## Friday, 6 April 2012

### Electrical Energy and Power

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:

 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)

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).
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 kilowatt-hour (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 past-year 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)