00:01
In this question, we are given a circuit v of t plus minus capacitor and here register r l and current is flowing like this.
00:33
Now, on applying kvl, we consider v of t will be equals to l of di over dt plus r i plus 1 over c integral of i dt.
00:55
Now, if we differentiate this with respect to t that is differentiating with respect to t.
01:04
So, dv over dt will be equals to l of d2i over dt2 plus r di over dt plus i over c that is equals to 0.
01:24
Now, d2i over dt2 plus r over l di over dt plus 1 over c.
01:34
Now, we can write this as v of t plus i over lc is also equal to 0.
01:38
We just divided this whole equation by l and we will get this result.
01:44
Now, we are given that r is 10 ohm, the capacitor value capacitance is 1 over 100 farad and the value of l is given as 1 over 200 v equals 12 volt.
01:59
So, putting all these values in this equation, we will get that d2i over dt2 plus r over l that is 20 di over dt plus 1 over lc will be 200 i equals 0.
02:22
Now, we can clearly see that this equation is similar to equation of m square plus 20 m plus 200 equals 0.
02:34
So, that is a quadratic equation and we know that the root of quadratic equation m will be given by the formula minus b plus minus under the root b square minus 4ac over 2a...