Specific objectives.

  1. Perform and describe experiments to illustrate the heating effect of electric current.
  2. State he factors affecting heating by an electric current.
  3. Derive the equations for electrical energy and power.
  4. Identify devices in which heating effect of an electric current is applied.
  5. Solve numerical problems involving electrical energy and electrical power.

 

  • Electric current asit flows or passes through a conductor,it creates electric energy. This energy can be converted to other forms e.g. heat energy,light energy,sound and mechanical energy.
  • The appliances that convert electric energy to heat energy have heating doment/resistor.
  • The heating appliances are
  1. Electric kettle.
  2. Electric heater
  3. Iron box
  4. Electric cooker.
  • The heating element contain a resistor wound on insulator or asbestos. The common one is nichrome wire.

                        Properties of heating elements.

  • High resistance. The heat energy is directly proportional to resistance.
  • High melting point.
  • Should not react with air to form an oxide.
  • Should not wear out easily.
  • High intensity. Should be able to withstand high voltage.

                        Factors affecting heating effect of electric current.

  • The electric current produced is determined by;
  1. Electric current (I).
  • The heat energy produced by an electric current is directly proportional to the square of e.m.f.

 

  1. The resistance of the conductor.
  • The heat energy produced is directly proportional to the resistance R

 

  1.  
  • The heat energy produced is directly proportional to the time taken to pass the current.

 

  • From the above factors.

                        Heat energy            H  I2RT

                        H = I2RT

  • This is called Joules law which states that energy developed in a heating element is directly proportional to the square of current, the resistance of the heating element and the time taken.

                         H = I2RT

                             = I2  T

The SI units is joules.

Example.

  1. A heating coil producing 3600J/min is required when the potential difference (p.d) across it is 24. Calculate the length of the wire making the coil if the cross- section of the wire is 1 x 10-7 mm and the resistivity of the wire is 1 x 10-6

            H =                       3600 =

            R =     R = 9.6 n  

  1. An electric heater is connected to a 200V supply. If the heating element has a resistance of 10n, calculate
  2. The electric energy supplied in 2 minutes.

H = 480,000

H = 4.8 X 104

  1. The power of the heater in kilo watts.

                                    Heating element.

  • It is an alloy of nickel, iron and chromium.
  • Has high resistance.
  • Has high melting point.
  • It does not form an oxide when hot.
  • It is used in electric fire and electric kettle.

                                    Lighting effect of electric current.

  • Electric current can be converted to electric energy by lighting appliances. These are filament lamb and fluorescent lamp/gas tube.
  • The appliances contain lighting element which is either a resistor and the resistor is tungsten or gas or sodium or silver vapor.
  • They convert electrical energy to light energy.

                                    Properties of lighting elements (tungsten).

  1. Have high melting point of 3500.
  2. Has high resistance.
  3. Becomes white hot when current flows through it.
  4. As the temperature increases, its ability to change electrical energy to light energy increases.
  5. Does not form an oxide easily.
  6. Does not wear out easily.
  7. Filament lamb.
  • Contains tungsten as the lighting element which is coiled.
  • The bulb is filled with inert gas, nitrogen or argon.
  • They do not react. This prevents vaporization of the metal filament and blackening of the flask.
  • The tungsten is coiled to increase its length and its surface area.
  • The length is to increase the resistance. Long thin wire has high resistance. A lot of heat is produced hence lowering efficiency of the bulb when a current flows through it.
  • Most of the heat is carried away from the filament by convection current of air and radiation.
  • Very little energy is converted to light energy hence the bulb is not efficient.
  • The filament is coiled double made by winding the filament on itself. This makes it more efficient than a coiled one because of increased surface area and resistance. It becomes more hotter than the coil, that is, it is compact and less affected by the cooling effect of the convection in gas.

                                    Disadvantage of filament bulb.

  1. Not efficient as most of the electrical energy is converted to heat energy. It consumes a lot of electrical energy.
  2. Very expensive to maintain as they do not last long.

                                    Advantage of filament bulb.

  1. It is cheap.
  2. Fluorescent bulb.
  • Contain mercury or sodium vapor and the tube is coated inside with fluorescent powder.
  • Different powder produces different colors.
  • When switched on, the sodium vapor emits ultra violet radiation which makes the powder on the walls of the tube to fluorescence or to glow or to emit light.

                                    Advantages of fluorescent bulb.

  1. More efficient as most of the electrical energy is converted to light energy.
  2. Cheap to maintain
  3. They last longer.

                                    Disadvantages of fluorescent bulb.

  1. They are expensive to install.

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