practical uses of electricity

8/8/2019 Practical Uses of Electricity

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PRACTICAL USES OF

ELECTRICITY


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Heating & Lighting Appliances


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Glass bulb

Coil of coiled filament made oftungsten

Inert argon gas at a low pressure

Tungsten high melting point, high resistivity value, high resistance to oxidation

Argon helps to minimise evaporation of tungsten filament

Coiled filament reduces energy losses through convection and increases

resistance of coil


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thermostat

Base plate

Heating element is made of

nichrome

Nichrome high melting point, high electrical resistance and high resistance to

oxidation

Thermostat regulates the temperature of the iron


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Watt is the Power of these appliances?


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Electrical Power and Electrical Resistance areconstant for an electrical appliance

Electrical Energy = Electrical Power x Time taken

Electrical Power in kilowatt ( kW)

Time taken in hours (h)

and electrical energy in kilowatt-hour (kWh)

One unit of electricity is 1 kWh

1 kWh is the amount of electrical energy

dissipated by a 1 kW appliance switched on for

1 hour.


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Cost ofusing electricity

1. A 2000 W heater is used for 10 hours to warm a room during winter.What is the cost of using the heater if 1 kWh of electricity costs 15cents ?

Energy used = Power (kW) x Time (h) = 2 kW x 10 h = 20 kWh

Cost = 20 x 15 = 300 cents = $ 3.00

2. A 2kW electric fire is used for 10 hours/week and a 100 W lamp isused for 10 hours/day. Find the total energy consumed per week, and

hence the total cost per week if 1 kWh of electricity costs 3 pence.

Total energy used = (2 x 10) + ( 0.10 x 10 x 7) = 27 kWh

Cost = 27 x 3 = 81 pence


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Electrical Power

Power = Voltage x Current

P = VI where P is Power in Watts (W)V is voltage in volts (V)

I is current in ampere (A)

Proof: E = VQ

P = ( E / t ) = ( V Q / t ) = V ( Q / t ) = VI


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Other alternative formulae for

electrical power

P = I 2 R

P = V2 / R


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240 V mains supply

Kettle rated at 960 W

resistance of 58 ;

cable of electrical resistance of 2.0 ;

Current supplied to the kettle

I = V / R = 240 / ( 58 + 2 ) = 4.0 A

Power Loss in cable

Power loss = I 2 R = (4.0)2 x 2.0 = 32 W

Potential difference (Voltage) across the kettle

V = IR = 4.0 x 58 = 232 V ( 8 V is lost to the cable )

Power output of the kettle

P = VI = 232 x 4.0 = 928 W ( = I2 R = 4.02 x 58 = 928 W ) = (V2/R = 2322/58 = 928W)

If cable has no resistance,

then P = VI,

I = P / V = 960 / 240

= 4.0 A


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More Questions

1. An electric kettle that is connected to a 240 V mains supply draws a current of 10 A.Calculate (a) the power of the kettle,

(b) the heat energy produced in 20 s.

(c) the rise in temperature if all this energy is given to 2 kg of water.

[Specific heat capacity of water = 4200 J/(kg0C) ]

(a) P = V I = 240 x 10 = 2400 W

(b) E = P t = 2400 x 20 = 48000 J

(c) E = 48000 = m c

= 48000 / ( 2 x 4200 ) = 5.7 0C


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2. The diagram shows a circuit which contains a bulb of resistance 3 ; connected to a

battery of e.m.f. of 12 V and connecting wires which has an internal resistance of 0.015

;. Calculate (a) the current in the circuit, (refer to notes for diagram)

(b) the power loss in the wires,

(c) the power of the bulb,

(d) the voltage across the bulb.

(a) I = V / R = 12 / 3.015 = 3.98 A

(b) Power loss in the wires = I 2 R = 3.982 x 0.015 = 0.238 W

(c) P = I2 R = 3.982 x 3 = 47.5 W

(d) V = I R = 3.98 x 3 = 11.9 V

[Note that p.d. across the bulb is less than 12 V because there is some p.d.

across the internal resistance of the wires.]

For part (c) using P = V I = 11.9 x 3.98 = 47.4 W

using P = V2 / R = 11.92 / 3 = 47.2 W


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3. Page 116, Question 4

(a) I = P / V = 72 / 12 = 6.0 A

(b) Q = I t = 6.0 x 20 x 60 = 7200 C

(c) E = V Q = 12 x 7200 = 86400 J or E = P t = 72 x 20 x 60 = 86400 J

(d) R = V / I = 12 / 6.0 = 2.0 ;

4. Page 116, Question 5

(a) P = V2 / R ; R = V2 / P = 4.02 / 12 = 1.3 ;

(b) I = V / R = 2 / 1.3 = 1.5 A

(c) P = V I = 2.0 x 1.5 = 3.0 W

[Note that when the operating voltage applied to the lamp is reduced by half; i.e.

from 4 V to 2 V; the current is also reduced by half, causing the power of the lamp to bereduced by a quarter ( Power = voltage x current )since the resistance of the lamp is

constant and is not affected by changes in voltage or current.


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Live, Earth, Neutral wires

LIVEWIRE

EARTHWIRE

NEUTRALWIRE

POWERSUPPLY


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3-pins plug


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Earth wire


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LIVEWIRE

CARRIES CURRENT FROMTHE

POWER SUPPLY TOTHEAPPLIANCE

- ATA HIGH VOLTAGEOF 240 V

NEUTRAL

WIRE

CARRIES CURRENT FROMTHE

APPLIANCETOTHE POWER SUPPLY- ATA LOW VOLTAGEOF 0 V

E

ARTHWIRE DOE

S NOT

CARRY ANY CURRE

NT

UNLESS AN ELECTRICAL HAZARD OF

A LEAKAGEOF CURRENTOCCURS

- ATA LOW VOLTAGEOF 0 V


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ELECTRICAL HAZARDS

SHORT CIRCUIT

OVERLOADING

LEAKAGEOF CURRENTCAUSES & EFFECTS ON APPLIANCEOR LIVES


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CAUSES & EFFECTOF SOMEELECTRICAL HAZARDS

CAUSES EFFECT

Short

Circuit

Electrical Contact between the barewires (live wire touches neutral wire)caused by torn or damaged insulationdue to deterioration from overusedwires.

Since the bare wires havealmost negligibleresistance, current will belarge. The amount of heatgenerated from this largecurrent will overheat thewires and appliance, andcan even start a fire.

Overloading Plugging too many appliances to onepower socket through the use of anadaptor.

Current drawn from thepower socket will be verylarge and this may result inoverheating of cable.

Electric

Shock

Commonly also known aselectrocution. This is caused either bydamp conditions (wet hands touchbare wires) or when the live wire isexposed and touches the metal partsof the appliance

The person who touchesthe live metal parts willreceive an electric shock,which may be fatal,depending on the hisresistance and current

through him.


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Safety Measures

Effects Safety

devices

How it works?

Short Circuit

& Overloading

Fuses and circuit-breakers

When a large current develops in the event of a

short circuit or overloading, this large current will

melt the fuse wire and when the fuse blows,

current flow to the cable and appliance is cut off.

Each fuse has a certain fuse rating which is thecurrent limit it can withstand before the fuse

blows. Thus the fuse protects the cable andthe appliance from overheating.

Electric Shock Earth wire connected tothe metal parts of the

appliance

When the appliance becomes live due to broken

insulation resulting in the bare live wire touching

the metal parts of the appliance, current from thelive wire will flow along the earth wire instead of

through the person. Thus the Earth wireprevents the user from an electric shock.

Damp

conditions

Hands must be dry and

appliances must not be

used in wet areas


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Switches and Fuses

Switches and fuses must be placed along the

live wire.Live wire

Neutral wire

fuseswitch


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