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Question: Two coils connected in series have resistances \(600 \Omega\) and \(300 Omega\) at 20°C and temperat...

Two coils connected in series have resistances 600Ω600 \Omega and 300Omega300 Omega at 20°C and temperature coefficient of resistivity 0.001K−10.001 {K}^{-1} and 0.004K−10.004 {K}^{-1} respectively. The effective temperature coefficient of the combination is
A. 11000degrees−1\dfrac {1}{1000}{degrees}^{-1}
B. 1250degrees−1\dfrac {1}{250}{degrees}^{-1}
C. 1500degrees−1\dfrac {1}{500}{degrees}^{-1}
D. 31000degrees−1\dfrac {3}{1000}{degrees}^{-1}

Explanation

Solution

To solve this problem, use the relationship between temperature and resistance of conductor which is given as Rt=R0(1+αΔt){R}_{t}= {R}_{0}(1 + \alpha \Delta t). Substitute the values in this formula for both the resistances and add them up as both the resistances are connected in series. Solve the obtained equation and find the effective temperature of the combination by comparing with the general formula mentioned above.
Formula used:
Rt=R0(1+αΔt){R}_{t}= {R}_{0}(1 + \alpha \Delta t)

Complete answer:
Given: R1=600Ω{R}_{1}= 600\Omega
R2=300Ω{R}_{2}= 300\Omega
α1=0.001K−1{\alpha}_{1}= 0.001 {K}^{-1}
α2=0.004K−1{\alpha}_{2}= 0.004 {K}^{-1}
The relationship between temperature and resistance of conductor is given by,
Rt=R0(1+αΔt){R}_{t}= {R}_{0}(1 + \alpha \Delta t) …(1)
Where, Rt{R}_{t} is the resistance at temperature t
R0{R}_{0} is the resistance at temperature usually 20°C
α\alpha is the temperature coefficient of resistance
Δt\Delta t is the difference between operating and reference temperature
Resistances are connected in series. So, their equivalent resistance is given by,
Req=Rt1+Rt2{R}_{eq}= {R}_{t1}+ {R}_{t2} …(2)
Using equation. (1), equation. (2) can be written as,
Req=R1(1+α1Δt)+R2(1+α2Δt){R}_{eq}= {R}_{1}(1 + {\alpha}_{1} \Delta t) + {R}_{2}(1 + {\alpha}_{2} \Delta t)
Substituting values in above equation we get,
Req=600(1+0.001Δt)+300(1+0.004Δt){R}_{eq}= 600 (1+ 0.001 \Delta t)+ 300(1+ 0.004 \Delta t)
⇒Req=600+0.6Δt+300+1.2Δt\Rightarrow {R}_{eq}= 600 + 0.6 \Delta t + 300 + 1.2 \Delta t
⇒Req=900+1.8Δt\Rightarrow {R}_{eq}= 900+ 1.8 \Delta t
⇒Req=900(1+0.002Δt)\Rightarrow {R}_{eq}= 900 (1 + 0.002 \Delta t) …(3)
Comparing equation. (3) with equation. (1) we get,
α=0.002K−1\alpha= 0.002 {K}^{-1}
⇒α=1500degrees−1\Rightarrow \alpha= \dfrac {1}{500} {degrees}^{-1}
Hence, the effective temperature of the combination is 1500degrees−1\dfrac {1}{500} {degrees}^{-1}.

So, the correct answer is option C i.e. 1500degrees−1\dfrac {1}{500}{degrees}^{-1}.

Note:
If the temperature coefficient of resistance is positive for a material then it means that the resistance increases with increase in temperature. Whereas, if the temperature coefficient of resistance is negative for a material then it means that the resistance decreases with increase in temperature. Pure metals have positive temperature coefficient of resistance whereas semiconductors and insulators have negative temperature coefficient of resistance.