
What are the essential requirements of a heater filament?
A. high resistivity, low melting point
B. low resistivity, low melting point
C. high resistivity, high melting point
D. low resistivity, high melting point
Answer
569.1k+ views
Hint: Firstly, you could recall the expression for Joule’s heating and the substitute for the resistance in terms of resistivity and determine how the resistivity of the heater filament should be. Also, remember that the filament has to withstand high heat and thus determine whether melting point should be high or low.
Formula used:
Joule’s Law of heating,
$H={{I}^{2}}Rt$
Complete answer:
In the question, we are asked to find the essential requirements of a heater filament from the given options.
Let us recall Joule’s law of heating which states that, for a current (I) passing through a wire of resistance R in t time interval, the heat developed will be the product of the square of the current, resistance and time. That is,
$H={{I}^{2}}Rt$ ………………………………………. (1)
But we know that resistance R is given by,
$R=\rho \dfrac{L}{A}$ …………………………….. (2)
Substituting (2) in (1),
$H={{I}^{2}}\left( \rho \dfrac{L}{A} \right)t$
So, we could say that greater the resistivity of the filament is, greater will be the heat produced by it. Also, from (2) we see that resistivity is inversely proportional to the length of the filament. So by increasing the resistivity, we could also minimize the length of the filament.
Now, for the melting point of the filament, it should definitely be high. This is because the filament has to withstand high heat, so, by having a high melting point it can bear this high heat without being destroyed.
Therefore, we could summarize that high resistivity and high melting point are the essential requirements of a heater filament. Hence, option C is found to be the correct answer.
Note:
The heater filament or the heating element basically works on the principle of Joule’s heating. It is converting electric energy into heat. This heating happens as the result of the encounter of electric current with resistance. The types of heating elements used may be metals, polymer, thick films, ceramic or semiconductors.
Formula used:
Joule’s Law of heating,
$H={{I}^{2}}Rt$
Complete answer:
In the question, we are asked to find the essential requirements of a heater filament from the given options.
Let us recall Joule’s law of heating which states that, for a current (I) passing through a wire of resistance R in t time interval, the heat developed will be the product of the square of the current, resistance and time. That is,
$H={{I}^{2}}Rt$ ………………………………………. (1)
But we know that resistance R is given by,
$R=\rho \dfrac{L}{A}$ …………………………….. (2)
Substituting (2) in (1),
$H={{I}^{2}}\left( \rho \dfrac{L}{A} \right)t$
So, we could say that greater the resistivity of the filament is, greater will be the heat produced by it. Also, from (2) we see that resistivity is inversely proportional to the length of the filament. So by increasing the resistivity, we could also minimize the length of the filament.
Now, for the melting point of the filament, it should definitely be high. This is because the filament has to withstand high heat, so, by having a high melting point it can bear this high heat without being destroyed.
Therefore, we could summarize that high resistivity and high melting point are the essential requirements of a heater filament. Hence, option C is found to be the correct answer.
Note:
The heater filament or the heating element basically works on the principle of Joule’s heating. It is converting electric energy into heat. This heating happens as the result of the encounter of electric current with resistance. The types of heating elements used may be metals, polymer, thick films, ceramic or semiconductors.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

