In a simple DC RL circuit, the instantaneous current $i(t)$ as a function of time is governed by the following differential equation. The circuit consists of a resistor $R$, an inductor $L$, and a voltage source $V$. The general form of the equation is derived from Kirchhoff's Voltage Law (KVL) applied to the circuit.

The equation for the voltage drops across the resistor and inductor is:

$V = L \frac{di(t)}{dt} + i(t) R$

Where:

• $i(t)$ is the current at time $t$,
• $L$ is the inductance of the coil,
• $R$ is the resistance,
• $V$ is the constant voltage of the source.

The solution to this differential equation depends on whether the circuit is being powered on or off (i.e., during the charging or discharging phase of the current).

Charging Phase (Switch is closed at $t = 0$):

When the switch is closed at $t = 0$, the current increases over time. The solution to the differential equation is:

$i(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right)$

Where:

• $\frac{V}{R}$ is the maximum current (also called the steady-state current),
• $e^{-\frac{R}{L} t}$ represents the time-dependent part of the current growth.

Key Characteristics:

• Initial current at $t = 0$ is 0, since the inductor initially resists any change in current.
• Steady-state current: As $t \to \infty$, $i(t)$ approaches $\frac{V}{R}$, which is the final current once the inductor no longer resists changes.

Discharging Phase (Switch is open at $t = 0$):

If the source is disconnected (i.e., the switch opens), the current decays over time according to:

$i(t) = i_0 e^{-\frac{R}{L} t}$

Where $i_0$ is the initial current at $t = 0$, and the exponential term describes the decay of the current due to the resistor dissipating energy.

Would you like further explanation on how to derive these equations or explore some examples?

---

Here are some related questions:

1. How does the inductance value affect the time constant in an RL circuit?
2. What is the time constant of an RL circuit and how is it calculated?
3. How does the behavior of the current change if the resistor value is doubled in a charging RL circuit?
4. What happens in an RL circuit when the voltage source is suddenly turned off?
5. Can the formula for the charging current be used for AC circuits as well?

Tip: In RL circuits, the time constant $\tau = \frac{L}{R}$ is crucial for understanding how quickly the current reaches its steady-state value.