RL Circuits

RL Circuits building off Inductance

• Overview

  • An RL (Resistor-Inductor) circuit contains both inductors and resistors.
  • These circuits are fundamental in understanding how the current changes when influenced by inductance and resistance.

• Inductors in a Circuit

  • Inductors oppose changes in current. When current changes, an electromotive force (emf) is induced.
  • The induced emf ($\epsilon$) across an inductor is given by:
    • $\epsilon =-L\frac{dI}{dt}$
    • Where $L$ is the inductance and $\frac{dI}{dt}$ is the rate of change of current.

• RL Circuit during Current Change

  • When an RL circuit is connected to a voltage source $V_0$ at $t=0$, the current through the inductor changes over time:
    • $I(t)=\frac{V_0}{R}\left(1-e^{-\frac{R}{L}t}\right)$
    • Voltage across the Resistor: $V_R(t)=I(t)R=V_0\left(1-e^{-\frac{R}{L}t}\right)$
    • Voltage across the Inductor: $V_L(t)=V_0{e}^{-\frac{R}{L}t}$
  • The time constant ($\tau$) of an RL circuit is defined as $\tau =\frac{L}{R}$, indicating how quickly the current reaches its maximum value.

• Example of RL Circuit during Current Change

  • Given:

    • Inductance: $L=1\text{H}$
    • Resistance: $R=10\Omega$
    • Voltage source: $V_0=5\text{V}$

  • Calculate:

    • Time constant ($\tau$): $\tau =\frac{L}{R}=\frac{1\text{H}}{10\Omega }=0.1\text{sec}$
    • Current at $t=0.2\text{sec}$: $I(t)=\frac{5\text{V}}{10\Omega }\left(1-e^{-\frac{10}{1}\cdot 0.2}\right)=0.5\left(1-e^{-2}\right)\text{A}$

• Discharging an Inductor

  • When the switch in an RL circuit is opened, the current through the circuit decreases exponentially.
  • The current and voltage at any time $t$ during discharging are given by:
    • $I(t)=I_0{e}^{-\frac{R}{L}t}$
    • $V_L(t)=-L\frac{dI}{dt}=I_{0}R{e}^{-\frac{R}{L}t}$
    • $I_0$ is the initial current at $t=0$.

• Example of Discharging RL Circuit

  • Given:

    • Inductance: $L=2\text{H}$
    • Resistance: $R=5\Omega$
    • Initial current: $I_0=1\text{A}$

  • Calculate:

    • Time constant ($\tau$): $\tau =\frac{L}{R}=\frac{2\text{H}}{5\Omega }=0.4\text{sec}$
    • Current at $t=0.5\text{sec}$: $I(t)=1\text{A}\times e^{-\frac{5}{2}\cdot 0.5}=e^{-1.25}\text{A}$
    • Voltage across inductor at $t=0.5\text{sec}$: $V_L(t)=-L\frac{dI}{dt}=-2\text{H}\times \left(-\frac{5}{2}{e}^{-1.25}\text{A/sec}\right)=5e^{-1.25}\text{V}$[1]``[2]``[3].