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Class 10, Science, Chapter-11, Lecture-4, Heating Effect of current (Notes)

Joule’s Law of Heating:

Heat produced in a resistor is
(i) directly proportional to the square of current for a given resistance,
(ii) directly proportional to resistance for a given current, and
(iii) directly proportional to the time for which the current flows through the resistor

HEAT PRODUCED BY ELECTRIC CURRENT

The quantity of heat produced when a current, I, flows through a wire of resistance R, in time t, is given by the formula H = I2 R t

DERIVATION:

The amount of work done in carrying a charge Q, through a potential difference V, is given by

$${\rm{W}} = {\rm{Q}} \cdot {\rm{V - - - (i)}}$$

And, the quantity of charge carried by a current I, in time t, is given by the relation

$${\rm{Q}} = {\rm{I}} \cdot {\rm{t - - - (ii)}}$$

And, the potential difference V, across a wire of resistance R, carrying a current I, is given by

$${\rm{V}} = {\rm{I}} \cdot {\rm{R - - - (iii)}}$$

Substituting the values of  ${\rm{Q}}$  from  ${\rm{(ii)}}$  and  ${\rm{V}}$  from  ${\rm{(iii)}}$  in equation  ${\rm{(i)}}$

$${\rm{W}} = \left( {{\rm{I}} \cdot {\rm{R}}} \right) \cdot \left( {{\rm{I}} \cdot {\rm{t}}} \right)$$

$$ \Rightarrow {\rm{W}} = {{\rm{I}}^2} \cdot {\rm{R}} \cdot {\rm{t}}$$

This work done is converted into heat,  ${\rm{H}}$ 

Therefore, ${\rm{H}} = {{\rm{I}}^2} \cdot {\rm{R}} \cdot {\rm{t}}$

 

FACTORS ON WHICH HEAT PRODUCED DEPENDS:

The heat produced is directly proportional to

  1. Square of current $\left( {{{\rm{I}}^2}} \right)$
  2. Resistance $\left( {{\rm{R}}} \right)$
  3. Time for which current is passed $\left( {{\rm{t}}} \right)$

ELECTRIC POWER:

The rate at which electrical energy is consumed with respect to time is called electrical power.
Mathematically, ${\rm{P}} = {{\rm{E}} \over {\rm{t}}}$ , where ${\rm{E}}$ is Electrical energy consumed, ${\rm{t}}$ is the time for which energy is consumed.
and

Formulae for power:

$${\rm{P}} = {{\rm{E}} \over {\rm{t}}}$$

$$ \Rightarrow {\rm{P}} = {{{{\rm{I}}^2}{\rm{Rt}}} \over {\rm{t}}} = {{\rm{I}}^2}{\rm{R}} = {\rm{IR}} \cdot {\rm{I}}$$

$$ = {\rm{V}} \cdot {\rm{I}} = {\rm{V}} \cdot \left( {{{\rm{V}} \over {\rm{R}}}} \right) = {{{{\rm{V}}^2}} \over {\rm{R}}}$$

Therefore,

$${\rm{P}} = {\rm{V}} \cdot {\rm{I}}$$

$${\rm{P}} = {{\rm{I}}^2} \cdot {\rm{R}}$$

$${\rm{P}} = {{{{\rm{V}}^2}} \over {\rm{R}}}$$

RELATION BETWEEN COMMERCIAL UNIT AND SI UNIT OF ELECTRICAL ENERGY:

SI unit of energy is joule ( J )

Commercial unit of energy is kilo watt hour ( kWh )

The amount of energy consumed in 1 hour by an appliance having a power of 1 kW is equal to 1 kWh

Therefore,

1 kWh = 1 kW $ \times $ 1 h

= 1000 W  $ \times $ 60 $ \times $ 60 s

= 1000 Js –1 $ \times $ 3600 s

= 3600000 J

$\therefore$  1 k W h = 3.6 $ \times $ 10 6 J

 

The current that makes the heater element very hot only slightly warms the connecting wires leading to the heater.

Reason:

When the heater element is connected to the power supply by insulated copper wires, a large amount of heat is produced in the heating coils because they have high resistivity, but negligible heat is produced in the connecting wires of copper because copper has very low resistivity.