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Question #43065

Difficulty: 8/10 | 🟢 Free

The Question

An isolated LCR circuit consists of an inductor of inductance LL, a capacitor of capacitance CC, and a resistor of mass mm, specific heat capacity ss, and initial resistance R0R_0 at ambient temperature T0T_0. The resistance varies with temperature TT as R(T)=R0[1+α(T−T0)]R(T) = R_0 [1 + \alpha (T - T_0)], where α\alpha is the temperature coefficient of resistance. At t=0t=0, the capacitor is charged to a voltage V0V_0 (giving an initial total energy E0=12CV02E_0 = \frac{1}{2} C V_0^2) and the circuit is closed. The resistor is perfectly thermally insulated from its surroundings. Assume the circuit is highly underdamped (Q≫1Q \gg 1) so that the fractional energy loss per cycle is very small, allowing the cycle-averaged power dissipation to be approximated as ⟨P⟩=ELR(T)\langle P \rangle = \frac{E}{L} R(T), where EE is the instantaneous total electrical energy of the circuit. Which of the following expressions gives the time tt at which the total electrical energy of the circuit reduces to E0/2E_0 / 2? (Let γ=αE0ms\gamma = \frac{\alpha E_0}{ms})

Options

  • A.

    t=LR0(1+γ)ln⁡(2+γ)t = \frac{L}{R_0 (1 + \gamma)} \ln (2 + \gamma)

  • B.

    t=LR0(1−γ)ln⁡(2−γ)t = \frac{L}{R_0 (1 - \gamma)} \ln (2 - \gamma)

  • C.

    t=LR0ln⁡(2+γ1+γ)t = \frac{L}{R_0} \ln \left( \frac{2 + \gamma}{1 + \gamma} \right)

  • D.

    t=LR0(1+γ/2)ln⁡2t = \frac{L}{R_0 (1 + \gamma/2)} \ln 2

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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

Quality Factor|LC Oscillations Energy Cycle|Thermal Properties of Matter (Resistance-Temperature relation)

Why This Question Is Tricky

Common Mistakes

This question tests the limits of the standard exponential decay model in damped LC oscillations.

Students typically memorize the energy decay formula E(t)=E0e−(R/L)tE(t) = E_0 e^{-(R/L)t} and the constant damping coefficient β=R/2L\beta = R/2L.

The trap lies in failing to recognize that resistive heating dynamically alters the damping coefficient.

To solve this, students must couple the First Law of Thermodynamics with AC circuit power dissipation, realizing that resistance is no longer a constant or explicitly time-dependent, but rather a function of the remaining electrical energy E(t)E(t).

This creates a non-linear separable differential equation, heavily penalizing those who blindly apply the standard constant-R half-life formula t=(L/R0)ln⁡2t = (L/R_0)\ln 2 or attempt to use an arithmetic mean resistance.

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Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptQuality Factor (Q-factor)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree