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

Hint: 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
$\begin{align}
  & {{R}_{1,T}}=R(1+{{\alpha }_{1}}T) \\
 & {{R}_{2,T}}=R(1+{{\alpha }_{2}}T) \\
 & {{R}_{s,T}}={{R}_{s}}(1+{{\alpha }_{s}}T) \\
\end{align}$
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
\[\begin{align}
  & {{R}_{1,T}}=R(1+{{\alpha }_{1}}T) \\
 & {{R}_{2,T}}=R(1+{{\alpha }_{2}}T) \\
 & {{R}_{p,T}}={{R}_{p}}(1+{{\alpha }_{p}}T) \\
\end{align}\]
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.