Electrical Energy and Power

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Welcome to class! 

In today’s class, we will be talking about electrical energy and power. Enjoy the class!

Electrical Energy and Power

Introduction

When current flows through a conductor, heat energy is generated in the conductor. The heating effect of an electric current depends on three factors:

1. The resistance, R of the conductor. A higher resistance produces more heat.
2. The time, t for which current flows. The longer the time the larger the amount of heat produced
3. The amount of current, I. the higher the current the larger the amount of heat generated.

Hence the heating effect produced by an electric current, I through a conductor of resistance, R for a time, t is given by H = I²Rt. This equation is called Joule’s equation of electrical heating.

Electrical energy and power

• Electrical energy:

Electrical Energy is the workplace in moving a charge from one point to another in an electric circuit.

W = QW

But Q = It

Hence, electrical energy, E = Ivt

Since V = IR

E = I²Rt ……………………………………… heating effect

The S.I unit is Joule. The commercial unit of electric energy is kilowatt-hour (kW h), commonly known as ‘unit’

• Electrical power:

The rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power. It is also defined as the electrical energy consumed per unit time. The power P is given by:

P = VI or

From Ohm’s Law

V = IR (substituting V = IR)

P = (IR) x (R)

P = I²R = V²/R

The SI unit of electric power is the watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.

Thus, 1W = 1volt × 1 ampere = 1 VA.

If a piece of electrical equipment is examined it will usually be found to have a label giving the working voltage and power consumption in watts. For example, an electric lamp may be marked 240V 60w or an electric stove 240v 3kw.

The electricity meter used by PHCN/N.E.P.A measure the electrical energy consumed in kWh. This stands for kilowatt-hour, which is the commercial unit of electric energy.

As its name implies, the kilowatt-hour is the energy supplied by a rate of working of 1000 watts for 1 hour.

i.e        kWh    = 1000 watts x 1 hour

= 1000 x 60 x 60 joules

= 3.6 x 10⁶ joules

When working out problems on the consumption of electric energy, it is unnecessary to convert to joules.

Worked examples

(1) Find the energy consumed in kWh when five 60W lamps and four 100w lamps are lit for 8 hours.

Solution

Total power consumption   =  (5 x 60w) + (4 x 100w) =  700 watts

Time =   8 hours

Energy consumed = 5600 watts hours =  5.6 kilowatts hours =  5.6 kWh

(2) If electricity is sod at N4 per kWh, find the cost of running the lamps in the example (1) given above for the time state.

Solution

Energy consumed   =  5.6 kWh

at N4 per kWh

Total cost  =  5.6 x 4 = N 22.40

(3) An electric lamp is marked 12v, 24w. Calculate the current in the lamps and the resistance of the lamp while in use.

Solution

To calculate for the current (I)

P = IV

P = 24w, V = 12v

24 = 12(I)

I = 2A

To calculate the resistance (R)

From Ohm’s law V = IR

V = 12v, I = 2A

Therefore, R =  = 6 Ω

(4) An electric palm and electric stove are rated 60W and 3000W respectively. Explain the meaning.

Solution

The electric lamp uses energy at the rate of 60 watts (60 joules per second) when operating on a 240 v supply. Likewise, the stove consumes 3000 joules of energy per second while operating on 240 v supply.

(5) What is the cost of operating two (2) – horsepower water pump and 1000 watts electricity cooker for 2hours if electricity is sold at N6.00 per kWh?

Solution

1 horse power = ¾ KW

Therefore, 2 hp = 1.5 KW

Two water pumps will have the power:  2 x 1.5 KW = 3KW

Cooker has: 100 KW  = 1KW

Total power = (3 + 1) KW = 4KW

When operated for two (2) hours, energy consumed is given as 4kwh x 2hrs = 8kWh.

At N4.00, cost = N4.00 x 8 = N32.00

(6) An electrical bulb is labelled 100W, 240V. Calculate:

(a)The current through the filament when the bulb works normally.

(b)The resistance of the filament used in the bulb.

Solution

(a) Given parameters: P = 100W, V = 240V

Therefore current I:

I = P/V = 100/240 = 0.4167A

(b) R = P/I²

100/ 0.4167² = 576.04Ω or

R = V²/P =240²/100 = 576Ω

(7) Find the energy dissipated in 5 minutes by an electric bulb with a filament of resistance of 500Ω connected to a 240V supply.

Solution

E = Pt =[(240 * 240)/500] * [5*60] = 115.2 * 300

E = 34,560J

(8) A 2.5 kW immersion heater is used to heat water. Calculate:

(a) The operating voltage of the heater if its resistance is 24Ω

(b) The electrical energy converted to heat energy in 2 hours.

Solution

(a) Given parameters: P = 2.5kW, R = 24Ω

Therefore, P=VI = I²R

I =  =10.2062A

Hence:          V=IR= 10.2062 x 24 = 244.9488V

 

(b) E = VIt = Pt = 2500 x 2 x 60 x 60 = 1.8 x 10⁷J

OR E= VIt = 244.9488 x 10.2062 x 2 x 60 x 60 = 1.8 x 10⁷J

(9) An electric bulb is labeled 100W, 240V. Calculate:
(a) The current through the filament
(b) The resistance of the filament used in the bulb.

Solution

Parameters given: P = 100W, V = 240V

(a) P = VI

I = P/V = 100/240 = 0.4167A

(b) From Ohm’s law, V =IR

R=V/I = 240/0.4167 = 575.95Ω

Applications of heating effect of electric current

Most household electrical appliances convert electrical energy into heat by this means. These include filament lamps, electric heater, electric iron, electric kettle, etc.

In lighting appliances

• Fluorescent lamps: these lamps are more efficient compared to filament lamps and last much longer. They have mercury vapour in the glass tube which emits ultraviolet radiation when switched on. This radiation causes the powder in the tube to glow (fluoresce) i.e. emits visible light. Different powders produce different colours. Note that fluorescent lamps are expensive to install but their running cost is much less.

In electrical heating

1. Electric cookers: electric cookers turn red hot and the heat energy produced is absorbed by the cooking pot through conduction.
2. Electric heaters: radiant heaters turn red at about 900⁰C and the radiation emitted is directed into the room by polished reflectors.
3. Electric kettles: the heating element is placed at the bottom of the kettle so that the liquid being heated covers it. The heat is then absorbed by water and distributed throughout the whole liquid by convection.
4. Electric irons: when current flows through the heating element, the heat energy developed is conducted to the heavy metal base raising its temperature. This energy is then used to press clothes. The temperature of the electric iron can be controlled using a thermostat (a bimetallic strip).

 

In our next class, we will be talking about Safety Devices.  We hope you enjoyed the class.

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