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Question #42744

Difficulty: 8/10 | 🟢 Free

The Question

A resistor of initial resistance $R_0$ at temperature $T_0$ is connected across an ideal DC voltage source of EMF $V$ . The resistor is thermally insulated from its surroundings and has a constant heat capacity $C$ . Its resistance varies with temperature as $R(T) = R_0[1 + \alpha(T - T_0)]$ , where $\alpha$ is the temperature coefficient of resistance. Find the time $t$ required for the current in the circuit to drop to exactly half of its initial value.

Options

A.
$\frac{3 C R_0}{2 \alpha V^2}$
B.
$\frac{C R_0}{\alpha V^2}$
C.
$\frac{2 C R_0}{\alpha V^2}$
D.
$\frac{C R_0 \ln 2}{\alpha V^2}$

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

Kirchhoff’s Loop Rule|Temperature Coefficient (Alpha)|First Law of Thermodynamics

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle non-linear electro-thermal coupling.

The primary trap is assuming that either the current $I$ or the power dissipation $P$ remains constant during the heating process.

Students often incorrectly write $P = I_0^2 R(T)$ (treating current as constant despite changing resistance, leading to Option D) or $P = V^2/R_0$ (ignoring the feedback loop entirely, leading to Option B).

By enforcing Kirchhoff's Loop Rule with a constant voltage source, the current must strictly obey $I(t) = V/R(T(t))$ , making the power $P(t) = V^2/R(T(t))$ inversely proportional to the temperature-dependent resistance.

This requires setting up and integrating the correct separable differential equation.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Kirchhoff’s Loop Rule
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free