- Sources of Mains Electricity
- Power Transmission
- Electrical Power and Energy
- Costing Electricity
- Domestic Wiring System
Sources of Mains Electricity
- Mains electricity comes from a power station and its current is the alternating current which can either be stepped up or down by a transformer.
- A.c is produced when a conductor is rotated in a magnetic field or when a magnetic field is rotated near a conductor.
- This method is known as electromagnetic induction.
- The source of energy for rotating the turbine is the actual source of electrical energy.
- Most of the electricity in East Africa is generated from water.
Power Transmission
- This is the bulk transfer of electric power from one place to another.
- A power transmission system in a country is referred to as the national grid.
- This transmission grid is a network of power generating stations, transmission circuits and sub-stations.
- It is usually transmitted in three phase alternating current .
Grid Input
- At the generating plant the power is produced at a relatively low voltage of up to 25 kV then stepped up by the power station transformer up to 400 kV for transmission.
- It is transmitted by overhead cables at high voltage to minimize energy losses. The cables are made of aluminium because it is less dense than copper.
- Metallic poles ( pylons ) carry four cables, one for each phase and the fourth is the neutral cable which is thinner and completes the circuit to the generator.
Grid Exit
- At sub-stations transformers are used to step down voltage to a lower voltage for distribution to industrial and domestic users.
- The combination of sub-transmission (33 kV to 132 kV) and distribution (11 kV to 33 kV) which is then finally transformed to a voltage of 240 V for domestic use.
Electricity Distribution
- This is the penultimate process of delivery of electric power.
- It is considered to include medium voltage (less than 50 kV) power lines, low voltage (less than 1,000 V) distribution, wiring and sometimes electricity meters.
Dangers of High Voltage Transmission
- They can lead to death through electrocution
- They can cause fires during upsurge
- Electromagnetic radiations from power lines elevate the risk of certain types of cancer
Electrical Power and Energy
- Work done = volts × coulombs = VQ, but Q = current × time = It. So work done = VIt
- Other expressions for work may be obtained by substituting V and I from Ohms law as below
V = IR and I = V/R, work done = IR × It = I2Rt, or work done = V × V × t/R = V2t/R . - The three expressions can be used to calculate work done.
- Electrical power may be computed from the definition of power.
Power = work/time = I2Rt/t = I2R or V2t/Rt = V2/R Using work done = VIt, then Power = VI. - These expressions are useful in solving problems in electricity.
- Work done or electrical energy is measured in joules (J) and power is measured in watts (W).
- 1W = 1 J/s.
Example
An electric heater running on 240 V mains has a current of 2.5 A.- What is its power rating?
- What is the resistance of its element?
Solution
a) Power = VI = 240 × 2.5 = 600 W. Rating is 600 W, 240 V.
b) Power = V/R = 600 W.
R = V/I.
R = 240/2.5
Costing Electricity
- The power company uses a unit called kilowatt hour (kWh) which is the energy transformed by a kW appliance in one hour .
1 kW = 1,000 W × 60 × 60 seconds = 3,600,000 J . - The meter used for measuring electrical energy uses the kWh as the unit and is known as joule meter.
Examples
- An electric kettle is rated at 2,500 W and uses a voltage of 240 V.
- If electricity costs Ksh 1.10 per kWh, what is the cost of running it for 6 hrs?
- What would be its rate of dissipating energy if the mains voltage was dropped to 120 V?
Solution
a) Energy transformed in 6 hrs = 2.5 × 6 = 15 kWh. Cost = 15 × 1.10 × 6 = Ksh 99.00
b) Power = V2/R = 2500.
R = (240 × 240)/2500 = Current = V/R = (240 × 2500)/(240 × 240) = 10.42 A
Power = VI = (2500 × 120)/240 = 1,250 W.
- An electric heater is made of a wire of re supply. Determine the;
- Power rating of the heater
- Current flowing in the circui
- Time taken for the heater to raise the temperature of 200 g of water from 23 0 C to 95 0 C. (specific heat capacity of water = 4,200 J Kg -1 K -1 )
- Cost of using the heater for two hours a day for 30 days if the power company charges Ksh 5.00 per kWh.
Solution
a) Power = V2/R = (240 × 240)/100 = 576 W
b) P = VI
I = P/V = 576/240 = 2.4 A
c) P × t = heat supplied = (mcθ) = 576 × t
Hence t = (0.2 × 4200 × 72)/576 = 105 seconds.
d) Cost = kWh × cost per unit = (0.576 × 2 × 30) × 5.0 = Ksh 172.80
- A house has five rooms each with a 60 W, 240 V bulb. If the bulbs are switched on from 7.00 pm to 10.30 pm, calculate the;
- Power consumed per day in kWh
- Cost per week for lighting those rooms if it costs 90 cents per unit.
Solution
a) Power consumed by 5 bulbs = 60 × 5 = 300 W = 0.3 kWh.
Time = 10.30 –7.00 = 3 ½ hrs.
Therefore for the time duration = 0.3 × 3 ½ = 1.05 kWh.
b) Power consumed in 7 days = 1.05 × 7 = 7.35 kWh. Cost = 7.35 × 0.9 = Ksh 6.62
Domestic Wiring System
- Power is supplied by two cables where one line is live wire (L) and the other is neutral (N). Domestic supply in Kenya is usually of voltage 240 V .
- The current alternates 50 times per second hence the frequency is 50 Hz . The neutral is earthed to maintain a zero potential.
- The main fuse is fitted on the live wire to cut off supply in case of a fault.
- A fuse is a short piece of wire which melts if current of more value flows through it.
- Supply to the house is fed to the joule meter which measures the energy consumed.
- From the meter both L and N cables go to the consumer box (fuse box) through the main switch which is fitted on the live cable.
- Consumer units within the house are fitted with circuit breakers which go off whenever there is a default in the system.
- Lights in the house are controlled by a single or double switch (two way).
- In most wiring systems the main sockets are connected to a ring main which is a cable which starts and end at the consumer unit.
- Plugs used are the three- pin type.
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