Question

66. II FIGURE P30.66 shows the current through a$10\text{mH}$ inductor. Draw a graph showing the potential difference$\Delta V_{\text{L}}$ across the inductor for these$6\text{ms}$ .

Short answer

The potential difference across the inductor is

$\begin{array}{l}\Delta V_L=-10\text{Vfor}0\text{ms}<t<2\text{ms}\\ \Delta V_L=0\text{V\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}for}2\text{ms}<t<3\text{ms}\\ \Delta V_L=20\text{V\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}for}3\text{ms}<t<5\text{ms}\\ \Delta V_L=0\text{V\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}for}t>5\text{ms}\end{array}$

Step-by-step solution

Step 1. Introduction

It is given that the value of inductor,$L=10\text{mH}=10\times {10}^{-3}\text{H}$

We have potential difference across the inductor as$\Delta V_L=-L\frac{dI}{dt}$

Step 2. Explanation

From $t=0\text{ms}$to $t=2\text{ms}$, current $I$changes from $0\text{\hspace{0.33em}A\hspace{0.33em}to\hspace{0.33em}2A}$.

$\begin{array}{l}\frac{dI}{dt}=\frac{(2\text{A}-0\text{A})}{(2\text{ms}-0\text{ms})}=\frac{2\text{A}}{(2\text{ms})\left({10}^{-3}\text{s}/\text{ms}\right)}=1000\text{A}/\text{s}\\ \Delta V_L=-\left(10\times {10}^{-3}\text{H}\right)(1000\text{A}/\text{s})=-10\text{V}\end{array}$

Step 3.

From $t=2\text{ms}$to role="math" $t=3\text{ms},$role="math" $I$remains constant

Therefore voltage becomes zero.

Step 3.

Now from $t=3\text{ms}$to $5\text{ms}$, the current decreases from $2\text{A}$to $-2\text{A}$

$\begin{array}{l}\frac{dI}{dt}=\frac{(-2\text{A}-2\text{A})}{(5\text{ms}-3\text{ms})}\\ =\frac{-4\text{A}}{(2\text{ms})\left({10}^{-3}\text{s}/\text{ms}\right)}\\ =-2000\text{A}/\text{s}\end{array}$

So, the inductor voltage is

$\Delta V_L=1\left(10\times {10}^{-3}\text{H}\right)(-2000\text{A}/\text{s})=20\text{V}$