12405942 chapter 7 electricity teachers guide 2009

8/8/2019 12405942 Chapter 7 Electricity Teachers Guide 2009

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JPN Pahang Physics Module Form 5Teachers 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 thatproduces andstore electric charges at high voltage on its dome

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dome

rubber belt

Metal dome

roller

roller

motor

+ ++

+

+

+

++

+

+


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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,

sparklingoccurs.

An electric currentflow.

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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Electric Current

1. Electric current consists of aflow of electrons

2. The more charges that flow through a cross

section within a given time, the largeris the

current.

3. Electric current is defined as the

rate of flow of electric charge

4. In symbols, it is given as:

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)

(iv) By rearranging the above formula, Q = (It /t

I/

I

t)

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 of1.6 x 10-19 C.

8. 1 C of charge is 6.25 x 1018 electrons.

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I = Q

Each second, 15 coulombs of charge cross theplane. The current is I = 15 amperes. Oneampere is one coulomb per second.


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Electric Field

a) An electric field is a region in which an electric chargeexperiences aforce.

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 apositively-chargedobject to a

negatively-chargedobject to infinity, or from infinity to a negatively-charged object,

(ii) electric field lines nevercross each other,

(iii) electric field lines are closerin astrongerelectric 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

DEMOA)

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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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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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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EFFECT OF AN ELECTRIC FIELD ON A POLYSTYRENE BALL

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 thetwo 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.

The ball will then move to the positively

charged plate.

When the ball touches the plate, the

ball loses some of its negative charges

to the plate and becomes positively

charged.

It then experiences a repulsive force.

This process continues.

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EFFECT OF AN ELECTRIC FIELD ON A CANDLE FLAME

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:

The heat of the flame ionizes the air

molecules to become positive and

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 repeleach other but opposite charges attracteach other.

3. Electric field lines are lines of force in an electric field. The direction of the field

lines is frompositive to negative.

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Exercise 7.1

1. 5 C of charge flows through a wire in 10 s. What is the current in the wire?

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)

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?

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Q = It

I = Q/t

= 5 / 10= 0.5 A

Q = It

I = Q/t = 300 / 120

= 2.5 A

Q = It= 0.2 (60 x 60)

= 720 C

Q = It

= 0.8 (60) Convert: 1 minute = 60s= 48 C

1.6 x 10 -19 C of charge 1 electron.

Hence, 48 C of charges is brought by 48 C = 3 x 1020 electrons

1.6 x 10 -19 C

a) Q = It

= 200 x 10-3 (3)

= 0.6 C

b) 1.6 x 10 -19 C of charge 1 electron.

Hence, 0.6 C of charges is 0.6 C = 3.75 x 1018 electrons

1.6 x 10 -19 C


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Ideas of Potential Difference

(a)

(b)

Pressure at point P isgreaterthan the pressure

at point Q

Water will flow fromPto Q when the valve is

opened.

This due to the difference in the pressure of

water

Gravitational potential energy at X isgreaterthan

the gravitational potential energy at Y.

The apple will fall fromXto Ywhen the apple is

released.

This due to the difference in the gravitational

potential energy.

(c) Similarly,

Point A is connected topositiveterminal

Point B is connected to negativeterminal

Electric potential at A is greater than the electric potential at

B.

Electric current flows from A to B, passing the bulb in the

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 lightand heatenergy.

The potential difference, Vbetween two points in a circuit is

defined as the amount of work done, Wwhen 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:

where W is work or energy in Joule (J)

Q is charge in Coulomb (C)

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A B

Bulb

V=echQuantityof

Work

arg=

Q

W

P Q


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Device and symbol

ammeter

Cells

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

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A

V


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1. Name the device used to measure

electrical current.

An ammeter

2. (a) What is the SI unit for current?

Amperes

(b) What is the symbol for the unit of

current?

3. How is an ammeter connected in an

electrical circuit?

In series

4. The positive terminal of an ammeter is

connected to which terminal of the dry

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.

1. Name the device used to measure

potential difference.

A voltmeter

2. (a) What is the SI unit for potential

difference?

Volts

(b) What is the symbol for the unit of

potential difference?

3. How is an voltmeter connected in an

electrical circuit?

In parallel

4. The positive terminal of a voltmeter is

connected to which terminal of the dry

cell?

Positive

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A

V


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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),

(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) HypothesisThe higher the current flows through a wire, the higher the potential difference across

it.

(c) AimTo determine the relationship between current and potential difference for a

constantan wire.

(d) Variables

(i) manipulated variable

(ii) responding variable

(iii) fixed variable

:current, I

:potential difference, V

: length of the wire // cross sectional area //

temperatureApparatus /

materials

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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.

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/V0.2 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 VagainstI

Discussion : 1. From the graph plotted.

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Potential difference, V /V

Current, I /A

3.0 -

2.0 -

1.0 -

0.2 0.4 0.6 0.8

4.0 -


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(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 proportionalto the

current, I.

(c) Does the gradient change as the current increases?

The gradient the ratio ofI

Vis a constantas current increases.

2. 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.

7.

5.3

o

= 5

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 conductorincreases when the current, I passing

through it increases as long as the conductor is kept at constant temperature.

Ohms Law

(a)

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Ohms law states

that the electric current, I flowing through a conductor is directly proportional to

the potential difference across the ends of the conductor,

iftemperature and other physical conditions remain constant

(b) By Ohms law: V I

= constant I

orI

V= constant

(c) The constant is known as resistance,Rof 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

R =I

Vand V = I R

(e) The unit of resistance is volt per ampere (V A-1) or ohm ()

(f) An ohmic conductoris one which obeys Ohms law, while a conductor which does not

obey Ohms law is known as a non-ohmic conductor

Factors Affecting Resistance

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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.

Factors Diagram Hypothesis Graph

Lengthofthe

ndu

ctor,

l

The longer the conductor, the

higher its resistance

Resistance is directlyproportional to the length of a

conductor

Thecross-sectional

tor,

The bigger the cross-sectional

area, the lower the its resistance

Resistance is inverselyproportional to the cross-

sectional area of a conductor

Thetypeofthe

alofthe

Different conductors with the

same physical conditions havedifferent resistance

Thetemperatureofthe

The higher temperature of

conductor, the higher theresistance

6. From, the following can be stated:

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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 theA A substance

Exercise 7.2

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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/t

I/

I

t)

ii) Work done, W= (QV/Q

V/

V

Q)

iii) Base on your answer in 2(i) and (ii) derive the work done, Win terms ofI, Vand t.

W = QV

= ItV

3. If a charge of 5.0 C flows through a wire and the amount of electrical energy convertedinto 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 chargespassed through it and 25 J of electrical energy is converted to light and heat energy. Whatis the potential difference across the bulb?

W = QV

25 = 5 (V)

V = 5 V

5. The potential difference of 10 V is used to operate an electric motor. How much work isdone in moving 3 C of electric charge through the motor?

W = QV

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= 3 (10)

= 30 J

6. When the potential difference across a bulb is20 V, the current flow is 3 A. How much workdone to transform electrical energy to light andheat energy in 50 s?

W = VIt

= 20 (3) (50)

= 3000 J

7. What is the potential difference across a light bulbof 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 of2.0 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, ifthe dry cells supply 2.0 V and the ammeterreading 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?

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3 A

20

Bulb


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V = IR

3.0 = I (6)

I = 0.5 A

11. If a current of 0.5 A flows through the resistor of3 in the figure, calculate the voltage supplied

by the dry cells?

V = IR

= 0.5 (3)

V = 1.5 V

12. The graph shows the result of an experiment todetermine the resistance of a wire. The resistanceof the wire is

From V-I graph, resistance = gradient

=5

2.1

= 0.24

13. An experiment was conducted to measure thecurrent, I flowing through a constantan wire whenthe 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 =1/ gradient

=1/ ( 4

10x8 3

)

=1/( 2.0 x

10-3)

= 500

14.Referring to the diagram on the right, calculate

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5 I

V/V

I/A0 5

1.2


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(a) The current flowing through the resistor.

V = IR

12 = I (5)

I = 2.4 A

(b) The amount of electric charge that passesthrough the resistor in 30 s

Q = It

= 2.4 (30)

= 72 C

(c) The amount of work done to transform theelectric energy to the heat energy in 30 s.

W = QV or W = VIt

= 72 (12) = 12(2.4)(30)

= 864 J = 864 J

15. Figure shows a torchlight that uses two 1.5 V drycells. The two dry cells are able to provide acurrent 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)

R = 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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1.5 V+ - 1.5 V+ -


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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.


=2

8

= 4

b) Calculate the resistance of conductor Y.


=22

= 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 ofIagainst Vfor three

conductors, P, Q and R.

i) QCompare the resistance of conductor P, Q and R.

RR > RQ >Rp

ii) Explain your answer in (a)

From I-V graph, resistance = 1/gradientThe greater the gradient, the lower the resistance

Gradient of P > Gradient of Q > Gradient of RThus, RR > RQ >Rp

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I/A

V/V

00

X

Y

2

8

2

V/V

I/A P

Q

R

R =A

l

=l

RA

=2.1

)10x0.5(4 6

= 1.67 x 10-5m


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19. Figure shows a wire P of length, lwith a cross-

sectional area, A and a resistance, R. Another

wire, Q is a conductor of the same material with

a length of 3land twice the cross-sectional area

of P. What is resistance of Q in terms of R?

Conductor P R =A

l

Conductor Q R ='A

'l(notes: P and R have the same resistivity, )

=A2

)l3(

=2

3R

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. Whatis 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 =

100

20(10)

R = 2

Total resistance

2+ 2= 4

Current, I =R

V

=4

3

= 0.75 A

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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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7.3 SERIES AND PARALLEL CIRCUITS

Current Flow and Potential Difference in Series and Parallel Circuit

SERIES CIRCUIT PARALLEL CIRCUIT

1 the current flows through each bulb/resistor isthe same

I= I1 = I2 = I3

2 the potential difference across each bulb /

resistor depends directly on its resistance. 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 Ohms 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, weget the effective resistance

R = R1 + R2 + + R3

1 the potential difference is the same across eachbulb/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

fromthe battery

3 If Ohms 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

VR

VR

VR

1R

1R

1R +

1R

I

V


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(a) (b) (c)

(d)

Series Parallel A, B - series Q, S - parallel

Ammeter reading Current

Voltmeter reading Potential difference Voltage

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Effective resistance, R

(a)R = 20 + 10 + 5= 35

(b)1/R = +1/5 + 1/10 = 4/5

Effective R = 1.25

(c)

1/R = 1/8 + 1/8= 1/8

R = 4

Effective R = 20 + 10 + 4 = 34

(d)1/R =1/16 + 1/8 + 1/8=5/16

Effective R = 3.2

(e)1/R = 1/4 + 1/2=3/4

R = 1.33

Effective R = 1.33 + 1 = 2.33

(f)1/R = 1/4 + 1/12=1/3

R = 3

Effective R = 3 + 2 = 5

(g)Effective R = 2+5+3+10

= 20

(h)1/R = 1/20 + 1/20=1/10

R = 10

Effective R = 10 + 10 + 5 =2 5

(i)1/R = 1/5 + 1/10=3/10

R = 3.33

(j)1/R = 1/10 + 1/10=2/10

R = 5

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Solve problems using V = IR

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.

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

- 30 -

V = IR9 =I(18)

= 0.5

V = IR240 = 6(R)

I =40 A

1/R = 1/5 + 1/20=1/4

R = 4


V = IR= 2(5) = 10 V

1/R = 1/10 + 1/10 =2/10

R = 5


V = IR

12 =I(5)= 2.4 A

0.5A

2 3


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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

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 ammeterwhen switch, S

(a) is open? (b) is closed?

Effective R = 6 Effective R = 4

V = IR V = IR12 =I(6) 12 =I(4)

I = 2 A I = 3 A

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6. Determine the voltmeter reading.

(a)

(b)

Determine the ammeter reading.

(a)

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

- 32 -

R = 12

I = 24/12= 2A

V= IR= (2)(8)

= 16 V

R = 12

I = 6/12

= 0.5A

V at 9: V= IR

= (0.5)(9)= 4.5 V

V readin : 6 4.5 = 1.5 V

R =9

I = 4.5/9

= 0.5A

A reading : 0.5/2= 0.25 A

Notes:Divide 2 because

the resistors have similar


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8.

The electrical components in our household appliances are connected in a combination of series andparallel circuits. The above figure shows a hair dryer which has components connected in series andparallel. 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 heldby 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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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)

- 34 -

No current flow

Voltmeter reading,e.m.f.

Voltmeter reading,potential difference, V < e.m.f.,

E , r

Current flowing


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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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3.

a) Why is the potential difference across the resistor not the same as the e.m.f. of thebattery?

The potential drops as much as 0.4 Vacross the internal resistance

b) 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

c) Determine the value of the external resistor.

Since V = IR

1.1 = 0.5RR = 2.2

Therefore, the value of the external resistance is 2.2

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Activity : To determine the values of the electromotive force (e.m.f.) and

the internal resistance, r of the cell

Interference

Hypothesis To investigate the relationshipbetween V and I

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, 0.5A and 0.6 A.

Tabulation of

data

:

Current,I/A Volt, V/V0.2 2.6

0.3 2.5

0.4 2.4

0.5 2.2

0.6 2.0

0.7 1.9

- 37 -

V

Dry cell

Internal resistance

+ -

Switch

Rheostat

Ammeter

Voltmeter


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39/60


= 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 thebattery and the ammeter reading is 0.3 A. Find the internal resistance of

the battery.

E = 1.5 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 resistence 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 , I = 0.25 A

Calculate internal resistance : E = I(R + r), 3.0 = 0.25(10+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

- 39 -

e.m.f.


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V = IR

11.4 = 0.5 (R)

R = 22.8

5 A 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

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)

- 40 -

R/

1.3

- 0.2

0.51 (A-1)I

/A

/ V

1/A

V / V

1.5

0.2

5


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Rearrange : R =I

E- 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 currentflows and the electrical energy

supplied by the source is transformedto other forms of energy.

3. Therefore, we can define electrical energyas : The energy carried by electrical charges

which can be transformed to other forms of energy by the operation of an electrical

appliance.

Electrical Energy and Electrical Power

- 41 -

battery(chemical energy)

Light and heat

currentcurrent

Energy Conversion:

Chemical energy Electrical energy

Light energy

+ Heat energy

battery(chemical energy)

currentcurrent

Energy Conversion:

Chemical energy Electrical energy

Kinetic energy


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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,

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

Hence, ; where I = VR

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

- 42 -

QV =

E = VQ

E = VI t

E = I2Rt

V2 tR

=

tP =

VQt

P =

P = VI

I2 RP =

P = I2 R

Electrical energy dissipated, E

Charge, Q

Potential difference, V =

E = VQ

Energy dissipated, Etime, t

Power P =


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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 andpower rating, P.

5. The energy consumption of an electrical appliance depends on thepower ratingand 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

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

EnergyConsumed

(kWh)

Bulb 5 60 0.06 8 hours 2.4

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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 2Electricity cost: RM 0.28 per kWh

Total energy consumed, E = (2.4 + 9.6 + 4.5 + 2.0)

= 18.5 kWh

Cost = 18.5 kWh x RM 0.28

= RM 5.18

Comparing Various Electrical Appliances in Terms of Efficient Use of Energy

1. A tungsten filament lamp changes electrical energy to

useful lightenergy and unwanted heatenergy

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 100

60

= 5 %

b) Efficiency of a fluorescent lamp and an energy

saving lamp

Efficiency = Output power x 100

Input power

= 3 x 10012

= 25 %

Exercise 7.5

1. How much power dissipated in the bulb?

(a)

- 44 -5 V

R = 10 P = V

2

R

= 52 / 10= 2.5 W


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(b)

2.

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?

- 45 -

E = Pt= 24 (1 x 60 x 60)

= 86 400 J

5 V

R = 10

R = 10

R1=2

R2=4 R3=4

V= 15V I

P = V2

R= 52 / 5

= 5 W

Total resistance, R = (2 + 4 + 4)

= 10

V = IR

I = V/R= 15 / 10

= 1.5 A

E = I 2Rt

= (1.5)2 (10)(10)

= 225 J

E = I 2Rt

= (1.5)2 (2)(10)= 45 J


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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.

(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

- 46 -

E = VI t= 24 (5) (2 x 60)

= 14 400 J

Q = I t

= (1.5) (2 x 60)= 180 C

E = QV

= 180 (240)= 43.2 kJ

P = IV

= 1.5 (240)= 360 W

P = IV R = V2/P

I = P/V

= 2000 / 240 R =28.8

= 8.3 A


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(b) What is the current flow through the kettle?

(c) Determine the suitable fuse to be used in the kettle.

13 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

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 = 15.2 kWh x 30 x RM 0.218

= RM 99.41

- 47 -

P = IV

3000 = I (240)

I = 12.5 A

P = I2 R

3000 = (12.5)2 R

R = 19.2


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49/60


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

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 isrun for 5 minutes?A. 90 J

B. 120 J

C. 1800 J

D. 7200 J

- 49 -

E =VIt= 12(2)(5x 60)= 7200J


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5. The electric current supplied by abattery in a digital watch is 3.0 x 10-5

A. What is the quantity of charge thatflows in 2 hours?

A. 2.5 x 10-7 CB. 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 beused to determine the resistance of the

bulb?

A.

B.

C.

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

- 50 -

Q=It= 3.0 X10-5 (2 X 60 X 60)= 0.216C= 2.2 X 10-1 C


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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?

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

- 51 -

P = IV= 10(240)= 2400 W

R= 1 R= 0.4

R= 2.5 R= 1

R = 6

I = V/R= 12/6

= 2 A


52/60


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.

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?

- 52 -

E = Pt= 2 x 103 x 20 x 60= 2400 x 103

= 2.4 x 106 J 80 = 600 x 4 x 100t (2.5 x 12 )

t = 6000 x 4 X 1002.5 x 12 80

= 100s


53/60


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

Part B: Structured Questions

1.

- 53 -

R = 24V = 12

P = V2/R= 144/24= 6.0W


54/60


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 Ohms law?

Conductor Y

(b) State Ohms 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

=11.0

1

= 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.

- 54 -

Gradient

=06.3

2.06.0

= 0.11 A V-1


55/60


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

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.

- 55 -


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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.

Part C: Essay Questions

1. Figure 1 shows the reading of the voltmeter in a simple electric circuit

- 56 -


57/60


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?

Suggested Answers

1. (a) The work done by a battery to move a unit charge around a complete circuit.

- 57 -


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(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)

e.m.f = intercept on the v-axis

internal resistance = -(gradient of the graph)(d)

(i) To improve the contact between the dry cells and the terminals of thetorchlight

(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 GridNetwork.

- 58 -

emf

0

Potential difference, V/V

Currrent, I/A

V

Dry cell

Internal resistance

+ -

Switch

Rheostat

Ammeter

Voltmeter


59/60


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 loadbefore breaking/

N

Density /kgm-3

Rate ofexpansion

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-10C-1]

(ii) What is the assumption made in the calculations above?

Suggested Answers

2.(a) Resistance is the ratio of potential difference to current flowing in an ohmic conductor.

(b)

- 59 -


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Characteristics Explanations

A low resistivity Energy loss during transmission is reduced

Max load beforebraking is high

Stronger and long lasting

A low densityMass or weight reduced. Can be supported by transmission

tower

Rate of expansion 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.

ac

(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 = mc

(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 CHAPTER 7

12405942 chapter 7 electricity teachers guide 2009

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