Chapter 1, Problem 16P

(a)

To determine

Sketch the power delivered to the element for $t>0$.

Explanation of Solution

Given data:

Refer to Figure 1.27 in the textbook for required data.

Formula used:

Write the expression for power delivered to the element as follows:

$p\left(t\right)=v\left(t\right)i\left(t\right)$ (1)

Here,

$i\left(t\right)$ is the current through the element and

$v\left(t\right)$ is the voltage across the element.

Calculation:

From the given current waveform, write the expression for current for 0 to 2 s as follows:

$i{\left(t\right)}_{0\text{s}-2\text{s}}=\left(\frac{i_{2\text{s}}-i_{0\text{s}}}{2\text{s}-0\text{s}}\right)t,\text{for 0}<t<2\text{s}$

From the given current waveform, substitute $60\text{mA}$ for $i_{2\text{s}}$ and $\text{0}\text{mA}$ for $i_{0\text{s}}$ as follows:

$\begin{array}{c}i{\left(t\right)}_{0\text{s}-2\text{s}}=\left(\frac{60\text{mA}-0\text{mA}}{2\text{s}-0\text{s}}\right)t,\text{for 0}<t<2\text{s}\\ =30t\text{mA}\text{for 0}<t<2\text{s}\end{array}$

From the given current waveform, write the expression for current for 2 to 4 s as follows:

$i{\left(t\right)}_{2\text{s}-4\text{s}}=120-\left(\frac{i_{4\text{s}}-i_{2\text{s}}}{4\text{s}-2\text{s}}\right)t,\text{for 2}<t<4\text{s}$

From the given current waveform, substitute $0\text{mA}$ for $i_{4\text{s}}$ and $\text{60}\text{mA}$ for $i_{2\text{s}}$ as follows:

$\begin{array}{c}i{\left(t\right)}_{2\text{s}-4\text{s}}=120-\left(\frac{0\text{mA}-60\text{mA}}{4\text{s}-2\text{s}}\right)t,\text{for 2}<t<4\text{s}\\ =\left(120-30t\right)\text{mA}\text{for 2}<t<4\text{s}\end{array}$

From the given voltage waveform, write the expression for voltage for 0 to 2 s as follows:

$v{\left(t\right)}_{0\text{s}-2\text{s}}=5\text{V},\text{for 0}<t<2\text{s}$

From the given voltage waveform, write the expression for voltage for 2 to 4 s as follows:

$v{\left(t\right)}_{2\text{s}-4\text{s}}=-5\text{V},\text{for 2}<t<4\text{s}$

Modify the expression in Equation (1) for the power delivered to the element for 0 to 2 s as follows:

$p{\left(t\right)}_{0\text{s}-2\text{s}}=v{\left(t\right)}_{0\text{s}-2\text{s}}i{\left(t\right)}_{0\text{s}-2\text{s}}$

Substitute 5 V for $v{\left(t\right)}_{0\text{s}-2\text{s}}$ and $30t\text{mA}$ for $i{\left(t\right)}_{0\text{s}-2\text{s}}$ to obtain the power delivered to the element from 0 to 2 s.

$\begin{array}{c}p{\left(t\right)}_{0\text{s}-2\text{s}}=\left(5\text{V}\right)\left(30t\text{mA}\right)\\ =150t\text{mW}\left\{\because 1\text{V}\cdot \text{A}=1\text{W}\right\}\end{array}$

Modify the expression in Equation (1) for the power delivered to the element for 2 to 4 s as follows:

$p{\left(t\right)}_{2\text{s}-4\text{s}}=v{\left(t\right)}_{2\text{s}-4\text{s}}i{\left(t\right)}_{2\text{s}-4\text{s}}$

Substitute $-5\text{V}$ for $v{\left(t\right)}_{2\text{s}-4\text{s}}$ and $\left(120-30t\right)\text{mA}$ for $i{\left(t\right)}_{2\text{s}-4\text{s}}$ to obtain the power delivered to the element from 2 to 4 s.

$\begin{array}{c}p{\left(t\right)}_{2\text{s}-4\text{s}}=\left(-5\text{V}\right)\left[\left(120-30t\right)\text{mA}\right]\\ =-600+150t\text{mW}\end{array}$

From the calculation, write the expression for power delivered to the element as follows:

$p\left(t\right)=\{\begin{array}{l}150t\text{mW},\text{for 0}<t<2\text{s}\\ -600+150t\text{mW},\text{for 2}<t<4\text{s}\end{array}$

From the expression of power, draw the waveform for power delivered to the element as shown in Figure 1:

Conclusion:

Thus, the waveform for power delivered to the element is sketched.

(b)

To determine

Find the amount of energy absorbed by the element for the period of $\text{0}<t<4\text{s}$.

Answer to Problem 16P

The amount of energy absorbed by the element for the period of $\text{0}<t<4\text{s}$ is $\underset{\_}{0\text{mJ}}$.

Explanation of Solution

Formula used:

Write the expression for energy as follows:

$w={\displaystyle \underset{t_1}{\overset{t_2}{\int }}p\left(t\right)dt}$ (2)

Here,

$p\left(t\right)$ is the power expended to charge the battery,

$t_1$ is the initial time, and

$t_2$ is the final time.

Calculation:

From Part (b), the expression for power delivered to the element is written as follows:

$p\left(t\right)=\{\begin{array}{l}150t\text{mW},\text{for 0}<t<2\text{s}\\ -600+150t\text{mW},\text{for 2}<t<4\text{s}\end{array}$

Substitute 0 s for $t_1$ and 4 s for $t_2$ in Equation (2) as follows:

$w={\displaystyle \underset{0\text{s}}{\overset{4\text{s}}{\int }}p\left(t\right)dt}$

Rewrite the expression as follows:

$w={\displaystyle \underset{0\text{s}}{\overset{2\text{s}}{\int }}p\left(t\right)dt}+{\displaystyle \underset{2\text{s}}{\overset{4\text{s}}{\int }}p\left(t\right)dt}$

Substitute $150t\text{mW}$ for $p\left(t\right)$ in the time interval of 0 to 2 s and $-600+150t\text{mW}$ for $p\left(t\right)$ in the time interval of 2 to 4 s to obtain the energy absorbed by the element as follows:

$\begin{array}{c}w={\displaystyle \underset{0\text{s}}{\overset{2\text{s}}{\int }}\left(150t\text{mW}\right)dt}+{\displaystyle \underset{2\text{s}}{\overset{4\text{s}}{\int }}\left(-600+150t\text{mW}\right)dt}\\ =\left(150\right){\left(\frac{t^2}{2}\right)}_0^2+\left[\left(-600\right){\left(t\right)}_2^4+\left(150\right){\left(\frac{t^2}{2}\right)}_2^4\right]\text{mJ}\left\{\because 1\text{mW}\cdot \text{s}=1\text{mJ}\right\}\\ =\left(150\right)\left(\frac{2^2}{2}-\frac{0^2}{2}\right)+\left[\left(-600\right)\left(4-2\right)+\left(150\right)\left(\frac{4^2}{2}-\frac{2^2}{2}\right)\right]\text{mJ}\\ =300+\left(-1200+900\right)\text{mJ}\end{array}$

Simplify the expression as follows:

$\begin{array}{c}w=\left(1200-1200\right)\text{mJ}\\ =0\text{mJ}\end{array}$

Conclusion:

Thus, the amount of energy absorbed by the element for the period of $\text{0}<t<4\text{s}$ is $\underset{\_}{0\text{mJ}}$.