Chapter 8, Problem 48E

(a)

To determine

Find the expression for $v_C$ valid for all values of $t$.

Answer to Problem 48E

The expression for $v_C$ valid for all values of $t$ is $1-e^{\frac{-t}{6\text{66}\text{.67 ns}}}\text{ V}$.

Explanation of Solution

Formula used:

The expression for the equivalent resistor when resistors are connected in parallel is as follows:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\mathrm{......}+\frac{1}{R_N}$ (1)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The expression for the time constant is as follows:

$\tau =R_{eq}C$ (2)

Here,

$\tau$ is the time constant,

$R_{eq}$ is the equivalent resistance across the capacitor and

$C$ is the capacitance of $\text{1 nF}$ capacitor.

The expression for the final response is as follows:

$v_C(t)=v_C(\infty )+\left(v_C(0)-v_C(\infty )\right)e^{\frac{-t}{\tau }}$ (3)

Here,

$v_C(t)$ is the voltage across the capacitor,

$v_C(\infty )$ is the voltage across the capacitor at $t=\infty$,

$v_C(0)$ is the voltage across the capacitor at $t=0$ and

$t$ is the time.

Calculation:

The redrawn circuit diagram is given in Figure 1.

Refer to the redrawn Figure 1:

The given circuit is connected for long time,

So, capacitor behaves as open circuit.

The independent voltage source is unit step function.

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 argument of unit step function becomes zero.

The independent voltage source is:

$v=3u\left(t\right)\text{ V}$ (4)

Substitute $0^{-}$ for $t$ in equation (4).

$\begin{array}{c}v=3u\left(0^{-}\right)\text{ V}\\ =\text{0 V}\end{array}$

Apply KCL at node 1.

$\frac{v_1-v}{R_1}+\frac{v_1}{R_2}=\text{0 A}$ (5)

Here,

$v_1$ is the voltage at node 1,

$v$ is the voltage of $3u\left(t\right)\text{ V}$ independent source,

$R_1$ is the resistance of $2\text{ k}\Omega$ resistor and

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

Substitute $\text{0 V}$ for $v$, $500\Omega$ for $R_1$ and $2\text{ k}\Omega$ for $R_2$ in equation (5).

$\begin{array}{l}\frac{v_1-0\text{ V}}{2\text{ k}\Omega }+\frac{v_1}{\text{1 k}\Omega }=0\text{ A}\\ \frac{v_1}{2\text{ k}\Omega }+\frac{v_1}{\text{1 k}\Omega }=0\text{ A}\\ \frac{v_1+2v_1}{\text{2 k}\Omega }=0\text{ A}\end{array}$

Rearrange for $v_1$.

$\begin{array}{l}{v}_1+2v_1=0\text{ V}\\ 3v_1=0\text{ V}\\ v_1=0\text{ V}\end{array}$

So, at $t<0$ the voltage $v_C(t)$ across the capacitor is $0\text{ V}$.

The voltage across capacitorat $t=0$ is,

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

So,

$v_C\left(0\right)=0\text{ V}$

Substitute $0^{+}$ for $t$ in equation (4).

$\begin{array}{c}v=3u\left(0^{+}\right)\text{ V}\\ =3\left(1\right)\text{ V}\\ =3\text{ V}\end{array}$

So, the value of the voltage across $3u\left(t\right)\text{ V}$ independent source for $t>0$ is

$3\text{ V}$.

Substitute $3\text{ V}$ for $v$, $2\text{ k}\Omega$ for $R_1$ and $1\text{ k}\Omega$ for $R_2$ in equation (5).

$\begin{array}{l}\frac{v_1-3\text{ V}}{2\text{ k}\Omega }+\frac{v_1}{\text{1 k}\Omega }=0\text{ A}\\ \frac{v_1-3\text{ V}}{\text{2}\times {10}^3\Omega }+\frac{v_1}{1\times {10}^3\Omega }=0\text{ A }\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ \frac{v_1-3\text{ V}+2v_1}{2\times {10}^3\Omega }=0\text{ A}\end{array}$

Rearrange for $v_1$.

$\begin{array}{l}{v}_1-3\text{ V}+2v_1=0\text{ V}\\ 3v_1-3\text{ V}=0\text{ V}\\ 3v_1=3\text{ V}\\ v_1=1\text{ V}\end{array}$

So, the voltage $v_C\left(\infty \right)$ across capacitor at $t=\infty$ is $1\text{ V}$.

The redrawn circuit diagram is given in Figure 2.

Refer to the redrawn Figure 2:

$2\text{ k}\Omega$ and $1\text{ k}\Omega$ resistor are connected in parallel.

So, form equation (1),

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{2\text{ k}\Omega }+\frac{1}{\text{1 k}\Omega }\\ =\frac{1}{2\times {10}^3\Omega }+\frac{1}{1\times {10}^3\Omega }\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =\frac{1+2}{2\times {10}^3\Omega }\\ =\frac{3}{2\times {10}^3\Omega }\end{array}$

Rearrange the equation for $R_{eq}$.

$\begin{array}{c}{R}_{eq}=\frac{2\times {10}^3}{3}\Omega \\ =\text{666}\text{.67 }\Omega \end{array}$

The simplified diagram is shown in Figure 3.

Refer to the redrawn Figure. 3:

Substitute $\text{666}\text{.67 }\Omega$ for $R_{eq}$ and $\text{1 nF}$ for $C$ in the equation (2).

$\begin{array}{c}\tau =\left(\text{666}\text{.67 }\Omega \right)\left(\text{1 nF}\right)\\ =\left(\text{666}\text{.67 }\Omega \right)\left(1\times {10}^{-9}\text{ F}\right)\left\{\because 1\mu \text{F}={10}^{-6}\text{ F}\right\}\\ =6.6667\times {10}^{-7}\text{ s}\\ =6\text{66}\text{.67}\times {10}^{-9}\text{ s }\end{array}$

$\tau =6\text{66}\text{.67 ns }\left\{\because {10}^{-9}\text{ s}=1\text{ ns }\right\}$

Substitute $6\text{66}\text{.67 ns}$ for $\tau$ and $\text{1 V}$ for $v_C(\infty )$ and $\text{0 V}$$v_C(0)$ in the equation (3).

$\begin{array}{c}{v}_C\left(t\right)=\text{1 V}+\left(\text{0 V}-1\text{ V}\right)e^{\frac{-t}{6\text{66}\text{.67 ns}}}\\ =\text{1 V}-\text{1 V}{e}^{\frac{-t}{6\text{66}\text{.67 ns}}}\end{array}$

$v_C\left(t\right)=1-e^{\frac{-t}{6\text{66}\text{.67 ns}}}\text{ V}$ (5)

So, the expression for the voltage across the capacitor valid for all values of $t$ is.

$v_C(t)=1-e^{\frac{-t}{6\text{66}\text{.67 ns}}}\text{ V}$

Conclusion:

Thus, the expression for $v_C$ valid for all values of $t$ is $1-e^{\frac{-t}{6\text{66}\text{.67 ns}}}\text{ V}$.

(b)

To determine

Sketch $v_C(t)$ over the range $0\le t\le 4\mu \text{s}$

Explanation of Solution

Given Data:

The range for the time is $0\le t\le 4\mu \text{s}$.

Calculation:

Substitute $0\text{ s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(0\text{ s}\right)=1-e^{\frac{-0\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ \text{=}\left(\text{1}-1\right)\text{}\text{ V}\\ =\text{0 V}\end{array}$

Substitute $0.25\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(0.25\mu \text{s}\right)=1-e^{\frac{-0.25\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-0.25\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-0.25\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.68729\right)\text{ V}\end{array}$

$v_C\left(0.25\mu \text{s}\right)=0.31271\text{ V}$

Substitute $0.5\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(0.5\mu \text{s}\right)=1-e^{\frac{-0.5\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-0.5\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-0.5\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.472368\right)\text{ V}\end{array}$

$v_C\left(0.5\mu \text{s}\right)=0.52763\text{ V}$

Substitute $0.75\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(0.75\mu \text{s}\right)=1-e^{\frac{-0.75\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-0.75\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-0.75\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.32465\right)\text{ V}\end{array}$

$v_C\left(0.75\mu \text{s}\right)\text{ = }0.67534\text{ V}$

Substitute $1\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(1\mu \text{s}\right)=1-e^{\frac{-1\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-1\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-1\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.223132\right)\text{ V}\end{array}$

$v_C\left(1\mu \text{s}\right)=0.77687\text{ V}$

Substitute $1.5\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(1.5\mu \text{s}\right)=1-e^{\frac{-1.5\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-1.5\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-1.5\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.1054\right)\text{ V}\end{array}$

$v_C\left(1.5\mu \text{s}\right)=0.894599\text{ V}$

Substitute $2\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(2\mu \text{s}\right)=1-e^{\frac{-2\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-2\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-2\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.0497878\right)\text{ V}\end{array}$

$v_C\left(2\mu \text{s}\right)=0.95021\text{ V}$

Substitute $3\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(3\mu \text{s}\right)=1-e^{\frac{-3\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-3\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-3\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(0.01111\right)\text{ V}\end{array}$

$v_C\left(3\mu \text{s}\right)=0.98889\text{ V}$

Substitute $4\mu \text{s}$ for $t$ in equation (5).

$\begin{array}{c}{v}_C\left(4\mu \text{s}\right)=1-e^{\frac{-4\mu \text{s}}{6\text{66}\text{.67 ns}}}\text{ V}\\ =1-e^{\frac{-4\times {10}^{-6}\text{ s}}{6\text{66}\text{.67 ns}}}\text{ V }\left\{\because 1\mu \text{s}={10}^{-6}\text{ s}\right\}\\ =1-e^{\frac{-4\times {10}^{-6}\text{ s}}{6\text{66}\text{.67}\times {\text{10}}^{-9}\text{ s}}}\text{ V }\left\{\because 1\text{ ns}={10}^{-9}\text{ s}\right\}\\ =1-\left(2.4788\times {10}^{-3}\right)\text{ V}\end{array}$

$v_C\left(4\mu \text{s}\right)=0.997521\text{ V}$

The different values of the $v_c\left(t\right)$ over the range $0\le t\le 4\mu \text{s}$ is given in the Table 1:

$\begin{array}{l}\begin{array}{cc}t\left(\mu \text{s}\right)& v_C\left(t\right)\left(\text{V}\right)\\ 0& 0\\ 0.25& 0.31271\\ 0.5& 0.52763\\ 0.75& 0.67534\\ 1& 0.77687\\ 1.5& 0.894599\\ 2& 0.95021\\ 3& 0.98889\\ 4& 0.99752\end{array}\\ \text{ Table 1}\end{array}$

The graph for $v_C(t)$ over the range $0\le t\le 4\mu \text{s}$ is shown in the Figure (4):

Conclusion:

Thus, the graph for $v_C(t)$ over the range $0\le t\le 4\mu \text{s}$ is plotted in the Figure 4.