How does RL circuit current behave at startup?

When voltage V applied to RL series circuit: I(t) = V/R·(1 - e^(-Rt/L)). Current starts at 0, rises exponentially to V/R steady state. Inductor opposes sudden current changes (back-EMF = L·dI/dt).

What happens when RL circuit voltage is suddenly removed?

Inductor maintains current: back-EMF = L·dI/dt can cause very high voltage spike if path opens. Freewheeling diode provides current path to prevent spike. This protects transistors switching inductive loads.

What energy is stored in an inductor?

W = ½LI². For L = 1H and I = 10A: W = 50 J. This energy must be absorbed safely when current is interrupted. Arc suppression and snubber circuits prevent voltage spikes from inductor energy release.

RL Time Constant Calculator

Calculate RL time constant τ = L/R. Current rises to 63.2% in 1τ after voltage step. I(t) = (V/R)·(1-e^(-t/τ)). Energy stored in inductor W = ½LI². Motor startup transient analysis.

2D stress state at a point. All stresses in MPa.

σx — Normal Stress X (MPa)

σy — Normal Stress Y (MPa)

τxy — Shear Stress (MPa)

⚙ Mechanical Note: Results are based on standard textbook formulas and ISO/AASHTO references. Always verify critical designs with applicable codes (ISO 898, AASHTO 1993, AISC, Eurocode 3) and a qualified engineer.

How to Use

Enter the values for the patient or scenario you are assessing. Calculate RL time constant τ = L/R. Current rises to 63.2% in 1τ after voltage step. I(t) = (V/R)·(1-e^(-t/τ)). Energy stored in inductor W = ½LI². Motor startup transient analysis. Use the rl time constant result to inform your clinical assessment.

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Frequently Asked Questions

How does RL circuit current behave at startup?

What happens when RL circuit voltage is suddenly removed?

What energy is stored in an inductor?