Question

67. II FIGURE P30.67 shows the potential difference across a potential difference across a $50\text{mH}$inductor. The current through the inductor at $t=0\text{s}$is $0.20\text{A}$. Draw a graph showing the current through the inductor from $t=0\text{s}$to $t=40\text{ms}$

Short answer

The following graph between current and time is

Step-by-step solution

Step 1. Given information

We have $\Delta r_L=-1{\text{V}}_{\text{for}}0\text{ms}<t<10\text{ms}$

$\begin{array}{l}\text{So}\\ -L\frac{dI}{dt}=-1\text{V}\\ \frac{dI}{dt}=\frac{1\text{V}}{50\text{mH}}=20\text{A}/\text{s}\end{array}$

$\begin{array}{l}dI=(20\text{A}/\text{s})dtI=(20\text{A}/\text{s})t+c\\ \text{Butat}t=0,I=0.2\text{A}\\ 0.2\text{A}=0+c\end{array}$

$\begin{array}{l}I=(20\text{A}/\text{s})t+0.2\text{A\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\\ \text{At}t=10\text{ms}=10\times {10}^{-3}\text{s}\\ I=(20\text{A}/\text{s})\left(10\times {10}^{-3}\text{s}\right)+0.2\text{A}\\ =0.2\text{A}+0.2\text{A}\\ =0.4\text{A}\end{array}$

Explanation

From $t=0\text{ms}$ to $t=10\text{ms}$current increase from$0.2\text{A}$ to $0.4\text{A}$.

From the figure below, for $\Delta V_L=0$

for $10\text{ms}<t<20\text{ms}$$I=0.4\text{A}$$10\text{ms}<t<20\text{ms}$

Now $\Delta V_L=2{\text{V}}_{\text{for}}20\text{ms}<t<30\text{ms}$

$\begin{array}{l}-L\frac{dI}{dt}=2\text{V}\\ \frac{dI}{dt}=\frac{-2V}{50\text{mH}}=-40\text{A}/\text{s}\left(\text{or}\right)\\ dI=(-40\text{A}/\text{s})dt\\ I=(-40\text{A}/\text{s})t+c\\ \\ \\ \text{At}t=20\text{ms},\\ I=0.4\text{A}\\ 0.4\text{A}=(-40\text{A}/\text{s})(20\text{ms})\left({10}^{-3}\text{s}/\text{ms}\right)+c\\ \Rightarrow 0.4\text{A}=-0.8\text{A}+c\\ \Rightarrow c=1.2\text{A}I\\ =(-40\text{A}/\text{s})t+1.2\text{A}\end{array}$

Now At

$t=30\text{ms\hspace{0.33em}=\hspace{0.33em}}30\times {10}^{-3}\text{s}$

$\begin{array}{l}I=(-40\text{A}/\text{s})\left(30\times {10}^{-3}\text{s}\right)+1.2\text{A}\\ =-1.2\text{A}+1.2\text{A}\\ =0\text{A}\end{array}$

Current decreases from$0.4{\text{A}}_{\text{to}}0{\text{A}}_{\text{in}}$ the time$20\text{ms}<t<30\text{ms}$

After$t=30\text{ms}$current remains at $0\text{A}$

So we can have the following graph:

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