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In steady state, a resistor is carrying a current $i$. The power lost to the surroundings is $b(T- T_0)$. Here $b$ is a constant, $T$ is the temperature of the resistance and $T_0$ is the surrounding temperature. If the coefficient of linear expansion is $ \alpha$. The strain in the wire is

$\begin {array} {1 1} (A)\;\alpha i^2 \large\frac{R}{\lambda} & \quad (B)\;\alpha i R \lambda \\ (C)\;\alpha^2i\large\frac{R}{\lambda} & \quad (D)\;\alpha i^2 \large\frac{R}{2\lambda} \end {array}$


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1 Answer

This question requires basic concepts from several chapters but can be solved simply by
realizing the principle of conservation of energy: power developed = power dissipated.
$ i^2R = \lambda ( T - T_0)$
or, $(T – T_0) = i^2 \large\frac{R}{ \lambda}$
This is also the increase in temperature of the resistance
$ \Delta l =l \alpha \Delta T$
Strain = $ \large\frac{ \Delta l}{l}$
Hence, the strain $= \alpha \times ( T-T_0)= \alpha i^2 \large\frac{R}{\lambda}$
Ans : (A)
answered Feb 8, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1

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