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Question: The temperature coefficient of resistance of tungsten is \(4.5 \times {10^{ - 3}}^\circ {C^{ - 1}}\)...

The temperature coefficient of resistance of tungsten is 4.5×10−3∘C−14.5 \times {10^{ - 3}}^\circ {C^{ - 1}} and that of germanium is −5×10−2∘C−1 - 5 \times {10^{ - 2}}^\circ {C^{ - 1}}. A tungsten wire of resistance 100Ω100\Omega is connected in series with a germanium wire of resistance RR. The value of RR for which the resistance of combination does not change with temperature is?
(A) 9Ω9\Omega
(B) 1111Ω1111\Omega
(C) 0.9Ω0.9\Omega
(D) 111.1Ω111.1\Omega

Explanation

Solution

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:
Req=R1+R2{R_{eq}} = {R_1} + {R_2}
Req{R_{eq}}Is the equivalent resistance, R1{R_1}andR2{R_2} are the two resistance connected in series.
⇒R′=R(1+αΔT)\Rightarrow R' = R(1 + \alpha \Delta T)
R′R'Is the new resistance after an increase in temperature,RR is the original resistance, α\alpha is the temperature coefficient of resistance, and ΔT\Delta Tis the change in temperature.

Complete Step-by-step answer
Let at initial temperature equivalent resistance be Req{R_{eq}} and resistance of tungsten be R1{R_1}and resistance of germanium beR2{R_2}.
Temperature coefficient of resistance for germanium beα2{\alpha _2} and temperature coefficient of resistance for tungsten beα1{\alpha _1}.
Given that,
α1=4.5×10−3∘C−1{\alpha _1} = 4.5 \times {10^{ - 3}}^\circ {C^{ - 1}}
α2=−5×10−2∘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 toReq{R_{eq}}.
We know that,
R′=R(1+αΔT)R' = R(1 + \alpha \Delta T)
R′R' Is the new resistance after an increase in temperature,RR is the original resistance, α\alpha is the temperature coefficient of resistance, and ΔT\Delta Tis the change in temperature.
Hence,
The resistance of tungsten and germanium after heating is R1′{R_1}'andR2′{R_2}' respectively.
Therefore
⇒R1′=R1(1+α1ΔT)\Rightarrow {R_1}' = {R_1}(1 + {\alpha _1}\Delta T)
⇒R2′=R2(1+α2ΔT)\Rightarrow {R_2}' = {R_2}(1 + {\alpha _2}\Delta T)
From the things we discussed above,
R1′+R2′=R1+R2{R_1}' + {R_2}' = {R_1} + {R_2}
⇒R1+R2=R1(1+α1ΔT)+R2(1+α2ΔT)\Rightarrow {R_1} + {R_2} = {R_1}(1 + {\alpha _1}\Delta T) + {R_2}(1 + {\alpha _2}\Delta T)
⇒R1+R2=R1+R1α1ΔT+R2+R2α2ΔT\Rightarrow {R_1} + {R_2} = {R_1} + {R_1}{\alpha _1}\Delta T + {R_2} + {R_2}{\alpha _2}\Delta T
⇒R1α1ΔT=−R2α2ΔT\Rightarrow {R_1}{\alpha _1}\Delta T = - {R_2}{\alpha _2}\Delta T
On substituting the values
R1(4.5×10−3∘C−1)ΔT=−R2(−5×10−2∘C−1)ΔT{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 R1=100Ω{R_1} = 100\Omega
⇒100(4.5×10−3)=R2(5×10−2)\Rightarrow 100(4.5 \times {10^{ - 3}}) = {R_2}(5 \times {10^{ - 2}})
⇒R2=100(4.5×10−3)5×10−2\Rightarrow {R_2} = \dfrac{{100(4.5 \times {{10}^{ - 3}})}}{{5 \times {{10}^{ - 2}}}}
On solving further,
R2=9Ω{R_2} = 9\Omega

Hence the correct answer to the question is (A) 9Ω9\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.