Chapter 9, Problem 16E

With regard to the circuit presented in Fig. 9.42, (a) obtain an expression for v(t) which is valid for all t > 0; (b) calculate the maximum inductor current and identify the time at which it occurs; (c) determine the settling time.

FIGURE 9.42

To determine

Find the equation for voltage $v\left(t\right)$ valid for all $t>0$.

Answer to Problem 16E

The equation of voltage is $-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\text{V}$.

Explanation of Solution

Formula used:

The expression for the exponential damping coefficient is as follows:

$\alpha =\frac{1}{2RC}$ (1)

Here,

$\alpha$ is the exponential damping coefficient,

$R$ is the resistance of a parallel $RLC$ circuit and

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

The expression for the resonating frequency is as follows:

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

Here,

${\omega }_0$ is the resonating frequency and

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

The expression for the two solutions of the characteristic equation of a parallel $RLC$ circuit is as follows:

$s_1=-\alpha +\sqrt{{\alpha }^2-{\omega }_0{}^2}$ (3)

$s_2=-\alpha -\sqrt{{\alpha }^2-{\omega }_0{}^2}$ (4)

Here,

$s_1$ and $s_2$ is the solutions of the characteristic equation of a parallel $RLC$ circuit.

Here,

$i_L\left(t\right)$ is the natural response of the parallel $RLC$ circuit,

$A_1$ and $A_2$ are arbitrary constant and

$t$ is the time.

Calculation:

The redrawn circuit is shown in Figure 1 as follows:

Refer to the Figure 1,

At $t=0^{-}\text{s}$ when the switch is opened the capacitor is open circuited and inductor gets short circuited because steady state is reached. The expression for current $i_L\left(0^{-}\right)$ is as follows:

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

The current across inductor is zero because inductor is not energized.

The expression for voltage $v_C\left(0^{-}\right)$ is as follows:

$v_C\left(0^{-}\right)=i_S\left(0^{-}\right)R$ (5)

Here,

$i_S\left(0^{-}\right)$ is current for $t=0^{-}\text{s}$ across branch $CD$ and

$R$ is the resistance across branch $CD$.

The redrawn circuit at $t=0^{-}\text{s}$ is shown in Figure 2 as follows:

Refer to the Figure 2:

The expression for the voltage across capacitor at $t=0^{-}\text{s}$ is:

$i_C\left(0^{-}\right)=C\frac{d{v}_C\left(0^{-}\right)}{dt}$

Substitute $5\mu \text{A}$ for $i_s\left(0^{-}\right)$ and $0.2\text{Ω}$ for $R$ in equation (5).

$\begin{array}{c}{v}_C\left(0^{-}\right)=\left(5\mu \text{A}\right)\left(0.2\text{Ω}\right)\\ =1\mu \text{V}\end{array}$

The expression for the natural response of voltage for capacitor in the parallel $RLC$ circuit is as follows:

$v_C\left(t\right)=A_1{e}^{s_{1}t}+A_2{e}^{s_{2}t}$ (6)

Substitute $0\text{s}$ for $t$ and $1\mu \text{V}$ for $v_C\left(0^{-}\right)$ in equation (6).

$\begin{array}{l}1\mu \text{V}=A_1{e}^{S_1\times 0^{-}\text{s}}+A_2{e}^{S_2\times 0^{-}\text{s}}\\ 1\mu \text{V}=A_1+A_2\end{array}$

Rearrange for $A_1$.

$A_1=1\mu \text{V}-A_2$ (7)

At $t=0^{+}$ the capacitor does not allow sudden change in the voltage.

Therefore,

$v_C(0^{+})=v_C(0^{-})$

So,

$v_C\left(0^{+}\right)=1\mu \text{V}$

The redrawn circuit is shown in Figure 3 as follows:

Refer to the Figure 3:

The current across resistor $R$ is same as $i_S\left(0^{-}\right)$, therefore, from equation (5):

$i_R\left(0^{+}\right)=5\mu \text{A}$

The expression for KCL at node $E$ is as follows:

$i_L\left(0^{+}\right)+i_C\left(0^{+}\right)+i_R\left(0^{+}\right)=0$ (8)

Here,

$i_L\left(0^{+}\right)$ is current flowing through inductor for $t>0$,

$i_C\left(0^{+}\right)$ is current flowing through capacitor for $t>0$ and

$i_R\left(0^{+}\right)$ is current flowing through resistor $R$.

Substitute $5\mu \text{A}$ for $i_R\left(0^{+}\right)$ and $0\text{A}$ for $i_L\left(0^{+}\right)$ in equation (8).

$\begin{array}{l}0+i_C\left(0^{+}\right)+5\mu \text{A}=0\\ i_C\left(0^{+}\right)=-5\mu \text{A}\end{array}$

The expression for current flowing through capacitor for $t>0$ is as follows:

$i_C\left(0^{+}\right)=C\frac{d{v}_C\left(0^{+}\right)}{dt}$ (9)

Here,

$i_C\left(0^{+}\right)$ is capacitor current for $t>0$,

$C$ is capacitor and

$v_C\left(0^{+}\right)$ is voltage across capacitor for $t>0$.

At $t=0\text{s}$, when switch is connected, then resistor $R$ and current $i_s$ gets open circuited because it is defined only for $t<0\text{s}$.

The circuit diagram is redrawn as shown in Figure 4 for $t>0$.

Refer to the redrawn Figure 4:

Substitute $0.2\text{ Ω}$ for $R$ and $4\text{mF}$ for $C$ in equation (1).

$\begin{array}{c}\alpha =\frac{1}{2\left(0.2\text{ Ω}\right)\left(4\text{mF}\right)}\\ =\frac{1}{2\left(0.2\text{ Ω}\right)\left(4\times {10}^{-3}\text{F}\right)}\left\{\because 1\text{mF}={10}^{-3}\text{F}\right\}\\ =\frac{1}{1.6\times {10}^{-3}\text{s}}\\ =625{\text{s}}^{-1}\end{array}$

Substitute $\frac{2}{13}\text{H}$ for $L$ and $250\text{mF}$ for $C$ in equation (2).

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(1\text{mH}\right)\left(4\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(1\times {\text{10}}^{-3}\text{H}\right)\left(4\times {10}^{-3}\text{F}\right)}}\left\{\begin{array}{l}\because 1\text{mF}={10}^{-3}\text{F}\\ \text{1 mH=1}\times {\text{10}}^{-3}\text{H}\end{array}\right\}\\ =\frac{1}{2\times {\text{10}}^{-3}}\frac{\text{rad}}{\text{s}}\\ =500\frac{\text{rad}}{\text{s}}\end{array}$

As value of exponential frequency $\alpha$ is greater than resonating frequency ${\omega }_0$ which means circuit is over-damped.

Substitute $625{\text{s}}^{-1}$ for $\alpha$ and $500\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (3).

$\begin{array}{c}{s}_1=-625{\text{s}}^{-1}+\sqrt{{\left(-625{\text{s}}^{-1}\right)}^2-{\left(500\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-625{\text{s}}^{-1}+\sqrt{390625{\text{s}}^{-2}-250000{\left(\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-625{\text{s}}^{-1}+375{\text{s}}^{-1}\\ =-250{\text{s}}^{-1}\end{array}$

Substitute $625{\text{s}}^{-1}$ for $\alpha$ and $500\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (4).

$\begin{array}{c}{s}_2=-625{\text{s}}^{-1}-\sqrt{{\left(-625{\text{s}}^{-1}\right)}^2-{\left(500\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-625{\text{s}}^{-1}-\sqrt{390625{\text{s}}^{-2}-250000{\left(\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-625{\text{s}}^{-1}-375{\text{s}}^{-1}\\ =-1000{\text{s}}^{-1}\end{array}$

Differentiate equation (6) both the sides with respect to time $t$.

$\frac{d{v}_C\left(t\right)}{dt}=A_1{s}_1{e}^{s_{1}t}+A_2{s}_2{e}^{s_{2}t}$ (10)

Substitute $-250{\text{s}}^{-1}$ for $s_1$, $-1000{\text{s}}^{-1}$ for $S_2$ and $0^{+}$ for $t$ in equation (10).

$\begin{array}{c}\frac{d{v}_C\left(0\right)}{dt}=A_1\left(-250{\text{s}}^{-1}\right)e^{\left(-250{\text{s}}^{-1}\right)\left(0\right)}+A_2\left(-1000{\text{s}}^{-1}\right)e^{\left(-1000{\text{s}}^{-1}\right)\left(0\right)}\\ =A_1\left(-250{\text{s}}^{-1}\right)\left(1\right)+A_2\left(-1000{\text{s}}^{-1}\right)\left(1\right)\end{array}$

$\frac{d{v}_C\left(0\right)}{dt}=-250A_1-1000A_2$ (11)

Substitute $4\text{mF}$ for $C$, $-5\mu \text{A}$ for $i_C\left(0^{+}\right)$ and $-250A_1-1000A_2$ for $\frac{d{v}_C\left(0^{+}\right)}{dt}$ in equation (9).

$\begin{array}{l}-5\mu \text{A}=\left(4\text{mF}\right)\left(-250A_1-1000A_2\right)\\ \frac{-5\times {10}^{-6}\text{A}}{4\times {10}^{-3}\text{F}}=-250A_1-1000A_2\end{array}$

$-1.25\times {10}^{-3}=-250A_1-1000A_2$ (12)

Substitute $1\mu \text{V}-A_2$ for $A_1$ in equation (12).

$\begin{array}{l}-1.25\times {10}^{-3}=-250\times \left(1\mu \text{V}-A_2\right)-1000A_2\\ -1.25\times {10}^{-3}=-250\mu \text{V}+250A_2-1000A_2\end{array}$

Rearrange for $A_2$.

$\begin{array}{l}-750A_2=250\mu \text{V}-1.25\times {10}^{-3}\\ -750A_2=-0.001\end{array}$

Rearrange for $A_2$.

$\begin{array}{c}{A}_2=\frac{-0.001}{-750}\\ =1.33\times {10}^{-6}\end{array}$

Substitute $1.33\times {10}^{-6}$ for $A_2$ in equation (7).

$\begin{array}{c}{A}_1=1\mu \text{V}-1.33\times {10}^{-6}\\ =-3.33\times {10}^{-7}\end{array}$

Substitute $-3.33\times {10}^{-7}$ for $A_1$, $1.33\times {10}^{-6}$ for $A_2$, $-250{\text{s}}^{-1}$ for $s_1$ and $-1000{\text{s}}^{-1}$ for $s_2$ in equation (6).

$\begin{array}{c}{v}_C\left(t\right)=-3.33\times {10}^{-7}{e}^{-250{\text{s}}^{-1}t}+1.33\times {10}^{-6}{e}^{-1000{\text{s}}^{-1}t}\\ =-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\text{V}\end{array}$

Therefore, voltage across inductor $v\left(t\right)$ is $-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\text{V}$.

Conclusion:

Thus, the voltage $v\left(t\right)$ is $-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\text{V}$.

To determine

Find the maximum inductor current and the time of occurrence.

Answer to Problem 16E

The maximum inductor current $i_L\left(t\right)$ is $6.28\times {10}^{-7}\text{A}$ and maximum current occurs at $1.84\times {10}^{-3}\text{s}$.

Explanation of Solution

Calculation:

Refer to the Figure 4:

The expression for inductor current is:

$i_L\left(t\right)=\frac{1}{L}{\displaystyle \int v\left(t\right)dt+i_L\left(0^{-}\right)}$ (13)

Substitute $-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\text{V}$ for $v\left(t\right)dt$, $0\text{A}$ for $i_L\left(0^{-}\right)$ and $1\text{mH}$ for $L$ in the equation (13).

$\begin{array}{c}{i}_L\left(t\right)=\frac{1}{1\text{mH}}{\displaystyle \int \left(-3.33\times {10}^{-7}{e}^{-250t}+1.33\times {10}^{-6}{e}^{-1000t}\right)dt+i_L\left(0^{-}\right)}\\ =\frac{1}{1\times {10}^{-3}\text{H}}\left[-3.33\times {10}^{-7}\frac{e^{-250t}}{-250}+1.33\times {10}^{-6}\frac{e^{-1000t}}{-1000}\right]+0\text{A}\\ =\text{1}\text{.33}\times {10}^{-6}{e}^{-250t}-1.33\times {10}^{-6}{e}^{-1000t}\\ =\text{1}\text{.33}\times {10}^{-6}\left(e^{-250t}-e^{-1000t}\right)\end{array}$

$i_L\left(t\right)=\text{1}\text{.33}\times {10}^{-6}\left(e^{-250t}-e^{-1000t}\right)\text{A}$ (14)

For the maximum positive value of inductor current, take the derivative of its equation and then, equate it to zero to find the time at which maximum value of current occurs.

Differentiate equation (14) both the sides with respect to time $t$.

$\begin{array}{c}\frac{d{i}_L\left(t\right)}{dt}=1.33\times {10}^{-6}\left(-250e^{-250t}\right)-1.33\times {10}^{-6}\left(-1000e^{-1000t}\right)\text{A}/\text{s}\\ =-332.5\times {10}^{-6}{e}^{-250t}+1330\times {10}^{-6}{e}^{-1000t}\text{A}/\text{s}\end{array}$

Equate the equation to zero.

$-332.5\times {10}^{-6}{e}^{-250t}+1330\times {10}^{-6}{e}^{-1000t}\text{A}/\text{s}=0$

Rearrange the equation.

$\begin{array}{l}\frac{e^{-250t}}{e^{-1000t}}=\frac{1330\times {10}^{-6}}{332.5\times {10}^{-6}}\\ e^{-250t+1000t}=4\\ e^{750t}=4\end{array}$

Take log both the sides.

$\begin{array}{l}\ln \left(e^{750t}\right)=\ln \left(4\right)\\ 750t=1.386\text{s}\end{array}$

Rearrange for $t$.

$\begin{array}{c}t=\frac{1.386}{750}\text{s}\\ =1.84\times {10}^{-3}\text{s}\end{array}$

Substitute $1.84\times {10}^{-3}\text{s}$ for $t$ in equation (14).

$\begin{array}{c}{i}_L\left(1.84\times {10}^{-3}\text{s}\right)=1.33\times {10}^{-6}{e}^{-250\times 1.84\times {10}^{-3}\text{s}}-1.33\times {10}^{-6}{e}^{-1000\times 1.84\times {10}^{-3}\text{s}}\text{A}\\ \text{=}1.33\times {10}^{-6}{e}^{-0.46\text{s}}-1.33\times {10}^{-6}{e}^{-1.84\text{s}}\text{A}\\ \text{=}1.33\times {10}^{-6}\times 0.6312-1.33\times {10}^{-6}\times 0.158\text{A}\\ \text{=}0.839\times {10}^{-6}-0.211\times {\text{10}}^{-6}\text{A}\end{array}$

$i_L\left(1.84\times {10}^{-3}\text{s}\right)=6.28\times {10}^{-7}\text{A}$ (15)

This is the maximum inductor current.

Conclusion:

Thus, the maximum inductor current is $6.28\times {10}^{-7}\text{A}$ and maximum current occurs at $1.84\times {10}^{-3}\text{s}$.

To determine

Find the settling time.

Answer to Problem 16E

The settling time of the voltage $t_s$ is. $21.4\times {10}^{-3}\text{s}$.

Explanation of Solution

Calculation:

The settling time is the time at which inductor current reaches to $1\text{}\text{\%}$ of its maximum value.

$i_L\left(t_s\right)=\frac{1}{100}\times i_L{\left(t\right)}_{\text{max}}$ (16)

Substitute $6.28\times {10}^{-7}\text{A}$ for $i_L{\left(t\right)}_{\text{max}}$ in equation (16).

$\begin{array}{c}{i}_L\left(t_s\right)=\frac{1}{100}\times 6.28\times {10}^{-7}\text{A}\\ =6.28\times {10}^{-9}\text{A}\end{array}$

The settling time is the time at which the inductor current is decreased to $6.28\times {10}^{-9}\text{A}$ and is expressed as follows:

$\text{1}\text{.33}\times {10}^{-6}\left(e^{-250t_s}-e^{-1000t_s}\right)\text{A = }6.28\times {10}^{-9}\text{A}$ (17)

Equation (17) is solved by scientific calculator which can determine the value of time $t_s$ at which current gets settled.

$t_s=21.4\times {10}^{-3}\text{s}$

Conclusion:

Thus, the settling time of the inductor current $t_s$ is $21.4\times {10}^{-3}\text{s}$.