Question

Two wires of resistance R1 and R2 have temperature coefficient of resistance Alpha 1 in Alpha 2 respectively, these are joined in series. The effective temperature coefficient of resistance is: ----

Answer 1

Answer: = $\frac{R_1\alpha_2+R_2\alpha_2}{R_1+R_2}$

Step by Step Solution:

We have,

Resistance of wires are R₁ and R₂  respectively.

Temperature  coefficient  is α₁ and α₂ respectively.

Let us take that their resistance at temperature t  are $R_{t1}$ and $R_{t2}$ respectively.

Now,

We know that,

Resistance of a wire t having temperature coefficient  α at temperature t° C is,

$R_t=R_0(1+\alpha t)$

Similarly,

For First wire,

$R_{t1}=R_1(1+\alpha_1t)\;\;\;\;..........i)$

For second wire,

$R_{t2}=R_2(1+\alpha_2t)\;\;\;\;.........ii)$

Now,

As per given that, resistance are connected in  the series,So their equivalent resistance is given by,

$R_{eq.}=R_{t1}+R_{t2}\\\;\\R_{eq.}=R_1(1+\alpha_1t})+R_2(1+\alpha_2t})\\\;\\R_{eq.}=R_1+R_1\alpha_1t+R_2+R_2\alpha_2t\\\;\\R_{eq.}=R_1+R_2+(R_1\alpha_1+R_2\alpha_2)t\\\;\\R_{eq.}=R_1+R_2+(R_1+R_2)[\frac{R_1\alpha_1+R_2\alpha_2}{R_1+R_2}]t\\\;\\R_{eq.}=(R_1+R_2)[1+(\frac{R_1\alpha_2+R_2\alpha_2}{R_1+R_2})t]\\\;\\\text{Comparing this equation with standard form of equation,}\\\;\\\text{Effective Temperature coefficient}=\frac{R_1\alpha_2+R_2\alpha_2}{R_1+R_2}$

Answer 2

Suppose at t°c, resistance of two wires become R1t & R2t & their equivalent is

Rt= R1t +R2t

.

.

. (open image)

∴ effective temp coefficient is {(R1α1 + R2α2)/(R1 + R2)}