Chapter 2, Problem 2.4HP

The charge cycle shown in Figure P2.4 is an example of a three-rate charge. The current is held constant at 30 mA for 6 h. Then it is switched to 20 mA for the next 3 h. Find:
a. The total charge transferred Lo the battery.
b. The energy transferred to the battery.
Hint: Recall that energy w is the integral of power, or $P=dw\text{/}dt$ .

To determine

(a)

The total charge transferred to the battery.

Answer to Problem 2.4HP

The total charge delivered to the battery is $\text{864}\text{C}$ .

Explanation of Solution

Calculation:

The given diagram for the battery voltage graph is shown below.

The given diagram is shown in Figure 1

  

The given diagram for the battery current is shown in Figure 2

  

The conversion of $\text{h}$ into $\text{sec}$ is given by,

  $\text{1}\text{h}=3600\text{sec}$

The conversion of $\text{6}\text{h}$ into $\text{sec}$ is given by,

  $\begin{array}{r}\text{6}\text{h}=12\times 3600\text{sec}\\ =6\times 60\times 60\text{sec}\end{array}$

The conversion of $\text{9}\text{h}$ into $\text{sec}$ is given by,

  $\begin{array}{r}\text{9}\text{h}=9\times 3600\text{sec}\\ =9\times 60\times 60\text{sec}\end{array}$

The conversion of $\text{mA}$ into $\text{A}$ is given by,

  $1\text{mA}={\text{10}}^{-3}\text{A}$

The conversion of $\text{40}\text{mA}$ into $\text{A}$ is given by,

  $\text{30}\text{mA}=30\times {\text{10}}^{-3}\text{A}$

The conversion of $\text{20}\text{mA}$ into $\text{A}$ is given by,

  $\text{20}\text{mA}=20\times {\text{10}}^{-3}\text{A}$

The expression for the current for the entire cycle is given by,

  $i=\{\begin{array}{l}30\text{mA}\text{0<t<6}\text{h}\\ 20\text{mA}6\text{h<t<9}\text{h}\end{array}$

The formula for the charge delivered to the battery is calculated as,

  $\begin{array}{c}Q={\displaystyle \underset{0}{\overset{6\times 60\times 60\text{sec}}{\int }}30\times {\text{10}}^{-3}dt}+{\displaystyle \underset{6\times 60\times 60\text{sec}}{\overset{9\times 60\times 60\text{sec}}{\int }}20\times {\text{10}}^{-3}dt}\\ =30\times {\text{10}}^{-3}{\left[t\right]}_0^{6\times 60\times 60}+20\times {\text{10}}^{-3}{\left[t\right]}_{6\times 60\times 60}^{9\times 60\times 60}\\ =\left(30\times {\text{10}}^{-3}\right)\left(6\times 60\times 60\right)+\left(20\times {\text{10}}^{-3}\right)\left(9\times 60\times 60-6\times 60\times 60\right)\\ =648+216\end{array}$

Solve further as,

  $Q=864\text{C}$

Conclusion:

Therefore, the total charge delivered to the battery is $\text{864}\text{C}$ .

To determine

(b)

The energy that is transferred to the battery.

Answer to Problem 2.4HP

The total power delivered to the battery is $\mathrm{70.6.452}\text{J}$ .

Explanation of Solution

Calculation:

The line equations for the points that joins the point $\left(0,0.5\right)\text{and}\left(3,0.6\right)$ in graph shown in Figure 2 is given by,

  $\begin{array}{l}v-0.5=\frac{0.6-0.5}{3\times 60\times 60-0}\left(t-0\right)\\ v-0.5=\frac{0.1}{10800}t\\ v=9.26\times {10}^{-6}t+0.5\text{V}\end{array}$

The line equations for the points that joins the point $\left(3,0.6\right)\text{and}\left(6,1.2\right)$ in graph shown in Figure 2 is given by,

  $\begin{array}{l}v-0.6=\frac{1.2-0.6}{6\times 60\times 60-3\times 60\times 60}\left(t-3\times 60\times 60\right)\\ v-0.6=\frac{0.6}{10800}\left(t-3\times 60\times 60\right)\\ v=5.55\times {10}^{-5}t-0.6\text{+0}\text{.6}\\ v=5.55\times {10}^{-5}t\text{V}\end{array}$

The line equations for the points that joins the point $\left(6,0.5\right)\text{and}\left(9,1.7\right)$ in graph shown in Figure 2 is given by,

  $\begin{array}{l}v-0.5=\frac{1.7-0.5}{9\times 60\times 60-6\times 60\times 60}\left(t-6\times 60\times 60\right)\\ v-0.5=\frac{1.7-0.5}{10800}\left(t-6\times 60\times 60\right)\\ v-0.5=1.11\times {10}^{-4}t-2.4+0.5\\ v=1.11\times {10}^{-4}t-1.9\text{V}\end{array}$

The evaluated value of the battery voltage for different cycles is shown below,

  $v=\{\begin{array}{l}9.26\times {10}^{-6}t+0.5\text{V}0\text{<}t<3\text{h}\\ 5.55\times {10}^{-5}t\text{V}3\text{h<}t<6\text{h}\\ 1.11\times {10}^{-4}t-1.9\text{V}6\text{h<}t<9\text{h}\end{array}$

The formula to calculate power supplied by the battery is given by,

  $p=vi$ ........ (1)

Substitute $30\times {10}^{-3}\text{A}$ for $i$ and $9.26\times {10}^{-6}t+0.5$ for $v$ in the above equation.

  $\begin{array}{c}p=\left(9.26\times {10}^{-6}t+0.5\right)\left(30\times {10}^{-3}\text{A}\right)\\ =2.778\times {10}^{-7}t+15\times {10}^{-3}\text{W}\end{array}$

Substitute $30\times {10}^{-3}\text{A}$ for $i$ and $5.55\times {10}^{-5}t\text{V}$ for $v$ in equation (1)

  $\begin{array}{c}p=\left(5.55\times {10}^{-5}t\text{V}\right)\left(30\times {10}^{-3}\text{A}\right)\\ =1.665\times {10}^{-6}t\text{W}\end{array}$

Substitute $30\times {10}^{-3}\text{A}$ for $i$ and $1.11\times {10}^{-4}t-1.9\text{V}$ for $v$ in the equation (1)

  $\begin{array}{c}p=\left(1.11\times {10}^{-4}t-1.9\text{V}\right)\left(30\times {10}^{-3}\text{A}\right)\\ =2.22\times {10}^{-6}t-38\times {10}^{-3}\text{W}\end{array}$

The formula for the relation between the power delivered and the energy stored is given by,

  $\omega ={\displaystyle \underset{0}{\overset{9\text{h}}{\int }}p\left(t\right)}dt$

The powered delivered to the battery is calculated as,

  $\begin{array}{c}\omega =\left[\begin{array}{l}{\displaystyle \underset{0}{\overset{3\times 60\times 60\text{sec}}{\int }}\left(2.778\times {10}^{-7}t+15\times {10}^{-3}\right)}dt+{\displaystyle \underset{3\times 60\times 60\text{sec}}{\overset{6\times 60\times 60\text{sec}}{\int }}\left(1.665\times {10}^{-6}t\right)}dt\\ +{\displaystyle \underset{6\times 60\times 60\text{sec}}{\overset{9\times 60\times 60\text{sec}}{\int }}\left(2.22\times {10}^{-6}t-38\times {10}^{-3}\text{W}\right)dt}\end{array}\right]\\ =\left[\begin{array}{l}2.778\times {10}^{-7}{\left[\frac{t^2}{2}\right]}_0^{3\times 60\times 60}+15\times {10}^{-3}{\left[t\right]}_0^{3\times 60\times 60}+1.665\times {10}^{-6}{\left[\frac{t^2}{2}\right]}_{3\times 60\times 60}^{6\times 60\times 60}\\ +2.22\times {10}^{-6}{\left[\frac{t^2}{2}\right]}_{6\times 60\times 60}^{9\times 60\times 60}-38\times {10}^{-3}{\left[t\right]}_{6\times 60\times 60}^{9\times 60\times 60}\end{array}\right]\\ =16.2+162+291.3+647.352-410.4\\ =\mathrm{70.6.452}\text{J}\end{array}$

Conclusion:

Therefore, the total power delivered to the battery is $\mathrm{70.6.452}\text{J}$ .