Two wires of resistance R1 and R2 have temperature coefficient of resistance α1 and α2, respectively. These are joined in series. The effective temperature coefficient of resistance is

Question

Two wires of resistance R1 and R2 have temperature coefficient of resistance α1 and α2, respectively. These are joined in series. The effective temperature coefficient of resistance is (A) (α1+α2/2) (B) √α1 α2 (C) (α1 R1+α2 R2/R1+R2) (D) (√R1 R2 α1 α2/√R12+R22)

Tardigrade Admin · Accepted Answer

Rt1=R1(1+α1 t) and Rt2=R2(1+α2 t) Also Re q .=Rt1+Rt2 ⇒ Re q=R1+R2+(R1 α1+R2 α2) t ⇒ Re q=(R1+R2) 1+((R1 α1+R2 α2/R1+R2)) ⋅ t So αe f f=(R1 α1+R2 α2/R1+R2)

Q. Two wires of resistance $R_{1}$ and $R_{2}$ have temperature coefficient of resistance $α_{1}$ and $α_{2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is

$\frac{α _{1} + α _{2}}{2}$

$α_{1} α_{2}$

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

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

Q. Two wires of resistance R1​ and R2​ have temperature coefficient of resistance α1​ and α2​, respectively. These are joined in series. The effective temperature coefficient of resistance is

2α1​+α2​​

α1​α2​​

R1​+R2​α1​R1​+α2​R2​​

R12​+R22​​R1​R2​α1​α2​​​

Rt1​​=R1​(1+α1​t) and Rt2​​=R2​(1+α2​t) Also Req.​=Rt1​​+Rt2​​ ⇒Req​=R1​+R2​+(R1​α1​+R2​α2​)t ⇒Req​=(R1​+R2​){1+(R1​+R2​R1​α1​+R2​α2​​)⋅t} So αeff​=R1​+R2​R1​α1​+R2​α2​​

Q. Two wires of resistance $R_{1}$ and $R_{2}$ have temperature coefficient of resistance $α_{1}$ and $α_{2}$ , respectively. These are joined in series. The effective temperature coefficient of resistance is

$\frac{α _{1} + α _{2}}{2}$

$α_{1} α_{2}$

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

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