Question

It is required to construct a resistance of 100 Ω having a temperature coefficient of 0.001 per °C. Wires of two materials of suitable cross-sectional area are available. For material A, the resistance is 97 Ω per 100 meters and for material B, the resistance is 40 Ω per 100 meters. The temperature coefficient of resistance for material A is 0.003 per °C and for material B is 0.0005 per °C. Determine suitable lengths of wires of materials A and B.

Answer 1

Let Ra and Rb be the resistances of suitable lengths of materials A and B respectively which when joined in series will have a combined temperature coeff. of 0.0021. Hence, combination resistance at any given temperature is (Ra + Rb). Suppose we heat these materials through t°C. When heated, resistance of A increases from Ra to Ra (1 + 0.003 t). Similarly, resistance of B increases from Rb to Rb (1 + 0.0015 t). ∴ combination resistance after being heated through t°C = Ra(1 + 0.003 t) + Rb(1 + 0.0015 t) The combination α being given, value of combination resistance can be also found directly as

$(R_a+R_b)(1+0.0021t)$

$R_a(1+0.003t)+R_b(1+0.0015t)$

Simplifying the above equation we get $\frac{R_b}{R_a}=\frac{3}{2} ...1$

$R_a+R_b=80Ω$

Substituting the values of $R_b$ from eqn 1 to eqn 2 we get

$R_a+\frac{3}{2}R_a=80Ω$

$R_a=32Ω$ $R_b=48Ω$

The required lengths $L_a$ and $L_b$ =

$L_a=\frac{100}{80}×32= 40M$

$L_b=\frac{100}{60}×48= 80M$