Two resistances R1 and R2 are made of different materials. The temperature coefficient of the material of R1 is α and of the material of R2 is -β . The resistance of the series combination of R1 and R2 will not change with temperature, if R1 / R2 equals

Question

Two resistances R1 and R2 are made of different materials. The temperature coefficient of the material of R1 is α and of the material of R2 is -β . The resistance of the series combination of R1 and R2 will not change with temperature, if R1 / R2 equals (A) (α/β) (B) (α+β/α-β) (C) (α2+β2/α β) (D) (β/α)

Tardigrade Admin · Accepted Answer

R1+R2=R1(1+α t)+R2(1-β t) ⇒ R1+R2=R1+R2+R1 α t-R2 β t ⇒ (R1/R2)=(β/α)

$\frac{α}{β}$

$\frac{α + β}{α - β}$

$\frac{α ^{2} + β ^{2}}{α β}$

$\frac{β}{α}$

$R_{1} + R_{2} = R_{1} (1 + α t) + R_{2} (1 - βt)$
$\Rightarrow R_{1} + R_{2} = R_{1} + R_{2} + R_{1} α t - R_{2} βt$
$\Rightarrow \frac{R _{1}}{R _{2}} = \frac{β}{α}$

Q. Two resistances R1​ and R2​ are made of different materials. The temperature coefficient of the material of R1​ is α and of the material of R2​ is −β. The resistance of the series combination of R1​ and R2​ will not change with temperature, if R1​/R2​ equals

βα​

α−βα+β​

αβα2+β2​

αβ​

R1​+R2​=R1​(1+αt)+R2​(1−βt) ⇒R1​+R2​=R1​+R2​+R1​αt−R2​βt ⇒R2​R1​​=αβ​

$\frac{α}{β}$

$\frac{α + β}{α - β}$

$\frac{α ^{2} + β ^{2}}{α β}$

$\frac{β}{α}$

$R_{1} + R_{2} = R_{1} (1 + α t) + R_{2} (1 - βt)$
$\Rightarrow R_{1} + R_{2} = R_{1} + R_{2} + R_{1} α t - R_{2} βt$
$\Rightarrow \frac{R _{1}}{R _{2}} = \frac{β}{α}$