Chapter 9, Problem 15E

(a) Assuming the passive sign convention, obtain an expression for the voltage across the 1 Ω resistor in the circuit of Fig. 9.41 which is valid for all t > 0. (b) Determine the settling time of the resistor voltage.

■ FIGURE 9.41

To determine

Obtain an expression for voltage across $1\Omega$ resistor valid for $t>0$.

Answer to Problem 15E

The expression for voltage across $1\Omega$ resistor valid for is $t>0$ is

$-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$

Explanation of Solution

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 $s_2$ are 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:

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

Here,

$v_R\left(t\right)$ 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 diagram is given in Figure 1 for $t>0$.

Refer to the redrawn Figure 1:

Substitute $1\text{ Ω}$ for $R$ and $2\text{mF}$ for $C$ in equation (1).

$\begin{array}{c}\alpha =\frac{1}{2\left(1\text{ Ω}\right)\left(2\text{mF}\right)}\\ =\frac{1}{2\left(1\text{ Ω}\right)\left(2\times {10}^{-3}\text{F}\right)}\left\{\because 1\text{mF}={10}^{-3}\text{F}\right\}\\ =\frac{1}{0.004\text{s}}\\ =250{\text{s}}^{-1}\end{array}$

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

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(250\text{mH}\right)\left(2\text{mF}\right)}}\\ =\frac{1}{\sqrt{\left(250\times {10}^{-3}\text{H}\right)\left(2\times {10}^{-3}\text{F}\right)}}\left\{\begin{array}{l}\because 1\text{mH}={10}^{-3}\text{H}\\ \because 1\text{mF}={10}^{-3}\text{F}\end{array}\right\}\\ =\frac{1}{\sqrt{0.0005}}\frac{\text{rad}}{\text{s}}\\ =44.72\frac{\text{rad}}{\text{s}}\end{array}$

As value of neper frequency $\alpha$ is greater than resonating frequency ${\omega }_0$ which means circuit is overdamped

Substitute $250{\text{s}}^{-1}$ for $\alpha$ and $44.72\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (3).

$\begin{array}{c}{s}_1=-250{\text{s}}^{-1}+\sqrt{{\left(250{\text{s}}^{-1}\right)}^2-{\left(44.72\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-250{\text{s}}^{-1}+\sqrt{62500{\text{s}}^{-2}-2000{\left(\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-250{\text{s}}^{-1}+245.96{\text{s}}^{-1}\\ =-4.03{\text{s}}^{-1}\end{array}$

Substitute $250{\text{s}}^{-1}$ for $\alpha$ and $44.72\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (4).

$\begin{array}{c}{s}_2=-250{\text{s}}^{-1}-\sqrt{{\left(250{\text{s}}^{-1}\right)}^2-{\left(44.72\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-250{\text{s}}^{-1}-\sqrt{62500{\text{s}}^{-2}-2000{\left(\frac{\text{rad}}{\text{s}}\right)}^2}\\ =-250{\text{s}}^{-1}-245.96{\text{s}}^{-1}\\ =-495.96{\text{s}}^{-1}\end{array}$

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 become zero.

So, at $t=0^{-}$ the $10u\left(-t\right)\text{mA}$ independent current source is equal to $10\text{ mA}$.

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 2 for $t=0^{-}$.

Refer to the redrawn Figure 2:

The expression for the current flowing in the $48\Omega$ resistor at $t=0^{-}$ is as follows:

$i_L\left(0^{-}\right)=i\left(\frac{R_2}{R_1+R_2}\right)$        (6)

Here,

$i_L\left(0^{-}\right)$ is the current flowing in the $48\Omega$ resistor at $t=0^{-}$,

$i$ is the current through $10u\left(-t\right)\text{mA}$ independent current source,

$R_1$ is the resistance of $48\Omega$ resistor and

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

Substitute $10\text{ mA}$ for $i$, $48\Omega$ for $R_1$ and $\text{1 }\Omega$ for $R_2$ in equation (6).

$\begin{array}{c}{i}_L\left(0^{-}\right)=\left(10\text{ mA}\right)\left(\frac{\text{1 }\Omega }{48\Omega +1\Omega }\right)\\ =\left(10\text{ mA}\right)\left(0.02041\right)\\ =0.2041\text{ mA}\end{array}$

The expression for the voltage across the $2\text{ mF}$ capacitor at $t=0^{-}$ is as follows:

$v_C\left(0^{-}\right)=i_L\left(0^{-}\right)R_1$        (7)

Here,

$v_C\left(0^{-}\right)$ is the voltage across the $2\text{ mF}$ capacitor at $t=0^{-}$.

Substitute $0.2041\text{ mA}$ for $i_L\left(0^{-}\right)$ and $48\Omega$ for $R_1$ in equation (7).

$\begin{array}{c}{v}_C\left(0^{-}\right)=\left(0.2041\text{ mA}\right)\left(48\Omega \right)\\ =9.796\text{ mV}\end{array}$

The switch closes at $t=0$.

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

So,

$v_C\left(0^{+}\right)=v_C\left(0^{-}\right)=9.796\text{ mV}$ and

$i_L\left(0^{+}\right)=i_L\left(0^{-}\right)=0.2041\text{ mA}$.

As parallel branches have same voltage across them, so, voltage across resistor $R_2$ at $t=0^{-}$ is same.

Therefore,

$v_R\left(0^{-}\right)=9.796\text{ mV}$

The redrawn circuit diagram is given in Figure 3 for $t>0$.

Refer to the redrawn Figure 3:

The expression for the current flowing in the resistor $R_2$ is as follows:

$i_R\left(0^{+}\right)=\frac{v_C\left(0^{+}\right)}{R_2}$        (8)

Here,

$i_R\left(0^{+}\right)$ is the current flowing in the resistor $R_2$ at $t=0^{+}$.

Substitute $9.796\text{ mV}$ for $v_C\left(0^{+}\right)$ and $\text{1 }\Omega$ for $R_2$ in equation (8).

$\begin{array}{c}{i}_R\left(0^{+}\right)=\frac{9.796\text{ mV}}{\text{1 }\Omega }\\ =9.796\text{ mA}\end{array}$

Apply KCL at node $A$.

$i_C\left(0^{+}\right)+i_L\left(0^{+}\right)+i_R\left(0^{+}\right)=0\text{ A}$        (9)

Here,

$i_C\left(0^{+}\right)$ is the current flowing through $2\text{ mF}$ capacitor at $t=0^{+}$.

Substitute $0.2041\text{ mA}$ for $i_L\left(0^{+}\right)$ and $9.796\text{ mA}$ for $i_R\left(0^{+}\right)$ in equation (9).

$\begin{array}{l}{i}_C\left(0^{+}\right)+0.2041\text{ mA+}9.796\text{ mA}=0\\ i_C\left(0^{+}\right)+10.0001\text{ mA}=0\end{array}$

Rearrange for $i_C\left(0^{+}\right)$.

$i_C\left(0^{+}\right)=-10.0001\text{ mA}$

Substitute $-4.03{\text{s}}^{-1}$ for $s_1$ and $-495.96{\text{s}}^{-1}$ for $s_2$ in equation (5).

$v_R\left(t\right)=A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$        (10)

Substitute $0$ for $t$ in equation (10).

$\begin{array}{c}{v}_R\left(0\right)=A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)\left(0\right)}+A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)\left(0\right)}\text{ V}\\ =A_1\left(1\right)+A_2\left(1\right)\text{ V}\end{array}$

$v_R\left(0\right)=A_1+A_2\text{ V}$        (11)

The voltage across the resistor $R_2$ at $t=0$ is $9.796\text{ mV}$.

Substitute $9.796\text{ mV}$ for $v_R\left(0\right)$ in equation (11).

$9.796\text{ mV}=A_1+A_2\text{ V}$

Rearrange for $A_1$.

$A_1=9.796\text{ mV}-A_2$        (12)

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

$i_C\left(t\right)=\frac{Cd{v}_C\left(t\right)}{dt}$        (13)

$v_C\left(t\right)$ is the voltage across the $2\text{ mF}$ capacitor for $t=0^{+}$.

Substitute $A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$ for $v\left(t\right)$ in equation (13).

$i_C\left(t\right)=\frac{Cd\left(A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}\right)}{dt}$

$i_C\left(t\right)=C\left(-4.03A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)t}-495.966A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)t}\right)$        (14)

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

$\frac{i_C\left(t\right)}{C}=-4.03A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)t}-495.966A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)t}$        (15)

Substitute $0$ for $t$ in equation (15).

$\begin{array}{c}\frac{i_C\left(0\right)}{C}=-4.03A_1{e}^{\left(-4.03{\text{s}}^{-1}\right)\left(0\right)}-495.966A_2{e}^{\left(-495.96{\text{s}}^{-1}\right)\left(0\right)}\\ =-4.03A_1\left(1\right)-495.966A_2\left(1\right)\end{array}$

$\frac{i_C\left(0\right)}{C}=-4.03A_1-495.966A_2$        (16)

The current flowing through the $2\text{ mF}$ capacitor at $t=0$ is $-10.0001\text{ A}$.

Substitute $-10.0001\text{ mA}$ for $i_C\left(0\right)$ and $2\text{ mF}$ for $C$ in equation (16).

$\begin{array}{c}\frac{-10.0001\text{ mA}}{2\text{ mF}}=-4.03A_1-495.966A_2\\ -5.00005=-4.03A_1-495.966A_2\end{array}$

Rearrange for $A_1$ and $A_2$.

$-4.03A_1-495.966A_2=-5.00005$        (17)

Substitute $9.796\text{ mV}-A_2$ for $A_1$ in equation (17).

$\begin{array}{l}-4.03\left(9.796\text{ mV}-A_2\right)-495.966A_2=-5.00005\\ -4.03\left(0.009796\text{ V}-A_2\right)-495.966A_2=-5.00005\left\{\because 1\text{ V}={10}^3\text{mV}\right\}\\ -0.0395+4.03A_2-495.966A_2=-5.00005\\ -0.0395-491.936A_2=-5.00005\end{array}$

Rearrange for $A_2$.

$\begin{array}{l}491.936A_2=5.00005-0.0395\\ A_2=\frac{4.96055}{491.936}\\ =0.01\end{array}$

Substitute $0.01$ for $A_2$ in equation (12).

$\begin{array}{c}{A}_1=9.796\text{ mV}-0.01\\ =9.796\times {10}^{-3}\text{ V}-0.01\left\{\because 1\text{V=}{10}^3\text{ mV}\right\}\\ =-2.04\times {10}^{-4}\end{array}$

Substitute $-2.04\times {10}^{-4}$ for $A_1$ and $0.01$ for $A_2$ in equation (10).

$v_R\left(t\right)=-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$

Conclusion:

Thus, expression for voltage across $1\Omega$ resistor valid for is $t>0$ is

$-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$.

To determine

Find settling time of the resistor voltage.

Answer to Problem 15E

The settling time is $\text{9}\text{.33}\text{ms}$.

Explanation of Solution

Calculation:

Function for voltage across resistor $R_2$ is as follows:

$v_R\left(t\right)=-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$        (18)

The maximum value of voltage $v_R\left(t\right)$ occurs at $t=0$.

Substitute $0$ for $t$ in equation (18).

$\begin{array}{c}{v}_R\left(t\right)=-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)\left(0\right)}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)\left(0\right)}\text{ V}\\ \text{=}-2.04\times {10}^{-4}\left(1\right)+0.01\left(1\right)\text{ V}\\ =9.796\times {10}^{-3}\text{V}\end{array}$

So, the maximum value of voltage $v_R\left(t\right)$ is $9.796\times {10}^{-3}\text{V}$.

Settling time is the time at which the value of the voltage $v_R\left(t\right)$ reaches $1\%$ of maximum value

The expression for the voltage $v_R\left(t\right)$ at $1\%$ of maximum value is as follows:

$v_R{\left(t\right)}_{\text{ at }1\%}=v_R{\left(t\right)}_{\text{ max}}\left(1\%\right)$        (19)

Here,

$v_R{\left(t\right)}_{\text{ at }1\%}$ is the $1\%$ of maximum value of the  voltage.

$v_R{\left(t\right)}_{\text{ max}}$ the maximum voltage.

Substitute $9.796\times {10}^{-3}\text{V}$ for $v_R{\left(t\right)}_{\text{ max}}$ in equation (19).

$\begin{array}{c}{v}_R{\left(t\right)}_{\text{ at }1\%}=9.796\times {10}^{-3}\text{V}\left(1\%\right)\\ =9.796\times {10}^{-3}\text{V}\left(\frac{1}{100}\right)\\ =9.796\times {10}^{-5}\text{V}\end{array}$

Substitute $9.796\times {10}^{-5}\text{V}$ for $v_R\left(t\right)$ in equation (18).

$9.796\times {10}^{-5}\text{V}=-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}+0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}\text{ V}$

Since the component $0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}$ dominates the component $-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}$, hence neglect the component $-2.04\times {10}^{-4}{e}^{\left(-4.03{\text{s}}^{-1}\right)t}$.

So, the new equation is:

$9.796\times {10}^{-5}\text{V}=0.01e^{\left(-495.96{\text{s}}^{-1}\right)t}$        (20)

Rearrange equation (20).

$\begin{array}{c}{e}^{-495.96t}=\frac{9.796\times {10}^{-5}}{0.01}\\ =9.796\times {10}^{-3}\end{array}$

Take natural logarithm both sides.

$\begin{array}{c}\ln e^{-495.96t}=\ln \left(9.796\times {10}^{-3}\right)\\ -495.96t=-4.626\\ t=9.33\times {10}^{-3}\text{s}\\ t=\text{9}\text{.33}\text{ms}\left\{\because 1\text{s}={10}^3\text{ms}\right\}\end{array}$

Conclusion:

Thus, the settling time is $\text{9}\text{.33}\text{ms}$.