Chapter 5, Problem 5.71HP

To determine

(a)

The capacitor voltage $v_C$ at $t=0^{+}$ .

Answer to Problem 5.71HP

The value of the capacitor voltage $v_C$ for time $t=0^{+}$ is $15\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

 

For time $t<0$ the inductor terminals are short circuited and the capacitor terminals are open circuited.

The required diagram is shown in Figure 2

 

From above, the expression for the initial voltage across the capacitor is given by,

  $v_C\left(0^{-}\right)=15\text{V}$

The expression for the voltage across the capacitor for time $t=0^{+}$ is given by,

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

Substitute $15\text{V}$ for $v_C\left(0^{-}\right)$ in the above equation.

  $v_C\left(0^{+}\right)=15\text{V}$

Conclusion:

Therefore, the value of the capacitor voltage $v_C$ for time $t=0^{+}$ is $15\text{V}$ .

To determine

(b)

The capacitor voltage $v_C$ at $t=0^{+}$ .

Answer to Problem 5.71HP

The value for the voltage across the capacitor at time $t=20\mu \text{sec}$ is $\text{13}\text{.48}\text{V}$ .

Explanation of Solution

Calculation:

The conversion from $\mu \sec$ into $\sec$ is given by,

  $1\mu \sec =1\times {10}^{-6}\text{sec}$

The conversion from $20\mu \sec$ into $\sec$ is given by,

  $20\mu \sec =20\times {10}^{-6}\text{sec}$

The expression for the current flowing through the inductor for time $t=0^{-}$ is given by,

  $i_L\left(0^{-}\right)=0\text{A}$

The inductor opposes sudden change in the current, thus the current $i_L\left(0^{+}\right)$ is given by,

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

Substitute $0\text{A}$ for $i_L\left(0^{-}\right)$ in the above equation.

  $i_L\left(0^{+}\right)=0\text{A}$

Mark the values and redraw the diagram for time $t>0$ .

The required diagram is shown in Figure 3

 

Apply KVL in the above circuit.

  $\begin{array}{l}{i}_L\left(t\right)R+\frac{Ld{i}_L\left(t\right)}{dt}+\frac{1}{C}{\displaystyle {\int }_0^t{i}_L\left(t\right)dt}=0\\ \frac{L{d}^2{i}_L\left(t\right)}{d{t}^2}+\frac{Rd{i}_L\left(t\right)}{dt}+\frac{i_L\left(t\right)}{C}=0\\ \frac{LC{d}^2{i}_L\left(t\right)}{d{t}^2}+\frac{CRd{i}_L\left(t\right)}{dt}+i_L\left(t\right)=0\end{array}$   ........... (1)

The standard second order equation for the differential equation.

  $\frac{1}{{\omega }_n^2}\frac{d^{2}x\left(t\right)}{d{t}^2}+\frac{2\varsigma }{{\omega }_n}\frac{dx\left(t\right)}{dt}+x\left(t\right)=0$

From above and from equation (1), the angular frequency is derived as,

  $\begin{array}{l}\frac{1}{{\omega }_n^2}=LC\\ {\omega }_n=\frac{1}{\sqrt{LC}}\end{array}$

Substitute $20\times {10}^{-3}\text{H}$ for $L$ and $0.1\times {10}^{-6}\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}{\omega }_n=\frac{1}{\sqrt{\left(1\times {10}^{-3}\text{H}\right)\left(0.1\times {10}^{-6}\text{F}\right)}}\\ =22.36\times {10}^3\text{rad}/\text{s}\end{array}$

The expression for the damping coefficient is given by,

  $\varsigma =\frac{{\omega }_{n}RC}{2}$

Substitute $22.36\times {10}^3\text{rad}/\text{s}$ for ${\omega }_n$, $0.1\times {10}^{-6}\text{F}$ for $C$ and $200\Omega$ for $R$ in the above equation.

  $\begin{array}{c}\varsigma =\frac{\left(22.36\times {10}^3\text{rad}/\text{s}\right)\left(200\Omega \right)\left(0.1\times {10}^{-6}\text{F}\right)}{2}\\ =0.2236\end{array}$

The value of $\varsigma$ is less then $1$, thus the circuit is under damped.

The expression to calculate the damping frequency of the circuit is given by,

  ${\omega }_d={\omega }_n\sqrt{1-{\varsigma }^2}$

Substitute $22.36\times {10}^3\text{rad}/\text{s}$ for ${\omega }_n$ and $0.2236$ for $\varsigma$ in the above equation.

  $\begin{array}{c}{\omega }_d=\left(=22.36\times {10}^3\text{rad}/\text{s}\right)\sqrt{1-{\left(0.2236\right)}^2}\\ =21.793\times {10}^3\text{rad}/\text{s}\end{array}$

The expression for the output response of the capacitor is given by,

  $v_C\left(t\right)=e^{-\varsigma {\omega }_{n}t}\left({\alpha }_1\cos \left({\omega }_{d}t\right)+{\alpha }_2\sin \left({\omega }_{d}t\right)\right)$

Substitute $22.36\times {10}^3\text{rad}/\text{s}$ for ${\omega }_n$, $21.793\times {10}^3\text{rad}/\text{s}$ for ${\omega }_d$ and $0.2236$ for $\varsigma$ in the above equation.

  $\begin{array}{c}{v}_C\left(t\right)=\left[e^{-\left(0.2236\right)\left(22.36\times {10}^3\text{rad}/\text{s}\right)t}\left(\begin{array}{l}{\alpha }_1\cos \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)t\right)\\ +{\alpha }_2\sin \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)t\right)\end{array}\right)\right]\\ =\left[e^{-5000t}\left(\begin{array}{l}{\alpha }_1\cos \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)t\right)\\ +{\alpha }_2\sin \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)t\right)\end{array}\right)\right]\end{array}$   .......... (1)

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

  $\begin{array}{c}{v}_C\left(0\right)=\left[e^{-5000\left(0\right)}\left(\begin{array}{l}{\alpha }_1\cos \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)\left(0\right)\right)\\ +{\alpha }_2\sin \left(\left(21.793\times {10}^3\text{rad}/\text{s}\right)\left(0\right)\right)\end{array}\right)\right]\\ ={\alpha }_1\end{array}$

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

  ${\alpha }_1=15\text{V}$

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

  $\frac{d{v}_C\left(t\right)}{dt}=\left[\begin{array}{l}{e}^{-5000t}\left[\begin{array}{l}\left(-21.793\times {10}^3\right){\alpha }_1\sin \left(\left(21.793\times {10}^3\right)t\right)+\\ \left(21.793\times {10}^3\right){\alpha }_2\cos \left(\left(21.793\times {10}^3\right)t\right)\end{array}\right]-\\ 5000e^{-5000t}\left[{\alpha }_1\cos \left(21.793\times {10}^{3}t\right)+{\alpha }_2\sin \left(21.793\times {10}^{3}t\right)\right]\end{array}\right]$

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

  $\begin{array}{c}\frac{d{v}_C\left(0\right)}{dt}=\left[\begin{array}{l}{e}^{-5000\left(0\right)}\left[\begin{array}{l}\left(-21.793\times {10}^3\right){\alpha }_1\sin \left(\left(21.793\times {10}^3\right)\left(0\right)\right)+\\ \left(21.793\times {10}^3\right){\alpha }_2\cos \left(\left(21.793\times {10}^3\right)\left(0\right)\right)\end{array}\right]-\\ 5000e^{-5000t}\left[{\alpha }_1\cos \left(21.793\times {10}^3\left(0\right)\right)+{\alpha }_2\sin \left(21.793\times {10}^3\left(0\right)\right)\right]\end{array}\right]\\ =21.793\times {10}^3{\alpha }_2-5000{\alpha }_1\end{array}$   .......... (2)

The expression for the current through the inductor and the capacitor is same and is given by,

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

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

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

Substitute $0$ for $i_L\left(0\right)$ in the above equation.

  $\begin{array}{l}0=C\frac{d{v}_C\left(0\right)}{dt}\\ \frac{d{v}_C\left(0\right)}{dt}=0\end{array}$

Substitute $0$ for $\frac{d{v}_C\left(0\right)}{dt}$ and $15$ for ${\alpha }_1$ in equation (2).

  $\begin{array}{l}0=21.793\times {10}^3{\alpha }_2-5000\left(15\right)\\ {\alpha }_2=3.44\end{array}$

Substitute $3.44$ for ${\alpha }_2$ and $15$ for ${\alpha }_1$ in equation (2) in equation (1).

  $v_C\left(t\right)=e^{-5000t}\left(15\cos \left(\left(21.793\times {10}^3\right)t\right)+3.44\sin \left(\left(21.793\times {10}^3\right)t\right)\right)\text{V}$   .......... (3)

Substitute $20\times {10}^{-6}\text{sec}$ for $t$ in the above equation.

  $\begin{array}{c}{v}_C\left(20\times {10}^{-6}\text{sec}\right)=e^{-5000\left(21.793\times {10}^3\right)}\left(15\cos \left(\left(21.793\times {10}^3\right)\left(20\times {10}^{-6}\text{sec}\right)\right)+3.44\sin \left(\left(21.793\times {10}^3\right)\left(20\times {10}^{-6}\text{sec}\right)\right)\right)\text{V}\\ =0.9\left(15\left(0.999\right)+3.4\left(0\right)\right)\text{V}\\ \text{=13}\text{.48}\text{V}\end{array}$

Conclusion:

Therefore, the value for the voltage across the capacitor at time $t=20\mu \text{sec}$ is $\text{13}\text{.48}\text{V}$ .

To determine

(c)

The capacitor voltage $v_C$ at $t=\infty$ .

Answer to Problem 5.71HP

The final voltage across the capacitor for the time $t=\infty$ is $0\text{V}$ .

Explanation of Solution

Calculation:

For time $t=\infty$ the circuit is in steady state and the capacitor acts as short circuit, mark the values and redraw the circuit.

The required diagram is shown in Figure 4

 

From the above circuit, the circuit is source free and the final voltage across the capacitor is given by,

  $v_C\left(\infty \right)=0\text{V}$

Conclusion:

Therefore, the final voltage across the capacitor for the time $t=\infty$ is $0\text{V}$ .

To determine

(d)

The value of maximum capacitor voltage.

Answer to Problem 5.71HP

The value of maximum capacitor voltage is $\text{16}\text{.93}\text{V}$ .

Explanation of Solution

Calculation:

The maximum capacitor voltage is obtained by evaluating the expression for voltage across the capacitor equal to zero and is given by,

  $\frac{d{v}_C\left(t\right)}{dt}=\left[\begin{array}{l}{e}^{-5000t}\left[\begin{array}{l}\left(-21.793\times {10}^3\right){\alpha }_1\sin \left(\left(21.793\times {10}^3\right)t\right)+\\ \left(21.793\times {10}^3\right){\alpha }_2\cos \left(\left(21.793\times {10}^3\right)t\right)\end{array}\right]-\\ 5000e^{-5000t}\left[{\alpha }_1\cos \left(21.793\times {10}^{3}t\right)+{\alpha }_2\sin \left(21.793\times {10}^{3}t\right)\right]\end{array}\right]$

Substitute $0$ for $\frac{d{v}_C\left(t\right)}{dt}$ in the above equation.

  $\begin{array}{l}0=\left[\begin{array}{l}{e}^{-5000t}\left[\begin{array}{l}\left(-21.793\times {10}^3\right){\alpha }_1\sin \left(\left(21.793\times {10}^3\right)t\right)+\\ \left(21.793\times {10}^3\right){\alpha }_2\cos \left(\left(21.793\times {10}^3\right)t\right)\end{array}\right]-\\ 5000e^{-5000t}\left[{\alpha }_1\cos \left(21.793\times {10}^{3}t\right)+{\alpha }_2\sin \left(21.793\times {10}^{3}t\right)\right]\end{array}\right]\\ e^{-5000t}\left(-344.1\times {10}^3\sin \left(21.793\times {10}^{3}t\right)-32.08\cos \left(21.793\times {10}^{3}t\right)\right)=0\\ -344.1\times {10}^3\sin \left(21.793\times {10}^{3}t\right)-32.08\cos \left(21.793\times {10}^{3}t\right)=32.08\cos \left(21.793\times {10}^{3}t\right)\\ \frac{\sin \left(21.793\times {10}^{3}t\right)}{\cos \left(21.793\times {10}^{3}t\right)}=\frac{32.08}{344.1\times {10}^3}\end{array}$

Solve further as,

  $\begin{array}{l}\tan \left(21.793\times {10}^{3}t\right)={\tan }^{-1}\left(-9.32\times {10}^{-3}\right)\\ 21.793\times {10}^{3}t=-0.534\\ t=-24.5\times {10}^{-6}\text{sec}\end{array}$

Substitute $-24.5\times {10}^{-6}\text{sec}$ for $t$ in the above equation.

  $\begin{array}{c}{v}_C\left(-24.5\times {10}^{-6}\text{sec}\right)=e^{-5000\left(\left(-24.5\times {10}^{-6}\text{sec}\right)\right)}\left(\begin{array}{l}15\cos \left(\left(21.793\times {10}^3\right)\left(-24.5\times {10}^{-6}\text{sec}\right)\right)\\ +3.44\sin \left(\left(21.793\times {10}^3\right)\left(-24.5\times {10}^{-6}\text{sec}\right)\right)\end{array}\right)\text{V}\\ =\left(1.13\right)\left(15\left(0.999\right)+3.44\left(0\right)\right)\text{V}\\ =\text{16}\text{.93}\text{V}\end{array}$

Conclusion:

Therefore, the value of maximum capacitor voltage is $\text{16}\text{.93}\text{V}$ .