What is the expression of current in an RL circuit?
2 Answers
In an RL circuit, the expression for the current \( I(t) \) flowing through the circuit as a function of time \( t \) can be derived from the differential equation governing the circuit. The basic form of the equation for a series RL circuit (with a resistor \( R \) and an inductor \( L \) in series connected to a voltage source \( V \)) is:
\[
V = L \frac{dI}{dt} + RI
\]
When the circuit is switched on at \( t = 0 \) (assuming the initial current is zero), the solution for the current as a function of time is:
\[
I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right)
\]
Here:
- \( V \) is the voltage of the power source,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( e \) is the base of the natural logarithm,
- \( t \) is the time after the circuit is closed.
As \( t \) approaches infinity, the current approaches its steady-state value \( I_{\text{max}} = \frac{V}{R} \). The time constant \( \tau \) of the circuit is given by:
\[
\tau = \frac{L}{R}
\]
This time constant indicates how quickly the current reaches its steady-state value.
\[
V = L \frac{dI}{dt} + RI
\]
When the circuit is switched on at \( t = 0 \) (assuming the initial current is zero), the solution for the current as a function of time is:
\[
I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} t}\right)
\]
Here:
- \( V \) is the voltage of the power source,
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( e \) is the base of the natural logarithm,
- \( t \) is the time after the circuit is closed.
As \( t \) approaches infinity, the current approaches its steady-state value \( I_{\text{max}} = \frac{V}{R} \). The time constant \( \tau \) of the circuit is given by:
\[
\tau = \frac{L}{R}
\]
This time constant indicates how quickly the current reaches its steady-state value.