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title: "An inductor of inductance \\(L\\) is initially carrying a steady current \\(I_0\\). At time \\(t = 0\\), the inductor is disconnected from its source and connected across a resistor of resistance \\(R_1\\) in Trial 1, allowing the current to decay to zero. The experiment is repeated in Trial 2 with the same inductor carrying the same initial current \\(I_0\\), but connected across a resistor of resistance \\(R_2 = 2R_1\\). Which of the following correctly compares the total thermal energy dissipated in the resistor in Trial 2 to that in Trial 1 as \\(t \\to \\infty\\), and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/121298/"
date_modified: "2026-08-23T04:59:37+00:00"
---

# An inductor of inductance \(L\) is initially carrying a steady current \(I_0\). At time \(t = 0\), the inductor is disconnected from its source and connected across a resistor of resistance \(R_1\) in Trial 1, allowing the current to decay to zero. The experiment is repeated in Trial 2 with the same inductor carrying the same initial current \(I_0\), but connected across a resistor of resistance \(R_2 = 2R_1\). Which of the following correctly compares the total thermal energy dissipated in the resistor in Trial 2 to that in Trial 1 as \(t \to \infty\), and provides the correct physical justification?

An inductor of inductance \(L\) is initially carrying a steady current \(I_0\). At time \(t = 0\), the inductor is disconnected from its source and connected across a resistor of resistance \(R_1\) in Trial 1, allowing the current to decay to zero. The experiment is repeated in Trial 2 with the same inductor carrying the same initial current \(I_0\), but connected across a resistor of resistance \(R_2 = 2R_1\). Which of the following correctly compares the total thermal energy dissipated in the resistor in Trial 2 to that in Trial 1 as \(t \to \infty\), and provides the correct physical justification?

![A single closed rectangular circuit loop. The left vertical branch contains an inductor labeled L. The right vertical branch contains a resistor labeled R. A curved arrow labeled i(t) indicates clockwise current flow through the loop. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461176-We4Rw2.jpg)

- **A.** The total energy dissipated in Trial 2 is greater than in Trial 1 because the instantaneous power dissipation \(P = I^2 R\) is proportional to the resistance, resulting in a higher rate of energy conversion throughout the discharge.
- **B.** The total energy dissipated in Trial 2 is less than in Trial 1 because the circuit time constant \(\tau = \dfrac{L}{R}\) is halved, causing the current to decay to zero more rapidly and leaving less time for energy transfer.
- **C.** The total energy dissipated in Trial 2 is greater than in Trial 1 because the initial induced back EMF across the inductor is doubled due to the larger initial \(\dfrac{di}{dt}\), doing more total work on the charges.
- **D.** The total energy dissipated in Trial 2 is equal to that in Trial 1 because the initial magnetic energy stored in the inductor is \(\dfrac{1}{2} L I_0^2\), and the time integral of the power \(\int_0^\infty I_0^2 R e^{-2Rt/L}\,dt\) evaluates to \(\dfrac{1}{2} L I_0^2\), independent of \(R\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121298/*