Decay of Current in RL Circuit

In the analysis of RC circuits, we found that the capacitor could hold its charge and store energy in the form of an electric field for a period of time determined by the leakage factors. In RL circuits, the energy is stored in the form of a magnetic field established by the current through the coil. Unlike the capacitor, however, an isolated inductor cannot continue to store energy since the absence of a closed path would cause the current to drop to zero, releasing the energy stored in the form of a magnetic field.
Demonstrating the effect of opening a switch in series with an inductor with a steady-state current.
Fig. 1: Demonstrating the effect of opening a switch in series with an inductor with a steady-state current.
If the series RL Transients circuit of [Fig. 1] had reached steady-state conditions and the switch were quickly opened, a spark would probably occur across the contacts due to the rapid change in current from a maximum of $E/R$ to zero amperes. The change in current $di/dt$ of the equation $v_L = L(di/dt)$ would establish a high voltage $v_L$ across the coil that in conjunction with the applied voltage $E$ appears across the points of the switch. This is the same mechanism as applied in the ignition system of a car to ignite the fuel in the cylinder. Some $25,000 V$ are generated by the rapid decrease in ignition coil current that occurs when the switch in the system is opened. (In older systems, the "points" in the distributor served as the switch.) This inductive reaction is significant when you consider that the only independent source in a car is a 12V battery.
If opening the switch to move it to another position will cause such a rapid discharge in stored energy, how can the decay phase of an RL circuit be analyzed in much the same manner as for the RC circuit? The solution is to use a network such as that appearing in [Fig. 2(a)]. When the switch is closed, the voltage across the resistor $R_2$ is $E$ volts, and the RL branch will respond in the same manner as described above, with the same waveforms and levels.
Fig. 2: Initiating the storage phase for the inductor L by closing the switch.
A thevenin network of E in parallel with $R_2$ would simply result in the source as shown in [Fig. 2(b)] since $R_2$ would be shorted out by the short-circuit replacement of the voltage source $E$ when the Thevenin resistance is determined.
Fig. 3: Network of Fig. 1 the instant the switch is opened.
After the storage phase has passed and steady-state conditions are established, the switch can be opened without the sparking effect or rapid discharge due to the resistor $R_2$, which provides a complete path for the current $i_L$. In fact, for clarity the discharge path is isolated in [Fig. 3]. The voltage $v_L$ across the inductor will reverse polarity and have a magnitude determined by
$$ \bbox[10px,border:1px solid grey]{v_L = -v_{R_1} - v_{R_2}=-(v_{R_1} + v_{R_2})} \tag{1}$$
Recall that the voltage across an inductor can change instantaneously but the current cannot. The result is that the current $i_L$ must maintain the same direction and magnitude as shown in [Fig. 3]. Therefore, the instant after the switch is opened, $i_L$ is still $I_m = E/R_1$, and
$$\begin{split} v_L &= -(v_{R1} + v_{R2}) = -(i_L R_1 + i_L R_2)\\ &= -i_L (R_1 + R_2) = {-E \over R_1} (R_1 + R_2) \\ &= - ({R_1 \over R_1} + {R_2 \over R_1})E\\ \end{split}$$
$$ \bbox[10px,border:1px solid grey]{v_L= - (1 + {R_2 \over R_1})E} \tag{2}$$
which is bigger than $E$ volts by the ratio $R_2 /R_1$. In other words, when the switch is opened, the voltage across the inductor will reverse polarity and drop instantaneously from E to $-(1 + (R_2 /R_1))E $ volts.
As an inductor releases its stored energy, the voltage across the coil will decay to zero in the following manner:
$$ \bbox[10px,border:1px solid grey]{v_L= - (1 + {R_2 \over R_1})E e^{-t/\tau ^\circ}} \tag{3}$$
$$\bbox[10px,border:1px solid grey]{\tau ^\circ = {L \over R_T} = {L \over R_1 + R_2}}$$
The current will decay from a maximum of $I_m = E/R_1$ to zero.
Therefore $I_i = E/R_1$ and $I_f = 0 A$ so that
$$i_L = I_f + (I_i - I_f)e^{-t/\tau ^ \circ}$$ $$ = 0 + ({E \over R_1} - 0)e^{-t/\tau ^ \circ}$$
$$ \bbox[10px,border:1px solid grey]{i_L={E \over R_1} e^{-t/\tau ^ \circ}} \tag{4}$$
Example 1: The resistor $R_2$ was added to the network of [Fig. 4], as shown in [Fig. 1].
a. Find the mathematical expressions for $i_L$, $v_L$, $v_{R_1}$, and $v_{R_2}$ for five time constants of the storage phase.
b. Find the mathematical expressions for $i_L$, $v_L$, $v_{R_1}$ , and $v_{R_2}$ if the switch is opened after five time constants of the storage phase.
c. Sketch the waveforms for each voltage and current for both phases covered by this example if five time constants pass between phases.
Fig. 4: For Example 1.
$$ \tau = {L \over R} = { 4H \over 2kΩ} = 2ms$$
$$ v_L = E e^{-t/\tau} = 50e^{-t/2\times 10^{-3}}$$
$$ \begin{split} i_L &= I_m(1-e^{-t/\tau}) = {E \over R_1}(1-e^{-t/\tau})\\ &= {50 \over 2kΩ}(1-e^{-t/2\times 10^{-3}})\\ i_L &= 25 \times 10^{-3}(1-e^{-t/2\times 10^{-3}})\\ \end{split} $$
$$\bbox[10px,border:1px solid grey]{i_L = 25 \times 10^{-3}(1-e^{-t/2\times 10^{-3}})}$$
$$ \bbox[10px,border:1px solid grey]{v_{R1} = E (1 - e^{-t/\tau}) = 50(1-e^{-t/2\times 10^{-3}})} $$
$$ \bbox[10px,border:1px solid grey]{v_{R2} = E = 50V}$$
b. $$ \bbox[10px,border:1px solid grey]{\tau ^{\circ} = {L \over R_1 + R_2} = {4H \over 2kΩ + 3kΩ} = 0.8 ms}$$
$$ v_L= - (1 + {R_2 \over R_1})E e^{-t/\tau ^\circ}$$ $$ v_L= - (1 + {5kΩ \over 2kΩ})50 e^{-t/0.8ms}$$ $$ \bbox[10px,border:1px solid grey]{v_L= - 125 e^{-t/0.8ms}}$$
$$i_L = I_m e^{-t/\tau} = {E \over R_1}e^{-t/\tau ^\circ}$$ $$i_L = {50 \over 2kΩ}e^{-t/0.8ms}$$ $$i_L = 25 \times 10^{-3} e^{-t/0.8ms}$$
$$\bbox[10px,border:1px solid grey]{v_{R1} = Ee^{-t/\tau ^\circ} = 50e^{-t/0.8ms}}$$
$$v_{R2} = -i_{R2} R_2 = -i_LR_2$$ $$v_{R2} = -{E \over R_1} R_2e^{-t/\tau ^\circ}=-{R_2 \over R_1} E e^{-t/\tau ^\circ}$$ $$v_{R2} = -{3kΩ \over 2kΩ} 50 e^{-t/0.8ms}$$
$$\bbox[10px,border:1px solid grey]{v_{R2} = -75 e^{-t/0.8ms}}$$
Fig. 5: The various voltages and the current for the network of Fig. 4

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