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Question #43371

Difficulty: 8/10 | 🟢 Free

The Question

A resistor of constant heat capacity ChC_h is connected across an ideal DC voltage source VV. The resistance varies with temperature TT as R(T)=R0[1+α(T−Tenv)]R(T) = R_0 [1 + \alpha (T - T_{env})], where R0R_0 is the resistance at the environment temperature TenvT_{env}, and α>0\alpha > 0 is the temperature coefficient of resistance. At t=0t=0, the resistor is at an initial temperature TiT_i (Ti>TenvT_i > T_{env}) and loses heat to the environment according to Newton's Law of Cooling. The rate of heat loss is given by Ploss=λ(T−Tenv)P_{loss} = \lambda (T - T_{env}), where λ\lambda is a constant. Let I0I_0 and RiR_i be the current and resistance at t=0t=0, respectively, and let ΔTi=Ti−Tenv\Delta T_i = T_i - T_{env}. Which of the following expressions correctly gives the initial rate of change of current, dIdt∣t=0\left. \frac{dI}{dt} \right|_{t=0}?

Options

  • A.

    αR0I02VCh(λΔTi−VI0)\frac{\alpha R_0 I_0^2}{V C_h} (\lambda \Delta T_i - V I_0)

  • B.

    αR0I02VCh(VI0−λΔTi)\frac{\alpha R_0 I_0^2}{V C_h} (V I_0 - \lambda \Delta T_i)

  • C.

    αR0I02λΔTiVCh\frac{\alpha R_0 I_0^2 \lambda \Delta T_i}{V C_h}

  • D.

    αRiI02VCh(λΔTi−VI0)\frac{\alpha R_i I_0^2}{V C_h} (\lambda \Delta T_i - V I_0)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Temperature Coefficient (Alpha)|First Law Application|Dimensional Consistency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple electrical and thermodynamic systems during a transient phase.

The 'Lethal Collision' occurs when students either ignore the non-conservative Joule heating term (assuming the resistor only cools passively, leading to Option C) or misapply the resistance derivative by incorrectly using the initial resistance RiR_i instead of the base resistance R0R_0 as the coefficient for α\alpha (leading to Option D).

Furthermore, students who fail to recognize that Joule heating opposes the cooling process will make a sign error in the thermal balance equation, resulting in Option B.

The correct approach requires a rigorous application of the First Law of Thermodynamics to find the instantaneous rate of temperature change, followed by the chain rule to link thermal and electrical rates of change.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptTemperature Coefficient (Alpha)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree