Initial Values of an Inductor
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Initial Values of an Inductor
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.
$$i_L = I_i + (I_f - I_i)(1 - e^{-t/\tau})$$
$$\begin{split}
i_L &= I_i + I_f - I_f e^{-t/\tau} - I_i + I_ie^{-t/\tau}\\
&= I_f - I_f e^{-t/\tau} + I_ie^{-t/\tau}\\
\end{split}
$$
$$\bbox[10px,border:1px solid grey]{i_L = I_f + (I_i - I_f)e^{-t/\tau}} \tag{1}$$
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.)
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:
The time constant is determined by
Applying Eq. (1):
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:
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:
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].
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.
and
When $t > 5 \tau$,
and
Fig. 2: Example 1.
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:
$$ \begin{split}
I_f &= {E \over R_1 + R_2} \\
&= {16 v\over 2.2kΩ + 6.8kΩ} \\
&= {16 v \over 9kΩ} = 1.78mA\\
\end{split}
$$
$$\tau = {L \over R_T}= {100 mH \over 9kΩ} = 11.11\mu s$$
$$ \begin{split}
i_L &= I_f + (I_i - I_f)e^{-t/\tau} \\
&= 1.78mA + (4 mA - 1.78mA)e^{-t/11.11 \mu s} \\
&= 1.78mA + 2.22mA \, e^{-t/11.11 \mu s} \\
\end{split}
$$
$$ \begin{split}
Vm &= E - V_{R1} - V_{R2} \\
&=16 V - (4 mA)(2.2 kΩ) - (4 mA)(6.8 kΩ)\\
&= -20V
\end{split}
$$
The equation for $v_L$ is therefore:
$$ \begin{split}
v_L &= E e^{-t/\tau}\\
&= -20 e^{-t/11.11 \mu s}\\
\end{split}
$$
Fig. 3: iL and vL for the network of [Fig. 2].
$$ e^{-t/\tau} = e^{-0/\tau} = e^0=1$$
$$ \begin{split}
i_L &= 1.78 mA + 2.22 mAe^{-0/\tau}\\
&= 1.78 mA + 2.22 mA\\
&= 4 mA\\
\end{split}
$$
$$ e^{-t/\tau} = e^{-5\tau/\tau} = e^{-5}=0$$
$$i_L = 1.78 mA + 2.22 mA (0) = 1.78 mA$$
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