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Question #43919

Difficulty: 8/10 | 🟢 Free

The Question

An ideal inductor of inductance $L = 2.0 \text{ H}$ carries an initial steady current $I_0 = 10 \text{ A}$ . At time $t=0$ , a switch is thrown such that the inductor begins to discharge through a resistor. The resistor has an initial resistance $R_0 = 5.0 \ \Omega$ at $t=0$ , mass $m = 10 \text{ g}$ , and specific heat capacity $s = 500 \text{ J kg}^{-1}\text{K}^{-1}$ . The temperature coefficient of resistance of the resistor material is $\alpha = 4.0 \times 10^{-3} \text{ K}^{-1}$ . Assuming all Joule heating is entirely retained by the resistor (adiabatic process), calculate the value of $\frac{d^2 I}{dt^2}$ at $t=0$ .

Options

A.
$52.5 \text{ A s}^{-2}$
B.
$62.5 \text{ A s}^{-2}$
C.
$72.5 \text{ A s}^{-2}$
D.
$10.0 \text{ A s}^{-2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Maxwell-Boltzmann Distribution|Transient Current in RL Circuit|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize that in a transient RL circuit, resistance is not strictly constant if Joule heating is significant.

Students typically memorize the standard exponential decay solution where $d^2I/dt^2 = I_0 R_0^2 / L^2$ (which yields $62.5 \text{ A s}^{-2}$ here).

The trap is failing to treat the system as a coupled differential equation where the thermodynamic state of the resistor dynamically alters the electrical time constant $\tau = L/R(T)$ .

By asking for the second derivative at $t=0$ , the problem forces the application of the chain rule to $R(T)$ and the integration of the non-conservative Joule heating rate $I^2 R$ without requiring a full, non-linear analytical solution.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	Maxwell-Boltzmann Distribution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free