Question

A resistance R of the thermal coefficient of resistance=α is connected in parallel with a resistance =3R, having a thermal coefficient of resistance=3α. The value ${{\alpha }_{eff}}$ is given as ${{\alpha }_{eff}}=\dfrac{x}{4}\alpha .$ Find x.

Answer 1

Hint: Recall what is resistance and the effective resistance of two resistors connected in parallel. Then define what thermal coefficient of resistance is and write the equation to find it. Equating the equations ${{R}_{eff(t)}}=\dfrac{{{R}_{1t}}{{R}_{2t}}}{{{R}_{1t}}+{{R}_{2t}}}$ and${{R}_{eff(t)}}={{R}_{eff(0)}}(1+{{\alpha }_{eff}}t)$, find the value of x.

Complete step-by-step answer:
Whenever electric current is allowed to pass through a conductor, it faces a sort of opposition which blocks the smooth flow of electrons through the conductor. It may be due to friction or any other thing. This opposition is generally termed as resistance. Its unit is ohm($\Omega$). Resistance is inversely proportional to current.

Parallel connection:
Parallel connection is a connection in which the two ends of the resistors are joined together and then it is connected to the circuit. Then the total resistance is given as
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$ or
$R=\dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}$………….(1)
Now thermal coefficient of a resistance is the measure of the change in resistance per unit temperature. It is generally denoted as α. At a temperature t, the relation between resistance and α is given by
${{R}_{t}}={{R}_{0}}(1+{{\alpha }_{0}}t)$……….(2)
Where, ${{R}_{t}}$- the resistance at t
${{R}_{0}}$- the initial resistance when t=0
${{\alpha }_{0}}$- the thermal coefficient of resistance.
In the given question,
The initial resistance when t=0 of the first resistor, ${{R}_{10}}=R$
The initial resistance when t=0 of the second resistor, ${{R}_{20}}=3R$
Thermal coefficient of resistor R, ${{\alpha }_{10}}$ = α
Thermal coefficient of resistor 3R,${{\alpha }_{20}}$ = 2α
The effective thermal coefficient, ${{\alpha }_{eff}}=\dfrac{x}{4}\alpha .$
The effective resistance in the parallel connection, when t=0
${{R}_{eff(0)}}=\dfrac{{{R}_{10}}{{R}_{20}}}{{{R}_{10}}+{{R}_{20}}}$
${{R}_{eff(0)}}=\dfrac{R\times 3R}{R+3R}$
${{R}_{eff(0)}}=\dfrac{3R}{4}$…………………(3)
Then the value of resistance when the temperature =t is
${{R}_{1t}}={{R}_{10}}(1+{{\alpha }_{10}}t)$. ie
${{R}_{1t}}=R(1+\alpha t)$…………..(4)
Similarly for the second resistor,
${{R}_{2t}}=3R(1+2\alpha t)$………..(5)
Therefore the ${{R}_{eff}}$ at temperature t is
${{R}_{eff(t)}}=\dfrac{{{R}_{1t}}{{R}_{2t}}}{{{R}_{1t}}+{{R}_{2t}}}$
${{R}_{eff(t)}}=\dfrac{R(1+\alpha t)\times 3R(1+2\alpha t)}{R(1+\alpha t)+3R(1+2\alpha t)}$
In order to simplify the equation, let us expand the numerator. Then we get,
${{R}_{eff(t)}}=\dfrac{3{{R}^{2}}(1+\alpha t+2\alpha t+2{{\alpha }^{2}}{{t}^{2}})}{R(1+\alpha t+3+6\alpha t)}$ or
${{R}_{eff(t)}}=\dfrac{3R(1+3\alpha t+2{{\alpha }^{2}}{{t}^{2}})}{4+7\alpha t}$
Since resistors are made of carbon, whose thermal coefficient is less. It is -0.0005/${{C}^{0}}$. If we take the square, it will be in the power of ${{10}^{-9}}$, which we approximately take as zero. So we will ignore the term that includes ${{\alpha }^{2}}$. Then the above equation becomes,
${{R}_{eff(t)}}=\dfrac{3R(1+3\alpha t)}{4+7\alpha t}$………………..(6)
We can also calculate the effective resistance of the total resistance at temperature t using the equation
${{R}_{eff(t)}}={{R}_{eff(0)}}(1+{{\alpha }_{eff}}t)$ i.e
${{R}_{eff(t)}}=\dfrac{3}{4}R(1+{{\alpha }_{eff}}t)$…………………..(7)
No what we have to do is simply equate the RHS of equation (6) n equation (7). i.e.
$\dfrac{3}{4}R(1+{{\alpha }_{eff}}t)=\dfrac{3R(1+3\alpha t)}{4+7\alpha t}$
Simplifying
$\dfrac{1}{4}(1+{{\alpha }_{eff}}t)=\dfrac{1+3\alpha t}{4+7\alpha t}$
On cross multiplication,
$(4+7\alpha t)(1+{{\alpha }_{eff}}t)=4(1+3\alpha t)$
$4+4{{\alpha }_{eff}}t+7{{\alpha }_{eff}}\alpha {{t}^{2}}+7\alpha t=4+12\alpha t$……………….(8)
Note that ${{\alpha }_{eff}}=\dfrac{x}{4}\alpha .$
Then the equation (8) becomes
$4+4\dfrac{x}{4}\alpha t+\dfrac{7}{4}x{{\alpha }^{2}}{{t}^{2}}+7\alpha t=4+12\alpha t$
Again ignore the term containing ${{\alpha }^{2}}$. Then our equation will be
$4+x\alpha t+7\alpha t=4+12\alpha t$
$x\alpha t+7\alpha t=12\alpha t$
$x\alpha t=12\alpha t-7\alpha t$ .ie
$x=12-7=5$
So the value of x=5and the effective thermal coefficient is ${{\alpha }_{eff}}=\dfrac{5}{4}\alpha$.

Note: The thermal coefficient of resistance in metals is positive. As the current passes through it, the resistance increases and the metal starts to dissipate more energy in the form of heat. The thermal coefficient of resistance is unique to every element.