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.
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=80\u03a9"
Substituting the values of "R_b" from eqn 1 to eqn 2 we get
"R_a+\\frac{3}{2}R_a=80\u03a9"
"R_a=32\u03a9" "R_b=48\u03a9"
The required lengths "L_a" and "L_b" =
"L_a=\\frac{100}{80}\u00d732= 40M"
"L_b=\\frac{100}{60}\u00d748= 80M"
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