Question

Two conductors have the same resistance at $0{}^\circ C$ but their temperature coefficients of resistances are ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$. The respective temperature coefficients of their series and parallel combinations are nearly:
$A)\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},{{\alpha }_{1}}+{{\alpha }_{2}}$
$B){{\alpha }_{1}}+{{\alpha }_{2}},\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$
$C){{\alpha }_{1}}+{{\alpha }_{2}},\dfrac{{{\alpha }_{1}}{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}$
$D)\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$

Answer 1

Even though it is given that both the conductors have the same resistance at ${{0}^{0}}C$, their equivalent resistance in series and parallel combination will be different. Equivalent/net resistance of a conductor at a particular temperature is related to the temperature coefficient of the resistance of the conductor as well as a reference resistance at a reference temperature. We proceed by taking into consideration the rules which govern series and parallel combinations of resistors.

Complete answer:
We know that the equivalent or net resistance of a conductor at a particular temperature is given by
${{R}_{net}}={{R}_{ref}}(1+\alpha (T-{{T}_{ref}}))$
where
${{R}_{net}}$ is the equivalent resistance of a conductor at temperature $T$
$\alpha$ is the temperature coefficient of resistance
${{R}_{ref}}$ is the conductor resistance at reference temperature ${{T}_{ref}}$
Let this be equation 1.
We also know that equivalent/net resistance of a series combination of resistors is equal to the sum of resistances of each resistor connected in series whereas the reciprocal of equivalent/net resistance of a parallel combination of resistors is equal to the sum of reciprocals of resistances of each resistor connected in parallel.
Relating the above concepts in the given question, we have:
- At ${{T}_{ref}}={{0}^{0}}C$, conductors have the same resistance, say ${{R}_{ref}}=R$.
- Temperature coefficients of resistances at ${{T}_{ref}}={{0}^{0}}C$ are ${{\alpha }_{1}}$and ${{\alpha }_{2}}$, respectively.
Now, if ${{R}_{1,T}}$and ${{R}_{2,T}}$ are the resistances of the conductors at a temperature $T$, equivalent resistance $({{R}_{s,T}})$ for series combination is given by
${{R}_{s,T}}={{R}_{1,T}}+{{R}_{2,T}}$
Let this be equation 2.
Using equation 1, we have
$\begin{aligned} & {{R}_{1,T}}=R(1+{{\alpha }_{1}}T) \\\ & {{R}_{2,T}}=R(1+{{\alpha }_{2}}T) \\\ & {{R}_{s,T}}={{R}_{s}}(1+{{\alpha }_{s}}T) \\\ \end{aligned}$
Here
- ${{R}_{s}}={{R}_{1}}+{{R}_{2}}=R+R=2R$ is the equivalent resistance of ${{R}_{1}}=R$ and ${{R}_{2}}=R$, in series combination
- ${{\alpha }_{s}}$ is the assumed temperature coefficient of series equivalent resistance
- we have substituted ${{R}_{ref}}=R$ and ${{T}_{ref}}=0$
Let this set of equations be denoted by M.
Substituting the set of equation denoted by M in equation 2, we have
${{R}_{s}}\left( 1+{{\alpha }_{s}}T \right)=R\left( 1+{{\alpha }_{1}}T \right)+R\left( 1+{{\alpha }_{2}}T \right)\Rightarrow 2R\left( 1+{{\alpha }_{s}}T \right)=2R\left( 1+\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}T \right)$
Comparing the left-hand side and the right-hand side of the above expression, we have
${{\alpha }_{s}}=\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$
where
${{\alpha }_{s}}$ is the temperature coefficient of equivalent resistance of series combination of resistors
${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ are the given temperature coefficients of the resistors at ${{T}_{ref}}={{0}^{0}}C$
Let this be equation 3.
Similarly, equivalent resistance $({{R}_{p,T}})$ for parallel combination is given by
$\dfrac{1}{{{R}_{p,T}}}=\dfrac{1}{{{R}_{1,T}}}+\dfrac{1}{{{R}_{2,T}}}$
Let this be equation 4.
Using equation 1, we have

& {{R}_{1,T}}=R(1+{{\alpha }_{1}}T) \\\ & {{R}_{2,T}}=R(1+{{\alpha }_{2}}T) \\\ & {{R}_{p,T}}={{R}_{p}}(1+{{\alpha }_{p}}T) \\\ \end{aligned}$$ Here -$\dfrac{1}{{{R}_{p}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}=\dfrac{1}{R}+\dfrac{1}{R}=\dfrac{2}{R}$ is the equivalent resistance of ${{R}_{1}}=R$ and ${{R}_{2}}=R$, in parallel combination \- ${{\alpha }_{p}}$ is the assumed temperature coefficient of parallel equivalent resistance \- we have substituted ${{R}_{ref}}=R$ and ${{T}_{ref}}=0$ Let this set of equations be denoted by N. Substituting the set of equations denoted by N in equation 4, we have $$\dfrac{1}{{{R}_{p}}(1+{{\alpha }_{p}}T)}=\dfrac{1}{R(1+{{\alpha }_{1}}T)}+\dfrac{1}{R(1+{{\alpha }_{2}}T)}\Rightarrow \dfrac{2}{R(1+{{\alpha }_{p}}T)}=\dfrac{2+({{\alpha }_{1}}+{{\alpha }_{2}})T}{R(1+({{\alpha }_{1}}+{{\alpha }_{2}})T+{{\alpha }_{1}}{{\alpha }_{2}}{{T}^{2}})}$$ Reducing the above expression further, we have $$\dfrac{2}{R(1+{{\alpha }_{p}}T)}=\dfrac{2+({{\alpha }_{1}}+{{\alpha }_{2}})T}{R(1+({{\alpha }_{1}}+{{\alpha }_{2}})T+{{\alpha }_{1}}{{\alpha }_{2}}{{T}^{2}})}\Rightarrow 1+{{\alpha }_{p}}T=1+\dfrac{({{\alpha }_{1}}+{{\alpha }_{2}}+2{{\alpha }_{1}}{{\alpha }_{2}}T)T}{2+({{\alpha }_{1}}+{{\alpha }_{2}})T}$$ In the above equation, the temperature $T$ taken to be small and negligible (order of $${{10}^{-3}}{{/}^{0}}C$$). So, the terms $$2{{\alpha }_{1}}{{\alpha }_{2}}T$$and $$\left( {{\alpha }_{1}}+{{\alpha }_{2}} \right)T$$in numerator and denominator respectively, can be eliminated. After doing the same, we have $$1+{{\alpha }_{p}}T\approx 1+\dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$$ Comparing the left-hand side and the right-hand side of the above expression, we have $${{\alpha }_{p}}\approx \dfrac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}$$ where ${{\alpha }_{p}}$ is the temperature coefficient of equivalent resistance of parallel combination of resistors ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ are the given temperature coefficients of the resistors Let this be equation 5. **Therefore, from equation 3 and equation 5, we can conclude that the correct answer is option $D$.** **Note:** Resistance of conductors/metals increases with increase in temperature (directly proportional) as they have positive temperature coefficient. For insulators/non metals and semiconductors, the resistance decreases with increase in temperature (inversely proportional) as they have negative temperature coefficient.