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Question #43914

Difficulty: 8/10 | 🟢 Free

The Question

An RL circuit contains a constant voltage source $V = 20 \text{ V}$ , a resistor of constant resistance $R = 5 \text{ \Omega}$ , and a variable inductor. The inductor's core is mechanically coupled to a bimetallic strip, such that the inductance varies with the strip's temperature $T$ (measured above the ambient temperature) as $L(T) = L_0 + \alpha T$ , where $L_0 = 1 \text{ H}$ and $\alpha = 0.2 \text{ H/K}$ . The resistor $R$ is embedded in the bimetallic strip such that all its Joule heating is transferred to the strip. The strip has a heat capacity $C = 2 \text{ J/K}$ and loses heat to the surroundings according to Newton's law of cooling with a cooling constant $\lambda = 2 \text{ W/K}$ . At a particular instant $t_1$ , the current in the circuit is $I = 2 \text{ A}$ and the temperature of the strip is $T = 15 \text{ K}$ above ambient. What is the rate of change of current $\frac{dI}{dt}$ at this instant?

Options

A.
$3.0 \text{ A/s}$
B.
$2.5 \text{ A/s}$
C.
$2.0 \text{ A/s}$
D.
$10.0 \text{ A/s}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Bimetallic Strip Curvature|Transient Current in RL Circuit|Differential Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle a coupled electro-thermal system where circuit parameters are time-dependent state variables.

The primary trap lies in the misapplication of Faraday's law for an inductor.

Students commonly memorize the voltage drop across an inductor as $V_L = L \frac{dI}{dt}$ , which is only valid for a constant inductance.

When the inductance varies with time (due to the thermally induced mechanical deformation of the bimetallic strip), the correct expression derived from the rate of change of flux linkage is $V_L = \frac{d(LI)}{dt} = L \frac{dI}{dt} + I \frac{dL}{dt}$ .

Failing to account for the 'motional back-EMF' term $I \frac{dL}{dt}$ leads directly to the incorrect option of $2.5 \text{ A/s}$ .

Furthermore, the problem requires correctly setting up the transient thermal balance equation to find the instantaneous rate of cooling, linking Joule heating feedback to the mechanical and electrical domains.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermal Properties of Matter - Class 11 Part 2
Concept	Bimetallic Strip Curvature
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free