Chapter 5, Problem 5.79HP

To determine

The time $t$ at which value of $v=7.5\text{V}$ .

Answer to Problem 5.79HP

The value of $t$ for which the voltage $v$ is $7.5\text{V}$ is $0.873\text{s}$ .

Explanation of Solution

Calculation:

The given circuit is shown in Figure 1

 

For time $t<0$, the circuit is in steady state and the capacitor terminals are open circuited, mark the value and redraw the circuit.

The required diagram is shown in Figure 2

 

The value for the current through the resistance of $3\Omega$ is calculated as,

  $\begin{array}{c}{I}_1=\left(\frac{5\Omega }{5\Omega +3\Omega }\right)\left(20\text{A}\right)\\ =12.5\text{A}\end{array}$

The value for the current through the resistance of $3\Omega$ is calculated as,

  $\begin{array}{c}{I}_2=\left(\frac{3\Omega }{3\Omega +5\Omega }\right)\left(20\text{A}\right)\\ =7.5\text{A}\end{array}$

The expression for the initial voltage across the capacitor $C_1$ is given by,

  $V_{C1}=I_1\left(3\Omega \right)$

Substitute $12.5\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{V}_{C1}=\left(12.5\text{A}\right)\left(3\Omega \right)\\ =37.5\text{V}\end{array}$

The expression for the initial voltage across the capacitor $C_2$ is given by,

  $V_{C2}=I_2\left(3\Omega \right)$

Substitute $7.5\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{V}_{C1}=\left(7.5\text{A}\right)\left(3\Omega \right)\\ =15\text{V}\end{array}$

For time $t>0$, the circuit is source free, mark the values and redraw the circuit

The required diagram is shown in Figure 3

 

Apply KCL to the node $V_{C1}$ of the circuit.

  $\begin{array}{l}-C_1\left[\frac{d{V}_{C1}}{dt}-V_{C1}\left(0\right)\right]-\frac{V_{C1}-V_{C2}}{3\Omega }=0\\ -\left(\frac{d{V}_{C1}}{dt}\right)+C_1\left[V_{C1}\left(0\right)\right]-\frac{V_{C1}-V_{C2}}{3\Omega }=0\end{array}$

Substitute $37.5\text{V}$ for $V_{C1}\left(0\right)$ and $\frac{1}{6}\text{F}$ for $C_1$ in the above equation.

  $\begin{array}{l}-\left(\frac{1}{6}\text{F}\right)\left(\frac{d{V}_{C1}}{dt}\right)+\left(\frac{1}{6}\text{F}\right)\left[37.5\text{V}\right]-\frac{V_{C1}-V_{C2}}{3\Omega }=0\\ \frac{1}{6}\frac{d{V}_{C1}}{dt}+\frac{V_{C1}-V_{C2}}{3\Omega }=6.25\end{array}$   .......... (1)

Apply KCL to the node $V_{C2}$ .

  $\begin{array}{l}\frac{V_{C1}-V_{C2}}{3\Omega }-C_2\left[\frac{d{V}_{C2}}{dt}-V_{C2}\left(0\right)\right]-\frac{V_{C2}}{2\Omega }=0\\ \frac{V_{C1}-V_{C2}}{3\Omega }-C_2\frac{d{V}_{C2}}{dt}+C_2\left[V_{C2}\left(0\right)\right]-\frac{V_{C2}}{2\Omega }=0\end{array}$

Substitute $15\text{V}$ for $V_{C2}\left(0\right)$ and $\frac{1}{6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{l}\frac{V_{C1}-V_{C2}}{3\Omega }-\left(\frac{1}{6}\text{F}\right)\frac{d{V}_{C2}}{dt}+\left(\frac{1}{6}\text{F}\right)\left[15\text{V}\right]-\frac{V_{C2}}{2\Omega }=0\\ \frac{V_{C1}}{3}-\frac{1}{6}\frac{d{V}_{C2}}{dt}-\frac{5V_{C2}}{6}=2.5\\ V_{C1}=7.5+0.5\frac{d{V}_{C2}}{dt}+2.5V_{C2}\end{array}$

Substitute $7.5+0.5\frac{d{V}_{C2}}{dt}+2.5V_{C2}$ for $V_{C1}$ in the above equation.

  $\begin{array}{l}\frac{1}{6}\frac{d}{dt}\left[7.5+0.5\frac{d{V}_{C2}}{dt}+2.5V_{C2}\right]+\frac{7.5+0.5\frac{d{V}_{C2}}{dt}+2.5V_{C2}-V_{C2}}{3\Omega }=6.25\\ \frac{1}{6}\left[0.5\frac{d^2{V}_{C2}}{dt}+2.5\frac{d{V}_{C2}}{dt}\right]+2.5+0.166\frac{d{V}_{C2}}{dt}+0.833V_{C2}-\frac{V_{C2}}{3}=6.25\\ 0.083\frac{d^2{V}_{C2}}{d{t}^2}+0.582\frac{d{V}_{C2}}{dt}+0.5V_{C2}=3.75\\ 0.166\frac{d^2{V}_{C2}}{dt}+1.166\frac{d{V}_{C2}}{dt}+V_{C2}=7.5\end{array}$   .......... (2)

The standard second order equation for the above equation is given by,

  $\frac{1}{{\omega }_n^2}\frac{d^{2}x\left(t\right)}{d{t}^2}+\frac{2\varsigma }{{\omega }_n\frac{dx\left(t\right)}{dt}}+x\left(t\right)=K_{S}f\left(t\right)$

From above and from equation (2), the natural frequency of the circuit is calculated as,

  $\begin{array}{l}\frac{1}{{\omega }_n^2}=0.66\\ {\omega }_n=2.45\text{rad}/\text{s}\end{array}$

The damping ratio of the circuit is calculated as,

  $\begin{array}{l}\frac{2\varsigma }{{\omega }_n}=1.166\\ 2\varsigma =\left(1.166\right)\left({\omega }_n\right)\end{array}$

Substitute $2.45\text{rad}/\text{s}$ for ${\omega }_n$ in the above equation.

  $\begin{array}{l}2\varsigma =\left(1.166\right)\left(2.45\text{rad}/\text{s}\right)\\ \varsigma =1.43\end{array}$

The general equation for the voltage across the capacitor is given by,

  $V_{C2}\left(t\right)={\alpha }_1{e}^{\left(-\varsigma {\omega }_n+{\omega }_n\sqrt{{\varsigma }^2-1}\right)t}+{\alpha }_2{e}^{\left(-\varsigma {\omega }_n-{\omega }_n\sqrt{{\varsigma }^2-1}\right)t}+V_{C2}\left(\infty \right)$

Substitute $1.43$ for $\varsigma$ and $2.45\text{rad}/\text{s}$ for ${\omega }_n$ in the above equation.

  $\begin{array}{c}{V}_{C2}\left(t\right)={\alpha }_1{e}^{\left(-\left(2.45\text{rad}/\text{s}\right)+\left(2.45\text{rad}/\text{s}\right)\sqrt{{\left(1.43\right)}^2-1}\right)t}+{\alpha }_2{e}^{\left(-\left(1.43\right)\left(2.45\text{rad}/\text{s}\right)-\left(2.45\text{rad}/\text{s}\right)\sqrt{{\left(1.43\right)}^2-1}\right)t}+V_{C2}\left(\infty \right)\\ ={\alpha }_1{e}^{-t}+{\alpha }_2{e}^{-6t}+V_{C2}\left(\infty \right)\end{array}$

Substitute $0$ for $V_{C2}\left(\infty \right)$ in the above equation.

  $V_{C2}\left(t\right)={\alpha }_1{e}^{-t}+{\alpha }_2{e}^{-6t}$   .......... (3)

Substitute $0$ for $t$ in the above equation.

  $\begin{array}{c}{V}_{C2}\left(0\right)={\alpha }_1{e}^{-0}+{\alpha }_2{e}^{-6\left(0\right)}\\ ={\alpha }_1+{\alpha }_2\end{array}$

Substitute $15\text{V}$ for $V_{C2}\left(0\right)$ in the above equation.

  $\begin{array}{l}15\text{V}={\alpha }_1+{\alpha }_2\\ {\alpha }_1=15\text{V}-{\alpha }_2\end{array}$   .......... (4)

Apply KCL to the node $V_{C1}$ in the circuit shown in Figure 3

  $\begin{array}{l}\frac{V_{C1}-V_{C2}}{3\Omega }-C_2\left[\frac{d{V}_{C2}\left(t\right)}{dt}\right]-\frac{V_{C2}}{2\Omega }=0\\ C_2\left(\frac{d{V}_{C2}\left(t\right)}{dt}\right)=\frac{V_{C1}}{3}-\frac{V_{C2}}{3}-\frac{V_{C2}}{2}\end{array}$

Substitute $\frac{1}{6}\text{F}$ for $C_2$ in the above equation.

  $\begin{array}{l}\left(\frac{1}{6}\text{F}\right)\left(\frac{d{V}_{C2}\left(t\right)}{dt}\right)=\frac{V_{C1}}{3}-\frac{V_{C2}}{3}-\frac{V_{C2}}{2}\\ \left(\frac{1}{6}\text{F}\right)\left(\frac{d{V}_{C2}\left(t\right)}{dt}\right)=\frac{V_{C1}}{3}-\frac{5V_{C2}}{6}\\ \frac{d{V}_{C2}\left(t\right)}{dt}=2V_{C1}-5V_{C2}\end{array}$

Substitute $0$ for $t$, $37.5\text{V}$ for $V_{C1}\left(0\right)$ and $15\text{V}$ for $V_{C2}\left(0\right)$ in the above equation.

  $\begin{array}{c}\frac{d{V}_{C2}\left(t\right)}{dt}=2\left(37.5\text{V}\right)-5\left(15\text{V}\right)\\ =0\end{array}$

The differentiation of equation (3) with respect to $t$ is given by,

  $\begin{array}{c}{V}_{C2}\left(t\right)={\alpha }_1{e}^{-t}+{\alpha }_2{e}^{-6t}\\ \frac{d{V}_{C2}\left(t\right)}{dt}=-{\alpha }_1{e}^{-t}-6{\alpha }_2{e}^{-6t}\end{array}$

Substitute $0$ for $t$ in the above equation.

  $\begin{array}{c}\frac{d{V}_{C2}\left(0\right)}{dt}=-{\alpha }_1{e}^{-0}-6{\alpha }_2{e}^{-6\left(0\right)}\\ =-{\alpha }_1-6{\alpha }_2\end{array}$

Substitute $0$ for $\frac{d{V}_{C2}\left(0\right)}{dt}$ in the above equation.

  $0=-{\alpha }_1-6{\alpha }_2$

Substitute $15-{\alpha }_2$ for ${\alpha }_1$ in the above equation.

  $\begin{array}{l}0=-15-{\alpha }_2-6{\alpha }_2\\ {\alpha }_2=-3\end{array}$

Substitute $-3$ for ${\alpha }_2$ in equation (4).

  $\begin{array}{c}{\alpha }_1=15+3\\ =18\end{array}$

Substitute $18$ for ${\alpha }_1$ and $-3$ for ${\alpha }_2$ in the above equation.

  $V_{C2}\left(t\right)=18e^{-t}-3e^{-6t}$

Substitute $7.5\text{V}$ for $V_{C2}\left(t\right)$ in the above equation.

  $7.5\text{V}=18e^{-t}-3e^{-6t}$

Substitute $0.873$ for $t$ in the above equation.

  $\begin{array}{l}7.5\text{V}=18e^{-0.873}-3e^{-6\left(0.873\right)}\\ 7.5\text{V}=7.5\text{V}\end{array}$

Thus, the value of $t$ for which the voltage $v$ is $7.5\text{V}$ is $0.873\text{s}$ .

Conclusion:

Therefore, the value of $t$ for which the voltage $v$ is $7.5\text{V}$ is $0.873\text{s}$ .