In our daily life we make use of different forms of energy. Of these electricity is the most useful and indispensable. It is the only form which can be conveniently converted into any other form of energy to suit our various needs. Electricity is so much intimately connected to our life. We can not imagine a world without it. All electrical devices depend upon one or more effects of electric current. They are chemical effect, heating effect, lighting effect, magnetic effect and mechanical effect. , Here study about heating effect of electricity.
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Let us examine through an experiment how we can find out the amount of heat energy produced in a current carrying conductor. And also the factors depending on it.Two small pieces of thin aluminium wire and nichrome wire of nearly equal length and thickness are connected in series to a 6V battery using copper conductor. Pass an electric current through the circuit for some time. What do you observe? Nichrome wire becomes red hot.
Why is the nichrome wire red hot while the aluminium wire is not so? Though the magnitude of current in each wire is the same why does the nichrome wire alone become red hot? The resistance of the nichrome wire (R) is greater than that of aluminium wire. So we understand that a conductor of high resistance can convert more electric energy into heat energy.
Discuss the observations in the above experiment and list the factors affecting the conversion of electric energy into heat energy.
Electric current (I)
Resistance of the conductor (R)
Time of flow of electric current (t)
If I is the current in ampere (A), R the resistance of the conductor in ohm (Ω), t the time in seconds (s) then the heat H produced in joule (J), is
H = I2Rt joule
A law stating that the current in an circuit is directly proportional to the electromotive force (voltage) in the circuit and inversely proportional to the resistance in the circuit.
H = VI t
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Ex:1 An electric heater works in a 230V supply which draws a current of 5A for 3 minutes. Find out the quantity of heat produced in the heater. Also find out the resistance of the coil in the heater.
Sol: V = 230V
I = 5A
t = 3 minute
= 3 × 60 second
= 180 second
H = VI t or V = IR
= 230V × 5A × 180s
The resistance of the coil
R = V/I = 230V/5A
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