Predict resistor–inductor current transients

Follow current buildup through a resistor and inductor

For the source-connected history, the voltage balance is 6 V = IR + L dI/dt. The source voltage is shared between the resistor drop and the inductor drop. At the initial instant, current is zero and its slope is largest. As current rises, IR grows, leaving less voltage across the inductor and a smaller slope. This feedback produces an exponential approach to a steady value, rather than a constant rate of increase.

The model fixes L = 2 H and source voltage V s = 6 V. It lets you vary R from 1 to 5 Ω and select time from 0 to 5 s. The time control chooses a point on an analytic history; it is not a running clock. Changing resistance starts a different hypothetical history, including a different initial current in the decay comparison.

Interpret the time constant with units

The steady current is I limit = V s/R and the time constant is τ = L/R. Since H = V s/A and Ω = V/A, the ratio H/Ω has units seconds. Tau describes the fractional approach to the limit, not the instant of exact completion.

Starting from zero current, I buildup(t) = I limit [1 − exp(−t/τ)]. The function exp is the exponential; its dimensionless input t/τ counts time constants. At one time constant, about 63.2% of the final current has been reached. The remaining gap is about 36.8% of the original gap. At five time constants, about 99.33% has been reached, still not exactly 100%.

At R = 2 Ω, τ = 1 s and I limit = 3 A. At t = 1 s, current is approximately 1.89636 A. The resistor drop is approximately 3.79272 V and the inductor drop 2.20728 V. Their sum is the source’s 6 V. Capture t = 0, 1 and 5 s and explain the changing division of voltage.

Compare a separate source-free decay

The second history begins at I₀ = 6/R A with the source removed and the resistor–inductor loop still closed. This assumes prior preparation at steady current; it is not the current reached by the buildup row at the selected finite time. Each row has its own time zero.

The loop balance is 0 = IR + L dI/dt, giving I decay(t) = I₀ exp(−t/τ). Current remains positive in the shared clockwise reference. At one time constant, it is 36.8% of its initial value. At R = 2 Ω and t = 1 s, it is approximately 1.10364 A, with resistor drop +2.20728 V and inductor drop −2.20728 V.

The negative inductor drop supplies the resistor while stored energy decreases. Unlike the capacitor comparison, current in these two inductor histories uses the same positive direction. Do not borrow a sign from an RC plot without checking its defined reference.

Change resistance without confusing rate and size

At R = 4 Ω, τ becomes 0.50 s and the steady current becomes 1.50 A. The normalized approach is faster but its final value is smaller. Compare R = 2 Ω at 1 s with R = 4 Ω at 0.50 s: both are one time constant, yet the currents differ by a factor of two because their limits differ.

For this RL circuit, increasing resistance decreases L/R. For the earlier RC circuit, increasing resistance increased RC. Memorizing “more resistance always means slower” would fail here. Explain which quantity is stored and which differential relation governs it.

Changing R also changes the initial current and energy used by the decay row. A decay-time comparison under a fixed initial current would require a different preparation than this browser uses. Report this condition before claiming what resistance alone does to total dissipated energy.

Use a threshold to infer a time scale

A pure exponential current decay reaches half its initial value at t half = τ ln 2, where ln denotes natural logarithm. If an independently observed current halves after 0.70 s, the ideal estimate is τ = 0.70/ln 2 ≈ 1.010 s. With independently known R = 2 Ω, that would imply L ≈ 2.020 H. A single time-scale estimate identifies L/R; it cannot independently identify both L and R or establish that the exponential model describes all observations. Reserve another time point to test it.

The browser’s half-second sampling steps need not land on an exact half-current time. Interpreting a model formula is different from claiming the interface measured that threshold.

Transfer to another RL circuit

A 9 V source connects to a 3 Ω resistor and an ideal 1.5 H inductor starting at zero current. Find τ, the limiting current and the current at 0.50 s. Predict the passive inductor drop at that time.

Check the RL transient transfer

τ = 1.5/3 = 0.50 s and I limit = 9/3 = 3 A. At one time constant, current is 3(1 − exp(−1)) ≈ 1.89636 A. Resistor drop is about 5.68909 V and inductor drop about 3.31091 V, summing to 9 V. These components differ from the browser's fixed source and inductance.

Report a voltage balance, a fractional time-scale comparison and a distinction between initial conditions. A plotted curve is an ideal prediction, not measured component performance.

Reference: OpenStax: resistor–inductor transients. Comparisons, inference example and transfer task are original.