Track inductor energy and sustained resistor heating

Stored energy is different from heating

An ideal inductor stores energy in a magnetic field; the resistor dissipates energy. For a constant linear inductance, stored energy is U = LI²/2, where U is in joules, L in henries and I in amperes. This formula assumes the ideal linear element used here. It is not a general model of core saturation, hysteresis or frequency-dependent losses.

At R = 2 Ω, the current limit is 3 A and L = 2 H. The corresponding stored energy is 0.5 × 2 × 3² = 9 J. The buildup history starts with zero stored energy. The separate decay history starts with 9 J. They therefore need different starting terms in their energy accounts.

Read power with the passive sign convention

Power absorbed by the inductor is P L = V L I. With V L = L dI/dt, this is the rate of change of LI²/2. Positive power adds stored energy; negative power means the inductor is returning energy to the circuit. A negative absorbed power is not negative heat.

During buildup, both current and passive inductor drop are positive, so energy grows. During decay, current stays positive but the drop is negative, so stored energy decreases. The resistor power I²R is nonnegative in both histories. At the initial instant of buildup, current and power are zero even though the inductor voltage is 6 V. Voltage alone does not establish nonzero instantaneous power.

Check the two complete energy accounts

For buildup, source work = stored energy now + resistor heat. For decay, initial stored energy = stored energy now + resistor heat. These are balances for each separate history since its own time zero. Do not add the rows together as if both processes were occurring simultaneously in one circuit.

At R = 2 Ω and t = 1 s, buildup current is about 1.89636 A. Its stored energy is 3.59619 J, source work 6.62183 J and accumulated resistor heat 3.02564 J. Add stored energy and resistor heat using unrounded values to recover the source work.

At that same selected time, the decay row has current 1.10364 A, retains 1.21802 J and has dissipated 7.78198 J, summing to the initial 9 J. Unlike the earlier RC paired example, these RL histories do not have equal current magnitudes or equal accumulated heat at the same time.

Accumulate power rather than multiplying by its last value

Imagine dividing the history into short time intervals, estimating local power times interval duration, then adding those energies. This is the idea of integrating power. In buildup, current and resistor power rise, so using the final resistor power for the whole interval overestimates the heat. In decay they fall, so the same shortcut underestimates it.

The source supplies instantaneous power V s I. With x = t/τ and I limit = V s/R, its accumulated buildup work is V s I limit τ [x − (1 − exp(−x))]. The browser evaluates accumulated source work and heat analytically. Learners can check the values independently by summing sufficiently small time slices, without needing to derive the integral first.

The long-time source account differs from RC charging

Buildup energy approaches a finite value, 9 J for R = 2 Ω. But current approaches a nonzero 3 A through the resistor, so source power and resistor power approach 18 W. Source work and heat therefore continue increasing without a finite long-time total while the source stays connected. Do not apply the capacitor-charging claim that source work is twice final stored energy to this RL circuit.

The source-free decay has a different limit: its finite initial energy is ultimately dissipated in R. At half the initial current, energy is one quarter of its initial value, because energy depends on current squared. Energy reaches half its initial value after (τ ln 2)/2, earlier than the half-current time τ ln 2.

Identify the comparison’s preparation

Increasing R in this browser lowers the initial decay current 6/R, so it also lowers initial energy. At R = 4 Ω, initial current is 1.5 A and initial energy 2.25 J. This is not a comparison at fixed stored energy. With the same independently prepared current in the same ideal inductor, changing the discharge resistance would instead change the decay rate while preserving the total energy eventually dissipated.

No open-circuit interruption, arc, protective diode or parasitic capacitance is simulated. The closed decay path is part of the energy account. Real switching behavior cannot be inferred by simply removing that path from this diagram.

Transfer to another stored-energy state

An ideal 0.80 H inductor initially carries 2.0 A, then decays through a resistor with no source. Find its initial stored energy, energy remaining at 1.0 A, and energy dissipated by that point. If the same initial current is used with a larger decay resistance, what changes?

Check the inductor-energy transfer

Initial energy is 0.5 × 0.80 × 2.0² = 1.6 J. At 1.0 A, energy is 0.5 × 0.80 × 1.0² = 0.40 J, so 1.2 J has been dissipated. A larger resistance decreases L/R and releases the energy faster under the same ideal assumptions. With initial current fixed, it does not change the eventual total of 1.6 J. The browser's resistance slider uses a different initial-current preparation, so that fixed-initial-energy comparison is a reasoning transfer.

Use the ledger to verify both energy boundaries. Distinguish power, accumulated heat and stored energy, and state why the source-connected resistor continues dissipating after current becomes nearly steady.

References: OpenStax: magnetic-field energy and RL circuits. Worked comparisons and transfer cases are original.