Step-by-Step Example of an Inductor in a DC Transient RL Circuit

For a 12V DC source switched into a series circuit with a 20Ω resistor and a 50mH inductor, the time constant (τ) is 2.5 ms. At exactly t = 2.5 ms, the current is 0.379 A, which represents 63.2% of the 0.6 A steady-state maximum. This direct answer anchors our walkthrough, but understanding the derivation and the physical realities of the component is where the real engineering value lies.

The Core Problem: DC Transient Response

When analyzing an example of an inductor in a transient DC state, we are looking at how the component resists changes in current by storing energy in its magnetic field. The governing physics dictate that current cannot change instantaneously across an inductor.

Step-by-Step Algebraic Solution

We will solve this using Kirchhoff’s Voltage Law (KVL) and first-order ordinary differential equations. No steps are skipped.

1. Write the KVL equation:
   The sum of the voltage drops across the resistor and inductor must equal the source voltage.
   V_s = iR + L(di/dt)
12 = 20i + 0.05(di/dt)
2. Isolate the derivative term:
   Divide the entire equation by the inductance (0.05 H) to normalize the derivative.
   12 / 0.05 = (20 / 0.05)i + di/dt
240 = 400i + di/dt
3. Rearrange for separation of variables:
   di/dt = 240 - 400i
   Factor out -400 to prepare for integration:
   di/dt = -400(i - 0.6)
di / (i - 0.6) = -400 dt
4. Integrate both sides:
   ∫ [1 / (i - 0.6)] di = ∫ -400 dt
ln|i - 0.6| = -400t + C
5. Apply initial conditions to find C:
   At t = 0, the current i = 0 (inductor opposes instant current change).
   ln|0 - 0.6| = -400(0) + C
C = ln(0.6) (Note: we drop the absolute value as we know the steady state approaches 0.6 from below, but mathematically ln|-0.6| resolves to the same constant offset in the final exponent).
6. Substitute C and solve for i(t):
   ln|i - 0.6| - ln(0.6) = -400t
   Using logarithm rules (ln(a) - ln(b) = ln(a/b)):
   ln((0.6 - i) / 0.6) = -400t
   Exponentiate both sides:
   (0.6 - i) / 0.6 = e^(-400t)
0.6 - i = 0.6 * e^(-400t)
i(t) = 0.6 - 0.6 * e^(-400t)
   Final Equation: i(t) = 0.6(1 - e^(-400t))
7. Evaluate at t = 2.5 ms (0.0025 seconds):
   i(0.0025) = 0.6(1 - e^(-400 * 0.0025))
i(0.0025) = 0.6(1 - e^(-1))
i(0.0025) = 0.6(1 - 0.367879)
i(0.0025) = 0.6(0.632121) = 0.37927 A

Sanity Check and The 'Flyback' Trap

Sanity Check: The units are Amperes. The order of magnitude is correct: 0.379 A sits logically between the initial 0 A and the steady-state maximum of 0.6 A (calculated via Ohm's law: 12V / 20Ω). Furthermore, the time constant τ = L/R = 0.05 / 20 = 0.0025 s (2.5 ms). By definition, at exactly one time constant (1τ), an RL circuit reaches 63.2% of its final value. 63.2% of 0.6 A is 0.379 A. The math holds up perfectly.

The Trap in this Problem: The algebraic solution assumes the switch stays closed. In a real-world bench scenario, the trap occurs when you open the switch. The inductor attempts to maintain the 0.379 A current flow. The voltage across an inductor is defined by V = L(di/dt). If the mechanical switch breaks the circuit in 1 microsecond (1μs), the rate of change of current is massive:

V = 0.05 H * (0.379 A / 0.000001 s) = 18,950 Volts

This 18.9 kV spike will instantly arc across the mechanical switch contacts, welding them shut over time, or if you used a MOSFET, it will punch through the drain-source breakdown voltage (V_DSS) and destroy the silicon. Always place a flyback diode (like a 1N4007) in reverse parallel across the inductor to clamp this spike to ~0.7V.

Decision Path: Selecting a Real-World Inductor

Theoretical 50mH inductors are easy to draw; physical ones require careful selection based on DC bias current and core saturation. Use this decision matrix to select the right physical component.

The Concrete Pick: For this specific 50mH, 0.6A circuit, specify the Bourns RLB9012-503KL. It is a 50mH radial power inductor rated for 1.1 A maximum DC current, ensuring the core will not saturate at our 0.6 A steady state. It has a DC resistance (DCR) of 0.32Ω, which is low enough that our assumption of a 20Ω dominant circuit resistance remains valid (total R = 20.32Ω, altering the final current by a negligible 1.6%).

Frequently Asked Questions

Which theorem or method applies here and why?
We applied Kirchhoff’s Voltage Law (KVL) combined with the constitutive equation of an inductor (V = L(di/dt)). Because the circuit contains only one energy-storage element (the inductor), it forms a first-order linear ordinary differential equation (ODE). KVL is the mandatory starting point for any lumped-parameter circuit analysis to establish the relationship between the source and the voltage drops.

How can I verify this answer independently without doing the algebra?
The fastest verification method is to use a free SPICE simulator like LTspice. Build the schematic with a 12V step source, a 20Ω resistor, and a 50mH inductor. Run a Transient (.tran) simulation for 10ms. Place a cursor at exactly 2.5ms on the current trace; the simulator will read out ~0.379 A, confirming the analytical derivation. Alternatively, you can review the foundational transient theory on All About Circuits or Electronics Tutorials to cross-reference the 63.2% time-constant rule.

Does the physical layout of the breadboard affect this result?
At DC and low-frequency transients (like our 2.5ms time constant), breadboard parasitic capacitance (typically ~2pF between rows) and trace inductance (~10nH per cm) are entirely negligible. However, if you were analyzing a switch-mode power supply operating at 500 kHz, those parasitics would introduce high-frequency ringing that our first-order ODE completely fails to predict.