CalcRL Circuit Calculator
Calculate the time constant of an RL circuit.
What is an RL Circuit?
An RL circuit combines a resistor (R) and an inductor (L) in series or parallel. When voltage is applied, current doesn't instantly reach its final value — the inductor resists changes in current, creating a gradual exponential rise. When power is removed, the inductor's stored magnetic energy keeps current flowing temporarily, decaying exponentially. This behavior is fundamental to motors, transformers, relay coils, and switching power supplies.
Imagine a circuit with a 12 V battery, a 100 Ω resistor, and a 50 mH inductor in series. When you close the switch, current doesn't jump to 12V/100Ω = 0.12 A immediately. Instead, it rises exponentially with time constant τ = L/R = 0.050 H / 100 Ω = 0.0005 s = 0.5 ms. After 0.5 ms, current reaches 63% of final value (0.076 A). After 2.5 ms (5 time constants), it's at 99.3% of 0.12 A — essentially steady state. Open the switch and current decays with the same time constant.
The inductor's opposition to current change comes from Faraday's law: changing current creates changing magnetic flux, which induces a voltage opposing the change. This "back EMF" is V_L = -L(dI/dt). At the instant power is applied, current tries to rise rapidly, inducing maximum back EMF that nearly equals the supply voltage. As current stabilizes, dI/dt approaches zero and the inductor behaves like a short circuit (just its small wire resistance).
How it Works: Formulas Explained
The time constant τ = L/R determines how quickly an RL circuit responds. Inductance L is in henries (H), resistance R in ohms (Ω), giving τ in seconds. After one time constant, current reaches 63.2% of final value during rise, or falls to 36.8% during decay. After 5τ, the transient is 99.3% complete — for practical purposes, steady state.
For a series RL circuit with DC voltage V applied at t=0, current rises as I(t) = (V/R)(1 - e^(-t/τ)). The final steady current is I_final = V/R (inductor acts as short circuit). The voltage across the inductor is V_L(t) = V × e^(-t/τ), starting at V and decaying to zero. Voltage across the resistor is V_R(t) = V(1 - e^(-t/τ)), rising from 0 to V.
When the source is removed (circuit opened or shorted), current decays as I(t) = I₀ × e^(-t/τ), where I₀ is the initial current just before disconnection. The inductor's stored energy E = ½LI² dissipates as heat in the resistor. If the circuit is opened abruptly, dI/dt becomes very large, inducing a huge voltage spike (V = L×dI/dt) — this is why relay coils need flyback diodes to protect switching transistors.
Working through a complete example: A 24 V source connects to R = 120 Ω and L = 60 mH in series. Time constant: τ = L/R = 0.060/120 = 0.0005 s = 500 μs. Final current: I_final = 24/120 = 0.20 A = 200 mA. Current at t = 250 μs (0.5τ): I = 0.20 × (1 - e^(-0.5)) = 0.20 × (1 - 0.607) = 0.20 × 0.393 = 0.0786 A = 78.6 mA. At t = 1 ms (2τ): I = 0.20 × (1 - e^(-2)) = 0.20 × 0.865 = 0.173 A = 173 mA. Inductor voltage at t = 0: V_L = 24 V; at t = 500 μs: V_L = 24 × e^(-1) = 8.83 V.
Step-by-Step Guide
- Identify circuit configuration and component values.** Determine if R and L are in series or parallel. Note the resistance R in ohms and inductance L in henries. Convert units: 1 mH = 0.001 H, 1 μH = 0.000001 H, 1 kΩ = 1000 Ω. Example: R = 2.2 kΩ = 2200 Ω, L = 10 mH = 0.010 H.
- Calculate the time constant τ = L/R.** For the example: τ = 0.010/2200 = 4.55×10⁻⁶ s = 4.55 μs. This is a fast circuit — typical of high-frequency switching applications. A larger inductor or smaller resistance gives longer time constants. Motor windings might have τ = 10-100 ms; RF chokes might have τ = nanoseconds.
- Determine the final steady-state current.** For DC, I_final = V/R (inductor is a short circuit at steady state). With V = 15 V and R = 2200 Ω: I_final = 15/2200 = 0.00682 A = 6.82 mA. This is the maximum current after transients die out. For AC circuits, use impedance Z = √(R² + X_L²) where X_L = 2πfL.
- Write the current equation for your scenario.** Rising current (power applied): I(t) = I_final(1 - e^(-t/τ)). Decaying current (power removed): I(t) = I₀ × e^(-t/τ). Choose based on whether you're analyzing turn-on or turn-off. Time t starts at 0 when the switch changes state.
- Calculate current or voltage at specific times.** At t = τ: I = I_final(1 - e⁻¹) = 0.632 × I_final. At t = 3τ: I = 0.950 × I_final. At t = 5τ: I = 0.993 × I_final — essentially complete. For decay: at t = τ, I = 0.368 × I₀. Example: After 13.65 μs (3τ) in our circuit: I = 0.950 × 6.82 mA = 6.48 mA.
- Find inductor voltage if needed.** During rise: V_L(t) = V × e^(-t/τ). During decay with a freewheeling path: V_L(t) = -I₀R × e^(-t/τ). The negative sign indicates polarity reversal — the inductor becomes a source, not a load. Peak voltage during abrupt opening can reach hundreds or thousands of volts, limited only by arcing or breakdown.
Real-World Examples
Example 1: Relay coil drive circuit. A 12 V relay has coil resistance R = 400 Ω and inductance L = 2 H. Time constant: τ = 2/400 = 0.005 s = 5 ms. Pull-in current is 20 mA; final current is 12/400 = 30 mA. Time to reach pull-in: 0.020 = 0.030(1 - e^(-t/0.005)). Solving: 0.667 = 1 - e^(-t/0.005), so e^(-t/0.005) = 0.333, giving t = -0.005 × ln(0.333) = 5.5 ms. The relay closes 5.5 ms after voltage is applied. When turned off, a flyback diode clamps the voltage spike, and current decays with τ = 5 ms through the diode loop.
Example 2: DC motor startup. A DC motor armature has R = 0.5 Ω and L = 10 mH. Applied voltage is 24 V. Initial time constant: τ = 0.010/0.5 = 0.02 s = 20 ms. Final current (if motor were stalled): I = 24/0.5 = 48 A — huge! But as the motor speeds up, back EMF reduces effective voltage. At t = 0, current rises at rate dI/dt = V/L = 24/0.010 = 2400 A/s. After 20 ms, current would reach ~30 A if not for back EMF. Motor starters use series resistance or PWM to limit inrush current.
Example 3: Switching power supply inductor. A buck converter uses L = 10 μH with effective series resistance R = 0.05 Ω. Time constant: τ = 10×10⁻⁶/0.05 = 200 μs. The converter switches at 500 kHz (period = 2 μs), much faster than τ. During the ON time (1 μs at 50% duty), current rises by ΔI = (V_in - V_out) × t_on / L = (12 - 5) × 1×10⁻⁶ / 10×10⁻⁶ = 0.7 A. During OFF time, current falls through the freewheeling diode. Average output current depends on load; ripple current is ±0.35 A around the average.
Example 4: Ignition coil spark generation. An automotive ignition coil has primary inductance L = 8 mH and resistance R = 2 Ω. With 12 V applied, τ = 0.008/2 = 4 ms. After 12 ms (3τ), current reaches 95% of 12/2 = 6 A. Stored energy E = ½LI² = 0.5 × 0.008 × 36 = 0.144 J. When the points or transistor open, current tries to stop instantly. The induced voltage V = L×dI/dt can exceed 200 V on the primary. The transformer steps this up 100:1 to 20,000+ V on the secondary, creating the spark plug arc.
Example 5: MRI magnet quench protection. An MRI superconducting magnet has L = 100 H operating at I = 500 A. Stored energy E = ½ × 100 × 500² = 12.5 MJ — equivalent to 3 kg of TNT. During normal operation, R = 0 (superconductor), so τ = ∞ (current flows forever). If the magnet "quenches" (goes normal), resistance suddenly appears (R ≈ 0.1 Ω). Time constant becomes τ = 100/0.1 = 1000 s. Decay is slow, but power dissipation P = I²R = 500² × 0.1 = 25,000 W — 25 kW of heat in the cryostat! Quench protection circuits rapidly dump energy into external resistors.
Common Mistakes to Avoid
Confusing RL and RC time constants. RL circuits use τ = L/R; RC circuits use τ = RC. They're reciprocals in form (L/R vs. RC). A common error: calculating τ = R/L for an RL circuit, which gives units of 1/s instead of seconds. Dimensional analysis catches this: henry/ohm = (V·s/A)/(V/A) = seconds. Check your units — if τ doesn't come out in seconds, you've inverted the formula.
Forgetting that inductors are short circuits at DC steady state. After transients die out (t > 5τ), a DC inductor acts as a wire (zero voltage across it, limited only by wire resistance). Students sometimes keep the inductor in calculations for steady-state DC, getting wrong answers. For DC steady state: inductor = short circuit, capacitor = open circuit. Use this simplification for final values.
Ignoring voltage spikes when opening inductive circuits. Opening an RL circuit forces dI/dt to be very large (current goes to zero instantly). Since V = L×dI/dt, this creates enormous voltage — easily hundreds or thousands of volts. This destroys transistors, causes arcing in switches, and creates electromagnetic interference. Always provide a current path (flyback diode, snubber circuit, varistor) when switching inductive loads.
Using the wrong initial condition for decay. When analyzing current decay, I₀ must be the current flowing just before the switch changes, not the final steady-state current from a different configuration. If a circuit has been on long enough, I₀ = V/R. But if power is removed before steady state is reached, I₀ is whatever current existed at that moment: I₀ = (V/R)(1 - e^(-t_on/τ)).
Pro Tips
Use the 63% rule for quick mental math. After one time constant, rising current reaches 63% of final; decaying current falls to 37% of initial. After 2τ: 86% rise or 14% decay. After 3τ: 95% rise or 5% decay. Memorize these benchmarks: 1τ = 63%, 2τ = 86%, 3τ = 95%, 4τ = 98%, 5τ = 99%. For most purposes, 3τ is "mostly done" and 5τ is "completely done."
Estimate inductance from physical parameters. For a solenoid: L ≈ μ₀μ_r × N² × A / l, where N is turns, A is cross-sectional area, l is length, μ₀ = 4π×10⁻⁷ H/m, and μ_r is core relative permeability. An air-core coil with N = 100, A = 1 cm², l = 5 cm: L ≈ (4π×10⁻⁷) × 10000 × 10⁻⁴ / 0.05 = 2.5×10⁻⁵ H = 25 μH. A ferrite core (μ_r = 2000) multiplies this by 2000: L = 50 mH.
Apply Thevenin equivalent for complex circuits. When an inductor connects to a complex resistor network, find the Thevenin equivalent resistance seen by the inductor. Replace everything else with V_th and R_th. Then τ = L/R_th and I_final = V_th/R_th. This reduces any linear circuit to a simple series RL problem. Turn off independent sources (voltage sources → shorts, current sources → opens) to find R_th.
Recognize critical damping in RLC circuits. Adding capacitance creates an RLC circuit. Damping depends on R relative to √(L/C). Critical damping (fastest response without oscillation) occurs at R = 2√(L/C). Underdamped (R smaller) rings/oscillates; overdamped (R larger) responds sluggishly. Snubber circuits choose R and C to critically damp voltage spikes from inductive switching.
FAQs
The inductor's voltage is V = L(dI/dt) — proportional to the rate of change of current, not the current itself. For sinusoidal AC, the derivative of sine is cosine, which leads by 90°. So voltage peaks occur when current is changing fastest (at current's zero crossing), and voltage is zero when current peaks (no change). In phasor notation, voltage leads current by 90°, or equivalently, current lags voltage by 90°.
For AC, use impedance instead of resistance. Inductive reactance X_L = 2πfL increases with frequency. Total impedance Z = √(R² + X_L²). Current I = V/Z. Phase angle φ = arctan(X_L/R) — current lags voltage by φ degrees. At high frequency, X_L dominates and current is small. At DC (f = 0), X_L = 0 and only R limits current. This frequency dependence makes RL circuits useful as filters.
The diode must handle the peak inductor current (I_peak = V/R) and block the supply voltage when conducting. Reverse voltage rating should exceed supply voltage by 20-50%. Forward current rating should exceed steady-state coil current. Speed matters: fast recovery or Schottky diodes minimize voltage overshoot during turn-off. For a 12 V, 100 mA relay, a 1N4148 or 1N4001 works. For higher power, use UF4007 or MURS series.
Yes — an RL low-pass filter takes output across the resistor. At low frequency, X_L is small, so most voltage appears across R. At high frequency, X_L is large, dropping most voltage. Cutoff frequency f_c = R/(2πL). An RL high-pass filter takes output across the inductor. RC filters are more common (capacitors are cheaper and smaller than inductors), but RL filters handle higher currents and work well in power applications.
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