Im having difficulty understanding the integral calculation in number 4. How does (Response) section of "V/L" convert into "V/R"? Where does exponents go?
The last equation is from the textbook.
*I wanted to show the each steps more clearly so I took a picture of the each step.
Best Answer
The solution approach is just a bog standard 1st order diff eq solution. Probably in chapter 1 of any diff eq book. But it sounds more like you don't see how to integrate an exponential. Which is even earlier -- like MTH252, I think.
The standard form is just as given:
$$\frac{\textrm{d}y}{\textrm{d}t} + P_x\cdot y = Q_x$$
If you can set things up like that, then your integrating factor (which is a nifty way to solve these) is:
$$\mu=e^{\int P_x\; \textrm{d}x}$$
Then then the solution is:
$$ y=\frac{1}{\mu}\int \mu\cdot Q_x\;\; \textrm{d}x$$
You can see the minor trouble that they went to in order to put it into standard form, through they called one of the terms \$r_x\$ instead. Doesn't matter. So let's just look at the solution:
$$\begin{align*} i_t &= \frac{1}{\mu}\int \mu\cdot Q_x\;\; \textrm{d}t,~~~~~\mu= e^{\int P_x \textrm{d}x}=e^{\int \frac{R}{L} \textrm{d}x}=e^{\frac{R}{L}x}\\ i_t &= \frac{1}{e^{\frac{R}{L}t}}\int e^{\frac{R}{L}x}\cdot \frac{V_s}{L}\;\; \textrm{d}x \\ i_t &= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{L}\int e^{\frac{R}{L}x} \;\; \textrm{d}x \\ i_t &= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{L}\left[\frac{L}{R}\left(e^{\frac{R}{L}x}\right)\right]\Biggr\vert_0^{t} \\ i_t &= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{R}\left[e^{\frac{R}{L}x}\right]\biggr\vert_0^{t} \\ i_t &= \left(e^{-\frac{R}{L}t}\right)\frac{V_s}{R}\left[e^{\frac{R}{L}t} - 1\right] \\ i_t &= \frac{V_s}{R}\left[1 - \left(e^{-\frac{R}{L}t}\right)\right] =\frac{V_s}{R} - \frac{V_s}{R}\left(e^{-\frac{R}{L}t}\right)~~~~~~\tau=\frac{L}{R} \\ \therefore i_t &= \frac{V_s}{R} - \frac{V_s}{R}\left(e^{-\frac{t}{\tau}}\right) \end{align*}$$
I hope that helps.