Chapter 4, Problem 4.9HP

To determine

(a)

The energy stored in the capacitor.

Answer to Problem 4.9HP

The energy stored by the capacitor for different time interval is $w_C\left(t\right)=\{\begin{array}{l}0-\infty <t<0\\ 1.6t^{2}0\le t<10\text{s}\\ 0.025t^2-0.2t+0.4\text{J}4\le t<10\text{s}\\ 010\text{s}\le t<\infty \end{array}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression for the energy stored by the capacitor is given by,

  $w_C\left(t\right)=\frac{1}{2}C{v}^2\left(t\right)$

Substitute $0.2\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}{w}_C\left(t\right)=\frac{1}{2}\left(0.2\text{F}\right)v^2\left(t\right)\\ =0.1v^2\left(t\right)\end{array}$ ........ (1)

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

  $\begin{array}{c}{w}_C\left(t\right)=0.1{\left(0\right)}^2\\ =0\text{J}\end{array}$

Substitute $4t$ for $v\left(t\right)$ in equation (1)

  $\begin{array}{c}{w}_C\left(t\right)=0.1{\left(4t\right)}^2\\ =1.6t^2\text{J}\end{array}$

Substitute $2-0.5t$ for $v\left(t\right)$ in equation (1)

  $\begin{array}{c}{w}_C\left(t\right)=0.1{\left(2-0.5t\right)}^2\\ =0.025t^2-0.2t+0.4\text{J}\end{array}$

Substitute $0$ for $v\left(t\right)$ in equation (1)

  $\begin{array}{c}{w}_C\left(t\right)=0.1{\left(0\right)}^2\\ =0\text{J}\end{array}$

The energy stored by the capacitor for different time interval is given by,

  $w_C\left(t\right)=\{\begin{array}{l}0-\infty <t<0\\ 1.6t^{2}0\le t<10\text{s}\\ 0.025t^2-0.2t+0.4\text{J}4\le t<10\text{s}\\ 010\text{s}\le t<\infty \end{array}$

Conclusion:

Therefore, the energy stored by the capacitor for different time interval is $w_C\left(t\right)=\{\begin{array}{l}0\text{J}-\infty <t<0\\ 1.6t^{2}0\le t<10\text{s}\\ 0.025t^2-0.2t+0.4\text{J}4\le t<10\text{s}\\ 0\text{J}10\text{s}\le t<\infty \end{array}$ .

To determine

(b)

The energy delivered by the source.

Answer to Problem 4.9HP

The expression for the energy delivered by the source for different time interval is $\{\begin{array}{l}0\text{J}-\infty <t<0\\ 1.6t^2+\frac{4t^3}{3}\text{J}0\le t<10\text{s}\\ 0.02t^3-0.025t^2+t+0.44\le t<10\text{s}\\ 0\text{J}10\text{s}\le t<\infty \end{array}$

Explanation of Solution

Calculation:

The expression for the energy delivered by the source is given by,

  $w\left(t\right)=w_C\left(t\right)+\frac{1}{R}{\displaystyle \int v^2\left(t\right)dt}$

Substitute $4\Omega$ for $R$ in the above equation.

  $w\left(t\right)=w_C\left(t\right)+\frac{1}{4\Omega }{\displaystyle \int v^2\left(t\right)dt}$ ........(2)

Substitute $0$ for $v\left(t\right)$ and $0\text{J}$ for $w_C\left(t\right)$ in the above equation.

  $\begin{array}{c}w\left(t\right)=0\text{J}+\frac{1}{4}{\displaystyle \int {\left(0\right)}^{2}dt}\\ =0\text{J}\end{array}$

Substitute $4t$ for $v\left(t\right)$ and $1.6t^2$ for $w_C\left(t\right)$ equation (2)

  $\begin{array}{c}w\left(t\right)=1.6t^2+\frac{1}{4}{\displaystyle \int {\left(4t\right)}^{2}dt}\\ =1.6t^2+\frac{4t^3}{3}\text{J}\end{array}$

Substitute $2-0.5t$ for $v\left(t\right)$ and $0.025t^2-0.2t+0.4\text{J}$ for $w_C\left(t\right)$ equation (2)

  $\begin{array}{c}w\left(t\right)=0.025t^2-0.2t+0.4\text{J}+\frac{1}{4}{\displaystyle \int {\left(2-0.5t\right)}^{2}dt}\\ =0.025t^2-0.2t+0.4\text{J}+\frac{1}{4}{\displaystyle \int {\displaystyle {\int }_{4\text{s}}^{10\text{s}}\left(0.25t^2-2t+4\right)}dt}\\ =0.025t^2-0.2t+0.4\text{J}+\frac{1}{4}\left(\frac{0.25t^2}{2}-t^2+4t\right)\\ =0.02t^3-0.025t^2+t+0.4\end{array}$

Substitute $0$ for $v\left(t\right)$ and $0\text{J}$ for $w_C\left(t\right)$ equation (2)

  $\begin{array}{c}w\left(t\right)=0\text{J}+\frac{1}{2}{\displaystyle \int {\left(0\right)}^{2}dt}\\ =0\text{J}\end{array}$

Conclusion:

Therefore, the expression for the energy delivered by the source for different time interval is $\{\begin{array}{l}0\text{J}-\infty <t<0\\ 1.6t^2+\frac{4t^3}{3}\text{J}0\le t<10\text{s}\\ 0.02t^3-0.025t^2+t+0.44\le t<10\text{s}\\ 0\text{J}10\text{s}\le t<\infty \end{array}$