SPM Physics

Wow! I told them u guys need to be prepared, i'm sure b4 your trial, your teacher might ask u to learn certain things on your own. I do not know about u guys, hopefully all of u have almost finished chapter 2.

Ok, lets finish our discussion, in last 3 outcome of chapter 2,
i. u must be able to compare power rating & energy consumption of various electrical appliances
( e.g aircond, cattle, heater etc)

ii. u must also compare various electrical appliances in terms of efficient use of energy.
Here u must know how to calculate efficiency of an electrical appliance
Calculating efficency is to find the percentage; By formula,

Efficiency= (Output/input) x 100

Lets do some analogy,
Just look at your physics result in exam, how efficient r u in studying physics?
For example in paper 1 u r given an input of 50 questions. Out of that 50, you got 30 correct answers. U've wasted 20 questions. So applying efficiency formula,
your Input =50
your Output= 30
Thus your efficiency = (30/50) x 100
= 60%
Well if that is your efficiency in paper 1, i think u need to do lots of EXTRA work :) U r not efficient enough. Where does the 40% go? Is it lost in the middle of your subconscious mind?
If u want to b efficient u shud get 90% & above :)

The same thing goes to eletrical appliances, usually there'll be an input power or energy being supplied. From that input, some of the enery will be used, and some will be lost. The energy being used is the output.
Thus, we can measure its efficiency by applying the formula discussed earlier.

I leave u guys with this comparison between 2 different electrical appliances.