Chapter 6, Problem 58P

The current waveform in Fig. 6.80 flows through a 3-H inductor. Sketch the voltage across the inductor over the interval 0 < t < 6 s.

Figure 6.80

For Prob. 6.58.

To determine

Sketch the voltage across the inductor over the interval $0\text{s}<t<6\text{s}$.

Explanation of Solution

Given data:

The value of the inductor $\left(L\right)$ is $3\text{H}$.

Refer to Figure 6.80 in the textbook.

Formula used:

Write the expression to calculate the straight line equation for two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

$\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ (1)

Refer to Figure 6.80 in the textbook.

From the given graph, substitute $t$ for $x$ and $i$ for $y$ in equation (1).

$\left(i-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(t-x_1\right)$ (2)

Write the expression to calculate the voltage across the inductor.

$v=L\frac{di}{dt}$ (3)

Here,

$L$ is the value of the inductor, and

$\frac{di}{dt}$ is the rate of change of current with time.

Calculation:

The given current waveform is redrawn as Figure 1.

Refer to Figure 1, split up the time period as six divisions $0\text{s}<t<1\text{s}$, $1\text{s}<t<2\text{s}$, $2\text{s}<t<3\text{s}$, $3\text{s}<t<4\text{s}$, $4\text{s}<t<5\text{s}$ and $5\text{s}<t<6\text{s}$ to find the respective current value.

Case (i): $0\text{s}<t<1\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(0\text{s},\text{0}\text{A}\right)$ and $\left(1\text{s},2\text{A}\right)$.

Substitute $0\text{s}$ for $x_1$, $\text{0}\text{A}$ for $y_1$, $1\text{s}$ for $x_2$ and $2\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-\text{0}\text{A}\right)=\frac{2\text{A}-\text{0}\text{A}}{1\text{s}-0\text{s}}\left(t-0\text{s}\right)\\ =\frac{2\text{A}}{1\text{s}}t\\ =2t\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=2t\text{A}+0\text{A}\\ =2t\text{A}\end{array}$

Case (ii): $1\text{s}<t<2\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(1\text{s},2\text{A}\right)$ and $\left(2\text{s},0\text{A}\right)$.

Substitute $1\text{s}$ for $x_1$, $2\text{A}$ for $y_1$, $2\text{s}$ for $x_2$ and $0\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-2\text{A}\right)=\frac{0\text{A}-2\text{A}}{2\text{s}-1\text{s}}\left(t-1\text{s}\right)\\ =\frac{-2\text{A}}{1\text{s}}\left(t-1\text{s}\right)\\ =-2\text{A}\left(t-1\text{s}\right)\\ =-2t\text{A}+\text{2}\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=-2t\text{A}+\text{2}\text{A}+\text{2}\text{A}\\ =-2t\text{A}+\text{4}\text{A}\\ =\left(-2t+4\right)\text{A}\end{array}$

Case (iii): $2\text{s}<t<3\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(2\text{s},0\text{A}\right)$ and $\left(3\text{s},2\text{A}\right)$.

Substitute $2\text{s}$ for $x_1$, $0\text{A}$ for $y_1$, $3\text{s}$ for $x_2$ and $2\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-0\text{A}\right)=\frac{2\text{A}-0\text{A}}{3\text{s}-2\text{s}}\left(t-2\text{s}\right)\\ =\frac{2\text{A}}{1\text{s}}\left(t-2\text{s}\right)\\ =2\text{A}\left(t-2\text{s}\right)\\ =2t\text{A}-\text{4}\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=2t\text{A}-\text{4}\text{A}+0\text{A}\\ =2t\text{A}-\text{4}\text{A}\\ =\left(2t-\text{4}\right)\text{A}\end{array}$

Case (iv): $3\text{s}<t<4\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(3\text{s},2\text{A}\right)$ and $\left(4\text{s},0\text{A}\right)$.

Substitute $3\text{s}$ for $x_1$, $2\text{A}$ for $y_1$, $4\text{s}$ for $x_2$ and $0\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-2\text{A}\right)=\frac{0\text{A}-2\text{A}}{4\text{s}-3\text{s}}\left(t-3\text{s}\right)\\ =-\frac{2\text{A}}{1\text{s}}\left(t-3\text{s}\right)\\ =-2\text{A}\left(t-3\text{s}\right)\\ =-2t\text{A}+\text{6}\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=-2t\text{A}+\text{6}\text{A}+2\text{A}\\ =-2t\text{A}+\text{8}\text{A}\\ =\left(-2t+\text{8}\right)\text{A}\end{array}$

Case (v): $4\text{s}<t<5\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(4\text{s},0\text{A}\right)$ and $\left(5\text{s},2\text{A}\right)$.

Substitute $4\text{s}$ for $x_1$, $0\text{A}$ for $y_1$, $5\text{s}$ for $x_2$ and $2\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-0\text{A}\right)=\frac{2\text{A}-0\text{A}}{5\text{s}-4\text{s}}\left(t-4\text{s}\right)\\ =\frac{2\text{A}}{1\text{s}}\left(t-4\text{s}\right)\\ =2\text{A}\left(t-4\text{s}\right)\\ =2t\text{A}-\text{8}\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=2t\text{A}-\text{8}\text{A}+0\text{A}\\ =2t\text{A}-\text{8}\text{A}\\ =\left(2t-\text{8}\right)\text{A}\end{array}$

Case (vi): $5\text{s}<t<6\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(5\text{s},2\text{A}\right)$ and $\left(6\text{s},0\text{A}\right)$.

Substitute $5\text{s}$ for $x_1$, $2\text{A}$ for $y_1$, $6\text{s}$ for $x_2$ and $0\text{A}$ for $y_2$ in equation (2).

$\begin{array}{c}\left(i-2\text{A}\right)=\frac{0\text{A}-2\text{A}}{6\text{s}-5\text{s}}\left(t-5\text{s}\right)\\ =-\frac{2\text{A}}{1\text{s}}\left(t-5\text{s}\right)\\ =-2\text{A}\left(t-5\text{s}\right)\\ =-2t\text{A}+10\text{A}\end{array}$

Simplify the equation to find $i$.

$\begin{array}{c}i=-2t\text{A}+10\text{A}+2\text{A}\\ =-2t\text{A}+12\text{A}\\ =\left(-2t+12\right)\text{A}\end{array}$

Therefore, the current function of the signal in Figure 1 is,

$i=\{\begin{array}{l}2t\text{A},0\text{s}<t<1\text{s}\\ \left(-2t+4\right)\text{A},1\text{s}<t<2\text{s}\\ \left(2t-\text{4}\right)\text{A,}2\text{s}<t<3\text{s}\\ \left(-2t+\text{8}\right)\text{A,}3\text{s}<t<4\text{s}\\ \left(2t-\text{8}\right)\text{A,}4\text{s}<t<5\text{s}\\ \left(-2t+12\right)\text{A,}5\text{s}<t<6\text{s}\end{array}$

For $0\text{s}<t<1\text{s}$:

Substitute $3\text{H}$ for $L$ and $2t\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(2t\text{A}\right)\\ =\left(3\times 2\right)\frac{d}{dt}\left(t\right)\text{H}\cdot \text{A}\\ =\left(6\right)\left(1\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

For $1\text{s}<t<2\text{s}$:

Substitute $3\text{H}$ for $L$ and $\left(-2t+4\right)\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(\left(-2t+4\right)\text{A}\right)\\ =\left(3\right)\frac{d}{dt}\left(-2t+4\right)\text{H}\cdot \text{A}\\ =\left(3\right)\left(-2\left(1\right)+0\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =-6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

For $2\text{s}<t<3\text{s}$:

Substitute $3\text{H}$ for $L$ and $\left(2t-\text{4}\right)\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(\left(2t-\text{4}\right)\text{A}\right)\\ =\left(3\right)\frac{d}{dt}\left(2t-4\right)\text{H}\cdot \text{A}\\ =\left(3\right)\left(2\left(1\right)-0\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

For $3\text{s}<t<4\text{s}$:

Substitute $3\text{H}$ for $L$ and $\left(-2t+\text{8}\right)\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(\left(-2t+\text{8}\right)\text{A}\right)\\ =\left(3\right)\frac{d}{dt}\left(-2t+\text{8}\right)\text{H}\cdot \text{A}\\ =\left(3\right)\left(-2\left(1\right)+0\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =-6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

For $4\text{s}<t<5\text{s}$:

Substitute $3\text{H}$ for $L$ and $\left(2t-\text{8}\right)\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(\left(2t-\text{8}\right)\text{A}\right)\\ =\left(3\right)\frac{d}{dt}\left(2t-\text{8}\right)\text{H}\cdot \text{A}\\ =\left(3\right)\left(2\left(1\right)-0\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

For $5\text{s}<t<6\text{s}$:

Substitute $3\text{H}$ for $L$ and $\left(-2t+12\right)\text{A}$ for $i$ in equation (3) to find $v$.

$\begin{array}{c}v=\left(3\text{H}\right)\frac{d}{dt}\left(\left(-2t+12\right)\text{A}\right)\\ =\left(3\right)\frac{d}{dt}\left(-2t+12\right)\text{H}\cdot \text{A}\\ =\left(3\right)\left(-2\left(1\right)+0\right)\frac{\text{H}\cdot \text{A}}{\text{s}}\\ =-6\text{V}\left\{\because 1\text{V}=\frac{\text{1H}\cdot 1\text{A}}{\text{1s}}\right\}\end{array}$

The expression of the voltage $v$ is,

$v=\{\begin{array}{l}6\text{V},0\text{s}<t<1\text{s}\\ -6\text{V},1\text{s}<t<2\text{s}\\ 6\text{V,}2\text{s}<t<3\text{s}\\ -6\text{V,}3\text{s}<t<4\text{s}\\ 6\text{V,}4\text{s}<t<5\text{s}\\ -6\text{V,}5\text{s}<t<6\text{s}\end{array}$

From the voltage expression, the signal is drawn in Figure 2.

Conclusion:

Thus, the voltage across the inductor over the interval $0\text{s}<t<6\text{s}$ is sketched.