5:01 thanks for the video, what does it mean that the inductor has an initial current I(0-)? Is it like there's an initial magnetic field on the coil without being connected to anything or what does it mean?
It means that, the initially inductor was connected to some other circuit. And because of that, there is some flow of current through the inductor. Now, as you know, the inductor opposes the instantaneous change in the current. That means even after disconnecting it from the previous circuit, and connecting it to the new circuit, immediately the same current will continue to flow through the inductor. So, in the newly connected circuit of the inductor, that current will be the initial current through the inductor. I hope, it will clear your doubt. For more info, you may check this video, which I have made some time back. ua-cam.com/video/3YinmbkU0DE/v-deo.htmlsi=RzjcX-HDj1lVEQ5e
The Laplace Transform of unit step function (u(t)) is 1/s. For more info, please check the Laplace Transform Playlist, where I have already covered the Laplace Transform of some basic functions. The link is already provided in the pinned comment.
In the loop while moving around, if there is a rise in potential across the element, then it will considered as positive voltage. For 1/S voltage source, as we move in the clock wise direction, then there is a rise in potential across it. So, that is considered as positive voltage. While for other elements there is a drop in potential. So, they are considered negative. The net sum is 0. so, all the negative terms have taken on the right side. Using Ohm's law ( V = I*R), the drop across each element is calculated. For the element 3/S, the net current flowing in the downward direction is (I1- I2). So, drop across is 3/s (I1 - I2). And that is how entire equation is written. I hope, it will clear your doubt. For more info, you can check this basic video on KVL: ua-cam.com/video/dlROTnDbULo/v-deo.htmlsi=e3U2BNUMvEcNTav0
While writing the s domail equivalence of inductor, when we get the term v(s) = Lsi(s) + Li(0-), why does that translate to an inductor of ls inductance? Wouldn't that have a different voltage?
In the previous videos of Laplace transform, I have already covered the Laplace transform of some basic functions . For more info, you can check this video: ua-cam.com/video/KWcRYnyWDCo/v-deo.htmlsi=Hb2GkfUPkbAYTp-V It would be good if you watch this playlist in a sequence for better understanding. ua-cam.com/play/PLwjK_iyK4LLD8Cdj0SKiMZFoK2d3eE6v2.html&si=uSTWFLnT1eAqa-fs
can i solve ac circuit this way instead of the phasor domain? man i really hate dealing with complex numbers a quick look and i cant seem to find how to solve the phase in s domain seems only for transient analysis so im stuck with phasor? will it actually be harder to solve ac using s domain than phasor domain?
Yes, for sinusoidal analysis, it is better to solve it using the phasors. If you deal it in the s-domain, then it involves a little more mathematics. (Finding the Inverse Laplace Transform is additional step). Moreover, if you deal it in the s-domain, then the response of the circuit contains both transient and steady state response. If you just want to consider steady state response, then first you need to remove the transient response. (The terms which contains decaying exponential terms) On the channel, through different videos and different solved examples, the ac analysis of the circuit using the phasors have been already covered. If you required, you can check the playlist of network analysis. ua-cam.com/play/PLwjK_iyK4LLBN9RIDQfl9YB4caBYyD_uo.html&si=RnvwjOP3Cc03xiZp
For other useful videos related to Laplace Transform, check this playlist:
ua-cam.com/play/PLwjK_iyK4LLD8Cdj0SKiMZFoK2d3eE6v2.html&si=zPi5_EwlASecoTIR
Sir coordinate system and transformation stady this chapter please
Just now I searched for the same topic and came the notification of yours.
Thankyou so much for this wonderful classs❤🎉
We really thank you you are helping us understand more than we would imagine if its possible call tell us On sit transistor
Very informative 🎉🎉
Have you also made lecture videos on Three phase circuits??
And I really appreciate your hardwork..
amazing video, pls keep going
Thank you so much sir
5:01 thanks for the video, what does it mean that the inductor has an initial current I(0-)? Is it like there's an initial magnetic field on the coil without being connected to anything or what does it mean?
It means that, the initially inductor was connected to some other circuit. And because of that, there is some flow of current through the inductor. Now, as you know, the inductor opposes the instantaneous change in the current. That means even after disconnecting it from the previous circuit, and connecting it to the new circuit, immediately the same current will continue to flow through the inductor. So, in the newly connected circuit of the inductor, that current will be the initial current through the inductor. I hope, it will clear your doubt.
For more info, you may check this video, which I have made some time back.
ua-cam.com/video/3YinmbkU0DE/v-deo.htmlsi=RzjcX-HDj1lVEQ5e
@@ALLABOUTELECTRONICS thank you, as always excellent explanations 🙏
In example 2 at 20:20 why vout = 4*i why 4 multiplied
It is a simple ohm's law. Voltage across the resistor is I*R.
Sir my dought is that in first question final answer was written without u(t)
But second Question final answer was written in terms of u(t), why?
In the first example, I forgot to write it. But while writing the output, one should always write it.
@@ALLABOUTELECTRONICS
Thanks
In the first example, why do we introduce root 2 squared?
At 9:23, how did u(t) become 1/s?
The Laplace Transform of unit step function (u(t)) is 1/s. For more info, please check the Laplace Transform Playlist, where I have already covered the Laplace Transform of some basic functions. The link is already provided in the pinned comment.
Can anyone explain how at 12:17 kvl eq is written
In the loop while moving around, if there is a rise in potential across the element, then it will considered as positive voltage. For 1/S voltage source, as we move in the clock wise direction, then there is a rise in potential across it. So, that is considered as positive voltage. While for other elements there is a drop in potential. So, they are considered negative. The net sum is 0. so, all the negative terms have taken on the right side. Using Ohm's law ( V = I*R), the drop across each element is calculated. For the element 3/S, the net current flowing in the downward direction is (I1- I2). So, drop across is 3/s (I1 - I2).
And that is how entire equation is written.
I hope, it will clear your doubt.
For more info, you can check this basic video on KVL:
ua-cam.com/video/dlROTnDbULo/v-deo.htmlsi=e3U2BNUMvEcNTav0
While writing the s domail equivalence of inductor, when we get the term v(s) = Lsi(s) + Li(0-), why does that translate to an inductor of ls inductance? Wouldn't that have a different voltage?
This is S-domain circuit. In the S-domain circuit I(s) is the current in S-domain. So, Ls represents the equivalent inductance in S-domain.
sir what is the frequency shifting property?
I have covered it in this video. Please check the link :
ua-cam.com/video/3oTsZwaLjE8/v-deo.htmlsi=KIo3zJHOcglOoNEH
Sir did you explained convolution theorem in this video series
no
At 10:57 in the first question why did u write u(t)=1/s i mean how😅
In the previous videos of Laplace transform, I have already covered the Laplace transform of some basic functions . For more info, you can check this video: ua-cam.com/video/KWcRYnyWDCo/v-deo.htmlsi=Hb2GkfUPkbAYTp-V
It would be good if you watch this playlist in a sequence for better understanding.
ua-cam.com/play/PLwjK_iyK4LLD8Cdj0SKiMZFoK2d3eE6v2.html&si=uSTWFLnT1eAqa-fs
16:05 sir how it is time shifting?
It is a frequency shifting property.
why did you multiply 3s on both side ? 13:34
On the right hand side, after simplification, in the denominator you will get 3s. So, that is why both sides have been multiplied by 3s.
in example 2 how does v0(s) become 4*i(s) where does 4 come from
It is as per the ohm's law. V = I x R. And here, R is 4 ohm. So, Vo (s) = 4 x I(s). I hope, it will clear your doubt.
can i solve ac circuit this way instead of the phasor domain? man i really hate dealing with complex numbers
a quick look and i cant seem to find how to solve the phase in s domain
seems only for transient analysis
so im stuck with phasor?
will it actually be harder to solve ac using s domain than phasor domain?
Yes, for sinusoidal analysis, it is better to solve it using the phasors. If you deal it in the s-domain, then it involves a little more mathematics. (Finding the Inverse Laplace Transform is additional step). Moreover, if you deal it in the s-domain, then the response of the circuit contains both transient and steady state response. If you just want to consider steady state response, then first you need to remove the transient response. (The terms which contains decaying exponential terms)
On the channel, through different videos and different solved examples, the ac analysis of the circuit using the phasors have been already covered.
If you required, you can check the playlist of network analysis.
ua-cam.com/play/PLwjK_iyK4LLBN9RIDQfl9YB4caBYyD_uo.html&si=RnvwjOP3Cc03xiZp
Sir can you perform this in programming for better explanation
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