00:01
Hi in the given problem we are given with the rlc circuit and which l is 0 .1, r is 0 .6 ohms, c is 0 .4 faradays and et is the voltage is 60 volt.
00:24
So initial charge in the capacitor is 0 coulombs and the initial current is 7 .5 ampere.
00:30
7 .5 ampere is the initial current.
00:33
Let's call this as i0.
00:35
So we have to find the charge at t.
00:37
So we are given with the initial problem that l d square q over dt square plus r dq over dt plus q over c is equal to the total voltage.
00:56
So now q0 is equal to q0.
00:59
Q0 is the initial charge and initial current is q prime at 0, the derivative of the charge.
01:13
So this is q0.
01:16
Q prime 0 is the initial current and qt is the charge at that capacitor.
01:22
So if we put the values so 0 .1 d square q over dt square plus 0 .6 dq over dt plus 2 .5 q is equal to 60.
01:40
Now when you put the initial condition so on solving this by using the auxiliary equation using auxiliary equation...