Chapter 9, Problem 20E

For the circuit represented by Fig. 9.44, the two resistor values are R₁ = 0.752 Ω and R₂ = 1.268 Ω, respectively. (a) Obtain an expression for the energy stored in the capacitor, valid for all t > 0; (b) determine the settling time of the current labeled i_A.

FIGURE 9.44

To determine

Find the expression for the energy stored in the capacitor, valid for all $t>0$.

Answer to Problem 20E

The expression for the energy stored in the capacitor, valid for all $t>0$ is

$2.5{\left(0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)}^2\text{ J}$.

Explanation of Solution

Given Data:

The value of the resistor $R_1$ is $\text{0}\text{.752 }\Omega$ and $R_2$ is $\text{1}\text{.268 }\Omega$.

Formula used:

The expression for the exponential damping coefficient or the neper frequency is as follows:

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

Here,

$\alpha$ is the exponential damping coefficient or the neper frequency,

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

$C$ 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.

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

$v\left(t\right)=A_1{e}^{S_{1}t}+A_2{e}^{S_{2}t}\text{ V}$ (5)

Here,

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

$A_1$ and $A_2$ are arbitrary constant and

$t$ is the time.

The expression for the energy stored in the capacitor is as follows:

$w=\frac{1}{2}C{v}^2$ (6)

Here,

$w$ is the energy stored in the capacitor,

$v$ is the voltage across the capacitor.

Calculation:

The capacitor and the inductor are connected in the circuit for long time.

So, the capacitor behaves as open circuit and the inductor behaves as short circuit.

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

Refer to the redrawn Figure 1:

As parallel branches have same voltage so voltage across $0.752\Omega$ resistor at $t=0^{+}$ is $1.5\text{ V}$.

The expression for the current flowing through $0.752\Omega$ resistor is as follows:

$i_A\left(0^{-}\right)=\frac{v_1}{R_1}$ (7)

Here,

$i_A\left(0^{-}\right)$ is the current flowing through the $0.752\Omega$ resistor,

$v_1$ is the voltage across the $0.752\Omega$ resistor and

$R_1$ is the resistance of the $0.752\Omega$ resistor.

Substitute $1.5\text{ V}$ for $v_1$ and $0.752\Omega$ for $R_1$ in equation (7).

$\begin{array}{c}{i}_A\left(0^{-}\right)=\frac{1.5\text{ V}}{0.752\Omega }\\ =1.9947\text{ A}\end{array}$

The expression for the current flowing through the $2\text{ H}$ inductor is as follows:

$i_L\left(0^{-}\right)=\frac{v_1-v_2}{R_2}$ (8)

Here,

$i_L\left(0^{-}\right)$ is the current flowing through $2\text{ H}$ inductor,

$v_2$ is the voltage of $2i_A$ dependent source and

$R_2$ is the resistance of the $\text{1}\text{.268 }\Omega$ resistor.

Substitute  $1.5\text{ V}$ for $v_1$, $2i_A$ for $v_2$ and $\text{1}\text{.268 }\Omega$ for $R_2$ in equation (8).

$i_L\left(0^{-}\right)=\frac{1.5\text{ V}-2i_A}{\text{1}\text{.268 }\Omega }$ (9)

Substitute $1.9947\text{ A}$ for $i_A$ in equation in equation (9).

$\begin{array}{c}{i}_L\left(0^{-}\right)=\frac{1.5\text{ V}-2\left(1.9947\right)}{\text{1}\text{.268 }\Omega }\\ =\frac{-2.4894}{\text{1}\text{.268}}\text{ A}\\ \text{=}-\text{1}\text{.963 A}\end{array}$

As parallel branches have same voltage and the voltage across the short circuit branch has $0\text{}\text{ V}$ so voltage across $0.752\Omega$ resistor at $t=0^{+}$ is $0\text{}\text{ V}$.

The capacitor does not allow sudden change in the voltage and the inductor does not allow sudden change in 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 $5\text{ F}$ capacitor at $t=0$ is $0\text{ V}$ and the current flowing through the $2\text{ H}$ inductor at $t=0$ is $-\text{1}\text{.963 A}$.

The redrawn circuit diagram is given in Figure 2 at $t=0^{+}$.

l

Refer to the redrawn Figure 2:

As the voltage across the he $5\text{ F}$ capacitor is $0\text{ V}$ it behaves as short circuit so, the current flowing through the $5\text{ F}$ capacitor is equal to negative of the current flowing through the $2\text{ H}$ inductor.

$\begin{array}{c}{i}_C\left(0\right)=-\left(-\text{1}\text{.963 A}\right)\\ =\text{1}\text{.963 A}\end{array}$

To find equivalent resistance across the capacitor, a $1\text{ V}$ test source is connected across the capacitor.

The redrawn circuit diagram is given in Figure 3.

Refer to the redrawn Figure 3:

Apply KVL in the circuit.

$v_0=i_A{R}_1+i_A{R}_2-v_2$ (10)

Here,

$v_0$ is the $1\text{ V}$ test source.

Substitute $1\text{ V}$ for $v_0$, $2i_A$ for $v_2$ and $\text{1}\text{.268 }\Omega$ for $R_2$ in equation (10).

$\begin{array}{l}1\text{ V}=\left(0.752\Omega \right)i_A+\left(\text{1}\text{.268 }\Omega \right)i_A-2i_A\\ 1=\text{2}\text{.02}{i}_A-2i_A\\ 1=0.02i_A\end{array}$

Rearrange for $i_A$.

$\begin{array}{l}0.02i_A=1\\ i_A=\frac{1}{0.02}\text{ A}\\ i_A=50\text{ A}\end{array}$

The expression for the equivalent resistance the circuit is as follows:

$R_{eq}=\frac{v_0}{i_A}$ (11)

Here,

$R_{eq}$ is the equivalent resistance across the $5\text{ F}$ capacitor.

Substitute $1\text{ V}$ for $v_0$ and $50\text{ A}$ for $i_A$ in equation (11).

$\begin{array}{c}{R}_{eq}=\frac{1\text{ V}}{50\text{ A}}\\ =0.02\Omega \end{array}$

Substitute $0.02\Omega$ for $R$ and $5\text{ F}$ for $C$ in the equation (1).

$\begin{array}{c}\alpha =\frac{1}{2\left(0.02\Omega \right)\left(5\text{ F}\right)}\\ =\frac{1}{2\left(0.02\Omega \right)\left(5\text{ F}\right)}\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =\frac{1}{0.2}{\text{ s}}^{-1}\\ ={\text{5 s}}^{-1}\end{array}$

Substitute $\text{2 H}$ for $L$ and $5\text{ F}$ for $C$ in the equation (2).

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(\text{2 H}\right)\times \left(5\text{ F}\right)}}\\ =\frac{1}{\sqrt{10}}\text{ rad/s}\\ =\text{0}\text{.31623 rad/s}\end{array}$

Here, the exponential damping coefficient is greater than the resonating frequency,

So, the response of the parallel $RLC$ circuit is over-damped.

Substitute $\text{5 s}$ for $\alpha$ and $\text{0}\text{.31623 rad/s}$ for ${\omega }_0$ in equation (3).

$\begin{array}{c}{s}_1=-\left({\text{5 s}}^{-1}\right)+\sqrt{{\left(\text{5 s}\right)}^2-{\left(\text{0}\text{.31623 rad/s}\right)}^2}\\ ={\text{5 s}}^{-1}+4.98998{\text{ s}}^{-1}\\ =-0.01{\text{ s}}^{-1}\end{array}$

Substitute $\text{5 s}$ for $\alpha$ and $\text{0}\text{.31623 rad/s}$ for ${\omega }_0$ in equation (4).

$\begin{array}{c}{s}_2=-\left({\text{5 s}}^{-1}\right)-\sqrt{{\left(\text{5 s}\right)}^2-{\left(\text{0}\text{.31623 rad/s}\right)}^2}\\ ={\text{5 s}}^{-1}-4.98998{\text{ s}}^{-1}\\ =-9.998{\text{ s}}^{-1}\end{array}$

Substitute $-0.01{\text{ s}}^{-1}$ for $s_1$ and $-9.998{\text{ s}}^{-1}$ for $s_2$ in equation (5).

$v\left(t\right)=A_1{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}+A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}$ (12)

Substitute $0$ for $t$ in equation (12).

$\begin{array}{c}v\left(0\right)=A_1{e}^{\left(-0.01{\text{ s}}^{-1}\right)\left(0\right)}+A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)\left(0\right)}\text{ V}\\ =A_1\left(1\right)+A_2\left(1\right)\text{ V}\end{array}$

$v\left(0\right)=A_1+A_2\text{ V}$ (13)

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

Substitute $\text{0 V}$ for $v\left(0\right)$ in equation (13).

$0\text{ V}=A_1+A_2\text{ V}$

Rearrange for $A_1$.

$A_1+A_2=0$

$A_1=-A_2$ (14)

The expression for the current flowing through the $\text{5 F}$ capacitor is as follows:

$i_C\left(t\right)=\frac{Cd{v}_C\left(t\right)}{dt}$ (15)

$v_C\left(t\right)$ is the voltage across the $\text{5 F}$ capacitor valid for all $t$.

Substitute $A_1{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}+A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}$ for $v\left(t\right)$ in equation (15).

$\begin{array}{c}{i}_C\left(t\right)=\frac{Cd\left(A_1{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}+A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}\right)}{dt}\\ =C\left(\left(-0.01{\text{ s}}^{-1}\right)e^{\left(-0.01{\text{ s}}^{-1}\right)t}+\left(-9.998{\text{ s}}^{-1}\right)A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)\end{array}$

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

$\frac{i_C\left(t\right)}{C}=-0.01e^{\left(-0.01{\text{ s}}^{-1}\right)t}-9.998A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}$ (16)

Substitute $0$ for $t$ in equation (12).

$\begin{array}{c}\frac{i_C\left(0\right)}{C}=-0.01e^{\left(-0.01{\text{ s}}^{-1}\right)\left(0\right)}-9.998A_2{e}^{\left(-9.998{\text{ s}}^{-1}\right)\left(0\right)}\\ =-0.01\left(1\right)-9.998A_2\left(1\right)\\ =-0.01-9.998A_2\end{array}$

The current flowing through the $\text{5 F}$ capacitor at $t=0$ is $\text{1}\text{.963 A}$.

Substitute $\text{1}\text{.963 A}$ for $i_C\left(0\right)$ and $\text{5 F}$ for $C$ in equation (16).

$\begin{array}{l}\frac{\text{1}\text{.963 A}}{\text{5 F}}=-0.01-9.998A_2\\ 0.393=-0.01-9.998A_2\end{array}$

Rearrange for $A_1$ and $A_2$.

$-0.01-9.998A_2=0.393$ (17)

Substitute $-A_2$ for $A_1$ in equation (17).

$\begin{array}{l}\left(-0.01\right)\left(-A_2\right)-9.998A_2=0.393\\ 0.01A_2-9.998A_2=0.393\\ -9.988A_2=0.393\end{array}$

Rearrange for $A_2$.

$\begin{array}{c}{A}_2=-\frac{0.393}{9.988}\\ =-0.0393\end{array}$

Substitute $-0.0393$ for $A_2$ in equation (14).

$\begin{array}{c}{A}_1=-\left(-0.0393\right)\\ =0.0393\end{array}$

Substitute $0.0393$ for $A_1$ and $-0.0393$ for $A_2$ in equation (12).

$v\left(t\right)=0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}$ (18)

Substitute $0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}$ for $v$ and $\text{5 F}$ for $C$ in equation (6).

$\begin{array}{c}w=\left(\frac{1}{2}\right)\left(\text{5 F}\right){\left(0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}\right)}^2\\ =2.5{\left(0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)}^2\text{ J}\end{array}$

So, the energy stored in the capacitor, valid for all $t>0$ is

$2.5{\left(0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)}^2\text{ J}$

Conclusion:

Thus, the expression for the energy stored in the capacitor, valid for all $t>0$ is

$2.5{\left(0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)}^2\text{ J}$.

To determine

Find the settling time of the current $i_A$.

Answer to Problem 20E

The settling time of current $i_A$ is $461.32\text{ s}$.

Explanation of Solution

Calculation:

The redrawn circuit diagram is given in Figure 4.

Refer to the redrawn Figure 4:

The expression for the current flowing in the left hand mesh at $t>0$ is as follows:

$i_A\left(t\right)=\frac{v\left(t\right)}{{\mathrm{R}}_{eq}}$ (18)

Here,

$i_A\left(t\right)$ is the current flowing in the left hand mesh at $t>0$

${\mathrm{R}}_{eq}$ is the  equivalent resistance across capacitor.

Substitute $0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}$ for $v\left(t\right)$ and $0.02\Omega$ for ${\mathrm{R}}_{eq}$ in equation (18).

$\begin{array}{c}{i}_A\left(t\right)=\frac{0.0393e^{\left(-0.01{\text{ s}}^{-1}\right)t}-0.0393e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ V}}{0.02\Omega }\\ =1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}-1.965e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ A}\end{array}$

$i_A\left(t\right)=1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}-1.965e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ A}$ (19)

Differentiate both side of the equation (19).

$\begin{array}{c}\frac{d{i}_A\left(t\right)}{dt}=\frac{d\left(1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}-1.965e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ A}\right)}{dt}\\ =\left(\left(1.965\right)\left(-0.01{\text{ s}}^{-1}\right)e^{\left(-0.01{\text{ s}}^{-1}\right)t}-\left(1.965\right)\left(-9.998{\text{ s}}^{-1}\right)e^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)\frac{\text{A }}{\text{s}}\end{array}$

The maximum value is obtained when derivative is equated to zero.

$\left(\left(1.965\right)\left(-0.01{\text{ s}}^{-}\right)e^{\left(-0.01{\text{ s}}^{-}\right)t}-\left(1.965\right)\left(-9.998{\text{ s}}^{-}\right)e^{\left(-9.998{\text{ s}}^{-}\right)t}\right)\frac{\text{A}}{\text{s}}=\text{0}$

$\left(1.965\right)\left(-0.01{\text{ s}}^{-1}{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}+9.998{\text{ s}}^{-1}{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\right)=\text{0}$ (20)

Rearrange equation (20).

$\begin{array}{l}-0.01{\text{ s}}^{-1}{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}+9.998{\text{ s}}^{-1}{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}=\text{0}\\ 0.01{\text{ s}}^{-1}{e}^{\left(-0.01{\text{ s}}^{-1}\right)t}=9.998{\text{ s}}^{-1}{e}^{\left(-9.998{\text{ s}}^{-1}\right)t}\\ \frac{e^{\left(-0.01{\text{ s}}^{-1}\right)t}}{e^{\left(-9.998{\text{ s}}^{-1}\right)t}}=\frac{9.998}{0.01}\\ e^{\left(-0.01{\text{ s}}^{-1}+9.998{\text{ s}}^{-1}\right)t}=999.8\end{array}$

$e^{\left(9.988{\text{ s}}^{-1}\right)t}=999.8$

Take natural logarithm both sides.

$\begin{array}{l}\ln \left(e^{\left(9.988{\text{ s}}^{-}\right)t}\right)=\ln \left(999.8\right)\\ 9.988t=6.9075\text{ s}\end{array}$

Rearrange for $t$.

$\begin{array}{c}t=\frac{6.9075}{9.988}\\ =0.69089\text{ s}\end{array}$

Substitute $0.69089\text{ s}$ for $t$ in equation (19).

$\begin{array}{c}{i}_A\left(t\right)=1.965e^{\left(-0.01{\text{ s}}^{-1}\right)\left(0.69089\text{ s}\right)}-1.965e^{\left(-9.998{\text{ s}}^{-1}\right)\left(0.69089\text{ s}\right)}\text{ A}\\ =\left(1.965\right)\left(0.993\right)-\left(1.965\right)\left(1.000237\times {10}^{-3}\right)\text{ A}\\ \text{=}1.9513-1.9654\times {10}^{-3}\text{ A}\\ \text{=1}\text{.9493 A}\end{array}$

So, the value of the maximum current flowing in the left hand mesh is $\text{1}\text{.9493 A}$.

Settling time is the time at which the value of the current reaches $1\%$ of its maximum value.

The expression for the current at $1\%$ of maximum value is as follows:

$i_{A\text{ at }1\%}=i_{\text{A max}}\left(1\%\right)$ (21)

Here,

$i_{\text{A at }1\%}$ is the value of the current at $1\%$ of maximum value,

$i_{\text{A max}}$ is the maximum current.

Substitute $\text{1}\text{.9493 A}$ for $i_{\text{A max}}$ in equation (21).

$\begin{array}{c}{i}_{\text{A at }1\%}=\left(\text{1}\text{.9493 A}\right)\left(1\%\right)\\ =\left(\text{1}\text{.9493 A}\right)\left(\frac{1}{100}\right)\\ =0.019493\text{ A}\end{array}$

The value of current flowing in the left hand mesh at $1\%$ of maximum value is

$0.019493\text{ A}$.

Substitute $0.019493\text{ A}$ for $i_R$ in equation (19).

$0.019493\text{ A}=1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}-1.965e^{\left(-9.998{\text{ s}}^{-1}\right)t}\text{ A}$

Since the component $e^{\left(-0.01{\text{ s}}^{-1}\right)t}$ dominates the component $e^{\left(-9.998{\text{ s}}^{-1}\right)t}$, hence neglect the component $e^{\left(-9.998{\text{ s}}^{-1}\right)t}$.

So, the new equation is:

$0.019493\text{ A}=1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}\text{ A}$ (22)

Rearrange equation (22).

$\begin{array}{l}1.965e^{\left(-0.01{\text{ s}}^{-1}\right)t}=0.019493\\ e^{\left(-0.01{\text{ s}}^{-1}\right)t}=\frac{0.019493}{1.965}\\ e^{\left(-0.01{\text{ s}}^{-1}\right)t}=9.9201\times {10}^{-3}\end{array}$

Take natural logarithm both sides.

$\begin{array}{l}\ln \left(e^{\left(-0.01{\text{ s}}^{-1}\right)t}\right)=\ln \left(9.9201\times {10}^{-3}\right)\\ -0.01t=-4.6132\text{ s}\end{array}$

Rearrange for $t$.

$\begin{array}{c}t=\frac{-4.6132}{-0.01}\text{ s}\\ =461.32\text{ s}\end{array}$

So, the settling time of the currentflowing in the left hand mesh $i_A$ is $461.32\text{ s}$.

Conclusion:

Thus, the settling time of current $i_A$ is $461.32\text{ s}$.