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# The temperature coefficient of resistance of tungsten is $4.5 \times {10^{ - 3}}^\circ {C^{ - 1}}$ and that of germanium is $- 5 \times {10^{ - 2}}^\circ {C^{ - 1}}$. A tungsten wire of resistance $100\Omega$ is connected in series with a germanium wire of resistance $R$. The value of $R$ for which the resistance of combination does not change with temperature is?(A) $9\Omega$(B) $1111\Omega$(C) $0.9\Omega$(D) $111.1\Omega$

Last updated date: 14th Jun 2024
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Hint In electric conductors as temperature increases, interatomic vibration also increases which creates difficulty for the electrons to travel inside the electric conductor. But in the case of a semiconductor as the temperature increases the number of charge carriers also increases so as there are more charge carrier resistivity decreases.
Formula used:
${R_{eq}} = {R_1} + {R_2}$
${R_{eq}}$Is the equivalent resistance, ${R_1}$and${R_2}$ are the two resistance connected in series.
$\Rightarrow R' = R(1 + \alpha \Delta T)$
$R'$Is the new resistance after an increase in temperature,$R$ is the original resistance, $\alpha$ is the temperature coefficient of resistance, and $\Delta T$is the change in temperature.

Let at initial temperature equivalent resistance be ${R_{eq}}$ and resistance of tungsten be ${R_1}$and resistance of germanium be${R_2}$.
Temperature coefficient of resistance for germanium be${\alpha _2}$ and temperature coefficient of resistance for tungsten be${\alpha _1}$.
Given that,
${\alpha _1} = 4.5 \times {10^{ - 3}}^\circ {C^{ - 1}}$
${\alpha _2} = - 5 \times {10^{ - 2}}^\circ {C^{ - 1}}$
As we want that after heating equivalent resistance remains the same.
Therefore equivalent resistance after heating is equal to${R_{eq}}$.
We know that,
$R' = R(1 + \alpha \Delta T)$
$R'$ Is the new resistance after an increase in temperature,$R$ is the original resistance, $\alpha$ is the temperature coefficient of resistance, and $\Delta T$is the change in temperature.
Hence,
The resistance of tungsten and germanium after heating is ${R_1}'$and${R_2}'$ respectively.
Therefore
$\Rightarrow {R_1}' = {R_1}(1 + {\alpha _1}\Delta T)$
$\Rightarrow {R_2}' = {R_2}(1 + {\alpha _2}\Delta T)$
From the things we discussed above,
${R_1}' + {R_2}' = {R_1} + {R_2}$
$\Rightarrow {R_1} + {R_2} = {R_1}(1 + {\alpha _1}\Delta T) + {R_2}(1 + {\alpha _2}\Delta T)$
$\Rightarrow {R_1} + {R_2} = {R_1} + {R_1}{\alpha _1}\Delta T + {R_2} + {R_2}{\alpha _2}\Delta T$
$\Rightarrow {R_1}{\alpha _1}\Delta T = - {R_2}{\alpha _2}\Delta T$
On substituting the values
${R_1}(4.5 \times {10^{ - 3}}^\circ {C^{ - 1}})\Delta T = - {R_2}( - 5 \times {10^{ - 2}}^\circ {C^{ - 1}})\Delta T$
It is given in the question that ${R_1} = 100\Omega$
$\Rightarrow 100(4.5 \times {10^{ - 3}}) = {R_2}(5 \times {10^{ - 2}})$
$\Rightarrow {R_2} = \dfrac{{100(4.5 \times {{10}^{ - 3}})}}{{5 \times {{10}^{ - 2}}}}$
On solving further,
${R_2} = 9\Omega$

Hence the correct answer to the question is (A) $9\Omega$

Note
Semiconductors are the material whose resistivity lies between the range of electric conductors and electric insulators. Thus they share common traits with both electric conductors and electric insulators. The negative temperature coefficient represents that the material is a semiconductor as its resistance decreases on increase in temperature.