Chapter 7, Problem 11E

(a)

To determine

Sketch the voltage waveform of capacitor.

Explanation of Solution

Given data:

Value of capacitor is $1\text{mF}$.

Formula used:

Refer to FIGURE 7.44 in the textbook.

The expression for the voltage across the capacitor is:

$v_c\left(t\right)=\frac{1}{C}{\displaystyle {\int }_{-T}^T{i}_c\left(t\right)dt}$ (1)

Here,

$v_c\left(t\right)$ is the voltage across capacitor and

$i_c\left(t\right)$ is the capacitor current.

Calculation:

Refer to the FIGURE 7.43 (a).

Substitute the limits $0\le t<\infty$ in equation (1).

$v_c(t)=\frac{1}{C}{\displaystyle {\int }_0^{\infty }{i}_c\left(t\right)dt}$ (2)

The equation for the current waveform for time period $-1\le T<\infty$ is:

$i_c\left(t\right)=\left\{\begin{array}{l}\text{4 A , 0}\le T<0.2\\ \text{4 A , 0}\text{.2}\le T<0.4\\ 0\text{A , 0}\text{.4}\le T<0.8\\ \text{4 A , 0}\text{.8}\le T<1.0\\ \text{4 A , 1}\text{.0}\le T<1.2\\ 0\text{A, }T\ge 1.2\end{array}\right\}$ (3)

Substitute $4\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $\text{0}\le T<0.2$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_1=\frac{1}{1\text{mF}}{\displaystyle \int \left(4\text{A}\right)dt}\\ =\frac{4\text{A}}{1\times {10}^{-3}\text{F}}{\displaystyle \int dt}\end{array}$

$v_c{\left(t\right)}_1=4000t+v_0$ (4)

Here,

$v_0$ is the constant.

Substitute $0\text{ s}$ for $t$ in equation (4).

$\begin{array}{c}{v}_c\left(0\right)=0+v_0\\ =v_0\end{array}$

Since initial value of voltage at $t=0\text{s}$ is zero,

Therefore,

$v_0=0\text{ V}$

Substitute $0\text{ V}$ for $v_0$ in equation (4).

$v_c{\left(t\right)}_1=4000t+0$

$v_c{\left(t\right)}_1=4000t\text{ V}$ (5)

Substitute $8\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $\text{0}\text{.2}\le T<0.4$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_2=\frac{1}{1\text{mF}}{\displaystyle \int \left(8\text{A}\right)dt}\\ =\frac{8\text{A}}{1\times {10}^{-3}\text{F}}{\displaystyle \int dt}\end{array}$

$v_c{\left(t\right)}_2=8000t+v_0$ (6)

Substitute $\text{0}\text{.2 s}$ for $t$ in equation (6).

$\begin{array}{l}{v}_c{\left(0.2\text{s}\right)}_2=8000\times 0.2\text{s}+v_0\\ v_c{\left(0.2\text{s}\right)}_2=1600+v_0\end{array}$

Since initial value of voltage at $t=0.2\text{s}$ is $4000t\text{ V}$,

Therefore,

$\begin{array}{l}4000\times 0.2\text{ V = 1600}+v_0\\ 800\text{V= 1600}+v_0\\ v_0=-800\text{V}\end{array}$

Substitute $-800\text{ V}$ for $v_0$ in equation (6).

$v_c{\left(t\right)}_2=8000t-800\text{V}$ (7)

Substitute $\text{0}\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $\text{0}\text{.4}\le T<0.8$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_3=\frac{1}{1\text{mF}}{\displaystyle \int \left(0\text{A}\right)dt}+v_0\\ =0\text{V}+v_0\end{array}$

Substitute $0.4\text{ s}$ for $t$ in equation (7).

$\begin{array}{c}{v}_c\left(0.4\right)=8000\times 0.4-800\text{V}\\ =2\text{400 V}\end{array}$

Substitute $2400\text{ V}$ for $v_0$ in equation for $v_C{\left(t\right)}_3$.

$v_C{\left(t\right)}_3=2400\text{ V}$

Substitute $4\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $\text{0}\text{.8}\le T<1.0$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_4=\frac{1}{1\text{mF}}{\displaystyle \int \left(4\text{A}\right)dt}\\ =\frac{4\text{A}}{1\times {10}^{-3}\text{F}}{\displaystyle \int dt}\end{array}$

$v_c{\left(t\right)}_4=4000t+v_0$ (8)

Substitute $\text{0}\text{.8 s}$ for $t$ in equation (8).

$v_c{\left(0.8\text{s}\right)}_4=4000\times 0.8\text{s}+v_0$

Since initial value of voltage at $t=0.8\text{s}$ is $2400\text{V}$,

$\begin{array}{l}2400\text{V}=4000\times 0.8\text{s}+v_0\\ 2400\text{V}=3200+v_0\\ v_0=-800\text{V}\end{array}$

Substitute $-800\text{ V}$ for $v_0$ in equation (8).

$v_c{\left(t\right)}_4=4000t-800\text{ V}$

Substitute $8\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $\text{1}\text{.0}\le T<1.2$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_5=\frac{1}{1\text{mF}}{\displaystyle \int \left(8\text{A}\right)dt}\\ =\frac{8\text{A}}{1\times {10}^{-3}\text{F}}{\displaystyle \int dt}\end{array}$

$v_c{\left(t\right)}_5=8000t+v_0$ (9)

Substitute $\text{1 s}$ for $t$ in equation (9).

$\begin{array}{c}{v}_c{\left(1\text{s}\right)}_5=8000\times 1\text{s}+v_0\\ =8000+v_0\end{array}$

Since initial value of voltage at $t=1\text{s}$ is $4000t-800\text{ V}$,

$\begin{array}{l}4000\times 1\text{s}-800\text{ V}=8000+v_0\\ v_0=-4800\text{V}\end{array}$

Substitute $-4800\text{V}$ for $v_0$ in equation (9).

$v_c{\left(t\right)}_5=8000t-4800\text{V}$ (10)

Substitute $\text{0}\text{A}$ for $i_c\left(t\right)$ and $1\text{mF}$ for $C$, for $T\ge 1.2$ in equation (2).

$\begin{array}{c}{v}_c{\left(t\right)}_6=\frac{1}{1\text{mF}}{\displaystyle \int \left(0\text{A}\right)dt}+v_0\\ =0\text{V}+v_0\end{array}$

Substitute $\text{1}\text{.2 s}$ for $t$ in equation (10).

$\begin{array}{c}{v}_c\left(\text{1}\text{.2 s}\right)=8000\times \text{1}\text{.2}-4800\text{V}\\ =48\text{00 V}\end{array}$

Substitute $48\text{00 V}$ for $v_0$ in equation for $v_C{\left(t\right)}_6$.

$v_C{\left(t\right)}_6=48\text{00 V}$

The expression for the voltage across capacitor for time period $0\le T<\infty$ is:

$v_C\left(t\right)=v_C{\left(t\right)}_1+v_C{\left(t\right)}_2+v_C{\left(t\right)}_3+v_C{\left(t\right)}_4+v_C{\left(t\right)}_5+v_C{\left(t\right)}_6$ (11)

Substitute $4000t\text{ V}$ for $v_c{\left(t\right)}_1$, $8000t-800\text{V}$ for $v_c{\left(t\right)}_2$, $2400\text{V}$ for $v_c{\left(t\right)}_3$, $4000t-800\text{ V}$ for $v_c{\left(t\right)}_4$, $8000t-4800\text{V}$ for $v_c{\left(t\right)}_5$ and $4800\text{V}$ for $v_c{\left(t\right)}_6$ in equation (11).

$\begin{array}{c}{v}_C\left(t\right)=4000t\text{ V}+8000t-800\text{ V}+2400\text{V}+4000t-800\text{ V}+8000t-4800\text{V}+4800\text{V}\\ =\text{3200}t+800\text{V}\end{array}$

The labeled sketch of voltage waveform for $0\le T<\infty$ time period and voltage $v_C\left(t\right)=3200t+800\text{ V}$ for $0\le T\le 1.2$$s$ is shown in Figure 1:

Conclusion:

Thus, the resulting voltage waveform is sketched.

(b)

To determine

Find the voltage of capacitor.

Answer to Problem 11E

The voltage of capacitor for given time duration is $800\text{V}$, $2400\text{V}$ and $4800\text{V}$.

Explanation of Solution

Given Data:

Value of capacitor is $1\text{mF}$ and

Value of time duration for capacitor voltage is $200\text{ms}$, $600\text{ms}$ and $1.2s$.

Formula used:

The expression for the voltage at given point of time is,

$v_c=\frac{i\left(t\right)\times t\left(\text{s}\right)}{C\left(\text{mF}\right)}$ (12)

Calculation:

Substitute $200\text{ms}$ for $T$ in equation (12).

$\begin{array}{c}{v}_c\left(t\right)=\frac{4\text{A}\times \text{0}\text{.2}s}{1\text{ms}}\\ =\frac{0.8}{1\times {10}^{-3}s}\\ =800\text{V}\end{array}$

Substitute $600\text{ms}$ for $T$ in equation (12).

$\begin{array}{c}{v}_c\left(t\right)=\frac{4\text{A}\times 0.6s}{1\text{ms}}\\ =\frac{2.4}{1\times {10}^{-3}s}\\ =2400\text{V}\end{array}$

Substitute $1.2\text{s}$ for $T$ in equation (12).

$\begin{array}{c}{v}_c\left(t\right)=\frac{4\text{A}\times 1.2s}{1\text{ms}}\\ =\frac{4.8}{1\times {10}^{-3}s}\\ =4800\text{V}\end{array}$

Conclusion:

Thus, the voltage of capacitor for given time duration is $800\text{V}$, $2400\text{V}$ and $4800\text{V}$.