chapter 7 - electricity(teacher's guide) 2009

8/12/2019 Chapter 7 - Electricity(Teacher's Guide) 2009

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Physics Module Form 5

Teachers Guide Chapter 7: Electricity

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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 thatproducesandstore electric chargesat high voltage on its dome

dome

rubber belt

Metal dome

roller

roller

motor

+ ++

+

+

+

++

+

+


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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

rubberbelt.

This causes the rubber belt to against the rollerand hence becomes charged.

The charge is then carried by the moving belt up to the metal dome where it is

collected.

A large amount of chargeis built up on the dome

Positivecharges 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 deflectionof the pointer of

the meter.

This indicates an electric current flow.

+++

++

++


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Electric Current

1. Electric current consists of a fl ow of electrons

2.

The more charges that flow through a cross

section within a given time, the largeris the

current.

3. Electric currentis defined as the

rate of fl ow of electri c charge

4. In symbols, it is given as:

where I = electri c 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 = (I t / t

I / I

t )

4. If one coulomb of charge flows past in one second, then the current is oneampere.

5.

15 amperes means in eachsecond, 15coulomb of charge through a cross section of aconductor.

6. In a metal wire, the charges are carried by electrons.

7.

Each electron carries a charge of 1.6 x 10-19C.

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

I Q

t

Each second, 15 coulombs of charge crossthe plane. The current is I = 15amperes.

One ampere is onecoulomb per second.


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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 oscil lated between thetwo plates, touching one plate after

another.

Explanation:

When the polystyrene ball touches the

negatively charged plate, the bal l

receives negative charges from the plate

and experiences a repul sive force.

The ball wil l then move to the positi vely

charged plate.

When the ball touches the plate, the ball

loses some of its negative char ges to the

plate and becomes positively char ged.

I t then experiences a repulsive force.

Th is process conti nues.


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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 spli ts in to two porti ons in

opposite dir ection. The porti on that is

attracted to the negative plate is very much

larger than the porti on of the fl ame that is

attracted to the positive plate.

Explanation:

The heat of the fl ame ionizes the air

molecules to become positive and

negative charges.

The positive charges are attracted to the

negative plate whi le the negative

charges are attracted to the positive

plate.

The flame is dispersed in two opposite

dir ections but more to the negative

plate.

The positi ve charges are heavier than

the negative charges. Th is causes the

uneven dispersion of the fl ame.

Conclusion

1. Electric field is a region where an electr ic char ge experiences a for ce.

2. Like charges repeleach other but opposite charges attracteach other.

3.

Electric field lines are li nes of forcein an electric field. The direction of the field

lines is from positiveto 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 10-19C)

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?

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 -19C of charge 1 electron.Hence, 48 C of charges is brought by 48 C = 3 x 1020electrons

1.6 x 10 -19C

a) Q = It

= 200 x 10-3(3)

= 0.6 C

b) 1.6 x 10 -19C of charge 1 electron.Hence, 0.6 C of charges is 0.6 C = 3.75 x 1018electrons

1.6 x 10 -19C


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Ideas of Potential Difference

(a) (b)

Pressure at point P isgreaterthan the pressure

at point Q

Water will flow fromPto Qwhen the valve is

opened.

This due to the differencein 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 differencein the gravitational

potential energy.

(c) Similarly,

Point A is connected topositiveterminal

Point B is connected to negativeterminal

Electric potential at A is greaterthan the electric potential at

B.

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

circuit and lights upthe bulb.

This is due to the electricpotential differencebetween 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)

A B

Bulb

V=echQuantityof

Work

arg= Q

W

P Q

X

Y


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Device and symbol

ammeterCells

voltmeter Switch

connecting wireConstantan wire //

eureka wire

resistance

bulb

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

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

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 //

temperature

Apparatus /

materials


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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 ofI

Vis a constant as 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 conductor increaseswhen the current, I passing

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


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Ohms Law

(a)

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,

if temperature 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 ratioof the potential difference

across the conductor to the current, I flowing through the conductor. That is

R = I

V

and 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-ohmicconductor


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Factors Affecting Resistance

1. The resistance of a conductor is a measure of the ability of the conductor to (resist/

allow)the flow of an electric current through it.

2.

From the formula V = IR, the current I is (directly / inversely) proportional to the

resistance, R.

3. When the value of the resistance, R is large, the current, I flowing in the conductor is

(small/ large)

4. What are the factors affecting the resistance of a conductor?

a) the length of the conductor

b) the cross-sectional area of the conductor

c) type of material of the conductor

d) the temperature of the conductor

5. Write down the relevant hypothesis for the factors affecting the resistance in the table

below.

Factors Diagram Hypothesis Graph

Len

gthofthe

conductor,l

The longer the conductor, the

higher its resistance

Resistance is directly proportional

to the length of a conductor

Thecross-sectional

areaofthe

conductor,A

The bigger the cross-sectional

area, the lower the its resistance

Resistance isinversely

proportional to the cross-sectional area of a conductor

Thetypeofth

e

materialofthe

conductor

Different conductors with the

same physical conditions have

different resistance

Th

etemperatureof

theconductor

The higher temperature ofconductor, the higher the

resistance


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6.

From, the following can be stated:

Resistance of a conductor, R length

Resistance of a conductor, R 1

cross-sectional area

Hence, resistance of a conductor, R length

cross-sectional area

Or R l or R = l where = resistivity of the

A A substance


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Exercise 7.2

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 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

25 = 5 (V)

V = 5 V


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5. The potential difference of 10 V is used to operate an electric motor. How much work is

done in moving 3 C of electric charge through the motor?

W = QV

= 3 (10)

= 30 J

6. When the potential difference across a bulb is

20 V, the current flow is 3 A. How much work

done to transform electrical energy to light and

heat energy in 50 s?

W = VIt

= 20 (3) (50)= 3000 J

7. What is the potential difference across a light bulb

of resistance 5 when the current that passesthrough 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

3 A

20 V

Bulb


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10. If the bulb in the figure has a resistance of 6 ,

what is the reading shown on the ammeter, if the

dry cells supply 3 V?

V = IR

3.0 = I (6)

I = 0.5 A

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)

V = 1.5 V

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

=5

2.1

= 0.24

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 =1/ gradient

=1/ (4

10x8 3

)

=1/( 2.0 x 10-3)

= 500

V/V

I/A0 5

1.2


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14. Referring to the diagram on the right, calculate

(a)The current flowing through the resistor.

V = IR

12 = I (5)

I = 2.4 A

(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 J = 864 J

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)

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

1.5 V+ - 1.5 V+ -

12 V

5 I


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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.


=2

2

= 1

c) If the cross sectional area of X is 5.0 x 10-6

m2, and the length of X is 1.2 m, calculate its

resistivity.

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/gradient

The greater the gradient, the lower the resistance

Gradient of P > Gradient of Q > Gradient of R

Thus, RR> RQ>Rp

I/A

V/V

0

0

X

Y

2

8

2

V/V

I/AP

Q

R

R =A

l

=l

RA

=2.1

)10x0.5(4 6

= 1.67 x 10-5

m


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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 anammeter, 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 =10020 (10)

R = 2

Total resistance

2+ 2= 4

Current, I =R

V

= 4

3


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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 is

the same

I= I1= I2= I3

2 the potential difference across eachbulb /

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 eachbulb /

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

fromthe battery

3 If Ohms law is applied separately to eachbulb /

resistor, we get :

I = I1 + I2 + I3

If each term in the equation is divided by V, we

get the effective resistance

V

RV

R1

V

R2+=

V

R3+

1

R1

R1

1

R2+= +

1

R3

I

V


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Identify series circuit or parallel circuit

(a) (b) (c) (d)

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

Ammeter reading Current

Voltmeter reading Potential difference Voltage

1

5

6

3

2 2

0.5

1

3

3

2

5

2

1.5


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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 = 1.33Effective R = 1.33+ 1 = 2.33

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

R = 3Effective 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 2and 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) = 1V2.5 =I(5) 3: V = IR = (0.5)(3) = 1.5 V

= 0.5 A

V = IR

9 =I(18)

= 0.5

V = IR

240 = 6(R)

I =40 A

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

R = 4Effective R = 1 + 4= 5

V = IR= 2(5) = 10 V

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

R = 5Effective R = 1 + 4= 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 R1has 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, I1and 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 5and 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 = IR3 =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


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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 8and 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

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 VV reading : 64.5 = 1.5 V

R =9 I = 4.5/9

= 0.5A

A reading : 0.5/2= 0.25 A

Notes: Di vide 2 because

the resistors have simil ar

resistance.


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8.

The electrical components in our household appliances are connected in a combination of series and

parallel circuits. The above figure shows a hair dryer which has components connected in series and

parallel. Describe how the circuit works.

Suggested answer

The hair dryer has three switches A, B and C

When switch A is switched on, the dryer will only blow air at ordinary room temperature When switches A and B are both switched on, the dryer will blow hot air. As a safety feature to prevent overheating, the heating element will not be switched on if the fan is

not switched on

The hair dryer has an energy saving feature. Switch C will switch on the dryer only when it is

held by the hand of user

The body of the hair dryer must be safe to hold and does not get hot easily


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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, VRand drop

in potential difference due to internal resistance, V r.

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 the

battery?

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

Aim

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

the internal resistance, r of the cell

Apparatus /

materials

Dry cells holder, ammeter (01 A), voltmeter(05 V), rheostat (015 ), 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

V

Dry cell

Internal resistance

+ -

Switch

Rheostat

Ammeter

Voltmeter


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0.6 2.0

0.7 1.9

Analysis of data

Draw a graph of

VagainstI

:

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 VI graphR = the gradient of the VI graph

Potential difference, V /V

Current, I /A

3.0 -

2.0 -

1.0 -

0.2 0.40.6

0.8


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(c) determine the values of E and r from the graph.

By extrapolating the graph until it cuts the vertical axis,

E = 2.9 V

r = - gradient

= 1.4

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 = 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


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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)

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 2Hence, from VI graph : E = c = intercept of V-axis = 2

= 6 V

6

The 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

e.m.f.

6

2

2/A

/ V

1/A

V / V

1.5

0.2

5


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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 =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.

R/

1.3

- 0.2

0.5

1 (A-1

)I

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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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

kilowatt-hour, kWh.

1 kWh = 1000 x 3600 J

= 3.6 x 106J

= 1 unit

3. Onekilowatt-hour is the electrical energy dissipated or transferred by a 1 kW device in

onehour

4. Household electrical appliances that work on the heating effect of current are usually

marked with voltage, V andpower rating, P.

5. The energyconsumption 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)2100

= 25 watts

R= 100, V=50 W, P=?

P = (V/R)2R

= 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


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Cost of energy

Appliance Quantity Power / W Power / kW Time

EnergyConsumed

(kWh)

Bulb 5 60 0.06 8 hours 2.4

Refrigerator 1 400 0.4 24 hours 9.6

Kettle 1 1500 1.5 3 hours 4.5

Iron 1 1000 1.0 2 hours 2

Electricity cost: RM 0.28 per kWh

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

= 18.5kWh

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 energyto

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 100

Input power

= 3 x 100

60

= 5 %

b) Efficiency of a fluorescent lamp and an energy

saving lamp

Efficiency = Output power x 100

Input power

= 3 x 100

12

= 25 %


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Exercise 7.5

1. How much power dissipated in the bulb?

(a)

(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.

5 V

R = 10

5 V

R = 10

R = 10

R1=2R2=4

R3=4

V= 15V I

P = VR

= 52/ 10

= 2.5 W

P = V

R

= 52/ 5

= 5 W

Total resistance, R = (2 + 4 + 4) = 10

V = IRI = V/R

= 15 / 10

= 1.5 A

E = I Rt

= (1.5)2(10)(10)

= 225 J

E = I Rt

= (1.5)2(2)(10)

= 45 J


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2. A lamp is marked 12 V, 24 W. Howmany 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.

(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?

E = Pt

= 24 (1 x 60 x 60)

= 86 400 J

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 = V /P

I = P/V

= 2000 / 240 R =28.8= 8.3 A


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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.

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

P = IV

3000 = I (240)

I = 12.5 A

P = I R

3000 = (12.5)2R

R = 19.2


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(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

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

(c) Efficiency of the motor

Efficiency = Output power x 100 %

Input power

= 400 x 100 %

1000

= 40 %

E = VIt

= 12 (5.0) (2.5)

= 150 J

U = mgh

= 2 (9.8) (5)

= 98 J

Efficiency = Output power x 100 %

Input power

= 98 x 100 %

150

= 65.3 %


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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

CB. 1.5 x 10-5C

C. 6.0 x 10-5C

D. 3.6 x 10-3C

E. 2.2 x 10-1C

6. Which of the following circuits can be

used 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

E =VIt

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

Q=It

= 3.0 X10-5(2 X 60 X 60)

= 0.216C

= 2.2 X 10-1C


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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?

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

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


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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 106J

B. 1.8 x 106J

C. 2.4 x 106J

D. 3.6 x 106J

E. 4.8 x 106J

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?

E = Pt

= 2 x 103x 20 x 60

= 2400 x 103

= 2.4 x 106J 80 = 600 x 4 x 100

t (2.5 x 12 )

t = 6000 x 4 X 100

2.5 x 12 80

= 100s


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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 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.

Gradient

=06.3

2.06.0

= 0.11 A V-1


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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

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.


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(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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Part C: Essay Questions

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

Figure 2 shows the reading of the same voltmeter

(a)

What is meant by electromotive force (e.m.f.) of a battery?

(b)Referring to figure (a) and figure (b), compare the state of the switch, S, and

the readings of the voltmeter. State a reason for the observation on the

readings of the voltmeter.

(c)

Draw a suitable simple electric circuit and a suitable graph, briefly explain

how the e.m.f. and the quantity in your reason in (b) can be obtained.

(d)

The figure above shows a dry cell operated torchlight with metal casing

(i) What is the purpose of the spring in the torchlight?

(ii) Why it is safe to use the torchlight although the casing is made of metal?

(iii) What is the purpose of having a concave reflector in the torchlight?


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Suggested Answers

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

(b) - Switch in figure 1 is turned off

- Switch in figure 2 in turned on

- Reading of voltmeter in figure 1 is higher than in figure 2

- This is due to the presence of an internal resistance in the battery

(c)

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.

emf

0

Potential difference, V/V

Currrent, I/A

V

Dry cell

Internal resistance

+ -

Switch

Rheostat

Ammeter

Voltmeter


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Suggested Answers

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

(b)

Characteristics Explanations

A low resistivity Energy loss during transmission is reduced

Max load before

braking is highStronger 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

chapter 7 - electricity(teacher's guide) 2009

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