Question

Two resistors of resistance r and 3r have thermal coefficients of resistance α and 2α respectively. The equivalent thermal coefficient of resistance for the combination connected in parallel is__?

• A. $\frac{5\alpha }{4}$
• B. $\frac{5\alpha }{3}$
• C. $\frac{7\alpha }{5}$
• D. $\frac{7\alpha }{3}$

Answer 1

The correct option is A

$\frac{5\alpha }{4}$

Thermal coefficient

Resistance in an intrinsic property of every material. Change in electrical resistance value of a material with respect to per degree change in temperature is represented by thermal coefficient of resistance.

Explanation

Correct option is A.

Given: Two resistors are in parallel connection.

Resistance at t= 0 are R₁ = r and R₂ = 3r.

Thermal coefficients are α and 2α. Let α_e equivalent thermal coefficient.

The equivalent resistance of parallel connection at t= 0 is

$R_{e0}=\frac{R_{10}{R}_{20}}{R_{10}+R_{20}}=\frac{Rx3R}{R+3R}=\frac{3}{4}R---\left(1\right)$

The equivalent resistance of parallel connection at t is

$R_{e0}(1+{\alpha }_{e}t)=\frac{R_{10}(1+{\alpha }_{10}t)x{R}_{20}(1+{\alpha }_{20}t)}{R_{10}(1+{\alpha }_{10}t)+R_{20}(1+{\alpha }_{20}t)}$

Substituting the given values we get,

$\frac{3}{4}R(1+{\alpha }_{e}t)=\frac{R(1+\alpha t)x3R(1+2\alpha t)}{R(1+\alpha t)+3R(1+2\alpha t)}$

On simplifying we get, neglecting α² terms, as thermal coefficients are very small values.

$\frac{1}{4}(1+{\alpha }_{e}t)=\frac{1+3\alpha t}{4+7\alpha t}$

Cross multiply and simplify to get α_e value.

$$\begin{gathered} (1+{\alpha }_{e}t)(4+7\alpha t)=4(1+3\alpha t) \\ 4+4{\alpha }_{e}t+7\alpha t+7\alpha {\alpha }_{e}t=4+12\alpha t \\ 4{\alpha }_{e}t=12\alpha t-7\alpha t \\ {\alpha }_e=\frac{5}{4}\alpha \end{gathered}$$

Therefore equivalent thermal coefficient is $\frac{5}{4}\alpha$. Hence option A is correct.