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JEE Main 2020 · Physics · Current Electricity · medium

A copper coil of resistance 5.0 Ω at 0^circC (temperature coefficient +4×10^-3 ^circC^-1) is joined in series with a carbon coil whose temperature…

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A copper coil of resistance 5.0 Ω5.0\ \Omega at 0∘C0^\circ\mathrm{C} (temperature coefficient +4×10−3 ∘C−1+4\times10^{-3}\ ^\circ\mathrm{C}^{-1}) is joined in series with a carbon coil whose temperature coefficient is −5×10−4 ∘C−1-5\times10^{-4}\ ^\circ\mathrm{C}^{-1}. The combination has a resistance that does not change with temperature (to first order). Its total resistance at 0∘C0^\circ\mathrm{C} is
  1. A. 45 Ω45\ \Omega
  2. B. 40 Ω40\ \Omega
  3. C. 50 Ω50\ \Omega
  4. D. 25 Ω25\ \Omega
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Answer: (A) 45 Ω45\ \Omega

For small changes, R(t)=RCu(1+α1t)+RC(1+α2t)R(t) = R_{Cu}(1+\alpha_1 t) + R_C(1+\alpha_2 t). The temperature-dependent part vanishes when RCuα1+RCα2=0R_{Cu}\alpha_1 + R_C\alpha_2 = 0: 5(4×10−3)=RC(5×10−4)5(4\times10^{-3}) = R_C(5\times10^{-4}) gives RC=40 ΩR_C = 40\ \Omega. Total at 0∘C0^\circ\mathrm{C} is 5+40=45 Ω5 + 40 = 45\ \Omega. The common error is to quote only the carbon coil's 40 Ω40\ \Omega.
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