PHYSICS S.S ONE(ELECTRICAL ENERGY AND POWER)

 SECOND TERM

WEEK: 7

ELECTRICAL ENERGY AND POWER

Work is done when electricity flows from one point to another of different potential. If Q coulombs of electricity flows between the two points whose differences in potential is V volts, then the work done (W) is given by W = Q V (joules).

The quantity of electricity Q coulombs is given by Q = I t; where I is the current in amperes and t is the time of flow of current in seconds.

W = QV = IVt

From ohm’s law, V =IR. Then W = I2Rt

This electrical energy can be converted into other forms of energy as follows:

Heat energy e.g.  electrical iron, electrical kettle and heater. Light energy as in bulb. Mechanical energy as in washing machine, blender, and electric fan. Sound player as in telephone earpiece.

Electrical power

Power is defined as the rate of doing work. P = .

Electrical Power is the amount of electrical work done per seconds.   P =  = IV. the unit of power is the watt. Hence watt is Joule/seconds.

One watt is the power consumed in electric circuit when one joule of work is done in one second.

Buying and Selling of electric Power

Electric power consumption is measured and sold by Power Holding Company of Nigeria (PHCN) in units of kilowatts-hour (kWh).

One kilowatt-hour (kWh) is the electrical energy consumed by an appliance when power of one kilowatt is used by the appliance for one hour. 1kWh = 3.6x106 Joules

Example

1.        What current is taken by 8kwh electrical element on the 240v mains? What resistance rate of fuse should be used in the plug?

Solution

P=8kwh = 8000wh; V=240V; I=?

P=IV; I=P/V = 8000/240= 33.3A.     V=IR ; R=V/R = 240/33.3 =7.21ohm

2.       A television (65W), a refrigerator (1.2kW), an electric kettle (650W) and 10 lamps (40W each) are connected in series in a house. How much will it cost the house keeper to switch on all the appliances for 24 hours if the cost of electricity is 50kobo per kwh?

Solution

Electronics

Unit  x rating (W)

Total (W)

Television

1 x65

65

Refrigerator

1 x 1200

1200

Electric kettle

1 x 650

650

Lamp

10 x 40

400

                                                                                                                                                Total =2315W = 2.315kW

                Total Electric power consumption =  2.315kW x 24h =55.56kwh

                1 kwh cost 15 kobo

                55.56kwh will cost 15k x 55.56 =833.4 kobo = #8.33k        

PRESENTATION

Step I: The teacher explains electrical energy and how to derive the equation.

Step II: The teacher explains electrical power and buying and selling of electrical power.

Step III: The teacher leads the students to solve the mathematical problems on electrical energy and power.

EVALUATION

The teacher evaluates the students by asking the following questions:

i.                     Define electrical energy

ii.                   Give examples of electrical appliances and their ratings

iii.                  Define electric power

iv.                 state the unit of electrical consumption

v.                   A potential difference of 6v is used to produce a current of 5A for 200s through a heating coil. The heat produced is?

ASSIGNMENT

1.       A stand-by generator is connected to fifteen 40W lamps and musician’s 600W amplifying system. How much energy is used if the generator runs for 6 hours?

2.       If the maximum voltage across a 100 ohm resistor is 20V, then the maximum power it can dissipate is?

                                                                                                                                               

 

 

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