -gradient = -(-0.2) = 0.2
b) the e.m.f. of the cell
e.m.f. = gradient = 3 V
7.5 ELECTRICAL ENERGY AND POWER
Electrical Energy
1. Energy Conversion
(a) (b)
2. When an electrical appliance is switched on, the flows and the .............................. energy
supplied by the source is ................................... to other forms of energy.
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R/
1.3
- 0.2
0.51 (A-1)I
battery(chemical energy)
Light and heat energy
currentcurrent
Energy Conversion:Electrical energy Light energy + Heat energy
battery(chemical energy)
currentcurrent
Energy Conversion:Electrical energy Kinetic energy
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
3. Therefore, we can define electrical energy as :
................................................................................................................................................
................................................................................................................................................
................................................................................................................................................
Electrical Energy and Electrical Power
1. Potential difference, V across two points is the ............................ dissipated or
transferred by a coulomb of charge, Q that moves across the two points.
2. Therefore,
3. Hence,
4. Power is defined as the rate of energy dissipated or transferred.
5. Hence,
Electrical Energy, E Electrical Power, P
From the definition of potential
difference, V
Power is the rate of transfer of electrical energy,
Electrical energy converted, E
; where Q = It
Hence, ; where V = IR
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E = VQ
E = VI t
VQt
P =
P = VI
Electrical energy dissipated, ECharge, Q
Potential difference, V =
E = VQ
Energy dissipated, Etime, t
Power, P =
V = VQ VQt
P =
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Hence, ; where I = V R
Hence,
SI unit : Joule (J) SI unit : Joule per second // J s-1 // Watt(W)
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 = Power rating x Time
E = Pt where energy, E is in Joules
power, P is in watts
time, t is in seconds
2. The unit of measurement used for electrical energy consumption is the
………………………………………...
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 ….. kW device in
……... hour
4. Household electrical appliances that work on the heating effect of current are usually
marked with, ……………… and …………………..
5. The energy consumption of an electrical appliance depends on the ……………… and
the………………………., , E = Pt
6. Power dissipated in a resistor, three ways to calculate:
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E = I2R
V2 tRE =
I2 RtP =
P = I2 R
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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
V=50 V, I=0.5 A, P=?
P = I2(V/I) = IV = (0.50)50 = 25 watts
Cost of energy
Appliance Quantity Power / W Power / kW Time
Energy Consumed
(kWh)
Bulb 5 60 8 hours
Refrigerator 1 400 24 hours
Kettle 1 1500 3 hours
Iron 1 1000 2 hours
Electricity cost: RM 0.28 per kWh
Total energy consumed, E = (0.48 + 9.6 + 4.5 + 2.0)
= 16.58 kWh
Cost = 16.58 kWh x RM 0.28
= RM 4.64
Comparing Various Electrical Appliances in Terms of Efficient Use of Energy
1. A tungsten filament lamp changes ...............................to
useful ................ energy and unwanted ................energy
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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 100Input power
= 3 x 10060
= 5 %
b) Efficiency of a fluorescent lamp and an ‘energy
saving lamp’
Efficiency = Output power x 100Input power
= 3 x 10012
Exercise 7.5
1. How much power dissipated in the bulb?
(a)
(b)
2.
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5 V
R = 10
5 V
R = 10
R = 10
R1=2 R2=4 R3=4
V= 15V I
JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Calculate
(a) the current, I in the circuit (b) the energy released in R 1 in 10 s.
(b) the electrical energy supplied by the battery in 10 s.
2. A lamp is marked “12 V, 24 W”. How many joules of electrical energy does it consume
in an hour?
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?
4. An electric kettle is rated 240 V 2 kW. Calculate the resistance of its heating element and
the current at normal usage.
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.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(b) How much energy will be transferred to the water in the kettle in 2 minutes?
(c) What is the power dissipated in the kettle?
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?
(c) Determine the suitable fuse to be used in the kettle.
12 A
(d) Determine the resistance of the heating elements in the kettle.
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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Lamp 5 60 8 hours
Electricity cost: RM 0.218 per kWh
Calculate
(a) Energy consumed in 1 day
Energy consumed = Quantity x Power rating (kW) x Time used
Kettle jug, = 1 x 2 x 1 = 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
(b) How much would it cost to operate the appliances for 1 month?
Cost = 16.58 kWh x 30 x RM 0.218
= 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?
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
(b) Useful energy output of the motor
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(c) Efficiency of the motor
Reinforcement Exercise Chapter 7
Part A: Objective Questions
1. What is the unit of electric charge?
A. Ampere, A
B. kelvin,K
C. Coulomb, C
D. Volt, V
2. Which of the following diagrams
shows the correct electric field?
A.
B.
C.
3. Which of the following graphs shows
the correct relationship between the
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
potential difference, V and current, I
for an ohmic conductor?
A.
B.
C.
D.
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
B. 120 J
C. 1800 J
D. 7200 J
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
C. 6.0 x 10-5 C
D. 3.6 x 10-3 C
E. 2.2 x 10-1 C
6. Which of the following circuits can be used to determine the resistance of the bulb?
A.
B.
C.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
D.
7. Why is the filament made in the
shape of a coil?
A. To increase the length and produce
a higher resistance.
B. To increase the current and produce
more energy.
C. To decrease the resistance and
produce higher current
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.
A. temperature
B. length
C. cross-sectional area
D. current flow through the wire
9. The potential difference between two
points in a circuit is
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. An electric kettle connected to the
240 V main supply draws a current
of 10 A. What is the power of the
kettle?
A. 200 W
B. 2000 W
C. 2400 W
D. 3600 W
E. 4800 W
11. An e.m.f. of a battery is defined as
A. the force supplied to 1 C of charge
B. the power supplied to 1 C of charge
C. the energy supplied to 1 C of
charge
D. the pressure exerted on 1 C of
charge
12. Which two resistor combinations have
the same resistance between X and Y?
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
A. P and Q
B. P and S
C. Q and R
D. R and S
E.
13. In the circuit above, what is the
ammeter reading when the switch S
is turned on?
A. 1.0 A
B. 1.5 A
C. 2.0 A
D. 9.0 A
E. 10.0 A
14. A 2 kW heater takes 20 minutes to
heat a pail of water. How much
energy is supplied by the heater to
the water in this period of time?
A. 1.2 x 106 J
B. 1.8 x 106 J
C. 2.4 x 106 J
D. 3.6 x 106 J
E. 4.8 x 106 J
15. All bulbs in the circuits below are
identical. Which circuit has the
smallest effective resistance?
A.
B.
C.
D.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
16. An electric motor lifts a load with a
potential difference 12 V and fixed
current 2.5 A. If the efficiency of the
motor is 80%, how long does it take
to lift a load of 600 N through a
vertical height of 4 m
A. 20 s
B. 40 s
C. 60 s
D. 80 s
E. 100 s
17. The kilowatt-hour (kWh) is a unit of
measurement of
A. Power
B. Electrical energy
C. Electromotive force
18. The circuit above shows four
identical bulbs to a cell 6 V. Which
bulb labeled A, B, C and D is the
brightest?
19. A 24 resistor is connected across
the terminals of a 12 V battery.
Calculate the power dissipated in the
resistor.
A. 0.5 W
B. 2.0 W
C. 4.0 W
D. 6.0 W
E. 8.0 W
20. Which of the following quantities can
be measured in units of JC-1
A. Resistance
B. Potential difference
C. Electric current
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
Part B: Structured Questions
1.
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.
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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
=
= 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.
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.
diameter of wire / mm maximum current / A
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
0.80 8
1.00 10
1.20 13
1.40 15
(a) What is the current flowing through the cable when the kettle is switched
on?
P = IV
I = P/V = 2500 / 240 = 10.4 A
(b) 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.
(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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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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?
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(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?
Answer
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)
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
e.m.f = intercept on the v-axisinternal resistance = -(gradient of the graph)
(d)(i) To improve the contact between the dry cells and the terminals of the
torchlight(ii) Current flowing through the torchlight is very small, will not cause
electric shock(iii) To converge the light rays to obtain increase the intensity of the light rays
projected by the torchlight.
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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
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?
Answer
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
braking is high
Mass or weight reduced. Can be supported by transmission
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
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JPN Pahang Physics Module Form 5Student’s Copy Chapter 7: Electricity
(d) (i) Pt = mcθ
(2000)(t) = (1.5)(4200)(100-28)
t = 226.8 s
(ii) No heat is lost to the surroundings and absorbed by the kettle
END OF MODULE
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