The resistance of a wire at temperatures \[{t^ \circ }C\] and \[{0^ \circ }C\] are related by
A. ${R_t} = {R_0}\left( {1 + \alpha t} \right)$
B. ${R_t} = {R_0}\left( {1 - \alpha t} \right)$
C. ${R_t} = {R_0}^2\left( {1 + \alpha t} \right)$
D. ${R_t} = {R_0}^2\left( {1 - \alpha t} \right)$
Answer
266.4k+ views
Hint: To get an expression relating the resistance at two different temperatures we use the relationship between resistance and temperature. The change in resistance with temperature is directly proportional to the resistance at that temperature and the constant of proportionality obtained is known as temperature coefficient of resistance.
Complete step by step solution:
The resistance is related to temperature as the rate of change of resistance with change in temperature is directly proportional to the resistance at that temperature and the proportionality constant is known as the temperature coefficient of resistance denoted by the symbol $\alpha $. That is,
$\dfrac{{dR}}{{dT}} = \alpha R$
If the change in resistance with temperature is small then this relation can be written as,
$\dfrac{{\vartriangle R}}{{\vartriangle T}} = \alpha R$
Given two temperatures are ${0^ \circ }C\,\& \,{t^ \circ }C$. Here, the change in temperature, that is, $\vartriangle T$ is ${\left( {t - 0} \right)^ \circ }C$. So the equation becomes,
$\vartriangle R = \alpha {R_0}\left( {t - 0} \right) \\ $
That is,
$\vartriangle R = \alpha {R_0}t \\ $ ...(1)
Also,
$\vartriangle R = {R_t} - {R_0} \\ $
This implies that,
${R_t} = {R_0} + \vartriangle R \\ $ ...(2)
Substituting equation (1) in equation (2), we get,
\[{R_t} = {R_0} + \alpha {R_0}t \\ \]
Taking ${R_0}$ common on right hand side of the equation, we get,
\[{R_t} = {R_0}\left( {1 + \alpha t} \right)\]
Hence, option A is the correct answer.
Note: Consider a basic model of resistance to better understand the temperature dependency. Atoms and molecules in a conductor obstruct the flow of electrons. It becomes more difficult for the electrons to pass by as these atoms and molecules bounce about more. As a result, resistance typically rises with temperature.
Complete step by step solution:
The resistance is related to temperature as the rate of change of resistance with change in temperature is directly proportional to the resistance at that temperature and the proportionality constant is known as the temperature coefficient of resistance denoted by the symbol $\alpha $. That is,
$\dfrac{{dR}}{{dT}} = \alpha R$
If the change in resistance with temperature is small then this relation can be written as,
$\dfrac{{\vartriangle R}}{{\vartriangle T}} = \alpha R$
Given two temperatures are ${0^ \circ }C\,\& \,{t^ \circ }C$. Here, the change in temperature, that is, $\vartriangle T$ is ${\left( {t - 0} \right)^ \circ }C$. So the equation becomes,
$\vartriangle R = \alpha {R_0}\left( {t - 0} \right) \\ $
That is,
$\vartriangle R = \alpha {R_0}t \\ $ ...(1)
Also,
$\vartriangle R = {R_t} - {R_0} \\ $
This implies that,
${R_t} = {R_0} + \vartriangle R \\ $ ...(2)
Substituting equation (1) in equation (2), we get,
\[{R_t} = {R_0} + \alpha {R_0}t \\ \]
Taking ${R_0}$ common on right hand side of the equation, we get,
\[{R_t} = {R_0}\left( {1 + \alpha t} \right)\]
Hence, option A is the correct answer.
Note: Consider a basic model of resistance to better understand the temperature dependency. Atoms and molecules in a conductor obstruct the flow of electrons. It becomes more difficult for the electrons to pass by as these atoms and molecules bounce about more. As a result, resistance typically rises with temperature.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Young’s Double Slit Experiment Derivation Explained

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Kinematics Mock Test for JEE Main 2025-26: Practice & Ace the Exam

Kinematics Mock Test for JEE Main 2025-26: Comprehensive Practice

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Marks vs Rank: Estimate IIT Rank from Your Score

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Derivation of Equation of Trajectory Explained for Students

