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Question: An aluminium (\({\alpha _{Al}} = 4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}\)) wire resistance ‘\...

An aluminium (αAl=4×10−3  l∘C{\alpha _{Al}} = 4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}) wire resistance ‘R1{R_1}’ and carbon wire (αC=0.5×10−3  l∘C{\alpha _C} = 0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}) resistance ‘R2{R_2}’ are connected in series to have a resultant resistance of 18  ohm18\;{\rm{ohm}} at all temperatures. The values of R1{R_1} and R1{R_1} in ohms
A.22, 1616
B. 1212, 66
C. 1313, 55
D. 1414, 44

Explanation

Solution

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, αAl=4×10−3  l∘C{\alpha _{Al}} = 4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}
The temperature coefficient of resistance of carbon, αC=0.5×10−3  l∘C{\alpha _C} = 0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}}
The resistance of aluminium wire isR1{R_1} .
The resistance of carbon wire is R2{R_2} .
When wires of resistances R1{R_1} and R2{R_2} are connected in series, we can write the resultant resistance of the wires as
R=R1+R2R = {R_1} + {R_2}
Here RR is the resultant resistance of wires of resistances R1{R_1} andR2{R_2}.
We can substitute 18  Ω18\;\Omega for RR in in the above equation, we get
R1+R2=18  Ω{R_1} + {R_2} = 18\;\Omega
Let the temperature of the wires are increased by ΔT\Delta T. Then
The new resistance of the aluminium wire, R11=R1(1+αAlΔT){{R_1}^1} = {R_1}\left( {1 + {\alpha _{Al}}\Delta T} \right)
The new resistance of the carbon wire, R21=R2(1+αCΔT){{R_2}^1} = {R_2}\left( {1 + {\alpha _C}\Delta T} \right)
When wires of resistances R11{{R_1}^1} and R21{{R_2}^1} are connected in series, we can write the resultant resistance of the wires as
R1=R11+R21{R^1} = {{R_1}^1} + {{R_2}^1}
Here R1{R^1} is the resultant resistance of R11{{R_1}^1} and R21{{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 RR and R1{R^1} to get
R=R1 R1+R2=R11+R21 R = {R^1}\\\ {R_1} + {R_2} = {{R_1}^1} + {{R_2}^1}
Substituting R1(1+αAlΔT){R_1}\left( {1 + {\alpha _{Al}}\Delta T} \right) for R11{{R_1}^1} and R2(1+αCΔT){R_2}\left( {1 + {\alpha _C}\Delta T} \right) forR21{{R_2}^1}, we get
R1+R2=R1(1+αAlΔT)+R2(1+αCΔT) ⇒R1+R2=R1+R1αAlΔT+R2+R2αCΔT ⇒R2αCΔT=−R1αAlΔT {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
R2R1=αAlαC\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{\alpha _{Al}}}}{{{\alpha _C}}}
Hence, we obtained an equation in the form of the ratio between R2{R_2} andR1{R_1} .
Substituting 4×10−3  l∘C4 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}} for αAl{\alpha _{Al}} and 0.5×10−3  l∘C0.5 \times {10^{ - 3}}\;{\rm{l}}^\circ {\rm{C}} for αC{\alpha _C} in the above equation, we get
R2R1=4×10−3  l∘C0.5×10−3  l∘C ⇒R2R1=8 ⇒R2=8R1 \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 8R18{R_1} for R2{R_2} in R1+R2=18  Ω{R_1} + {R_2} = 18\;\Omega , we get
R1+8R1=18  Ω ⇒9R1=18  Ω ⇒R1=2  Ω {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 as2  Ω2\;\Omega .
Substituting 2  Ω2\;\Omega for R1{R_1} in the equationR1+R2=18  Ω{R_1} + {R_2} = 18\;\Omega , we obtain
R1+R2=18  Ω ⇒2  Ω+R2=18  Ω ∴R2=16  Ω {R_1} + {R_2} = 18\;\Omega \\\ \Rightarrow 2\;\Omega + {R_2} = 18\;\Omega \\\ \therefore{R_2} = 16\;\Omega
Therefore, we obtained that the resistance R1{R_1} of the aluminium wire is 2  Ω2\;\Omega and the resistance R2{R_2} of the carbon wire is 16  Ω16\;\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.