Chapter 2, Problem 21E

The current waveform depicted in Fig. 2.29 is characterized by a period of 4 s. (a) What is the average value of the current over a single period? (b) Compute the average current over the interval 1 < t < 3 s. (c) If q(0) = 1C, sketch q(t), 0 < t < 4 s.

FIGURE 2.29 An example of a time-varying current.

To determine

Find the average value of the given current waveform in Figure 2.29 over the period of 4 s.

Answer to Problem 21E

The average value of the given current waveform over the period of 4 s is $\underset{\_}{i_{avg}=0.75\text{ A}}$.

Explanation of Solution

Given data:

Refer to Figure 2.29 in textbook for the time-varying current with the period of 4 s.

Formula used:

Write the formula to find the average value of a function of current over a period as,

$i_{avg}=\frac{1}{T}{\displaystyle {\int }_0^{T}i\left(t\right)}dt$        (1)

Here,

T is the time period.

$i\left(t\right)$ is the function of current in time t.

Calculation:

From the given current waveform in Figure 2.29, the function of current $i\left(t\right)$ is written as,

For the period of $0\le t\le 1$:

$i\left(t\right)=3\text{A}$

For the period of $1\le t\le 2$:

$i\left(t\right)=-1\text{A}$

For the period of $2\le t\le 3$:

$i\left(t\right)=1\text{A}$

For the period of $3\le t\le 4$:

$i\left(t\right)=0\text{A}$

Therefore, the final function of current $i\left(t\right)$ is,

$i\left(t\right)=\{\begin{array}{l}3\text{ A }\text{ for }0\le t\le 1\\ -\text{1 A}\text{ for 1}\le t\le 2\\ 1\text{ A }\text{for 2}\le t\le 3\\ 0\text{A }\text{ for 3}\le t\le 4\end{array}$        (2)

The average value of current $i\left(t\right)$ is written as follows using the function of $i\left(t\right)$ in equation (2) and the formula in equation (1) over the period of 4 s.

$\begin{array}{l}{i}_{avg}=\frac{1}{4}\left[{\displaystyle {\int }_0^{1}3}dt+{\displaystyle {\int }_1^2\left(-1\right)}dt+{\displaystyle {\int }_2^3\left(1\right)}dt+{\displaystyle {\int }_2^3\left(0\right)}dt\right]\text{ A }\left\{\because T=4\text{ s}\right\}\\ i_{avg}=\frac{1}{4}\left[{\left[3t\right]}_0^1+{\left[-t\right]}_1^2+{\left[t\right]}_2^3\right]\text{ A}\\ i_{avg}=\frac{1}{4}\left[\left(3-0\right)+\left(-2+1\right)+\left(3-2\right)\right]\text{ A}\end{array}$

Reduce the equation as follows.

$\begin{array}{l}{i}_{avg}=\frac{1}{4}\left(3-1+1\right)\text{ A}\\ i_{avg}=\frac{3}{4}\text{ A}\\ i_{avg}=0.75\text{ A}\end{array}$

Conclusion:

Thus, the average value of the given current waveform over the period of 4 s is $\underset{\_}{i_{avg}=0.75\text{ A}}$.

To determine

Find the average current of the waveform over the period of $1<t<3\text{ s}$.

Answer to Problem 21E

The average current of the waveform over the period of $1<t<3\text{ s}$ is $\underset{\_}{i_{avg}=0\text{ A}}$.

Explanation of Solution

Given data:

Refer to part (a).

Formula used:

Write the formula to find the average value of a function of current over a period of $t_1<t<t_2$ as,

$i_{avg}=\frac{1}{t_2-t_1}{\displaystyle {\int }_{t_1}^{t_2}i\left(t\right)}dt$        (3)

Here,

$t_1,t_2$ are lower and upper time limits of the period.

Calculation:

Write the function of current $i\left(t\right)$ for the period of $1<t<3\text{ s}$ using the current function in equation (2).

$i\left(t\right)=\{\begin{array}{l}-\text{1 A}\text{ for 1}<t\le 2\\ 1\text{ A }\text{for 2}\le t<3\end{array}$        (4)

Now, the average current of $i\left(t\right)$ using the formula in equation (3) the function of $i\left(t\right)$ in equation (4) over the period of $1<t<3\text{ s}$ is written as follows.

$\begin{array}{l}{i}_{avg}=\frac{1}{3-1}\left[{\displaystyle {\int }_1^2\left(-1\right)}dt+{\displaystyle {\int }_2^3\left(1\right)}dt\right]\text{ A}\\ i_{avg}=\frac{1}{2}\left[{\left[-t\right]}_1^2+{\left[t\right]}_2^3\right]\text{ A}\\ i_{avg}=\frac{1}{2}\left[\left(-2+1\right)+\left(3-2\right)\right]\text{ A}\\ i_{avg}=0\text{ A}\end{array}$

Conclusion:

Thus, the average current of the waveform over the period of $1<t<3\text{ s}$ is $\underset{\_}{i_{avg}=0\text{ A}}$.

To determine

Sketch the waveform for the charge $q\left(t\right)$ of a given waveform over the period of $0<t<4\text{ s}$.

Explanation of Solution

Given data:

Refer to part (a).

The initial charge $q\left(0\right)=\text{ 1 C}$.

Formula used:

Write the expression for the relation between charge $q\left(t\right)$ and current $i\left(t\right)$.

$i\left(t\right)=\frac{dq\left(t\right)}{dt}$

Integrate the above equation on both sides.

$\begin{array}{l}{\displaystyle \int i\left(t\right)}={\displaystyle \int \frac{dq\left(t\right)}{dt}}\\ {\displaystyle \int i\left(t\right)}dt={\displaystyle \int dq\left(t\right)}\end{array}$

$q\left(t\right)={\displaystyle \int i\left(t\right)dt}$        (5)

Calculation:

Using the function of $i\left(t\right)$ in equation (2) and the formula in equation (5), the function of change $q\left(t\right)$ is written as follows.

For $0\le t\le 1$:

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

Substitute 3 A for $i\left(t\right)$.

$\begin{array}{c}q\left(t\right)={\displaystyle {\int }_0^t\left(3\text{ A}\right)dt}\\ =3t+q\left(0\right)\text{C}\end{array}$

Substitute 1 C for $q\left(0\right)$.

$q\left(t\right)=3t+1\text{C}\text{ for }0\le t\le 1\text{s}$        (6)

Substitute 1 for $t$ in equation (6) to find initial condition $q\left(1\right)$ for the period of $1\le t\le 2$.

$\begin{array}{c}q\left(1\right)=3\left(1\right)+1\text{C}\\ =4\text{C}\end{array}$

For $1\le t\le 2$:

For the period of $1\le t\le 2$ the initial charge is $q\left(1\right)=4\text{C}$.

$q\left(t\right)={\displaystyle {\int }_1^{t}i\left(t\right)dt}$

Substitute –1 A for $i\left(t\right)$ in above equation.

$\begin{array}{c}q\left(t\right)={\displaystyle {\int }_1^t\left(-1\text{ A}\right)dt}\\ =-t+1+q\left(1\right)\text{C}\end{array}$

Substitute 4 C for $q\left(1\right)$.

$q\left(t\right)=-t+1+4\text{C}$

$q\left(t\right)=-t+5\text{C}\text{for}1\le t\le 2\text{s}$        (7)

Substitute 2 for $t$ in equation (7) to find initial condition $q\left(2\right)$ for the period of $2\le t\le 3$.

$\begin{array}{c}q\left(2\right)=-\left(2\right)+5\text{C}\\ =3\text{C}\end{array}$

For $2\le t\le 3$:

For the period of $2\le t\le 3$ the initial charge is $q\left(2\right)=3\text{C}$.

$q\left(t\right)={\displaystyle {\int }_2^{t}i\left(t\right)dt}$

Substitute 1 A for $i\left(t\right)$ in above equation.

$\begin{array}{c}q\left(t\right)={\displaystyle {\int }_2^t\left(1\text{ A}\right)dt}\\ =t-2+q\left(2\right)\text{C}\end{array}$

Substitute 3 C for $q\left(2\right)$.

$q\left(t\right)=t-2+3\text{C}$

$q\left(t\right)=t+1\text{C}\text{for}2\le t\le 3\text{s}$        (8)

Substitute 3 for $t$ in equation (8) to find initial condition $q\left(3\right)$ for the period of $3\le t\le 4$.

$\begin{array}{c}q\left(3\right)=3+1\text{C}\\ =4\text{C}\end{array}$

For $3\le t\le 4$:

For the period of $3\le t\le 4$ the initial charge is $q\left(3\right)=4\text{C}$.

$q\left(t\right)={\displaystyle {\int }_3^{t}i\left(t\right)dt}$

Substitute 0 A for $i\left(t\right)$ in above equation.

$\begin{array}{c}q\left(t\right)={\displaystyle {\int }_3^t\left(\text{0 A}\right)dt}\\ =0+q\left(3\right)\text{C}\end{array}$

Substitute 4 C for $q\left(3\right)$.

$q\left(t\right)=4\text{C}\text{for}3\le t\le 4\text{s}$        (9)

Therefore, the charge $q\left(t\right)$ can be written as follows.

$q\left(t\right)=\{\begin{array}{l}3t+1\text{ C, }0\le t\le 1\text{ s}\\ -t+5\text{ C, 1}\le t\le 2\text{ s}\\ t+1\text{ C, 2}\le t\le 3\text{ s}\\ 4\text{ C, 3}\le t\le 4\text{ s}\end{array}$

Table 1 shows for $q\left(t\right)$ in coulombs for various values of t in seconds.

Table 1

t in seconds	$q\left(t\right)$ in C
0	1
1	4
2	2
3	4
4	4

Figure 1 shows the charge (C) vs time (s) waveform.

Conclusion:

Thus, the waveform for the charge $q\left(t\right)$ is sketched over the period of $0<t<4\text{ s}$.