Chapter 2, Problem 2.8HP

A car battery kept in storage in the basement needs recharging, lf he voltage and the current provided bythe charger during a charge cycle are shown in Figure P2.8,
a. Find the total charge transferred to the battery.
b. Find the total energy transferred to the battery.

To determine

(a)

The total charge transferred to the battery.

Answer to Problem 2.8HP

The total charge delivered to the battery is $-174.862\times {10}^6\text{C}$ .

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 current equation for the time interval $\left(0<t<0.5\text{hr}\right)$ is calculated as,

  $\begin{array}{l}\frac{0.5-0}{t-0}=\frac{6-10}{i_1\left(t\right)-10}\\ i_1\left(t\right)=10-8t\end{array}$

The current equation for the time interval $\left(0.5\text{hr}<t<2\text{hr}\right)$ is calculated as,

  $\begin{array}{l}\frac{2-0.5}{t-0.5}=\frac{4-6}{i_2\left(t\right)-6}\\ i_2\left(t\right)=\frac{20}{3}-\frac{4t}{3}\end{array}$

The current equation for the time interval $\left(2\text{hr}<t<3\text{hr}\right)$ is calculated as,

  $\begin{array}{l}\frac{3-2}{t-2}=\frac{0-4}{i_3\left(t\right)-4}\\ i_3\left(t\right)=-4t+12\end{array}$

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

  $\text{1}\text{h}=3600\text{sec}$

The conversion of $0.5\text{h}$ into $\text{sec}$ is given by,

  $\begin{array}{c}0.5\text{h}=0.5\times 3600\text{sec}\\ =1800\text{sec}\end{array}$

The conversion of $2\text{h}$ into $\text{sec}$ is given by,

  $\begin{array}{c}2\text{h}=2\times 3600\text{sec}\\ =7200\text{sec}\end{array}$

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

  $\begin{array}{c}3\text{h}=3\times 3600\text{sec}\\ =10800\text{sec}\end{array}$

The charge delivered to the battery is given by,

  $Q={\displaystyle \underset{0}{\overset{1800}{\int }}{i}_1\left(t\right)dt}+{\displaystyle \underset{1800}{\overset{7200}{\int }}{i}_2\left(t\right)dt}+{\displaystyle \underset{7200}{\overset{10800}{\int }}{i}_3\left(t\right)dt}$

Substitute $10-8t$ for $i_1\left(t\right)$, $\frac{20}{3}-\frac{4t}{3}$ for $i_2\left(t\right)$ and $\left(12-4t\right)$ for $i_3\left(t\right)$ in the above equation.

  $\begin{array}{c}Q={\displaystyle \underset{0}{\overset{1800}{\int }}\left(10-8t\right)dt}+{\displaystyle \underset{1800}{\overset{7200}{\int }}\left(\frac{20}{3}-\frac{4t}{3}\right)dt}+{\displaystyle \underset{7200}{\overset{10800}{\int }}\left(12-4t\right)dt}\\ ={\left[10t-4t^2\right]}_0^{1800}+{\left[\frac{20}{3}-\frac{4t}{3}\right]}_{1800}^{7200}+{\left[12-4t\right]}_{7200}^{10800}\\ =\frac{2}{3}\left({7200}^2-{1800}^2\right)+12\left(10800-7200\right)-2\left({10800}^2-{7200}^2\right)\\ =-174.862\times {10}^6\text{C}\end{array}$

Conclusion:

Therefore, the total charge delivered to the battery is $-174.862\times {10}^6\text{C}$ .

To determine

(b)

The energy that is transferred to the battery.

Answer to Problem 2.8HP

The total energy delivered to the battery is $-1.3618\times {10}^{12}\text{J}$ .

Explanation of Solution

Calculation:

The voltage equation of $v\left(t\right)$ as derived from Figure 1 is calculated as,

  $\begin{array}{l}\frac{12-9}{v\left(t\right)-9}=\frac{3-0}{t-0}\\ v\left(t\right)=t+9\end{array}$

The formula to calculate the energy is given by,

  $\omega ={\displaystyle {\int }_{t_1}^{t_2}i\left(t\right)v\left(t\right)dt}$ ..... (1)

The energy delivered to the battery for the time period $0<t<0.5\text{hr}$ is calculated as,

Substitute ${\omega }_1$ for $\omega$, $10-8t$ for $i\left(t\right)$$t+9$ for $v\left(t\right)$ and $1800\text{sec}$ for $t_2$ and $0$ for $t_1$ in equation (1)

  $\begin{array}{c}{\omega }_1={\displaystyle {\int }_0^{1800}\left(t+9\right)\left(10-8t\right)dt}\\ ={90t-30t^2-\frac{8}{3}{t}^3|}_0^{1800}\\ =90\left(1800-0\right)-30\left({1800}^2-0\right)-\frac{8}{3}\left({1800}^3-0\right)\\ =-1.5632\times {10}^{10}\text{J}\end{array}$

Substitute ${\omega }_2$ for $\omega$, $\frac{20}{3}-\frac{4t}{3}$ for $i\left(t\right)$$t+9$ for $v\left(t\right)$ and $7200\text{sec}$ for $t_2$ and $1800\text{sec}$ for $t_1$ in equation (1)

  $\begin{array}{c}{\omega }_2={\displaystyle {\int }_{1800}^{7200}\left(\frac{20}{3}-\frac{4t}{3}\right)\left(10-8t\right)dt}\\ =\frac{4}{3}{\left[45-\frac{t^3}{3}-2t^2\right]}_{1800}^{7200}\\ =-1.634\times {10}^{11}\text{J}\end{array}$

Substitute ${\omega }_3$ for $\omega$, $t-3$ for $i\left(t\right)$$t+9$ for $v\left(t\right)$ and $10800\text{sec}$ for $t_2$ and $7200\text{sec}$ for $t_1$ in equation (1)

  $\begin{array}{c}{\omega }_3={\displaystyle {\int }_{7200}^{10800}\left(t-3\right)\left(10-8t\right)dt}\\ =-4{\left[\frac{t}{3}+3t^2-27t\right]}_{7200}^{10800}\\ =-1.1827\times {10}^{12}\text{J}\end{array}$

The formula for the total energy transferred to the battery is given by,

  $\omega ={\omega }_1+{\omega }_2+{\omega }_3$

Substitute $-1.1827\times {10}^{12}\text{J}$ for ${\omega }_3$, $-1.634\times {10}^{11}\text{J}$ for ${\omega }_2$ and $-1.5632\times {10}^{10}\text{J}$ for ${\omega }_1$ in the above equation.

  $\begin{array}{c}\omega =-1.5632\times {10}^{10}\text{J}-1.634\times {10}^{11}\text{J}-1.1827\times {10}^{12}\text{J}\\ =-1.3618\times {10}^{12}\text{J}\end{array}$

Conclusion:

Therefore, the total energy delivered to the battery is $-1.3618\times {10}^{12}\text{J}$ .