Q 3.5-7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An industrial electromagnet can be modeled as an RL circuit, while it is being energized with a voltage source. If the inductance is 10 H and the wire windings contain   of resistance, how long does it take a constant applied voltage to energize the electromagnet to within 90% of its final value (that is, the current equals 90% of its asymptotic value)?

 

Step-by-Step Solution

Answer

The value or time is 7.68 sec.

Step 1: Determine the value of I(t)

Here given  $R=3\Omega ,L=10H$

 

We have the equation 

$$\begin{gathered} \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-\mathrm{R}}{\mathrm{L}}\mathrm{t}}(\int {\mathrm{e}}^{\frac{\mathrm{R}}{\mathrm{L}}}\frac{\mathrm{V}}{\mathrm{L}}\mathrm{dt}+\mathrm{K}) \\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-3}{10}\mathrm{t}}(\int {\mathrm{e}}^{\frac{3}{10}}\frac{\mathrm{V}}{10}\mathrm{dt}+\mathrm{K}) \\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-3}{10}\mathrm{t}}(\int {\mathrm{e}}^{\frac{3}{10}}\frac{\mathrm{V}}{10}\mathrm{dt}+\mathrm{K}) \\ \mathrm{I}\left(\mathrm{t}\right)={\mathrm{e}}^{\frac{-3}{10}\mathrm{t}}({\mathrm{e}}^{\frac{3}{10}\mathrm{t}}\frac{\mathrm{V}}{3}+\mathrm{K}) \\ \mathrm{I}\left(\mathrm{t}\right)=\frac{\mathrm{V}}{3}+{\mathrm{Ke}}^{\frac{-3}{10}\mathrm{t}} \end{gathered}$$

Apply the given condition then we get the value, $\mathrm{k}=-\frac{\mathrm{v}}{3}$ 

$\mathrm{I}\left(\mathrm{t}\right)=\frac{\mathrm{V}}{3}+(1-{\mathrm{e}}^{\frac{-3}{10}\mathrm{t}})$.

 

Step 2: Apply the asymptotic value on the equation

$\begin{array}{rcl}\underset{t\to \infty }{\lim }l\left(t\right)& =& \underset{t\to \infty }{\lim }\frac{V}{3}+\left(1-e^{\frac{-3}{10}t}\right)\\ & =& \frac{V}{3}\end{array}$

Step 3: Evaluate the value of time

Since the 90% 0f the asymptotic value of current is $0.9\frac{\mathrm{V}}{3}$ .

 

$$\begin{gathered} \mathrm{I}\left(\mathrm{t}\right)=0.9\frac{\mathrm{V}}{3} \\ \mathrm{I}\left(\mathrm{t}\right)=\frac{\mathrm{V}}{3}(1-{\mathrm{e}}^{\frac{-3}{10}\mathrm{t}}) \\ =0.9\frac{\mathrm{V}}{3} \\ (1-{\mathrm{e}}^{\frac{-3}{10}\mathrm{t}})=0.9 \end{gathered}$$

 

By solving for t = 7.68 sec.

 

Hence, the value or time is 7.68 sec.