Question

Two wires of resistances ${R_1}$ and ${R_2}$ at $0^\circ C$ have temperature coefficient of resistance ${\alpha _1}$ and ${\alpha _2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is:
A) $\dfrac{{{\alpha _1} + {\alpha _2}}}{2}$
B) \[\sqrt {{\alpha _1}{\alpha _2}} \]
C) $\dfrac{{{\alpha _1}{R_1} + {\alpha _2}{R_2}}}{{{R_1} + {R_2}}}$
D) $\sqrt {\dfrac{{{R_1}{R_2}{\alpha _1}{\alpha _2}}}{{R_1^2 + R_2^2}}} $

Answer 1

Hint: Let us suppose that the temperature of the wire has risen to $t$ . Now calculate the resistances of the wires at this temperature and add them to get the effective resistance of the resultant wire when the two wires have been joined together. Compare it with the equation used above to calculate resistances to get h=the value of the resultant temperature coefficient.

Formula Used:
${R_t} = {R_i}(1 + \alpha t)$ where ${R_t}$ is the resistance of the wire at temperature $t$ , ${R_i}$ is the resistance of wire at temperature $0^\circ C$ , $\alpha $ is the temperature coefficient of the wire.

Complete Step by Step Solution:
Wire 1 has resistance ${R_1}$ at $0^\circ C$ and has temperature coefficient of resistance ${\alpha _1}$
Similarly, wire 2 has resistance ${R_2}$ at $0^\circ C$ and has temperature coefficient of resistance ${\alpha _2}$
Now let us assume that after the rise in temperature, the temperature of the wires is $t$
Therefore, resistance for wire 1 at this temperature will be ${R_{t(1)}} = {R_1}(1 + {\alpha _1}t)$ (as per the given values)
And, resistance for wire 2 at this temperature will be ${R_{t(2)}} = {R_2}(1 + {\alpha _2}t)$ (as per the given values)
Now these wires are joined therefore their resistances will also be added to get the resultant resistance.
Therefore, ${R_{final}} = {R_1}(1 + {\alpha _1}t) + {R_2}(1 + {\alpha _2}t)$
Further solving it, we get ${R_{final}} = {R_1} + {R_1}{\alpha _1}t + {R_2} + {R_2}{\alpha _2}t$
Rearranging, ${R_{final}} = ({R_1} + {R_2}) + t({R_1}{\alpha _1} + {R_2}{\alpha _2})$
Taking common terms together, \[{R_{final}} = ({R_1} + {R_2})(1 + (\dfrac{{{R_1}{\alpha _1} + {R_2}{\alpha _2}}}{{{R_1} + {R_2}}})t)\]
Now compare this equation with the formula we used above, and we get
$\Rightarrow$ ${R_i} = {R_1} + {R_2}$ , $\alpha = \dfrac{{{R_1}{\alpha _1} + {R_2}{\alpha _2}}}{{{R_1} + {R_2}}}$

Therefore, option (C) is the correct answer.

Note: We often ignore the step where we have compared the equations and end up solving them further and ending up leaving the question unanswered. Pay attention to what is asked in the question. If it has a direct formula, that’s good but if not, these methods must be used otherwise you will never get the answer. More practice will make you sharper for questions like these.