Transient Response of RL Circuit

The changing voltages and current that result during the storing of energy in the form of a magnetic field by an inductor in a dc circuit can best be described using the circuit of [Fig. 1].
Fig. 1: Basic R-L transient network
At the instant the switch is closed, the inductance of the coil will prevent an instantaneous change in current through the coil. The potential drop across the coil, $v_L$, will equal the impressed voltage E as determined by Kirchhoff's voltage law since $v_R = iR = (0)R = 0 V$. The current $i_L$ will then build up from zero, establishing a voltage drop across the resistor and a corresponding drop in $v_L$.
The current will continue to increase until the voltage across the inductor drops to zero volts and the full impressed voltage appears across the resistor. Initially, the current $i_L$ increases quite rapidly, followed by a continually decreasing rate until it reaches its maximum value of $E/R$.
Fig. 2: Circuit of Fig. 1 the instant the switch is closed.
Fig. 3: Circuit of Fig. 1 under steady-state conditions
You will recall from the discussion of capacitors that a capacitor has a short-circuit equivalent when the switch is first closed and an open-circuit equivalent when steady-state conditions are established. The inductor assumes the opposite equivalents for each stage. The instant the switch of the RL transient circuit in [Fig. 1] is closed, the equivalent network will appear as shown in [Fig. 2]. The inductor obviously meets all the requirements for an open-circuit equivalent: $v_L = E$ volts, and $i_L = 0 A$.
When steady-state conditions have been established in the RL transient circuit and the storage cycle (phase) is complete, the "equivalent" network will appear as shown in [Fig. 3]. The network clearly reveals the following:
An ideal inductor ($R_l = 0 Ω$) assumes a short-circuit equivalent in a dc network once steady-state conditions have been established in the RL transient circuit.
Fortunately, the mathematical equations for the voltages and current for the storage phase are similar in many respects to those encountered for the RC network. The experience gained with these equations in chapter 9 will undoubtedly make the analysis of RL networks somewhat easier to understand.
The equation for the current $i_L$ during the storage phase is the following:
$$ \bbox[10px,border:1px solid grey]{i_L = I_m(1 - e^{-t/\tau}) = {E\over R}(1 - e^{-t/\tau})}\tag{1}$$
Note the factor ($1 - e^{-t/\tau}$), which also appeared for the voltage $v_C$ of a capacitor during the charging phase. A plot of the equation is given in [Fig. 4], clearly indicating that the maximum steady-state value of $i_L$ is $E/R$, and that the rate of change in current decreases as time passes.
Fig. 4: Plotting the waveform for $i_L$ during the storage cycle.
The abscissa is scaled in time constants, with $\tau$ for inductive circuits defined by the following:
$$ \begin{split} v_L &= L {di \over dt}\\ \end{split} $$
Taking the derivative of $i_L$ given in eq.(1) we get
$$ \begin{split} v_L &= L {di \over dt} = L {E \over R} {d(1-e^{-t/\tau}) \over dt}\\ &= {L\,E \over R} (0-({-1 \over \tau} e^{-t/\tau}))\\ v_L &= {L\,E \over R} ({1 \over \tau}) e^{-t/\tau}\\ \end{split} $$
At time $t=0$, inductor remains open circuit and voltage $v_L$ across inductor is equal to the applied voltage $E$. Therefore
$$ \begin{split} E&= {L\,E \over R} ({1 \over \tau}) e^{-0/\tau}\\ 1&= {L \over R} ({1 \over \tau}) (1)\\ \tau &= {L \over R} \\ \end{split} $$
$$\bbox[10px,border:1px solid grey]{\tau = {L \over R}} \, (seconds,s)\tag{2}$$
Our experience with the factor $(1 - e^{-t/\tau})$ verifies the level of 63.2% after one time constant, 86.5% after two time constants, and so on. For convenience, [Fig. 5] is to evaluate the functions $(1 - e^{-t/\tau})$ and $e^{-t/\tau}$ at various values of t.
Fig. 5: Plotting the functions $y = 1 - e^{-t/\tau}$ and $ y = e^{-t/\tau}$
If we keep R constant and increase L, the ratio L/R increases and the rise time increases. The change in transient behavior for the current $i_L$ is plotted in [Fig. 6] for various values of L.
Effect of L on the shape of the iL storage
Fig. 6: Effect of L on the shape of the iL storage waveform.
For most practical applications, we will assume that the storage phase has passed and steady-state conditions have been established once a period of time equal to five time constants has occurred.
In addition, since L/R will always have some numerical value, even though it may be very small, the period $5\tau$ will always be greater than zero, confirming the fact that the current cannot change instantaneously in an inductive network. In fact,
the larger the inductance, the more the circuit will oppose a rapid buildup in current level.
[Fig. 2] and [3] clearly reveal that the voltage across the coil jumps to $E$ volts when the switch is closed and decays to zero volts with time. The decay occurs in an exponential manner, and $v_L$ during the storage phase can be described mathematically by the following equation:
$$ v_L = E e^{-t/\tau}$$
A plot of $v_L$ appears in [Fig. 7] with the time axis again divided into equal increments of t. Obviously, the voltage $v_L$ will decrease to zero volts at the same rate the current presses toward its maximum value.
Fig. 7: Plotting the voltage vR versus time for the network of [fig. 1].
In five time constants, $iL = E/R$, $v_L = 0 V$, and the inductor can be replaced by its short-circuit equivalent.
$$v_R = i_RR = i_LR$$
$$v_R = ({E \over R}(1 - e^{-t/\tau}))R$$
$$v_R = E(1 - e^{-t/\tau}) \tag{4}$$
and the curve for $v_R$ will have the same shape as obtained for $i_L$.

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