00:01
Here we have to find out the transform function that is g8s which is equals to v0 of s which is divided by vi of s in the flowing electrical circuit.
00:11
One henry inductor is given which is connected to the one henry inductor which is further connected to the one ohm resistor which is connected in parallel with the inductor and with the capacitor that is one henry this is one farad capacitor.
00:28
So again another capacitor of one farad capacitance is connected from here.
00:34
So this is the system which we are given here we are considering that is vm of s from here is equals to v1 of s minus vh of s that from here is equals to v of s minus 0 that is equals to v1 of s and v0 of s from here is equals to v3 of s minus v4 of s that become equals to v3 of s minus 0 that is equals to v3 of s.
01:00
Now we have to we are considering about the node 2 and we have to apply the kcl there.
01:05
So this from here is equals to v2 of s minus v1 of s which is divided by 1 of s plus v2 of s minus v4 of s which is divided by 1 of s plus v2 of s minus v4 of s is equals to 0.
01:30
So this is the term which we are considering here so we are considering that v2 of s divided by 1 of s minus v1 of s which is divided by s plus v2 of s divided by s plus v2 of s which is divided by 1 divided by s plus v2 of s which is divided by 1 minus v3 of s which is divided by 1 that is equals to 0.
01:50
Simplifying the term we get the value of v2 of s that is s raised to the power 2 plus s plus 2 that from here is equals to v1 of s plus s of v3 of s let's say this is the equation number 1...