Initial Values

Since the current through a coil cannot change instantaneously, the current through a coil will begin the transient phase at the initial value established by the network (note Fig. 1) before the switch was closed.
Fig. 1: Defining the three phases of a transient waveform.
It will then pass through the transient phase until it reaches the steady-state (or final) level after about five time constants. The steady-state level of the inductor current can be found by simply substituting its short-circuit equivalent (or $R_l$ for the practical equivalent) and finding the resulting current through the element. Using the transient equation developed in the previous section, an equation for the current $i_L$ can be written for the entire time interval of Fig. 1; that is, $$i_L = I_i + (I_f - I_i)(1 - e^{-t/\tau})$$ with ($I_f - I_i$) representing the total change during the transient phase.
However, by multiplying through and rearranging terms: $$\begin{split} i_L &= I_i + I_f - I_f e^{-t/\tau} - I_i + I_ie^{-t/\tau}\\ &= I_f - I_f e^{-t/\tau} + I_ie^{-t/\tau}\\ \end{split}$$ we find $$\bbox[10px,border:1px solid grey]{i_L = I_f + (I_i - I_f)e^{-t/\tau}} \tag{1}$$ If you are required to draw the waveform for the current $i_L$ from initial value to final value, start by drawing a line at the initial value and steady-state levels, and then add the transient response (sensitive to the time constant) between the two levels. The following example will clarify the procedure.
Example 1: The inductor of Fig. 2 has an initial current level of $4 mA$ in the direction shown. (Specific methods to establish the initial current will be presented in the sections and problems to follow.)
Fig. 2: Example 1.
a. Find the mathematical expression for the current through the coil once the switch is closed.
b. Find the mathematical expression for the voltage across the coil during the same transient period.
c. Sketch the waveform for each from initial value to final value.
Solution:
a. Substituting the short-circuit equivalent for the inductor will result in a final or steady-state current determined by Ohm's law: $$\begin{split} I_f &= {E \over R_1 + R_2} \\ &= {16 v\over 2.2kΩ + 6.8kΩ} \\ &= {16 v \over 9kΩ} = 1.78mA\\ \end{split}$$ The time constant is determined by $$\tau = {L \over R_T}= {100 mH \over 9kΩ} = 11.11\mu s$$ Applying Eq. (1): $$\begin{split} i_L &= I_f + (I_i - I_f)e^{-t/\tau} \\ &= 1.78mA + (4 mA - 1.78mA)e^{-t/11.11 \mu s} \\ &= 1.78mA + 2.22mA \, e^{-t/11.11 \mu s} \\ \end{split}$$ b. Since the current through the inductor is constant at $4 mA$ prior to the closing of the switch, the voltage (whose level is sensitive only to changes in current through the coil) must have an initial value of $0 V$. At the instant the switch is closed, the current through the coil cannot change instantaneously, so the current through the resistive elements will be $4 mA$. The resulting peak voltage at $t = 0 s$ can then be found using Kirchhoff's voltage law as follows: $$\begin{split} Vm &= E - V_{R1} - V_{R2} \\ &=16 V - (4 mA)(2.2 kΩ) - (4 mA)(6.8 kΩ)\\ &= -20V \end{split}$$ Note the minus sign to indicate that the polarity of the voltage $v_L$ is opposite to the defined polarity of Fig. 2. The voltage will then decay (with the same time constant as the current $i_L$) to zero because the inductor is approaching its short-circuit equivalence.
The equation for $v_L$ is therefore: $$\begin{split} v_L &= E e^{-t/\tau}\\ &= -20 e^{-t/11.11 \mu s}\\ \end{split}$$
c. See Fig. 3. The initial and final values of the current were drawn first, and then the transient response was included between these levels. For the voltage, the waveform begins and ends at zero, with the peak value having a sign sensitive to the defined polarity of $v_L$ in Fig. 2.
Fig. 3: iL and vL for the network of Fig. 2.
Let us now test the validity of the equation for $i_L$ by substituting t = 0 s to reflect the instant the switch is closed. $$e^{-t/\tau} = e^{-0/\tau} = e^0=1$$ and $$\begin{split} i_L &= 1.78 mA + 2.22 mAe^{-0/\tau}\\ &= 1.78 mA + 2.22 mA\\ &= 4 mA\\ \end{split}$$ When $t > 5 \tau$, $$e^{-t/\tau} = e^{-5\tau/\tau} = e^{-5}=0$$ and $$i_L = 1.78 mA + 2.22 mA (0) = 1.78 mA$$