Two resistances equal at 0° C with temperature coefficient of resistance α1 and α2 joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

Question

Two resistances equal at 0° C with temperature coefficient of resistance α1 and α2 joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be (A) α1 + α2 (B) ( α1 α2/ α1 + α2)  (C) ( α1 - α2/2) (D) ( α1 + α2/2)

Tardigrade Admin · Accepted Answer

R 1= R 0(1+α1 t )+ R 0(1+α2 t ) =2 R 0(1+(α1+α2/2) t ) Comparing with R=R0(1+α t) α=(α1+α2/2) .

Q. Two resistances equal at $0^{\circ} C$ with temperature coefficient of resistance $α_{1}$ and $α_{2}$ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

$α_{1} + α_{2}$

$\frac{α _{1} α _{2}}{α _{1} + α _{2}}$

$\frac{α _{1} - α _{2}}{2}$

$\frac{α _{1} + α _{2}}{2}$

$R_{1} = R_{0} (1 + α_{1} t) + R_{0} (1 + α_{2} t)$
$= 2 R_{0} (1 + \frac{α _{1} + α _{2}}{2} t)$
Comparing with $R = R_{0} (1 + α t)$
$α = \frac{α _{1} + α _{2}}{2} .$

Q. Two resistances equal at 0∘C with temperature coefficient of resistance α1​ and α2​ joined in series act as a single resistance in a circuit. The temperature coefficient of their single resistance will be

α1​+α2​

α1​+α2​α1​α2​​

2α1​−α2​​

2α1​+α2​​