Explain self-inductance and voltage reference signs

A changing current changes its own magnetic field

The earlier induction models prescribed an external field. A coil also produces a field from its own current. If that current changes, the coil’s own flux linkage changes and an induced emf opposes the change. This is self-inductance. It is a circuit property associated with magnetic energy, not an additional resistor that permanently consumes power.

The browser uses an ideal linear inductor of L = 2 H, where H is the henry. For a fixed linear geometry, its self-flux linkage can be written NΦ self = LI. L is inductance and I is current. Thus one henry corresponds to one weber-turn per ampere, equivalently one volt-second per ampere. Coil geometry and magnetic materials determine actual inductance; the displayed coil symbol does not calculate it from dimensions.

Start at R = 2 Ω and t = 0. Predict the buildup current and the voltage across the inductor. Compare that row with the separate decay row. Their initial conditions are different; the diagram is not switching a single saved current between circuits.

Use a clear voltage convention

Faraday’s law gives the self-induced emf emf self = −L dI/dt for a constant linear inductance. Here dI/dt means the current’s rate of change in amperes per second. The minus sign expresses opposition to the change. It does not mean the coil always opposes the existing current: during decay the induced effect helps sustain that current.

Circuit analysis often reports the passive voltage drop V L = +L dI/dt instead. The current reference enters the terminal marked “+ ref”; V L is that terminal’s potential relative to the “− ref” terminal. The passive drop is the negative of the induced emf along the current direction. These are two sign conventions for the same relation, not conflicting laws. The browser consistently displays the passive drop.

At the start of buildup, current is zero and the 6 V source appears across the inductor: V L = +6 V. Its initial current slope is 6 V / 2 H = 3 A/s. The resistor drop is initially zero because I = 0. The current need not already be large to be changing rapidly.

For the decay case at R = 2 Ω, the independently prepared initial current is 3 A. The source is absent but the loop remains closed. At t = 0, resistor drop is +6 V and inductor drop is −6 V, summing to zero. The slope is −3 A/s. Current is still positive while decreasing; negative voltage does not imply reversed current.

Finite voltage limits how fast current changes

If inductance is finite and the voltage is bounded, the current cannot jump instantaneously. Integrating the relation over a short interval gives change in current = (1/L) × accumulated voltage-time area. A finite current jump would require an unbounded ideal voltage impulse as the interval tends to zero. Real circuit capacitance, switching paths and breakdown affect actual interruptions; this model does not predict their voltages.

The decay diagram retains a closed resistor–inductor path. It does not model opening a current-carrying inductor into an otherwise empty circuit. Do not treat its bounded −6 V initial drop as the voltage of an arbitrary disconnected coil, or use the browser as a physical switching procedure.

Compare the two kinds of transient storage

A capacitor stores electric-field energy and resists an instantaneous voltage change under finite current. An ideal inductor stores magnetic-field energy and resists an instantaneous current change under finite voltage. This is a useful comparison, but the circuit equations differ: the RL time constant is L/R, while the earlier RC time constant was RC.

At a long time in the ideal source-connected RL circuit, current becomes nearly constant, so V L becomes nearly zero even though current and stored magnetic energy are nonzero. The ideal inductor then has no resistive voltage drop. A real winding can have resistance and core losses; those are not produced by the ideal L element. Here the stated loop resistance accounts for all modeled dissipation.

Transfer to a different inductor

An ideal 0.40 H inductor carries current that rises uniformly from 0 to 2.0 A in 0.10 s. Calculate its passive voltage drop and self-induced emf. Then consider a uniform decrease from 2.0 to 0 A over the same interval, with the same reference.

Check the self-inductance transfer

The rising slope is +20 A/s, so passive drop is +0.40 × 20 = +8.0 V and self-induced emf is −8.0 V. During the decrease, the slope is −20 A/s, passive drop −8.0 V and induced emf +8.0 V. These are prescribed linear current changes, unlike the browser's exponential histories. A positive current can therefore coexist with a negative passive drop while energy is released.

Use the two initial-state rows to explain the signs of current, voltage and current slope. Identify the closed path in each diagram and state why an open-circuit interruption would need another model.

Reference for the circuit voltage relation: OpenStax: RL circuits. The sign-convention comparisons and transfer exercise are original.