0:00
Let's solve this question.
00:01
We have ri plus l di by dt is equal to a and we have given condition i of 0 is equal to 0.
00:12
So, we will get our equation as di by dt plus r by li is equal to e by l.
00:23
So, when using laplace theorem in both side, we will get it as l di by dt plus r by li is equal to e by ls.
00:42
So, when we solve this equation, we are going to get it as s is plus r by li is equal to e by l into 1 by s.
01:05
So, from here we will get i of s is equal to e upon l of s into 1 upon s plus r by l.
01:21
So, after doing some or more substitutions and all, i of s is equal to e upon s into ls plus r.
01:36
Then if we change it in some different form, we will get a upon s plus b upon ls plus of r.
01:46
So, at final, we will be getting this as e upon s into ls plus r is equal to a ls plus a r plus b s divided by s into ls plus r is equal to a l plus b s plus a r upon s into ls plus r.
02:28
Okay...