Question

An aluminium (${\alpha _{Al}} = 4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$) wire resistance ‘${R_1}$’ and carbon wire (${\alpha _C} = 0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$) resistance ‘${R_2}$’ are connected in series to have a resultant resistance of $18\;{\rm{ohm}}$ at all temperatures. The values of ${R_1}$ and ${R_1}$ in ohms
A.$2$, $16$
B. $12$, $6$
C. $13$, $5$
D. $14$, $4$

Answer 1

Hint:In this question, we can use the theory that the resistance of a combination of wires is independent of the variation in temperature. Using the expression for the resultant resistance of the wires in series and the expression for the resultant resistance after a variation in temperature we can find the solution.

Complete step by step answer:
Given:
The temperature coefficient of resistance of aluminium, ${\alpha _{Al}} = 4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$
The temperature coefficient of resistance of carbon, ${\alpha _C} = 0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$
The resistance of aluminium wire is${R_1}$ .
The resistance of carbon wire is ${R_2}$ .
When wires of resistances ${R_1}$ and ${R_2}$ are connected in series, we can write the resultant resistance of the wires as
$R = {R_1} + {R_2}$
Here $R$ is the resultant resistance of wires of resistances ${R_1}$ and${R_2}$.
We can substitute $18\;\Omega $ for $R$ in in the above equation, we get
${R_1} + {R_2} = 18\;\Omega $
Let the temperature of the wires are increased by $\Delta T$. Then
The new resistance of the aluminium wire, ${{R_1}^1} = {R_1}\left( {1 + {\alpha _{Al}}\Delta T} \right)$
The new resistance of the carbon wire, ${{R_2}^1} = {R_2}\left( {1 + {\alpha _C}\Delta T} \right)$
When wires of resistances ${{R_1}^1}$ and ${{R_2}^1}$ are connected in series, we can write the resultant resistance of the wires as
${R^1} = {{R_1}^1} + {{R_2}^1}$
Here ${R^1}$ is the resultant resistance of ${{R_1}^1}$ and ${{R_2}^1}$ connected in series.
The resistances of a combination of wires remain the same even if the temperatures of the wires are increased. Hence, we can equate the resistances $R$ and ${R^1}$ to get
$
R = {R^1}\\
{R_1} + {R_2} = {{R_1}^1} + {{R_2}^1}
$
Substituting ${R_1}\left( {1 + {\alpha _{Al}}\Delta T} \right)$ for ${{R_1}^1}$ and ${R_2}\left( {1 + {\alpha _C}\Delta T} \right)$ for${{R_2}^1}$, we get
$
{R_1} + {R_2} = {R_1}\left( {1 + {\alpha _{Al}}\Delta T} \right) + {R_2}\left( {1 + {\alpha _C}\Delta T} \right)\\
\Rightarrow{R_1} + {R_2} = {R_1} + {R_1}{\alpha _{Al}}\Delta T + {R_2} + {R_2}{\alpha _C}\Delta T\\
\Rightarrow{R_2}{\alpha _C}\Delta T = - {R_1}{\alpha _{Al}}\Delta T
$
We can avoid the negative sign in the equation.
So, rewriting the above equation, we get
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{\alpha _{Al}}}}{{{\alpha _C}}}$
Hence, we obtained an equation in the form of the ratio between ${R_2}$ and${R_1}$ .
Substituting $4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$ for ${\alpha _{Al}}$ and $0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}$ for ${\alpha _C}$ in the above equation, we get
$
\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{4 \times {{10}^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}}}{{0.5 \times {{10}^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}}}\\
\Rightarrow\dfrac{{{R_2}}}{{{R_1}}} = 8\\
\Rightarrow{R_2} = 8{R_1}
$
Substituting $8{R_1}$ for ${R_2}$ in ${R_1} + {R_2} = 18\;\Omega $, we get
$
{R_1} + 8{R_1} = 18\;\Omega \\
\Rightarrow 9{R_1} = 18\;\Omega \\
\Rightarrow{R_1} = 2\;\Omega
$
Hence, we obtained the value of the resistance of the aluminium wire as$2\;\Omega $.
Substituting $2\;\Omega $ for ${R_1}$ in the equation${R_1} + {R_2} = 18\;\Omega $, we obtain
$
{R_1} + {R_2} = 18\;\Omega \\
\Rightarrow 2\;\Omega + {R_2} = 18\;\Omega \\
\therefore{R_2} = 16\;\Omega
$
Therefore, we obtained that the resistance ${R_1}$ of the aluminium wire is $2\;\Omega $ and the resistance ${R_2}$ of the carbon wire is $16\;\Omega $.

Hence, the option (A) is correct.

Note:The resistance of the individual wires generally changes with the variation in temperature. Since, the resultant resistance of wires in series are considered, we took it as independent of temperature.