Question

What is the formula for current in a RL circuit?

Answer 1

To provide the ultimate answer, are you looking for the formula during transient response or steady-state conditions?

Answer 2

In an RL circuit, which consists of a resistor (R) and an inductor (L) connected in series, the formula for the current depends on the nature of the voltage applied and the time since the circuit was energized. Here’s a detailed breakdown of the formulas used for different scenarios:

### 1. **DC Voltage Source (Steady-State Condition)**

When a direct current (DC) voltage source \( V \) is applied to the RL circuit and has been connected long enough for the circuit to reach a steady state (after a long time), the inductor behaves like a short circuit (its reactance becomes zero). Therefore, the current in the circuit can be found using Ohm's Law:

\[ I_{\text{steady-state}} = \frac{V}{R} \]

### 2. **DC Voltage Source (Transient Response)**

When the circuit is first energized, the current does not immediately reach its steady-state value due to the inductor’s property of opposing changes in current. Instead, it increases gradually. The transient response of the current \( I(t) \) as a function of time \( t \) is given by:

\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]

Here’s what each term represents:
- \( I(t) \) is the current at time \( t \).
- \( V \) is the voltage of the DC source.
- \( R \) is the resistance.
- \( L \) is the inductance.
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

This formula describes how the current grows from 0 to its steady-state value \( \frac{V}{R} \) over time. The term \( \frac{R}{L} \) is known as the time constant \( \tau \) of the RL circuit:

\[ \tau = \frac{L}{R} \]

### 3. **AC Voltage Source**

For an alternating current (AC) voltage source, the analysis is a bit more complex. The current \( I(t) \) in response to an AC voltage \( V(t) = V_0 \sin(\omega t) \), where \( V_0 \) is the peak voltage and \( \omega \) is the angular frequency of the AC source, can be expressed using phasor notation. The impedance \( Z \) of the RL circuit is:

\[ Z = \sqrt{R^2 + (\omega L)^2} \]

The magnitude of the current \( I \) is:

\[ I = \frac{V_0}{Z} \]

And the phase angle \( \phi \) between the voltage and current is:

\[ \phi = \arctan\left(\frac{\omega L}{R}\right) \]

Thus, the current in the circuit can be expressed as:

\[ I(t) = \frac{V_0}{Z} \sin(\omega t - \phi) \]

### Summary

- **Steady-State DC**: \( I = \frac{V}{R} \)
- **Transient DC**: \( I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \)
- **AC**: \( I(t) = \frac{V_0}{\sqrt{R^2 + (\omega L)^2}} \sin(\omega t - \phi) \)

These formulas cover the behavior of current in an RL circuit under various conditions.