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Question #43370

Difficulty: 8/10 | 🟢 Free

The Question

A circuit consists of an ideal battery of electromotive force $V$ , a fixed resistor of resistance $R_0$ , and a variable resistor connected in series. At time $t=0$ , the variable resistor is at an initial temperature $T_0$ and has a resistance $R_0$ . The variable resistor has a temperature coefficient of resistance $\alpha$ such that its resistance at temperature $T$ is given by $R(T) = R_0[1 + \alpha(T - T_0)]$ . The heat capacity of the variable resistor also varies with temperature as $C(T) = C_0[1 + \alpha(T - T_0)]$ , where $C_0$ is a positive constant. Assuming that all the Joule heat generated in the variable resistor is retained by it (adiabatic process) and the fixed resistor maintains a constant resistance $R_0$ , find the time $t$ at which the current in the circuit drops to $\frac{V}{3R_0}$ .

Options

A.
$\frac{19 C_0 R_0}{3 \alpha V^2}$
B.
$\frac{7 C_0 R_0}{3 \alpha V^2}$
C.
$\frac{6 C_0 R_0}{\alpha V^2}$
D.
$\frac{19 C_0 R_0}{6 \alpha V^2}$

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

Temperature Coefficient (Alpha)|Kirchhoff’s Loop Rule|Power Dissipation Ratio

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to couple non-linear circuit dynamics with thermodynamics.

The primary trap lies in the misapplication of Joule heating and heat capacity.

Students often incorrectly assume a steady state, use a constant initial power dissipation $P_0 = V^2/(4R_0)$ , or forget that the current $I(t)$ is dynamically dependent on the changing resistance $R(t)$ via Kirchhoff's Loop Rule (ignoring the fixed series resistor).

Furthermore, the problem introduces a temperature-dependent heat capacity $C(T)$ which scales linearly with resistance.

Failing to express $dQ$ strictly in terms of $R$ and $dR$ leads to unsolvable or incorrect differential equations.

The correct approach requires setting up a first-order differential equation linking $dt$ and $dR$ by continuously updating the power $P = I^2R$ as the series loop resistance increases.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Temperature Coefficient (Alpha)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free