Unnecessary complication while calculating the initial current through the 12 ohm resistor (initial inductor current). No need to calculate all three currents. Consider that one can simply join the 24 and 8 ohm resistors to form a 24*8/32 = 6 ohm resistor and then by current division I2 = 15 amps * (6/(6 + 12)) = 5 amps.
Thank you so much for these videos. I love how they start from the dead and dumb basics and then gradually become sophisticated as it progresses. It has been instrumental in me developing confidence in these. Though, that current division thing did throw me off lol.
in case anyone is wondering the exercicises present in this video and on other from this series are from the book "Fundamentals of Electric Circuits- Charles K. Alexander,Mathew N.O. Sadiku 5th Edition"
Please mention your use of Current divider in solving branch circuits currents. Also, simplify Resistors in // to make it easier to understand. Other than that, its a good reference. Thanks.
I may have shown how that equation is developed but with 2 branches (R1 and R2) the equation is: i1 = I total ( (R2) / (R1 + R2)) with 3 resistors the equation is: i1 = I total (R2)(R3) / ((R1)(R2) + (R1)(R3) + (R2)(R3))
When currents have settled in and no longer change, an inductor doesn't oppose a steady state current. But as soon as the swith is opened, the current begins to change and the inductor opposes that change, initially not allowing the current to change for the first fraction of a second.
It’s opposite of capacitor When switch is closed across cap it acts as a short because opening or closing a switch is high frequency phenomenon happens quickly so caps are shorted at high frequency Similarly inductor acts as open at high frequency and acts as open when switch is closed
After the switch opens up, no current will flow from the current source to the right part of the circuit, since that part of the circuit is not a complete loop.
Before the switch is opened, the incuctor will indeed act like a short circuit. But just after the switch is opended the inductor will not act like a short circuit
During steady state condition, IN THIS CIRCUIT, no current will flow through the inductor. After the switch opens, current cannot flow from the left to the right part of the circuit, since the circuit is now open.
During steady state (t < 0) the inductor has no reactance and there is no resistor between the 8 ohm resistor and the bottom of the circuit. Therefore essentially those 2 parts of the circuit are shorts and all charges will "choose" the path of least resistance. They will not go through the 5 ohm resistor so, that part of the circuit will act as an open.
After the current passes through the 8 ohm resistor, it will not go through the 5 ohm resistor because it will take the path with no resistor instead. (the path of least resistance).
He's referring to the beginning you said that it was an open and you meant short, you even do the calculations as if it were a short figuratively. On a side note I am subscribed and I love your channel.
Unnecessary complication while calculating the initial current through the 12 ohm resistor (initial inductor current). No need to calculate all three currents. Consider that one can simply join the 24 and 8 ohm resistors to form a 24*8/32 = 6 ohm resistor and then by current division I2 = 15 amps * (6/(6 + 12)) = 5 amps.
likewise you can find the equiv circuit: 15A and 4ohm, get Veq = 60, then use Ohms law and divide 60V by each resistor value to get all current values
Thank you so much for these videos. I love how they start from the dead and dumb basics and then gradually become sophisticated as it progresses. It has been instrumental in me developing confidence in these.
Though, that current division thing did throw me off lol.
We have videos on current division as well. Glad you find these helpful
i think the inductor under the DC condition will be short circuit not an open circuit
He didn’t say so but he had correct diagram :)
yeah i thought i was the only one who noticed that
sir you deserve a nobel prize, thank you for your service to society
in case anyone is wondering the exercicises present in this video and on other from this series are from the book "Fundamentals of Electric Circuits- Charles K. Alexander,Mathew N.O. Sadiku 5th Edition"
That is correct.
You do an amazing job explaining the breakdown and recapping afterwards. Thank you so much for the quality videos and for the help!!
Please mention your use of Current divider in solving branch circuits currents. Also, simplify Resistors in // to make it easier to understand. Other than that, its a good reference. Thanks.
Sir, at 07:24 isn't some of the 5A current going to the 5 ohm resistor? Why did we take it as if it was all current going through the inductor?
At the moment the switch opens, there is no current flowing through the 5 ohm resistor.
@@MichelvanBiezen Thank you so much.
at 3:06, where does that formula calculating i1 come from? is there any way to solve for i1?
I may have shown how that equation is developed but with 2 branches (R1 and R2) the equation is: i1 = I total ( (R2) / (R1 + R2)) with 3 resistors the equation is: i1 = I total (R2)(R3) / ((R1)(R2) + (R1)(R3) + (R2)(R3))
Why it becomes e^-2t instead of -0.5? 8:11
1/0.5 = 2
at the previous video,the inductor were short circuit but now open circut, whats the difference,also is there any video about how inductors work ?
Inductors only oppose a CHANGE in current. For steady state current the inductor acts as if is not there and can be replaced by a short.
Thanks a lot! I've learned so much. Please make video tutorial more about Electrical Engineering. More power to you Sir Michel!
at the beginning sir, you said at steady state the inductor will act as an open, I guess you meant short
Immediately before the switch is openend, the inductor will act like a short. Immediately after the switch is openend, the incuctor acts as an open.
@@MichelvanBiezen can you please explain why
When currents have settled in and no longer change, an inductor doesn't oppose a steady state current. But as soon as the swith is opened, the current begins to change and the inductor opposes that change, initially not allowing the current to change for the first fraction of a second.
It’s opposite of capacitor
When switch is closed across cap it acts as a short because opening or closing a switch is high frequency phenomenon happens quickly so caps are shorted at high frequency
Similarly inductor acts as open at high frequency and acts as open when switch is closed
Cant seem to understand when it at
acts as short and open circuit , what should be the conditions ?
Richie when a component is connected to the same node. U can give me ur social media or iMessage so that I can draw some diagrams n show u
@@abpdev can u explain to me how to determine the current flow of the inductor after current of source cut out
Thank you very much sir, I follow you from IRAQ
Welcome to the channel!
Why does the inductor act as an open circuit, at steady state current isn't it meant to act as a short circuit?
An inductor at steady state acts as a "short" circuit.
I like your videos. In this example for the Final current is not supposed to be included in the final equation?
The final current through the 12 ohm resistor is zero
should the 24ohm resister be considered as part of the resister networks visible to the inductance?
After the switch opens up, no current will flow from the current source to the right part of the circuit, since that part of the circuit is not a complete loop.
0:49 correction the inductor will act like a short circuit and the 5 ohm resistor will act like an open circuit
Before the switch is opened, the incuctor will indeed act like a short circuit. But just after the switch is opended the inductor will not act like a short circuit
During steady state condition inductor should act like short circuit instead of open circuit as you mistakenly said.Am I right ?
During steady state condition, IN THIS CIRCUIT, no current will flow through the inductor. After the switch opens, current cannot flow from the left to the right part of the circuit, since the circuit is now open.
Why is 24 ohm resistor not used to calculate R eq?
I don't understand why we didn't use the 5 ohm Resistance
During steady state (t < 0) the inductor has no reactance and there is no resistor between the 8 ohm resistor and the bottom of the circuit. Therefore essentially those 2 parts of the circuit are shorts and all charges will "choose" the path of least resistance. They will not go through the 5 ohm resistor so, that part of the circuit will act as an open.
Sir, why we don't take the current through inductor as (5+7.5) i.e. the current passing through 12 ohm and
8 ohm respectively?
After the current passes through the 8 ohm resistor, it will not go through the 5 ohm resistor because it will take the path with no resistor instead. (the path of least resistance).
Sir why the initial current taken is 5 ohm
That is the current through the inductor for t < 0
For t
Thanks
Sir give pls give simple steps
Have you seen the videos in the earlier part of the playlist?
❤️
Thanks!
Lol, wrong . when inductor in steady stage it should be short circuit , not open circuit
video is correct
He's referring to the beginning you said that it was an open and you meant short, you even do the calculations as if it were a short figuratively. On a side note I am subscribed and I love your channel.
You also say it's a short shortly after you call it an open.
+1
0:48
you said open,but you are solving it like a short circuit.And why is the 5ohm resistor cut out.you do mistakes man,not a reliable channel
why you didnt take 5ohm i got it
@@fayeezullah9628 Stop thumbing yourself up. Look at the video again and pay attention to what he says in the beginning. He did it right.
@@burgerking220 hey stop fucking around
@@fayeezullah9628 meet me outside bro
@@burgerking220 sure where do you stay bitch?