Question

Two conductors have the same resistance at ${{0}^{0}}C$ but their temperature coefficients of resistance are ${{\alpha }_{1}}\text{ and }{{\alpha }_{2}}$. The respective temperature coefficients of their series and parallel combinations are nearly –

& \text{A) }\dfrac{{{\alpha }_{1}}\text{+}{{\alpha }_{2}}}{2},{{\alpha }_{1}}+{{\alpha }_{2}} \\\ & \text{B) }{{\alpha }_{1}}+{{\alpha }_{2}},\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2} \\\ & \text{C) }{{\alpha }_{1}}+{{\alpha }_{2}},\dfrac{{{\alpha }_{1}}{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}} \\\ & \text{D) }\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2} \\\ \end{aligned}$$

Answer 1

We can use the formulas for the series and parallel combination of the resistors to find the solution. A slight modification of the formulae with the equation for resistance of a resistor at any temperature involving the temperature coefficient.

Complete answer:
We know that each resistor has a unique temperature coefficient of resistance. The equation relating the resistance to the temperature coefficient is given by –
${{R}_{T}}={{R}_{o}}(1+\alpha T)$
Where, ${{R}_{T}}$is the resistance at temperature T,
${{R}_{o}}$ is the resistance at ${{0}^{0}}C$,
$\alpha$is the temperature coefficient of resistance.
Now, let us consider the situation given to us. It is said that the resistance at ${{0}^{0}}C$ is the same for both the resistors. Let us assume it to be ${{R}_{o}}$. Now, the let ${{\alpha }_{1}}\text{ and }{{\alpha }_{2}}$ be the temperature coefficients of the resistors ${{R}_{T1}}\text{ and }{{R}_{T2}}$respectively at a temperature T.
Then, we can derive the resistances as –

& {{R}_{T1}}={{R}_{o}}(1+{{\alpha }_{1}}T) \\\ & \text{and, } \\\ & {{R}_{T2}}={{R}_{o}}(1+{{\alpha }_{2}}T) \\\ \end{aligned}$$ Now, let us consider the different combinations of these two resistors. 1) Series combination – The equivalent resistance is the sum of the individual resistances. i.e., $$\begin{aligned} & R={{R}_{T1}}+{{R}_{T2}} \\\ & \Rightarrow \text{ }{{\text{R}}_{s}}(1+\alpha T)={{R}_{o}}(1+{{\alpha }_{1}}T)+{{R}_{o}}(1+{{\alpha }_{2}}T) \\\ & \Rightarrow \text{ }{{R}_{s}}(1+\alpha T)=2{{R}_{0}}(1+(\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2})T) \\\ \end{aligned}$$ From comparing the L.H.S and R.H.S of the above equation gives us the temperature coefficient of the combination of resistances as – $$\Rightarrow \text{ }{{\alpha }_{s}}=\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$$ 2) Parallel Combination: We can find the temperature coefficient of the combination of resistors in parallel combination as – $$\begin{aligned} & \dfrac{1}{R}=\dfrac{1}{{{R}_{T1}}}+\dfrac{1}{{{R}_{T2}}} \\\ & \Rightarrow \text{ }R=\dfrac{{{R}_{T1}}{{R}_{T2}}}{{{R}_{T1}}+{{R}_{T2}}} \\\ & \Rightarrow \text{ }{{R}_{p}}(1+\alpha T)=\dfrac{{{R}_{o}}(1+{{\alpha }_{1}}T){{R}_{o}}(1+{{\alpha }_{2}}T)}{{{R}_{o}}(1+{{\alpha }_{1}}T)+{{R}_{o}}(1+{{\alpha }_{2}}T)} \\\ & \Rightarrow \text{ }{{R}_{p}}(1+\alpha T)=\dfrac{{{R}_{o}}(1+{{\alpha }_{1}}T)(1+{{\alpha }_{2}}T)}{(1+{{\alpha }_{1}}T)+(1+{{\alpha }_{2}}T)} \\\ & \Rightarrow \text{ }{{R}_{p}}(1+\alpha T)=\dfrac{{{R}_{0}}(1+({{\alpha }_{1}}+{{\alpha }_{2}})T+{{\alpha }_{1}}{{\alpha }_{2}}{{T}^{2}})}{2+({{\alpha }_{1}}+{{\alpha }_{2}})T} \\\ & \text{but, } \\\ & {{R}_{p}}=\dfrac{{{R}_{o}}}{2} \\\ & \Rightarrow \text{ }(1+\alpha T)=\dfrac{2(1+({{\alpha }_{1}}+{{\alpha }_{2}})T+{{\alpha }_{1}}{{\alpha }_{2}}{{T}^{2}})}{2+({{\alpha }_{1}}+{{\alpha }_{2}})T} \\\ & \text{We can neglect the 2}{{\alpha }_{1}}{{\alpha }_{2}}\text{T in numerator and }({{\alpha }_{1}}+{{\alpha }_{2}})T\text{ term in the denominator as} \\\ & \text{it is very small}\text{.} \\\ & \Rightarrow \text{ }(1+\alpha T)=1+\dfrac{({{\alpha }_{1}}+{{\alpha }_{2}})}{2}T \\\ & \Rightarrow \text{ }{{\alpha }_{p}}=\dfrac{({{\alpha }_{1}}+{{\alpha }_{2}})}{2} \\\ \end{aligned}$$ So, we get the temperature coefficients of the resistors in both series and parallel combinations as – $$\text{ }\alpha =\dfrac{({{\alpha }_{1}}+{{\alpha }_{2}})}{2}$$ **The correct answer is option D.** **Note:** The above example illustrates that the temperature coefficient of a combination resistor is independent of the type of combination they are involved in. The quantity is dependent only on the resistance at zero Celsius and the temperature. The temperature coefficient of resistance is independent of the combination of resistances.