SPM Physics

1. Diagram 26 shows an electric circuit. The reading of the ammeter is 0.2 A
and the reading of the voltmeter is 2.8 V

Calculate the electrical energy released by the bulb in 2 minutes.
A 0.56 J B. 1.12J C. 67.20J D. 28.00 J E. 1680.00 J

Apply formula E = V I t , but don’t forget to change time into second
E = 2.8 x 0.2 x ( 2 x 60) = 67.2 J

2. Paper 2 :SPM 04

(a) Namakan satu bahan yang sesuai digunakan sebagai unsur pemanas dalam

pemanas rendam itu. [1 markah] Tungsten

(b) Pemanas rendam itu disambungkan kepada bekalan 240 V. Hitungkan;

(i) arus yang melalui pemanas rendam itu, [2 markah]
P = V x I
1000 = 240 x I
1000/240 = I ,
Thus I = 4.17 A

(ii) rintangan pemanas rendam itu. [2 markah]

V = I x R
R = V/ I
= 240 / 4.17
= 57.55 ohm

(i) State the energy change that occurs when the immersion heater switched on. [1 mark]

Electrical energy is converted to heat energy

(ii) Calculate the energy supplied by each of the immersion heaters P, Q and R to start boiling the water. [4 marks]

In this question, u need to calculate all the energy; use formula E= V I t but please convert the time to second.

• heater P: E = 240 x 6 x ( 8 x 6o) = 691200 J

• heater Q: E = 240 x 5 x (10 x 60) = 720000 J

• heater R: E = 240 x 4 x (9 x 60) = 518400 J

  
  

(iii) Using your answer in (c)(ii), suggest which immersion heater is the most suitable to
heat water. Give one reason for your answer. [2 marks ]

Clue, choose a heater with less energy consumption & time efficient. Most appropriate answer is heater R coz it uses lowest energy.