Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
(N/A) The resistivity of an alloy is generally much higher than that of its constituent pure metals.
Due to this high resistivity,alloys produce more heat when an electric current passes through them.
Furthermore,alloys do not oxidize (burn) or melt readily even at high temperatures.
Therefore,alloys are preferred for heating elements in appliances like electric toasters and electric irons.
Due to this high resistivity,alloys produce more heat when an electric current passes through them.
Furthermore,alloys do not oxidize (burn) or melt readily even at high temperatures.
Therefore,alloys are preferred for heating elements in appliances like electric toasters and electric irons.
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Similar Questions
What is meant by "electrical resistance" of a conductor? State how the resistance of a conductor is affected when $(i)$ a low current passes through it for a short duration; $(ii)$ a heavy current passes through it for about $30$ seconds.
Medium
View SolutionExplain the following.
$(a)$ Why is tungsten used almost exclusively for the filament of electric lamps?
$(b)$ Why are the conductors of electric heating devices,such as bread-toasters and electric irons,made of an alloy rather than a pure metal?
$(a)$ Why is tungsten used almost exclusively for the filament of electric lamps?
$(b)$ Why are the conductors of electric heating devices,such as bread-toasters and electric irons,made of an alloy rather than a pure metal?
Medium
View SolutionAn electric iron has a rating of $750\, W; 200\, V$. Calculate:
$(i)$ The current required.
$(ii)$ The resistance of its heating element.
$(iii)$ Energy consumed by the iron in $2$ hours.
$(i)$ The current required.
$(ii)$ The resistance of its heating element.
$(iii)$ Energy consumed by the iron in $2$ hours.
Medium
View SolutionThe electrical resistivity of a few materials is given below in ohm-metre. Which of these materials can be used for making the element of a heating device?
$A$. $6.84 \times 10^{-8} \, \Omega m$ $B$. $1.60 \times 10^{-8} \, \Omega m$ $C$. $1.00 \times 10^{-4} \, \Omega m$ $D$. $2.50 \times 10^{12} \, \Omega m$ $E$. $4.40 \times 10^{-5} \, \Omega m$ $F$. $2.30 \times 10^{17} \, \Omega m$
| $A$. $6.84 \times 10^{-8} \, \Omega m$ |
| $B$. $1.60 \times 10^{-8} \, \Omega m$ |
| $C$. $1.00 \times 10^{-4} \, \Omega m$ |
| $D$. $2.50 \times 10^{12} \, \Omega m$ |
| $E$. $4.40 \times 10^{-5} \, \Omega m$ |
| $F$. $2.30 \times 10^{17} \, \Omega m$ |
Easy
View SolutionIf the current $I$ through a resistor is increased by $100\%$ (assume that temperature remains unchanged), the increase in power dissipated will be ........ $\%$
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