Tag: electric current and its effects

Electric bulb works by heating up a small wire called a filament. Electric current heats the filament so much that it glow. In fact, $90\%$ of the electricity used by a electric bulb is converted to heat and this conversion will produce maximum wastage of electricity.

Treating initially that there is no radiation and on applying voltage to the heater, the temperature of the heater rises. As the temperature rises, the radiation also increases.
As the input power($i^2r$) is constant, the output power should also be constant.
Initially temperature will be increasing and  radiation as well. When the heater reaches a certain temperature, the radiation power equals the input power and the temperature of the heater stops rising.
Thus temperature becomes constant after some time.  

As  heat produced depends upon resistance which has no relation with AC or DC but heat produced  also depends upon current as $H \propto\  I^2$ so for heating effect both A.C. and D.C. can be used, that's why appliances based on heating effect can work on AC as well as DC. hence correct option is C.

The right matches are given below.
Heat Generated$\rightarrow $Propotional to the square of current$\rightarrow $ $\frac { VIt }{ 4.18 } cal$  
Resistance in parallel$\rightarrow $Is used to reduce effective resistance in a circuit$\rightarrow $  $\frac { 1 }{ { R } _{ p } } =\frac { 1 }{ { R } _{ 1 } } +\frac { 1 }{ { R } _{ 2 } } $
Resistivity$\rightarrow $Depends on the material of the conductor$\rightarrow $  $\rho =\frac { RA }{ l } $
Ohm's law$\rightarrow $Gives relation between V and I$\rightarrow $V=IR

An electric iron does not use an electric motor .

$\displaystyle I = \frac {m}{Zt} = \frac {297 \times 10^{-6}}{3.3 \times 10^7 \times 15 \times 60}=1A$

$H = I^2Rt = 1 \times 10 \times 15 \times 60$
$=9000J = 9kJ$

The differential form of the Joule heating equation gives the power per unit volume.