Chapter 1, Problem 9P

The current through an element is shown in

1. (a) t = 1 s
2. (b) t = 3 s
3. (c) t = 5 s

Figure 1.26

For Prob. 1.9.

Fig. 1.26. Determine the total charge that passed through the element at:

To determine

Find the total charge flows through the element at $t=1\text{s}$.

Answer to Problem 9P

The total charge flows through the element at $t=1\text{s}$ is $\underset{\_}{10\text{C}}$.

Explanation of Solution

Given data:

Refer to Figure 1.26 in the textbook to solve the required objective.

Formula used:

Write the expression for charge entering to an element from $t_1$ to $t_2$ time as follows:

$Q={\displaystyle \underset{t_1}{\overset{t_2}{\int }}idt}$ (1)

Here,

$i$ is the current through an element,

$t_1$ is the initial time at which the charge entering to the element, and

$t_2$ is the final time at which the charge entering to the element.

Calculation:

The given figure is a waveform of electric current versus time.

As the time $t=1\text{s}$ is in the interval $0\le t\le 1\text{s}$, the charge at $t=1\text{s}$ is equal to the total charge flows through the element in the interval 0 to 1 s.

From the given figure, the current for the interval $0\le t\le 1\text{s}$ is 10 A.

Substitute 10 A for $i$, 0 s for $t_1$, and 1 s for $t_2$ in Equation (1) to obtain the charge at $t=1\text{s}$.

$\begin{array}{c}Q={\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}\left(10\text{A}\right)dt}\\ =\left(10\text{A}\right){\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}\left(1\right)dt}\\ =\left(10\text{A}\right){\left(t\right)}_{0\text{s}}^{1\text{s}}\left\{\because {\displaystyle \int t^{n}dt}=\frac{t^{n+1}}{n+1}\right\}\\ =\left(10\text{A}\right)\left(1\text{s}-0\text{s}\right)\end{array}$

Simplify the expression as follows:

$\begin{array}{c}Q=10\text{A}\cdot \text{s}\\ =10\text{C}\left\{\because 1\text{A}\cdot \text{s}=1\text{C}\right\}\end{array}$

Conclusion:

Thus, the total charge flows through the element at $t=1\text{s}$ is $\underset{\_}{10\text{C}}$.

To determine

Find the total charge flows through the element at $t=3\text{s}$.

Answer to Problem 9P

The total charge flows through the element at $t=3\text{s}$ is $\underset{\_}{22.5\text{C}}$.

Explanation of Solution

Calculation:

As the time $t=3\text{s}$ is in the interval $0\le t\le 3\text{s}$, the charge at $t=3\text{s}$ is equal to the total charge flows through the element in the interval 0 to 3 s.

Substitute 0 s for $t_1$, and 3 s for $t_2$ in Equation (1) as follows:

$Q={\displaystyle \underset{0\text{s}}{\overset{3\text{s}}{\int }}idt}$

Rewrite the expression as follows:

$Q={\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}idt}+{\displaystyle \underset{1\text{s}}{\overset{2\text{s}}{\int }}idt}+{\displaystyle \underset{2\text{s}}{\overset{3\text{s}}{\int }}idt}$

Calculate the expression for current from 1 s to 2 s as follows:

$\begin{array}{c}i=\left(\frac{10\text{A}-5\text{A}}{2\text{s}-1\text{s}}\right)t\\ =\left(\frac{5\text{A}}{1\text{s}}\right)t\\ =5t\frac{\text{A}}{\text{s}}\end{array}$

From the given figure, the current is 10 A during 0 to 1 s, $5t\frac{\text{A}}{\text{s}}$ for 1 to 2 s, and 5 A for 2 to 4 s. Therefore, substitute 10 A for $i$ during 0 to 1 s time limit, $5t\frac{\text{A}}{\text{s}}$ for $i$ during 1 to 2 s time limit, and 5 A for $i$ during 2 to 3 s time limit to obtain the required amount of charge at $t=3\text{s}$ as follows:

$\begin{array}{c}Q={\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}\left(10\text{A}\right)dt}+{\displaystyle \underset{1\text{s}}{\overset{2\text{s}}{\int }}\left(5t\frac{\text{A}}{\text{s}}\right)dt}+{\displaystyle \underset{2\text{s}}{\overset{3\text{s}}{\int }}\left(5\text{A}\right)dt}\\ =\left(10\text{A}\right){\left(t\right)}_{0\text{s}}^{1\text{s}}+\left(5\frac{\text{A}}{\text{s}}\right){\left(\frac{t^2}{2}\right)}_{1\text{s}}^{2\text{s}}+\left(10\text{A}\right){\left(t\right)}_{2\text{s}}^{3\text{s}}\\ =\left(10\text{A}\right)\left(1\text{s}-0\text{s}\right)+\left(5\frac{\text{A}}{\text{s}}\right)\left[\frac{{\left(2\text{s}\right)}^2}{2}-\frac{{\left(1\text{s}\right)}^2}{2}\right]+\left(5\text{A}\right)\left(3\text{s}-2\text{s}\right)\\ =10\text{C}+\left(5\frac{\text{A}}{\text{s}}\right)\left(1.5{\text{s}}^2\right)+5\text{C}\end{array}$

Simplify the expression as follows:

$\begin{array}{c}Q=15\text{C}+7.5\text{C}\\ =22.5\text{C}\end{array}$

Conclusion:

Thus, the total charge flows through the element at $t=3\text{s}$ is $\underset{\_}{22.5\text{C}}$.

To determine

Find the total charge flows through the element at $t=5\text{s}$.

Answer to Problem 9P

The total charge flows through the element at $t=5\text{s}$ is $\underset{\_}{50\text{C}}$.

Explanation of Solution

Calculation:

As the time $t=5\text{s}$ is in the interval $0\le t\le 5\text{s}$, the charge at $t=5\text{s}$ is equal to the total charge flows through the element in the interval 0 to 5 s.

Substitute 0 s for $t_1$, and 5 s for $t_2$ in Equation (1) as follows:

$Q={\displaystyle \underset{0\text{s}}{\overset{5\text{s}}{\int }}idt}$

Rewrite the expression as follows:

$Q={\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}idt}+{\displaystyle \underset{1\text{s}}{\overset{2\text{s}}{\int }}idt}+{\displaystyle \underset{2\text{s}}{\overset{4\text{s}}{\int }}idt}+{\displaystyle \underset{4\text{s}}{\overset{5\text{s}}{\int }}idt}$

From Part (b), the current for 1 s to 2 s is $5t\frac{\text{A}}{\text{s}}$.

Calculate the expression for current from 4 s to 5 s as follows:

$\begin{array}{c}i=\left(\frac{5\text{A}-0\text{A}}{5\text{s}-4\text{s}}\right)t\\ =\left(\frac{5\text{A}}{1\text{s}}\right)t\\ =5t\frac{\text{A}}{\text{s}}\end{array}$

From the given figure, the current is 10 A during 0 to 1 s, $5t\frac{\text{A}}{\text{s}}$ for 1 to 2 s, 5 A for 2 to 4 s, and $5t\frac{\text{A}}{\text{s}}$ for 4 to 5 s. Therefore, substitute 10 A for $i$ during 0 to 1 s time limit, $5t\frac{\text{A}}{\text{s}}$ for $i$ during 1 to 2 s time limit, 5 A for $i$ during 2 to 4 s time limit, and $5t\frac{\text{A}}{\text{s}}$ for 4 to 5 s time limit to obtain the required amount of charge at $t=5\text{s}$ as follows:

$\begin{array}{c}Q={\displaystyle \underset{0\text{s}}{\overset{1\text{s}}{\int }}\left(10\text{A}\right)dt}+{\displaystyle \underset{1\text{s}}{\overset{2\text{s}}{\int }}\left(5t\frac{\text{A}}{\text{s}}\right)dt}+{\displaystyle \underset{2\text{s}}{\overset{4\text{s}}{\int }}\left(5\text{A}\right)dt}+{\displaystyle \underset{4\text{s}}{\overset{5\text{s}}{\int }}\left(5t\frac{\text{A}}{\text{s}}\right)dt}\\ =\left(10\text{A}\right){\left(t\right)}_{0\text{s}}^{1\text{s}}+\left(5\frac{\text{A}}{\text{s}}\right){\left(\frac{t^2}{2}\right)}_{1\text{s}}^{2\text{s}}+\left(5\text{A}\right){\left(t\right)}_{2\text{s}}^{4\text{s}}+\left(5\frac{\text{A}}{\text{s}}\right){\left(\frac{t^2}{2}\right)}_{4\text{s}}^{5\text{s}}\\ =\left(10\text{A}\right)\left(1\text{s}-0\text{s}\right)+\left(5\frac{\text{A}}{\text{s}}\right)\left[\frac{{\left(2\text{s}\right)}^2}{2}-\frac{{\left(1\text{s}\right)}^2}{2}\right]+\left(5\text{A}\right)\left(4\text{s}-2\text{s}\right)+\left(5\frac{\text{A}}{\text{s}}\right)\left[\frac{{\left(5\text{s}\right)}^2}{2}-\frac{{\left(4\text{s}\right)}^2}{2}\right]\\ =10\text{C}+\left(5\frac{\text{A}}{\text{s}}\right)\left(1.5{\text{s}}^2\right)+10\text{C}+\left(5\frac{\text{A}}{\text{s}}\right)\left(4.5{\text{s}}^2\right)\end{array}$

Simplify the expression as follows:

$\begin{array}{c}Q=20\text{C}+7.5\text{C}+22.5\text{C}\\ =50\text{C}\end{array}$

Conclusion:

Thus, the total charge flows through the element at $t=5\text{s}$ is $\underset{\_}{50\text{C}}$.