DC circuits — energy dissipation sources: Is it true that in a direct-current (DC) circuit the only conversion of electrical energy to heat occurs in the inductance of a coil?

Introduction / Context:
Power dissipation in circuits is tied to resistance (or other lossy mechanisms), not ideal reactance. In DC conditions, an ideal inductor behaves as a short circuit with zero average power dissipation; practical heat arises in its winding resistance and other resistive components. Understanding where heat comes from helps with thermal design, efficiency, and reliability.

Given Data / Assumptions:

• DC steady state after any transients have decayed.
• Inductor modeled as ideal L in series with winding resistance R_w.
• Other circuit elements may include resistors and semiconductors with conduction losses.

Concept / Approach:
Instantaneous power p = v * i. For an ideal inductor at steady DC, di/dt = 0, so v_L = L * di/dt = 0; therefore, p_L = v_L * i = 0. Real heating occurs in resistive parts: P = I^2 * R_w in windings, I^2 * R in series resistors, and conduction/switching losses in semiconductors. During transients, energy is temporarily stored or released by the inductor (E = 0.5 * L * I^2), but ideal inductors do not dissipate that energy—resistive paths do.

Step-by-Step Solution:

Verification / Alternative check:
Measure coil temperature with and without current: heating correlates with I^2 * R_w. If the same coil had zero resistance (superconducting), DC heating would vanish despite finite inductance, confirming that resistance—not inductance—causes heat.

Why Other Options Are Wrong:
“Correct” contradicts power relations. Transient-only and core-type qualifiers miss the steady-state point. Superconductors illustrate the opposite: inductance remains, but DC heating disappears because R ≈ 0.

Common Pitfalls:
Confusing reactive energy exchange with dissipation; only resistive elements dissipate real power in ideal analyses.

Final Answer:
Incorrect