Electrical
energy is the energy carried by electric charges which can be
converted or transformed to other forms of energy by the operation of an electrical
appliance. For examples:
(Updated on 11/4/2012 by: tutortan1@gmail.com)
Type of
Appliance

Energy
Conversion

a) Radio

Electrical energy
> Sound energy

b) Electric Fan

Electrical energy
>Kinetic energy

c) Electric kettle,
hot plate, iron

Electrical energy
> Heat energy

d) Hair dryer

Electrical energy
>Heat
+ Kinetic
energy

e) Light bulb

Electrical energy
> Light
+ Heat energy

2.
The SI unit of electrical energy is the joule, J.
3.
Electrical Power
is defined as the rate of transfer of
electrical energy.
(Yr 2009 SPM P1 Q36 at pg. 194)
4.
The SI unit of electrical power is joule per
second J/s, or watt W (1 W = 1 J/s).
5.
Formulae for electrical energy E and power P:
a.
Electrical
Energy E  From the definition
of potential difference or voltage V across 2 points:
Potential difference = Work done / Charge;
V = W / Q
V = E / Q (since Work done W = Energy E)
E = VQ
Or, E = VIt (since Q = It)…(1): (Yr 2006 SPM P1 Q40 at pg. 58)
E = I^2Rt (since V =
IR)…(2)
E = V^2t /R (since I =
V/R)…(3)
b. Power P from its own definition:
Power = Electrical Energy / Time
P = E/t
P = VIt/t (since E = VIt)
Or, P = VI…(1)
P = I^2R (since V = IR)…(2):
(Yr 2008 SPM P1 Q38 at pg. 148)
P = V^2/R (since I =
V/R)…(3): (Yr 2011 SPM P1 Q38 at pg. 289)
6.
Household electrical appliances are usually
marked with voltage V and power P ratings:
a.
The voltage
V rating indicates the voltage suitable to operate the appliance concerned;
and
b.
The power
P rating indicates the electrical power that will be consumed when operating
at the rated voltage:
i.
A light bulb rated 12 V, 24 W means:
1.
it will release 24 J of light and heat energy in
1 second when connected to a 12 V supply;
2.
its resistance R = V^2/P (since P = V^2/R) …R is
inversely proportional to P
R = 144/24 ohms = 6 ohms.
C. The voltage and power ratings also tell us these:
1.
Resistance: R = V^2/P
Higher power P rating means lower resistance R.
For normal performance of the lamp or appliances, the voltage V across a parallel circuit must be the same as or near to the voltage V rating of the appliance (that explains why the light bulb with higher power P rating but same V rating (in Example 21 at pg 386) has lower resistance R) (2010 Paper2 Q12(b) Circuit X at pg. 265) .
Questions:
* What if the voltage across the parallel circuit > the voltage rating?
(2010 Paper2 Q12(b) Circuit Y at pg. 265)
The appliance / lamp would be damaged (due to Actual I > Permitted I)
* What if the voltage across the parallel circuit < the voltage rating?
(2010 Paper2 Q12(b) Circuits W & Z at pg. 265)
The appliance would be performing at below normal (e.g. dimmer light)
Higher power P rating means lower resistance R.
For normal performance of the lamp or appliances, the voltage V across a parallel circuit must be the same as or near to the voltage V rating of the appliance (that explains why the light bulb with higher power P rating but same V rating (in Example 21 at pg 386) has lower resistance R) (2010 Paper2 Q12(b) Circuit X at pg. 265) .
Questions:
* What if the voltage across the parallel circuit > the voltage rating?
(2010 Paper2 Q12(b) Circuit Y at pg. 265)
The appliance / lamp would be damaged (due to Actual I > Permitted I)
* What if the voltage across the parallel circuit < the voltage rating?
(2010 Paper2 Q12(b) Circuits W & Z at pg. 265)
The appliance would be performing at below normal (e.g. dimmer light)
2.
Current: I = P/V (since P = VI)
In parallel circuit, appliances or lamps with higher power P rating will receive higher current I since V is a constant (that explains why, in Example 21 at pg. 386, when 2 light bulbs of different power ratings are connected in parallel, the one with higher power rating will receive higher current).
In parallel circuit, appliances or lamps with higher power P rating will receive higher current I since V is a constant (that explains why, in Example 21 at pg. 386, when 2 light bulbs of different power ratings are connected in parallel, the one with higher power rating will receive higher current).
iii.
Electrical appliances with higher power ratings
consume more electrical energy for the same usage time (E = Pt).
iv.
The power ratings of appliances involving heat
are comparatively higher.
v.
One (1) horse power (h.p.) = 746 watts (W) (1
h.p. = 746 W)
7.
The energy
consumption of any electrical appliance depends on the power rating and the usage
time (E = P t):
a.
Thus, a 1300 W hair dryer uses 1300 joules of
electrical energy per second.
b.
The amount payable for the use of electricity
depends on the units of electrical energy consumed and the applicable
tariff (rate payable per unit of electrical energy consumed).
c.
In Malaysia, 1 unit of electrical
energy = 1 kilowatthour (kWh): which is the electrical energy used by an
appliance with a power rating of 1000W for 1 hour – it is equivalent to 3.6 MJ:
1 kWh = 1 kW x 1 hr
= 1000 W x 60 x 60 s
= 1000 W x 3600 s
= 3 600 000 J
d.
The amount payable is the product of number of units
consumed and the applicable rates (in sen per unit) in the tariff. The tariffs
for residential and commercial/industrial use will be different.
8.
Choosing energy efficient appliances will give
significant savings in electrical energy. Household appliances should always be
chosen and used on the basis of their electrical operating efficiency:
Efficiency = Useful Energy Output Eo / Energy Input Ei x 100%
= Eo / Ei x 100%
Efficiency = Useful Power Output Po /
Input Power Pi x 100%
= Po / Pi x 100%
9. Interesting SPM pastyear questions on
“Electricity”:
a. Yr 2011 SPM P1 Q39 at pg. 289
b. Yr 2010 SPM P2 CQ12 at pg. 264  Very Interesting Question!
c. Yr 2009 SPM P2 CQ12 at pg. 216
(Updated on 11/4/2012 by: tutortan1@gmail.com)
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