Chapter 9, Problem 62E

Determine expressions for v_C(t) and i_L(t) in Fig. 9.59 for the time windows (a) 0 < t <2 μs and (b) t > 2 μs.

FIGURE 9.59

To determine

Determine expressions for $v_C\left(t\right)$ and $i_L\left(t\right)$ in Fig. 9.59 for the time windows $0<t<2\mu \text{s}$.

Answer to Problem 62E

The expressions for $v_C\left(t\right)$ is $1+e^{-\left({10}^6\right)t}\left(-{10}^{6}t-1\right)\text{ V}$ and the expressions for $i_L\left(t\right)$ is $\left({10}^6\right)t{e}^{-{10}^{6}t}\text{ A}$ for the time windows $0<t<2\mu \text{s}$.

Explanation of Solution

Given Data:

The range of the time is $0<t<2\mu \text{s}$.

Formula used:

The expression for the exponential damping coefficient or the neper frequency for series $RLC$ circuit is as follows:

$\alpha =\frac{R}{2L}$ (1)

Here,

$\alpha$ is the exponential damping coefficient or the neper frequency of series $RLC$ circuit,

$R$ is the resistance of a series $RLC$ circuit,

$L$ is the inductance of a series $RLC$ circuit.

The expression for the resonant frequency for series $RLC$ circuit is as follows:

${\omega }_0=\frac{1}{\sqrt{LC}}$ (2)

Here,

${\omega }_0$ is the resonant frequency of series $RLC$ circuit and

$C$ is the capacitance of a series $RLC$ circuit.

The expression for complete natural response for source free series $RLC$ circuit is:

$v_C\left(t\right)=v_f+v_n$ (3)

Here,

$v_C\left(t\right)$ is voltage across capacitor for $t>0$,

$v_f$ is forced response and

$v_n$ is the natural response.

The expression for the critically damped natural response of the series $RLC$ circuit is as follows:

$v_n=e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}$ (4)

Here,

$v_n$ is the natural response of the parallel $RLC$ circuit,

$A_1$ and $A_2$ are arbitrary constants and

$t$ is the time.

Calculation:

The unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument.

$u\left(t-t_0\right)=\{\begin{array}{l}0t<t_0\\ 1t>t_0\end{array}$

Here,

$t_0$ is the time at which the argument of the function becomes zero.

So, at $t=0^{-}$ the voltage $V_{i\text{n}}\left(t\right)=\left[u\left(t\right)+2u\left(t-t_1\right)\right]\text{ V}$ independent current source is equal to $0\text{ V}$.

Since the series $RLC$ circuit is source free at $t<0$ so, the voltage across the $1\mu \text{F}$ capacitor is $0\text{ V}$ and the current flowing through $\text{1 }\mu \text{H}$ is $0\text{ A}$.

The capacitor does not allow sudden change in the voltage and the inductor does not allow sudden change in the current.

So,

$v_C\left(0^{+}\right)=v_C\left(0^{-}\right)$

And,

$i_L\left(0^{+}\right)=i_L\left(0^{-}\right)$

Therefore, the voltage across the capacitor at $t=0$ is $0\text{ V}$ and the current flowing through inductor at $t=0$ is $0\text{ A}$.

At $t>0$ the voltage $V_{i\text{n}}\left(t\right)=\left[u\left(t\right)+2u\left(t-t_1\right)\right]\text{ V}$ independent current source is equal to $1\text{ V}$.

The redrawn circuit diagram is given in Figure 1 at $t=0^{+}$.

Refer to the redrawn Figure 1:

Substitute $2\Omega$ for $R$ and $\text{1 }\mu \text{H}$ for $L$ in the equation (1).

$\begin{array}{c}\alpha =\frac{2\Omega }{\left(2\right)\left(\text{1 }\mu \text{H}\right)}\\ =\frac{\text{2 }\Omega }{\left(2\right)\left(1\times {10}^{-6}\text{ H}\right)}\left\{\because 1\mu \text{H}={10}^{-6}\text{ H}\right\}\\ ={\text{10}}^6{\text{ s}}^{-1}\end{array}$

Substitute $\text{1 }\mu \text{H}$ for $L$ and $1\mu \text{F}$ for $C$ in equation (2).

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(\text{1 }\mu \text{H}\right)\times \left(1\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(\text{1}\times {10}^{-6}\text{ H}\right)\times \left(1\times {10}^{-6}\text{ F}\right)}}\left\{\begin{array}{l}\because 1\mu \text{H}={10}^{-6}\text{ H}\\ \because 1\mu \text{F}={10}^{-6}\text{ F}\end{array}\right\}\\ =\frac{1}{\sqrt{1\times {10}^{-12}}}{\text{ s}}^{-1}\\ ={10}^6\text{ rad/s}\end{array}$

Here, the resonant frequency is equal to the exponential damping coefficient.

Therefore, the response of the circuit is critically damped.

At $t>0$ the voltage $V_{i\text{n}}\left(t\right)=\left[u\left(t\right)+2u\left(t-t_1\right)\right]\text{ V}$ independent current source is equal to $1\text{ V}$.

Therefore, the value of forced response $v_f$ is $1\text{ V}$.

Substitute $1\text{ V}$ for $v_f$ and $e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}$ for $v_n$ in equation (3).

$v_C\left(t\right)=1\text{ V}+e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}$

$v_C\left(t\right)\text{=}1+e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}$ (5)

Substitute $0$ for $t$ in equation (3).

$\begin{array}{c}{v}_C\left(0\right)=1+e^{-\alpha \left(0\right)}\left(A_1\left(0\right)+A_2\right)\text{ V}\\ \text{=}1+\left(1\right)\left(A_2\right)\text{ V}\end{array}$

$v_C\left(0\right)=1+A_2\text{ V}$ (6)

The voltage across the capacitor at $t=0$ is $0\text{ V}$.

Substitute $0\text{ V}$ for $v_C\left(0\right)$ in equation (6).

$0\text{ V}=1+A_2\text{ V}$

Rearrange for $A_2$.

$A_2=-1$

The expression for the current flowing through $1\mu \text{F}$ capacitor in the series $RLC$ circuit is as follows:

$i_C=i_L$

So, the current flowing through $1\mu \text{F}$ capacitor is $0\text{ A}$.

The expression for the current flowing through the $1\mu \text{F}$ capacitor is as follows:

$i_C\left(t\right)=\frac{Cdv\left(t\right)}{dt}$ (7)

$v\left(t\right)$ is the voltage across the $1\mu \text{F}$ capacitor for $t=0^{+}$.

Substitute $1+e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}$ for $v\left(t\right)$ in equation (7).

$i_C\left(t\right)=\frac{Cd\left(1+e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}\right)}{dt}$

Rearrange for $\frac{i_C\left(t\right)}{C}$.

$\frac{i_C\left(t\right)}{C}=\frac{d\left(1+e^{-\alpha t}\left(A_{1}t+A_2\right)\text{ V}\right)}{dt}$

$\frac{i_C\left(t\right)}{C}=\left(0+A_1{e}^{-\alpha t}-A_1\alpha t{e}^{-\alpha t}-\alpha e^{-\alpha t}{A}_2\right)\text{ V}$ (8)

Substitute $0$ for $t$ in equation (8).

$\begin{array}{c}\frac{i_C\left(0\right)}{C}=\left(A_1{e}^{-\alpha \left(0\right)}-A_1\alpha \left(0\right)e^{-\alpha \left(0\right)}-\alpha e^{-\alpha \left(0\right)}{A}_2\right)\text{ V}\\ =A_1\left(0\right)-0-\alpha \left(1\right)A_2\text{ V}\end{array}$

$\frac{i_C\left(0\right)}{C}=A_1-\alpha A_2\text{ V}$ (9)

The current flowing through $1\mu \text{F}$ capacitor is $0\text{ A}$.

Substitute $-1$ for $A_2$ and ${10}^6{\text{ s}}^{-1}$ for $\alpha$ in equation (9).

$\frac{i_C\left(0\right)}{C}=A_1-\left({10}^6{\text{ s}}^{-1}\right)\left(-1\right)\text{V}$ (10)

Substitute $0\text{ A}$ for $i_C\left(0\right)$ and $1\mu \text{F}$ for $C$ in equation (10).

$\begin{array}{l}\frac{0\text{ A}}{1\mu \text{F}}=A_1-\left({10}^6{\text{ s}}^{-1}\right)\left(-1\right)\\ \text{0}=A_1+{10}^6{\text{ s}}^{-1}\end{array}$

Rearrange for $A_1$.

$A_1=-{10}^6{\text{ s}}^{-1}$

Substitute $-{10}^6{\text{ s}}^{-1}$ for $A_1$, $-1$ for $A_2$ and ${10}^6{\text{ s}}^{-1}$ for $\alpha$ in equation (10).

$v_C\left(t\right)=1+e^{-\left({10}^6{\text{ s}}^{-1}\right)t}\left(-\left({10}^6{\text{ s}}^{-1}\right)t-1\right)\text{ V}$ (11)

The expression for the current flowing through the $\text{1 }\mu \text{H}$ inductor in the circuit is as follows:

$i_L\left(t\right)=C\frac{d{v}_C\left(t\right)}{dt}$ (12)

Substitute $1+e^{-\left({10}^6{\text{ s}}^{-1}\right)t}\left(-\left({10}^6{\text{ s}}^{-1}\right)t-1\right)\text{ V}$ for $v_C\left(t\right)$ in equation (12).

$\begin{array}{c}{i}_L\left(t\right)=C\frac{d\left(1+e^{-\left({10}^6{\text{ s}}^{-1}\right)t}\left(-\left({10}^6{\text{ s}}^{-1}\right)t-1\right)\text{ V}\right)}{dt}\\ =C\frac{d\left(1-\left({10}^6{\text{ s}}^{-1}\right)t{e}^{-{10}^{6}t}-e^{-{10}^{6}t}\right)}{dt}\text{ V/s}\\ =C\left(0-\left({10}^6{\text{ s}}^{-1}\right)e^{-{10}^{6}t}+\left({10}^6{\text{ s}}^{-1}\right)\left({10}^6\right)t{e}^{-{10}^{6}t}+\left({10}^6\right)e^{-{10}^{6}t}\right)\text{ V/s}\end{array}$

$i_L\left(t\right)=C\left(\left({10}^6{\text{ s}}^{-1}\right)\left({10}^6\right)t{e}^{-{10}^{6}t}\right)\text{ V/s}$ (13)

Substitute $1\mu \text{F}$ for $C$ in equation (13).

$\begin{array}{c}{i}_L\left(t\right)=\left(1\mu \text{F}\right)\left(\left({10}^6{\text{ s}}^{-1}\right)\left({10}^6\right)t{e}^{-{10}^{6}t}\right)\text{ V/s}\\ =\left(1\times {10}^{-6}\text{ F}\right)\left(\left({10}^6{\text{ s}}^{-1}\right)\left({10}^6\right)t{e}^{-{10}^{6}t}\right)\text{ V/s}\end{array}$

$i_L\left(t\right)=\left({10}^6{\text{s}}^{-1}\right)t{e}^{-{10}^{6}t}\text{ A}$ (14)

Conclusion:

Thus, the expressions for $v_C\left(t\right)$ is $1+e^{-\left({10}^6{\text{ s}}^{-1}\right)t}\left(-\left({10}^6{\text{ s}}^{-1}\right)t-1\right)\text{ V}$ and the expressions for $i_L\left(t\right)$ is $\left({10}^6{\text{ s}}^{-1}\right)t{e}^{-{10}^{6}t}\text{ A}$ for the time windows $0<t<2\mu \text{s}$.

To determine

Determine expressions for $v_C\left(t\right)$ and $i_L\left(t\right)$ in Fig. 9.59 for the time windows $t>2\mu \text{s}$.

Answer to Problem 62E

The expressions for $v_C\left(t\right)$ is:

$\left[3\text{V}+e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left(\begin{array}{l}-2.406\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ -0.8645\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right)\text{ V}\right]$ andthe expressions for $i_L\left(t\right)$ is:

$e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left\{\begin{array}{l}0.2707\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ +2.542\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right\}\text{ A}$ for the time window $t<2\mu \text{s}$.

Explanation of Solution

Formula used:

The expression for the damped natural frequency in series $RLC$ circuit is:

${\omega }_d=\sqrt{{\omega }_0{}^2-{\alpha }^2}\text{rad/s}$ (15)

Here,

${\omega }_d$ is the damped natural frequency.

The expression for natural response for series $RLC$ circuit is:

$v_n=e^{-\alpha t}\left(B_1\text{cos}{\omega }_{d}t+B_2\text{sin}{\omega }_{d}t\right)\text{ V}$ (16)

Here,

$B_1$ and $B_2$ is arbitrary constant for underdamped response,

$\alpha$ is the exponentially decaying constant and

${\omega }_d$ is the damped resonant frequency.

Calculation:

The redrawn circuit diagram is given in Figure 2 at $t=2\mu \text{s}$.

Refer to the redrawn Figure 2:

The initial condition of the voltage across the $\text{1 }\mu \text{F}$ capacitor and the current through the $\text{1 }\mu \text{H}$ inductor is calculated at $t=2\mu \text{s}$.

Substitute $2\mu \text{s}$ for $t$ in equation (11).

$\begin{array}{c}{v}_C\left(2\mu \text{s}\right)=1+e^{-\left({10}^6{\text{ s}}^{-1}\right)\left(2\mu s\right)}\left(-\left({10}^6{\text{ s}}^{-1}\right)\left(2\mu s\right)-1\right)\text{ V}\\ =1+e^{-\left({10}^6{\text{ s}}^{-1}\right)\left(2\times {10}^{-6}\text{ s}\right)}\left(-\left({10}^6{\text{ s}}^{-1}\right)\left(2\times {10}^{-6}{\text{ s}}^{-1}\right)-1\right)\text{ V }\left\{\because 1\mu s={10}^{-6}s\right\}\\ =1+\left(0.135335\right)\left(-2-1\right)\text{ V}\\ =1+\left(-0.4060\right)\text{ V}\end{array}$

$v_C\left(2\mu \text{s}\right)=0.594\text{ V}$

So, the initial condition of the voltage across the $\text{1 }\mu \text{F}$ capacitor at $t=2\mu \text{s}$ is $0.594\text{ V}$.

Substitute $2\mu \text{s}$ for $t$ in equation (14).

$\begin{array}{c}{i}_L\left(2\mu \text{s}\right)=\left({10}^6{\text{ s}}^{-1}\right)\left(2\mu \text{s}\right)e^{-{10}^6\left(2\mu \text{s}\right)}\text{ A}\\ =\left({10}^6{\text{ s}}^{-1}\right)\left(2\times {10}^{-6}\text{ s}\right)e^{-{10}^6\left(2\times {10}^{-6}\text{ s}\right)}\text{ A }\left\{\because 1\mu s={10}^{-6}s\right\}\\ =\left(2\right)\left(0.135335\right)\text{ A }\\ =\text{0}\text{.2707 A}\end{array}$

So, the initial condition ofthe current through the $\text{1 }\mu \text{H}$ inductor at $t=2\mu \text{s}$ is $\text{0}\text{.2707 A}$.

At $t>2\mu \text{s}$ the voltage $V_{i\text{n}}\left(t\right)=\left[u\left(t\right)+2u\left(t-t_1\right)\right]\text{ V}$ independent current source is equal to $\text{3 V}$.

The expression for the equivalent resistor when resistors are connected in parallel is as follows:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\mathrm{......}+\frac{1}{R_N}$ (17)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistance of the resistors connected in parallel.

$\text{2 }\Omega$ and $\text{1}\text{.5454 }\Omega$ resistor are connected in parallel.

So, form equation (15),

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{2 }\Omega }+\frac{1}{\text{1}\text{.5454 }\Omega }\\ =1.14708{\Omega }^{-1}\end{array}$

Rearrange for $R_{eq}$.

$\begin{array}{c}{R}_{eq}=\frac{1}{1.14708}\Omega \\ =\text{0}\text{.8718 }\Omega \end{array}$

The redrawn circuit diagram is given in Figure 3.

Refer to the redrawn Figure 3:

Substitute $\text{0}\text{.8718 }\Omega$ for $R$ and $\text{1 }\mu \text{H}$ for $L$ in equation (1).

$\begin{array}{c}\alpha =\frac{\text{0}\text{.8718 }\Omega }{\left(2\right)\left(\text{1 }\mu \text{H}\right)}\\ =\frac{\text{0}\text{.8718 }\Omega }{\left(2\right)\left(1\times {10}^{-6}\text{ H}\right)}\left\{\because 1\mu \text{H}={10}^{-6}\text{ H}\right\}\\ =4.3589\times {\text{10}}^5{\text{ s}}^{-1}\end{array}$

Substitute $\text{1 }\mu \text{H}$ for $L$ and $1\mu \text{F}$ for $C$ in equation (2).

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(\text{1 }\mu \text{H}\right)\times \left(1\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(\text{1}\times {10}^{-6}\text{ H}\right)\times \left(1\times {10}^{-6}\text{ F}\right)}}\left\{\begin{array}{l}\because 1\mu \text{H}={10}^{-6}\text{ H}\\ \because 1\mu \text{F}={10}^{-6}\text{ F}\end{array}\right\}\\ =\frac{1}{\sqrt{1\times {10}^{-12}}}{\text{ s}}^{-1}\\ ={10}^6\text{ rad/s}\end{array}$

Here, the resonant frequency is greater than the exponential damping coefficient.

Therefore, the response of the circuit is under-damped damped.

Substitute $4.3589\times {\text{10}}^5{\text{ s}}^{-1}$ for $\alpha$ and ${10}^6\text{ rad/s}$ for ${\omega }_0$ in equation (4).

$\begin{array}{c}{\omega }_d=\sqrt{{\left({10}^6\text{ rad/s}\right)}^2-{\left(4.3589\times {\text{10}}^5{\text{ s}}^{-1}\right)}^2}\\ =\sqrt{8.1\times {10}^{11}}\text{rad}/\text{s}\\ =9\times {10}^5\text{rad}/\text{s}\end{array}$

At $t>2\mu \text{s}$ the voltage $V_{i\text{n}}\left(t\right)=\left[u\left(t\right)+2u\left(t-t_1\right)\right]\text{ V}$ independent current source is equal to $3\text{ V}$.

Therefore, the value of forced response $v_f$ is $3\text{ V}$.

Substitute $3\text{V}$ for $v_f$ and $e^{-\alpha t}\left(B_1\text{cos}{\omega }_{d}t+B_2\text{sin}{\omega }_{d}t\right)$ for $v_n$ in equation (3).

$v_C\left(t\right)=3\text{V}+e^{-\alpha t}\left(B_1\text{cos}{\omega }_{d}t+B_2\text{sin}{\omega }_{d}t\right)\text{ V}$ (18)

Apply time shift in equation (18).

$v_C\left(t\right)=3\text{V}+e^{-\alpha \left(t-2\mu \text{s}\right)}\left(B_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+B_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}$ (19)

Substitute $2\mu \text{s}$ for $t$ in equation (18).

$\begin{array}{c}{v}_C\left(2\mu \text{s}\right)=\left[3\text{V}+e^{-\alpha \left(2\mu \text{s}-2\mu \text{s}\right)}\left(\begin{array}{l}{B}_1\text{cos}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\\ +B_2\text{sin}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\end{array}\right)\text{ V}\right]\\ =3\text{V}+\left(1\right)\left(B_1\text{cos}{\omega }_d\left(0\right)+B_2\text{sin}{\omega }_d\left(0\right)\right)\text{ V}\\ =3\text{V}+\left(1\right)\left(B_1\left(1\right)+\left(0\right)\right)\text{ V}\end{array}$

$v_C\left(2\mu \text{s}\right)\text{=}3\text{V}+B_1\text{ V}$ (20)

The initial condition of the voltage across the $\text{1 }\mu \text{F}$ capacitor at $t=2\mu \text{s}$ is $0.594\text{ V}$.

Substitute $0.594\text{ V}$ for $v_C\left(2\mu \text{s}\right)$ in equation (20).

$0.594\text{ V}=3\text{V}+B_1\text{ V}$

Rearrange for $B_1$.

$\begin{array}{c}{B}_1=-3+0.594\\ =-2.406\end{array}$

The expression for the current flowing through $1\mu \text{F}$ capacitor in the series $RLC$ circuit is as follows:

$i_C=i_L$

So, the current flowing through $1\mu \text{F}$ capacitor at $t=2\mu \text{s}$ is $\text{0}\text{.2707 A}$.

The expression for the current flowing through the $1\mu \text{F}$ capacitor at $t=2\mu \text{s}$ is as follows:

$i_C\left(2\mu \text{s}\right)=C{\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}$

Here,

$v\left(t\right)$ is the voltage across the $1\mu \text{F}$ capacitor.

Rearrange for $\frac{i_C\left(2\mu \text{s}\right)}{C}$

${\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}=\frac{i_C\left(2\mu \text{s}\right)}{C}$ (21)

Substitute $\text{0}\text{.2707 A}$ for $i_C\left(2\mu \text{s}\right)$ and $1\mu \text{F}$ for $C$ in equation (21).

$\begin{array}{c}{\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}=\frac{\text{0}\text{.2707 A}}{1\mu \text{F}}\\ =\frac{\text{0}\text{.2707 A}}{1\times {10}^{-6}\text{ F}}\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =270700\text{ V/s}\end{array}$

Substitute $3\text{V}+e^{-\alpha \left(t-2\mu \text{s}\right)}\left(B_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+B_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}$ for $v\left(t\right)$ in equation (20).

$\begin{array}{c}\frac{d{v}_C\left(t\right)}{dt}=\frac{d}{dt}\left(3\text{V}+e^{-\alpha \left(t-2\mu \text{s}\right)}\left(B_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+B_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}\right)\\ =\frac{d}{dt}\left(3+e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}\end{array}$

$\frac{d{v}_C\left(t\right)}{dt}=\left\{\begin{array}{l}0+\left(\begin{array}{l}-\alpha e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)\\ -{\omega }_d{e}^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\end{array}\right)\\ +\left(\begin{array}{l}-\alpha e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\\ +{\omega }_d{e}^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)\end{array}\right)\end{array}\right\}\text{ V/s}$ (21)

Substitute $2\mu \text{s}$ for $t$ in equation (21).

$\begin{array}{c}{\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}=\left\{\begin{array}{l}\left(\begin{array}{l}-\alpha e^{-\alpha \left(2\mu \text{s}-2\mu \text{s}\right)}{B}_1\text{cos}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\\ -{\omega }_d{e}^{-\alpha \left(2\mu \text{s}-2\mu \text{s}\right)}{B}_1\text{sin}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\end{array}\right)\\ +\left(\begin{array}{l}-\alpha e^{-\alpha \left(2\mu \text{s}-2\mu \text{s}\right)}{B}_2\text{sin}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\\ +{\omega }_d{e}^{-\alpha \left(2\mu \text{s}-2\mu \text{s}\right)}{B}_2\text{cos}{\omega }_d\left(2\mu \text{s}-2\mu \text{s}\right)\end{array}\right)\end{array}\right\}\text{ V/s}\\ =\left\{\begin{array}{l}\left(-\alpha e^{-\alpha \left(0\right)}{B}_1\text{cos}{\omega }_d\left(0\right)-{\omega }_d{e}^{-\alpha \left(0\right)}{B}_1\text{sin}{\omega }_d\left(0\right)\right)\\ +\left(-\alpha e^{-\alpha \left(0\right)}{B}_2\text{sin}{\omega }_d\left(0\right)+{\omega }_d{e}^{-\alpha \left(0\right)}{B}_2\text{cos}{\omega }_d\left(0\right)\right)\end{array}\right\}\text{ V/s}\\ =\left\{\begin{array}{l}\left(-\alpha B_1\left(1\right)-{\omega }_d{e}^{-\alpha \left(0\right)}{B}_1\left(0\right)\right)\\ +\left(-\alpha e^{-\alpha \left(0\right)}{B}_2\left(0\right)+{\omega }_d\left(1\right)B_2\left(1\right)\right)\end{array}\right\}\text{ V/s}\end{array}$

${\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}=\left(-\alpha B_1+{\omega }_d{B}_2\right)\text{ V/s}$ (22)

Substitute $270700\text{ V/s}$ for ${\frac{d{v}_C\left(t\right)}{dt}|}_{t=2\mu \text{s}}$, $-2.406$ for $B_1$, $4.3589\times {10}^5{\text{ s}}^{-1}$ for $\alpha$ and $9\times {10}^5\text{ rad/s}$ for ${\omega }_d$ in equation (22).

$\begin{array}{l}270700=-\left(4.3589\times {10}^5\right)\left(-2.406\right)+\left(9\times {10}^5\right)B_2\\ 270700=1048751.34+\left(9\times {10}^5\right)B_2\end{array}$

Rearrange for $B_2$.

$\begin{array}{l}270700-1048751.34=\left(9\times {10}^5\right)B_2\\ -778051.34=\left(9\times {10}^5\right)B_2\\ B_2=\frac{-778051.34}{9\times {10}^5}\\ =-0.8645\end{array}$

Substitute $-2.406$ for $B_1$, $-0.8645$ for $B_2$, $4.3589\times {10}^5{\text{ s}}^{-1}$ for $\alpha$ and $9\times {10}^5\text{ rad/s}$ for ${\omega }_d$ in equation (19).

$v_C\left(t\right)=\left[3\text{V}+e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left(\begin{array}{l}-2.406\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ -0.8645\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right)\text{ V}\right]$ (23)

Substitute $3\text{V}+e^{-\alpha \left(t-2\mu \text{s}\right)}\left(B_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+B_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}$ for $v\left(t\right)$ in equation (13).

$\begin{array}{c}{i}_L\left(t\right)=C\frac{d}{dt}\left(3\text{V}+e^{-\alpha \left(t-2\mu \text{s}\right)}\left(B_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+B_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}\right)\\ =C\frac{d}{dt}\left(3+e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)+e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\right)\text{ V}\end{array}$

$i_L\left(t\right)=C\left\{\begin{array}{l}\left(\begin{array}{l}-\alpha e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)\\ -{\omega }_d{e}^{-\alpha \left(t-2\mu \text{s}\right)}{B}_1\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\end{array}\right)\\ +\left(\begin{array}{l}-\alpha e^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{sin}{\omega }_d\left(t-2\mu \text{s}\right)\\ +{\omega }_d{e}^{-\alpha \left(t-2\mu \text{s}\right)}{B}_2\text{cos}{\omega }_d\left(t-2\mu \text{s}\right)\end{array}\right)\end{array}\right\}\text{ V/s}$ (24)

Substitute $1\mu \text{F}$ for $C$, $-2.406$ for $B_1$, $-0.8645$ for $B_2$, $4.3589\times {10}^5{\text{ s}}^{-1}$ for $\alpha$ and $9\times {10}^5\text{ rad/s}$ for ${\omega }_d$ in equation (24).

$\begin{array}{c}{i}_L\left(t\right)=\left(1\mu \text{F}\right)\left\{\begin{array}{l}\left(\begin{array}{l}\left[\begin{array}{l}-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\\ \left(-2.406\right)\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\\ \left[\begin{array}{l}-\left(9\times {10}^5\text{ rad/s}\right)e^{-V\left(t-2\mu \text{s}\right)}\\ \left(-2.406\right)\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\end{array}\right)\\ +\left(\begin{array}{l}\left[\begin{array}{l}-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\\ \left(-0.8645\right)\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\\ \left[\begin{array}{l}+\left(9\times {10}^5\text{ rad/s}\right)e^{-\alpha \left(t-2\mu \text{s}\right)}\\ \left(-0.8645\right)\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\end{array}\right)\end{array}\right\}\text{ V/s}\\ \text{=}\left(1\mu \text{F}\right)e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left\{\begin{array}{l}\text{1048751}\text{.34cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ +2165400\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ +376826.905\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ -778050\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right\}\text{ V/s}\\ =\left(1\times {10}^{-6}\text{ F}\right)e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left\{\begin{array}{l}\left[\begin{array}{l}270701.34\\ \text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\\ +\left[\begin{array}{l}2542226.905\\ \text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right]\end{array}\right\}\text{ V/s }\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\end{array}$$i_L\left(t\right)=e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left\{\begin{array}{l}0.2707\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ +2.542\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right\}\text{ A}$

Conclusion:

Thus, the expressions for $v_C\left(t\right)$ is:

$\left[3\text{V}+e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left(\begin{array}{l}-2.406\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ -0.8645\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right)\text{ V}\right]$ and the expressions for $i_L\left(t\right)$ is:

$e^{-\left(4.3589\times {10}^5{\text{ s}}^{-1}\right)\left(t-2\mu \text{s}\right)}\left\{\begin{array}{l}0.2707\text{cos}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\\ +2.542\text{sin}\left(9\times {10}^5\text{ rad/s}\left(t-2\mu \text{s}\right)\right)\end{array}\right\}\text{ A}$ for the time window $t<2\mu \text{s}$.