Applications of laplace transform for circuit analysis.RLC circuit analysis by laplace transform.

#applicationOfLaplaceTransform
the laplace transform can be used to analyse the electrical circuits in very easy manner.
definition of laplace transform link: • Definition,application...
RLC circuit analysis by laplace transform

[@Newnawn]

Thank you so much, your explanations are far clearer than my university professor's. Teachers often try to overcomplicate their explanations instead of formulating it in an easy to understand way, like you did here. Good job!

[@PunmasterSTP]

Hey how'd the rest of your class go?

[@frost5182]

Outstanding explanation and very clear to the point.

[@manaspandey2355]

Amazing explanation 👍👍👍👍👏👏👏👏👏

[@PunmasterSTP]

This was a really useful explanation and derivation. Thank you for making and sharing this video!

[@ashlw]

Thank you so much. Great explanation. Crunching for exams

[@ebarbie5016]

The standard Laplace model for an inductor is a current source i(o-)/s in parallel to the inductor's impedance. The model used here is correct but is actually a source-converted equivalent.

[@billpulvera8817]

Thank you so much, well explained

[@jaimemanuelgarciadominguez1037]

Very well explained, thank you

[@alesfigue]

This has been extremely helpful!! thank you

[@crediblehuman9753]

This was perfect thank you so much.

[@bitdhbvhcbbhx7979]

Perfect explanation Thanks❤️

[@tayyabahmad5291]

you make a blunder t0 inductor act as open circuit and t-->infinity which means at steady state inductor act as short circuit VL=Ldi/dt if i is constant in steady state the derivative of a constant is 0 and the VL=0 which means that short circuit and reverse order for capacitor

[@zeyaley13]

thank you

[@faris4363]

Thanks

[@typing6568]

I need more plz

[@arbindgiri5617]

also teach us how to find voltage

[@pradyumnanayak9844]

Namaste🙏.

[@CC.cinematics]

In 0 condition L is open circuit and C is short circuit..you have written reverse

[@PunmasterSTP]

I think he was saying that the switch had already been closed for a long time, but then was *opened* right at time t = 0. Therefore the inductor would function as a short circuit and the capacitor would function as an open.

[@g-tensolution8527]

At t=0-, Circuit is energised, so L should be open circuit and C should sort cicuit.

[@PunmasterSTP]

@@g-tensolution8527 I may be misunderstanding something, but I thought that if voltage and current can reach steady state, then an inductor would offer no resistance to current (hence, act as a short circuit) and a capacitor would be fully charged and resist the buildup of any more charge (act as an open circuit). Does that sound reasonable?

[@Dalziel45]

Why is the equivalent circuit not explained? Feels like it's pulled out of thin air, so the whole video ends up void :c

[@frost5182]

Because explanation and where this came from is quite lengthy topic since it involves integrating and so on.