dry cells



Diagram 10.1 shows a photograph of a circuit 
with six new identical dry cells with 1.5 V , 
while Diagram 10.2 shows a photograph of a 
circuit consists of a new dry cell with 9 V. 
Each circuit is connected to a bulb.
Rajah 10.1 menunjukkan fotograf bagi satu litar 
yang  mengandungi enam 
sel kering baru yang serupa setiap satunya 1.5V, 
 sementara  Rajah 10.2 
menunjukkan fotograf bagi satu litar yang  mengandungi 
satu sel kering baru 9V, 
masing-masing disambungkan kepada sebiji mentol 
yang serupa.


(a)
What is the meaning of 9 V battery ?


Apakah maksud 9V pada bateri tersebut ?



[1 mark / markah]

(b)
(i)
State the energy transformation that 
take place in the bulb.



Nyatakan perubahan tenaga yang berlaku 
pada mentol.



[1 mark / markah]


(ii)
Using Diagram 10.1 and Diagram 10.2,
 compare the bulb brightness with the types 
of the dry cell used.

Relate the bulb brightness with the amount of 
current flow to make a deduction regarding the 
relationship between the bulb brightness and the 
 internal resistance.



Menggunakan Rajah 10.1 dan Rajah 10.2, 
bandingkan kecerahan mentol dengan jenis sel kering 
yang digunakan.

Hubungkaitkan kecerahan mentol dengan 
jumlah arus yang mengalir untuk membuat 
kesimpulan tentang hubungan antara 
kecerahan mentol dengan rintangan dalam bagi sel kering.



[4 marks / markah]










(c)
Diagram 10.3 shows a 12V accumulator used to start a car engine.


Diagram 10.3 menunjukkan akumulator  12V yang digunakan
 untuk meghidupkan enjin sebuah kereta.


What is the type of this accumulator?
Apakah jenis  akumulator ini?
[1 mark / markah]
Diagram 10.4 shows a 12V accumulator ( internal resistance, r = 0.01Ω ) 
is replaced  with eight 1.5V dry cells ( internal resistance, r = 0.5Ω ).
Rajah 10.4 menunjukkan akumulator 12V (rintangan dalam, r = 0.01Ω)
 digantikan dengan lapan sel kering 1.5 V(rintangan dalam, r = 0.5Ω).






 

Can the car be started ?. Explain your answer
Adakah kereta tersebut dapat dihidupkan ?. Terangkan jawapan anda.



Diagram 10.5 shows the condition in a closed room with unsuitable installation of lamp.
Rajah 10.5 menunjukkan suasana dalam sebuah bilik yang tertutup dan  pemasangan 
 lampu yang tidak sesuai.
 


Using appropriate physics concepts, explain suitable modification to the room and 
the lamp so that the room condition becomes brighter and more comfortable.
Your answer should include the following aspects :
Menggunakan konsep fizik yang sesuai, terangkan pengubahsuaian
 pada bilik dan mentol tersebut supaya suasana bilik menjadi lebih terang dan selesa.
Jawapan anda hendaklah merangkumi aspek-aspek berikut:
( i )
Type of the lamp.

Jenis  lampu  tersebut.
( ii )
Position of the lamp.

Kedudukan  lampu  itu.
( iii )
The energy efficiency of the lamp.

Kecekapan tenaga lampu  itu.
( iv )
Safety feature of the lamp.

Ciri keselamatan untuk lampu itu.
( v )
 Wiring system for the lamp.

Sistem pendawaian untuk lampu itu..

[10 marks / markah]


Supply 9 Joule electric energy for each coulomb of charge transfer from one point to another point within a material

Electrical energy à Light energy + Heat energy

Diagram 4.2 is brighter // Diagram 4.1 is dimmer

The brighter bulb has larger amount of current //
The dimmer bulb has smaller amount of current

The brighter the bulb , the higher the current flows
and
Internal resistance of dry cell is lower //otherwise

Lead-acid accumulator

No
8 dry cells are arranged in serial so the total internal resistance is
8 x 0.5 = 4 Ω
Current flow, I = V/R = 12/4 = 3 A

This small current cannot start the engine.

Modification
Explanation
Use the fluorescent lamp
Consume less power and economic
Use a more efficient fluorescent lamp
It brightens the room more clearly and thoroughly and prevents wastage
Use a more energy efficient fluorescent lamp
Helps the temperature in the room to be not too hot due to less energy loss in form of heat energy.
Fuses should be connected to the fluorescent lamp
To prevent overheating of lamps might cause the fluorescent lamp to blow out.
Two or more fluorescent lamp are arrange in parallel.
If one lamp blows the rest are still functioning.