Question

Two conductors have the same resistance at 0°C but their temperature coefficients of resistances are α₁ and α₂ The respective temperature coefficients of their series and  parallel combinations are nearly....

(A) α₁ + α₂, {(α₁α₂)/(α₁ + α₂)}

(B) {(α₁ + α₂)/2}, {(α₁ + α₂)/2}

(C) {(α₁ + α₂)/2}, α₁ + α₂

(D) α₁ + α₂, {(α₁ + α₂)/2}

Answer 1

The correct option (B) {(α₁ + α₂)/2}, {(α₁ + α₂)/2}

Explanation:

R = R_o[1 + αt] ---- R_o is resistance at 0°c

R₁­ = R_o[1 + α₁t]

 R₂ = R_o[1 + α₂t]

series combination:

R_s = R₁ + R₂

= R_o[1 + α₁t + 1 + α₂t]

R_s = R_o[2 + t(α₁ + α₂)]

at          t = 0°c, R_s = R_o × 2 = 2R_o

Hence R_s = 2R_o[1 + α_st] where αs is temp. coefficient for series connection.

Also R_s = R_o(1 + α₁t) + R_o(1 + α₂t)

hence comparing we get, 2 α_s = α₁ + α₂

α_s = {(α₁ + α₂) / 2}

For parallel combination,

(1/R_p) = (1/R₁) + (1/R₂)

At 0°c, (1/R_p) = (1/R_o) + (1/R_o) = (2/R_o) ∴ R_p = (R_o/2)

∴ [1/{(R_o/2) (1 + α_pt)}] = [1/{R_o(1 + α₁t)}] + [1/{R_o(1 + α₂t)}]

hence 2(1 + α_pt)^–1 = (1 + α₁t)^–1 + (1 + α₂ t)^–1

∴ 2(1 – α_pt) = (1 – α₁t) + (1 – α₂t) ---- (1 + xn)^–1 = 1 – xn

∴ α_P = {(α₁ + α₂)/2}