Why does the current not instantly reach its maximum value when the switch is closed?

The inductor generates a back emf that opposes the change in current, as described by Faraday's and Lenz's laws. This induced voltage across the inductor is maximum at the instant the switch closes, gradually decreasing as the current changes more slowly, allowing the current to rise exponentially to its final value.

What happens to the energy stored in the inductor when the switch is opened?

The energy (½ L I²) stored in the inductor's magnetic field must be dissipated. In a practical circuit, opening the switch can cause a large voltage spike (v_L = -L di/dt) as the current tries to stop abruptly. Simulators often include a path (like a freewheeling diode) for the current to decay safely through the resistor, converting the magnetic energy into heat.

How does the time constant τ = L/R actually affect the circuit's behavior?

The time constant is a measure of how quickly the circuit responds to changes. A larger L (more inductance) or a smaller R (less resistance) increases τ, making the current take longer to rise or fall. After one time constant, the current reaches about 63% of its final value during rise, or decays to about 37% of its initial value.

Is the inductor voltage always opposite to the battery voltage?

No, only during the initial current rise. Kirchhoff's law states V_battery = v_R + v_L. During rise, v_L is positive (opposing the battery) as it fights the increase. In steady state, di/dt=0 so v_L=0. During decay (after switch opening), the inductor's polarity reverses to try to maintain current, making v_L negative relative to its rise polarity.

RL Circuit

Series RL: τ = L/R, i(t) rise & decay, v_L; DC steady i = V/R.

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About this model

An RL circuit simulator visualizes the dynamic behavior of a simple series circuit containing a resistor (R) and an inductor (L) connected to a DC voltage source, often via a switch. The core physics is governed by Kirchhoff's voltage law and the fundamental properties of the inductor. When the switch is closed, the inductor opposes the change in current through its self-induced electromotive force (emf), given by v_L = -L(di/dt). This leads to a characteristic exponential rise in current from zero to its final steady-state value, described by the equation i(t) = (V/R)(1 - e^{-t/τ}). The key parameter is the time constant, τ = L/R, which dictates the speed of the transient response. After the current reaches steady state (where di/dt = 0 and the inductor behaves like a short circuit), the simulator typically allows the user to open the switch, modeling the decay of current through a freewheeling path. This decay follows i(t) = (V/R)e^{-t/τ}. The model simplifies real-world components by assuming ideal resistors and inductors with zero internal resistance, a perfect DC source, and an instantaneous switch. By manipulating R, L, and V, students directly observe how these parameters affect the time constant, the shape of the current and inductor voltage curves, and the final steady current. This interaction solidifies understanding of electromagnetic induction, transient analysis, and exponential functions in a physical context.

Who it's for: Undergraduate physics and electrical engineering students studying circuit theory, transient analysis, and electromagnetic phenomena.

Key terms

Time Constant (τ)
Inductor
Transient Response
Exponential Decay
Kirchhoff's Voltage Law
Electromotive Force (emf)
RL Circuit
Steady State

How it works

A resistor and inductor in series contrast with RC Circuit: here the current is continuous, while the inductor voltage can jump. With a DC source, L di/dt + R i = V gives i(t) = (V/R)(1 − e^(−t/τ)) from rest, τ = L/R. Removing the source with current flowing yields exponential decay of i with the same τ.

Key equations

τ = L/R, rising: L di/dt + R i = V

decay in RL loop: L di/dt + R i = 0 → i ∝ e^(−t/τ)

Frequently asked questions

Why does the current not instantly reach its maximum value when the switch is closed?: The inductor generates a back emf that opposes the change in current, as described by Faraday's and Lenz's laws. This induced voltage across the inductor is maximum at the instant the switch closes, gradually decreasing as the current changes more slowly, allowing the current to rise exponentially to its final value.
What happens to the energy stored in the inductor when the switch is opened?: The energy (½ L I²) stored in the inductor's magnetic field must be dissipated. In a practical circuit, opening the switch can cause a large voltage spike (v_L = -L di/dt) as the current tries to stop abruptly. Simulators often include a path (like a freewheeling diode) for the current to decay safely through the resistor, converting the magnetic energy into heat.
How does the time constant τ = L/R actually affect the circuit's behavior?: The time constant is a measure of how quickly the circuit responds to changes. A larger L (more inductance) or a smaller R (less resistance) increases τ, making the current take longer to rise or fall. After one time constant, the current reaches about 63% of its final value during rise, or decays to about 37% of its initial value.
Is the inductor voltage always opposite to the battery voltage?: No, only during the initial current rise. Kirchhoff's law states V_battery = v_R + v_L. During rise, v_L is positive (opposing the battery) as it fights the increase. In steady state, di/dt=0 so v_L=0. During decay (after switch opening), the inductor's polarity reverses to try to maintain current, making v_L negative relative to its rise polarity.

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