Chapter 2, Problem 2.9HP

Suppose the current through a wire is given by the curve shown in Figure P2.9.
a. Find the amount of charge q that flows through the wire between $t_1=0$ and $t_2=1\text{s}$ .
b. Repeatpartafor $t_2=2,3,4,5,6,7,8,9,$ and 10 s.
c. Sketch $q\left(t\right)$ for $0\le t\le 10\text{s}$ .

To determine

(a)

The amount of current that flows through the wire for the given time.

Answer to Problem 2.9HP

The current that flows though the wire for the time interval $0$ to $1\text{sec}$ is $2\times {10}^{-3}\text{C}$ .

Explanation of Solution

Calculation:

The given time is $t_1=0$ and $t_2=1\text{s}$

The conversion from $1\text{mA}$ into $\text{A}$ is calculated as,

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

The conversion from $4\text{mA}$ to $\text{A}$ is given by,

  $4\text{mA}=4\times {\text{10}}^{-3}\text{A}$

The conversion from $2\text{mA}$ to $\text{A}$ is given by,

  $2\text{mA}=2\times {\text{10}}^{-3}\text{A}$

The conversion from $8\text{mA}$ to $\text{A}$ is given by,

  $8\text{mA}=8\times {\text{10}}^{-3}\text{A}$

The given diagram is shown in Figure 1.

  

The expression for the current as a function of time is given by,

  $i\left(t\right)=mt$

The slope of the line that represents the current from $0$ to $1\text{sec}$ is calculated as,

  $\begin{array}{c}m=\frac{4\text{mA}-0\text{mA}}{1\text{sec}-0}\\ =4\times {10}^{-3}\text{A}/\text{sec}\end{array}$

The current for the time interval $0\le t\le 1\text{s}$ is given by,

  $\begin{array}{c}i\left(t\right)=4\times {10}^{-3}t\text{A}\\ =4t\text{mA}\end{array}$

The expression for the charge that flows through the wire for the time interval $0$ to $1\text{sec}$ is given by,

  $q={\displaystyle {\int }_0^{1}i\left(t\right)}dt$

Substitute $4\times {10}^{-3}tA$ for $i\left(t\right)$ in the above equation.

  $\begin{array}{c}q={\displaystyle {\int }_0^{1}4\times {10}^{-3}t}dt\\ =4\times {10}^{-3}{\left[\frac{t^2}{2}\right]}_0^1\\ =4\times {10}^{-3}\left[\frac{1}{2}-0\right]\\ =2\times {10}^{-3}\text{C}\end{array}$

Conclusion:

Therefore, the charge that flows though the wire for the time interval $0$ to $1\text{sec}$ $2\times {10}^{-3}\text{C}$ .

To determine

(b)

The charge that flows through the given time interval.

Answer to Problem 2.9HP

The charge that will flow through the wire form $0\text{s}$ to $2\text{s}$ is $4\text{mC}$, from $0\text{s}$ to $3\text{s}$ is $2\text{mC}$, $0\text{s}$ to $\text{5}\text{sec}$ is $0$, from $0$ to $6\text{sec}$ is $2\text{mC}$, from $0$ to $7\text{sec}$ is $4\text{mC}$, from $0\text{s}$ to $\text{8}\text{sec}$ is $4\text{mC}$, from $0\text{s}$ to $\text{9}\text{sec}$ is $4\text{mC}$ and from $0\text{s}$ to $\text{10}\text{sec}$ is $4\text{mC}$ .

Explanation of Solution

Calculation:

The given time interval is $t_2=2,3,4,5,6,7,8,9$ and $\text{10}\text{s}$ .The expression for the current for the straight line that represents the current from $1\text{s}$ to $2\text{s}$ is given by,

$i\left(t\right)=mt+c$

The slope of the line that represents the current from $1\text{sec}$ to $2\text{sec}$ is calculated as,

  $\begin{array}{c}m=\frac{0\text{mA}-4\text{mA}}{2\text{sec}-1\sec }\\ =-4\times {10}^{-3}\text{A}/\text{sec}\end{array}$

The conversion of $\text{C}$ into $\text{mC}$ is given by,

  $\text{1}\text{C}={\text{10}}^3\text{mC}$

The conversion of $\text{4}\times {10}^{-3}\text{C}$ into $\text{mC}$ is given by,

  $\text{4}\times {10}^{-3}\text{C}=\text{4}\text{mC}$

The conversion of $\text{2}\times {10}^{-3}\text{C}$ into $\text{mC}$ is given by,

  $2\times {10}^{-3}\text{C}=2\text{mC}$

The current for the time interval $1\text{sec}\le t\le 2\text{sec}$ is given by,

  $i\left(t\right)=-4\times {10}^{-3}t+c$.......(1)

Substitute $1\text{sec}$ for $t$ and $-4\times {10}^{-3}\text{A}$ for $i\left(1\text{sec}\right)$ in the above equation.

  $\begin{array}{l}4\times {10}^{-3}\text{A}=-4\times {10}^{-3}\left(1\right)\text{A}+c\\ c=8\times {10}^{-3}\text{A}\end{array}$

Substitute $8\times {10}^{-3}$ for $c$ in the equation (1)

  $\begin{array}{c}i\left(t\right)=-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}\\ \text{=}4t+8\text{mA}\end{array}$

The expression for the charge stored from $0$ to $2\text{sec}$ is given by,

  $q_2={\displaystyle {\int }_0^{1}i\left(t\right)}dt+{\displaystyle {\int }_1^{2}i\left(t\right)}dt$

Substitute $-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}$ for $i\left(t\right)$ and $q_1$ for ${\displaystyle {\int }_0^{1}i\left(t\right)}dt$ in the above equation.

  $q_2=q_1+{\displaystyle {\int }_1^2\left[-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}\right]}dt$

Substitute $2\times {10}^{-3}\text{C}$ for $q_1$ in the above equation.

  $\begin{array}{c}{q}_2=2\times {10}^{-3}\text{C}+{\displaystyle {\int }_1^2\left[-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}\right]}dt\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}{\left[\frac{t^2}{2}\right]}_1^2+8\times {10}^{-3}{\left[t\right]}_1^2\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}{\left[\frac{4}{2}-\frac{1}{2}\right]}_1^2+8\times {10}^{-3}{\left[2-1\right]}_1^2\\ =2\times {10}^{-3}\text{C}-6\times {10}^{-3}\text{C}+8\times {10}^{-3}\text{C}\end{array}$

Solve further as,

  $\begin{array}{c}{q}_2=4\times {10}^{-3}\text{C}\\ =\text{4}\text{mC}\end{array}$

The expression for the charge stored from $0\text{sec}$ to $3\text{sec}$ is given by,

  $q_3={\displaystyle {\int }_0^{1}i\left(t\right)}dt+{\displaystyle {\int }_1^{3}i\left(t\right)}dt$

Substitute $-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}$ for $i\left(t\right)$

  $q_1$ for ${\displaystyle {\int }_0^{1}i\left(t\right)}dt$ in the above equation.

  $q_3=q_1+{\displaystyle {\int }_1^3\left[-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}\right]}dt$

Substitute $2\times {10}^{-3}\text{C}$ for $q_1$ in the above equation.

  $\begin{array}{c}{q}_3=2\times {10}^{-3}\text{C}+{\displaystyle {\int }_1^3\left[-4\times {10}^{-3}t+8\times {10}^{-3}\text{A}\right]}dt\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}{\left[\frac{t^2}{2}\right]}_1^3+8\times {10}^{-3}{\left[t\right]}_1^3\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}\left[\frac{9}{2}-\frac{1}{2}\right]+8\times {10}^{-3}\left[3-1\right]\\ =2\times {10}^{-3}\text{C}-16\times {10}^{-3}\text{C}+16\times {10}^{-3}\text{C}\end{array}$

Solve further as,

  $\begin{array}{c}{q}_3=2\times {10}^{-3}\text{C}\\ =\text{2}\text{mC}\end{array}$

The mathematical expression for the straight line that represents the current waveform from $3\text{s}$ to $\text{4}\text{s}$ is given by,

  $\begin{array}{c}i\left(t\right)=-4\times {10}^{-3}\text{A}\\ =-4\text{mA}\end{array}$

The expression for the charge stored from $0\text{sec}$ to $4\text{sec}$ is given by,

  $q_4={\displaystyle {\int }_0^{3}i\left(t\right)}dt+{\displaystyle {\int }_3^{4}i\left(t\right)}dt$

Substitute $-4\times {10}^{-3}\text{A}$ for $i\left(t\right)$ and $2\times {10}^{-3}\text{C}$ for ${\displaystyle {\int }_0^{3}i\left(t\right)}dt$ in the above equation.

  $\begin{array}{c}{q}_4=2\times {10}^{-3}\text{C}+{\displaystyle {\int }_3^4\left[-4\times {10}^{-3}\right]}dt\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}{\left[t\right]}_3^4\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}\left[4-3\right]\\ =2\times {10}^{-3}\text{C}-4\times {10}^{-3}\text{C}\end{array}$

Solve further as,

  $\begin{array}{c}{q}_4=-2\times {10}^{-3}\text{C}\\ =\text{2}\text{mC}\end{array}$

The mathematical expression for the straight line that represents the current waveform from $\text{4}\text{s}$ to $\text{5}\text{s}$ is given by,

  $\begin{array}{c}i\left(t\right)=2\times {10}^{-3}\text{A}\\ =2\text{mA}\end{array}$

The expression for the charge stored from $0\text{sec}$ to $4\text{sec}$ is given by,

  $q_5={\displaystyle {\int }_0^{4}i\left(t\right)}dt+{\displaystyle {\int }_4^{5}i\left(t\right)}dt$

Substitute $2\times {10}^{-3}\text{A}$ for $i\left(t\right)$ and $-2\times {10}^{-3}\text{C}$ for ${\displaystyle {\int }_0^{4}i\left(t\right)}dt$ in the above equation.

  $\begin{array}{c}{q}_4=-2\times {10}^{-3}\text{C}+{\displaystyle {\int }_4^5\left[2\times {10}^{-3}\right]}dt\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}{\left[t\right]}_4^5\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}\left[5-4\right]\\ =\text{0}\end{array}$

The mathematical expression for the straight line that represents the current waveform from $\text{4}\text{s}$ to $\text{6}\text{s}$ is given by,

  $\begin{array}{c}i\left(t\right)=2\times {10}^{-3}\text{A}\\ =2\text{mA}\end{array}$

The expression for the charge stored from $4\text{sec}$ to $6\text{sec}$ is given by,

  $q_6={\displaystyle {\int }_0^{4}i\left(t\right)}dt+{\displaystyle {\int }_4^{6}i\left(t\right)}dt$

Substitute $2\times {10}^{-3}\text{A}$ for $i\left(t\right)$ and $-2\times {10}^{-3}\text{C}$ for ${\displaystyle {\int }_0^{4}i\left(t\right)}dt$ in the above equation.

  $\begin{array}{c}{q}_6=-2\times {10}^{-3}\text{C}+{\displaystyle {\int }_4^6\left[2\times {10}^{-3}\right]}dt\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}{\left[t\right]}_4^6\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}\left[6-4\right]\\ =2\text{mC}\end{array}$

The mathematical expression for the straight line that represents the current waveform from $\text{4}\text{s}$ to $\text{7}\text{s}$ is given by,

  $\begin{array}{c}i\left(t\right)=2\times {10}^{-3}\text{A}\\ =2\text{mA}\end{array}$

The expression for the charge stored from $4\text{sec}$ to $\text{7}\text{s}$ is given by,

  $q_7={\displaystyle {\int }_0^{4}i\left(t\right)}dt+{\displaystyle {\int }_4^{7}i\left(t\right)}dt$

Substitute $2\times {10}^{-3}\text{A}$ for $i\left(t\right)$ and $-2\times {10}^{-3}\text{C}$ for ${\displaystyle {\int }_0^{4}i\left(t\right)}dt$ in the above equation.

  $\begin{array}{c}{q}_7=-2\times {10}^{-3}\text{C}+{\displaystyle {\int }_4^7\left[2\times {10}^{-3}\right]}dt\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}{\left[t\right]}_4^7\\ =-2\times {10}^{-3}\text{C}+2\times {10}^{-3}\left[7-4\right]\\ =4\text{mC}\end{array}$

The current is zero for $t\ge 7\text{sec}$ as the charge flow from $7\text{sec}$ to $10\text{sec}$ will also be zero.

The expression for the charge stored from $\text{0}\text{s}$ to $8\text{sec}$ is given by,

  $q_8={\displaystyle {\int }_0^{7}i\left(t\right)}dt+{\displaystyle {\int }_7^{8}i\left(t\right)}dt$

Substitute $0$ for $i\left(t\right)$ and $4\times {10}^{-3}\text{C}$ for ${\displaystyle {\int }_0^{7}i\left(t\right)}d$ in the above equation.

  $\begin{array}{c}{q}_8=4\times {10}^{-3}\text{C}+{\displaystyle {\int }_7^8\left(0\right)}dt\\ =4\times {10}^{-3}\text{C}\\ =\text{4}\text{mC}\end{array}$

The charge stored from $0\text{s}$ to $\text{9}\text{sec}$ is given by,

  $q_9=q_7$

Substitute $4\text{mC}$ for $q_7$ in the above equation.

  $q_9=4\text{mC}$

The charge stored from $0\text{s}$ to $\text{10}\text{sec}$ is given by,

  $q_{10}=q_7$

Substitute $4\text{mC}$ for $q_7$ in the above equation.

  $q_{10}=4\text{mC}$

Conclusion:

Therefore, the charge that will flow through the wire form $0\text{s}$ to $2\text{s}$ is $4\text{mC}$, from $0\text{s}$ to $3\text{s}$ is $2\text{mC}$, $0\text{s}$ to $\text{5}\text{sec}$ is $0$, from $0$ to $6\text{sec}$ is $2\text{mC}$, from $0$ to $7\text{sec}$ is $4\text{mC}$, from $0\text{s}$ to $\text{8}\text{sec}$ is $4\text{mC}$, from $0\text{s}$ to $\text{9}\text{sec}$ is $4\text{mC}$ and from $0\text{s}$ to $\text{10}\text{sec}$ is $4\text{mC}$ .

To determine

(c)

To sketch:

The graph for $q\left(t\right)$ at $0\le t\le 10\text{s}$ .

Answer to Problem 2.9HP

The sketch for $q\left(t\right)$ at $0\le t\le 10\text{s}$ is shown in Figure 2.

Explanation of Solution

Calculation:

The expression for the current for different time interval is given by,

  $i_t=\{\begin{array}{l}4t\text{mA}\text{for}\text{0}\le t\le 1\\ -4t+8\text{mA}\text{for}1\le t\le 3\\ -\text{4}\text{mA}\text{for}3\le t\le 4\\ \text{2}\text{mA}\text{for}\text{4}\le t\le 7\\ \text{0}\text{for}7\le t\le 10\end{array}$

The integration of the above equation with respect to time to obtain the charge expression is given by,

  $q_t=\{\begin{array}{l}{\displaystyle \int 4tdt}\text{for}\text{0}\le t\le 1\\ {\displaystyle \int -4t+8}\text{for}1\le t\le 3\\ {\displaystyle \int -4tdt}\text{for}3\le t\le 4\\ {\displaystyle \int 2tdt}\text{for}\text{4}\le t\le 7\\ \text{0}\text{for}7\le t\le 10\end{array}$

The evaluated form for the expression of charge for the different time interval is given by,

  $q_t=\{\begin{array}{l}2t^2+k_1\text{mC}\text{for}\text{0}\le t\le 1\\ -2t^2+\text{8}t+k_2\text{mC}\text{for}1\le t\le 3\\ -4t+k_3\text{mC}\text{for}3\le t\le 4\\ 2t+k_4\text{mC}\text{for}\text{4}\le t\le 7\\ 0+k_5\text{mC}\text{for}7\le t\le 10\end{array}$ ....... (2)

Substitute $1$ for $t$ in the above equation for time interval $\text{0}\le t\le 1$ .

  $q_t=2{\left(1\right)}^2+k_1\text{mC}$

Substitute $2\text{mC}$ for $q_1$ in the above equation.

  $\begin{array}{l}2\text{mC}=2{\left(1\right)}^2+k_1\text{mC}\\ k_1=0\text{mC}\end{array}$

Substitute $2$ for $t$ in equation (2)for time interval $1\le t\le 3$ .

  $q_2=-2{\left(2\right)}^2+\text{8}\left(2\right)+k_2\text{mC}$

Substitute $4\text{mC}$ for $q_2$ in the above equation.

  $\begin{array}{l}4\text{mC}=-2{\left(2\right)}^2+\text{8}\left(2\right)+k_2\text{mC}\\ 4\text{mC}=-8+16+k_2\text{mC}\\ k_2=-4\text{mC}\end{array}$

Substitute $3$ for $t$ in equation (2) for time interval $3\le t\le 4$ .

  $q_3=-4\left(3\right)+k_3\text{mC}$

Substitute $2\text{mC}$ for $q_3$ in the above equation.

  $\begin{array}{l}2\text{mC}=-4\left(3\right)+k_3\text{mC}\\ k_3=-12\text{mC}+k_3\\ k_3=14\text{mC}\end{array}$

Substitute $5$ for $t$ equation (2) for time interval $\text{4}\le t\le 7$ .

  $q_5=2\left(5\right)+k_4\text{mC}$

Substitute $0$ for $q_5$ in the above equation.

  $\begin{array}{l}0=2\left(5\right)+k_4\text{mC}\\ k_4=-10\text{mC}\end{array}$

Substitute $8$ for $t$ in equation (2)for time interval $\text{4}\le t\le 7$ .

  $q_8=0+k_5\text{mC}$

Substitute $4\text{mC}$ for $0$ in the above equation.

  $\begin{array}{l}4\text{mC}=0+k_5\text{mC}\\ k_5=4\text{mC}\end{array}$

Substitute $0\text{mC}$ for $k_1$, $-4\text{mC}$ for $k_2$, $14\text{mC}$ for $k_3$. $-10\text{mC}$ for $k_4$ and $4\text{mC}$ for $k_5$ in equation (2).

  $q_t=\{\begin{array}{l}2t^2+0\text{mC}\text{for}\text{0}\le t\le 1\\ -2t^2+\text{8}t-4\text{mC}\text{for}1\le t\le 3\\ -4t+14\text{mC}\text{for}3\le t\le 4\\ 2t-10\text{mC}\text{for}\text{4}\le t\le 7\\ 4\text{mC}\text{for}7\le t\le 10\end{array}$

The sketch of $q\left(t\right)$ versus time $\left(t\right)$ for the time interval $7\le t\le 10$ is shown below.

The required diagram is shown in Figure 2.