An electric wire is stretched to increase its length by 25%.By what % will the resistance be increased and what will be increase in its resistivity?
Answers
Answer:
Explanation:
Formula used, R= PL/A
As length is increased by stretching the wire , therefore the volume of the wire won't change. Therefore, final volume=Initial volume
Hence, AL=A'L'
Final length= {1+(25/100)} Initial length
Hence, final length L'={125/100}L...(1)
R=pL/A
Multiply den and num by L
Hence, R= pL^2/V. (AL=V)
Finally, R'= pL'^2/V
Put the value of L' from eq1
R'= 1.5625R
Or resistance increase by
(156.25)R-R/(R)×100%= 156.25%
Resistivity is the resistance of per unit length of material and is independent of the dimensions of wire .
Therefore it will remain same
Answer:
Explanation:
Explanation:
Formula used, R= PL/A
As length is increased by stretching the wire , therefore the volume of the wire won't change. Therefore, final volume=Initial volume
Hence,
LA = L`A`
LA= 1.25L A`
A`= A/1.25
R / R` = PL/A whole divided by P 1.25L/A/1.25
R'= 1.5625R
Or resistance increase by
1.5625R - R / (R)×100%= 56.25%