you said at around 2:53 minutes non of the current that comes from the source passes through the conductor... i dont understand how after this you calculate that the current through the conductor equals 2*e^-t.... i dont get this part.... espicially the part where you take the equivalent circuit and calculate from it the time constant ....
We are only drawing the circuit to the right of the short. No current will flow through that part of the circuit (after passing through the 2 ohm resistor) coming from the voltage supply. So we drew the right part of the circuit separately.
Great series on RC and RL. One thing: in this example, the switch is closed at t=0, so the current starts increasing eventually becoming steady at 5tao. When using the equation i(t)=Io*e^-tR/L (as used above to find i(t) and also used in the previous example where the switch is opened at t=0) this would put the current near zero at 5 time constants which doesn't make sense. Shouldn't we be using the equation i(t) = Io(1-(e^-tRL)) for this example? This would result in having the same answer and sign for i4ohmR(t) whether you do the above method or use a current divider. Otherwise these two answers give opposite signs for current across the same resistor which doesn't make sense to me
As always, very helpful. I feel like it would have been good to show one more example (before going to #15), or even just briefly explain here the “what if” scenario, how it would be different if there were a 4th resistor between the 6Ω and 2H, to show how the 6Ω would NOT be short. Some people might realize this already from doing your previous example (#13) but I think it would be clearer for some who are still trying to figure out exactly what’s going on, because for me, it’s little things like that, that I get stuck on when it’s something very basic but just slipped my mind and has me second guessing myself.
Great vid! However, if all the current will take the short path shouldn't the 3 ohm resistor be discluded because no current will be passing through it
How come when t>0 the inductor is no longer acting like a short, thus preventing current from going into the 6ohm resistor? I.e, why doesn't the current just go through the inductor, then the 3ohm resistor, then back to the inductor ignoring the 6ohm? Is it because the inductor is no longer in a steady-state, thus opposing current movement?
First question: An inductor only affects the circuit when the current through the inductor changes (during the transient state). Second question: Current will flow through every possible path. If there is a path through the 3 ohm resistor and another path through the 6 ohm resistor, current will flow through both paths. (but more through the 3 ohm resistor).
Sir im a little bit confused. In the previous video (video #13) when your t is greater than 0 the switch was open. But here when your t is greater than 0 the switch is closed, am i missing something?
@ t>0, how did you know that current only flow across 10V and 2 Ohms ? How did you know that there is no current flowing across the 3 and 6 ohms ? Just curious.. Ty for reply.
After the switch closes, there branch with the switch offers no resistance (essentially a short circuit), so all of the current will flow through than branch and none of the current will flow through the branches with resistance.
It depends on the situation, what other components are there, if the switch opens or closes, etc. It is better to think about it like this: "If there is a steady state current flowing through an inductor, the reactance is zero and the inductor acts as a short.
at around 318 he said that there would still be current throught the resistive path after the short circuit. can somone explain why any current would pass through anywhere beside the path of least resistance namely the short
Good Day sir! What reference book do you use for this discussion? I have watched your video discussions about RL circuit and I Aced my test yesterday. Thank you very much
The inductor essentially become the current source (for a short while) and the current can either flow through the 6 ohm resistor OR the 3 ohm resistor. That is why they are in parallel. If the current must flow through both to get back to the inductor, then they would be in series.
After the switch closed, all of the current driven by the 10 V source will take the path along the "short", and the energy in the inductor will drive the current temporarily down the "short" path as well until the current through the inductor reaches zero. The 2 ohm resistor does not affect the inductor after the switch closes.
Your question is a fundamental part for solving RL circuits so if anyone else sees this, and has the same question, and doesn’t understand why he did this (especially after his reply) please go back and either watch his videos and/or read the book or your notes from class. Learning RC and RL took me many practice problems and TH-cam videos (hints why I’m here) to finally understand but it will click. Just give it time.
At steady state, (current doesn't change), the inductor does not oppose the current. (only when the current is changing). Therefore at steady state, the inductor looks like a wire with no resistance.
Because in a steady state condition, the inductor acts like a short (inductors only oppose a CHANGE in current) and thus all of the current will flow through the inductor and none through the 6 ohm resistor.
Once the switch is closed, there is a path with no resistance, which is the preferred path of the current. The inductor will have a current for a little while longer until steady state is reached.
In the problem given, the switch closes at t = 0 sec. That means there will be a transient period, before the current settles into a steady state condition. You will have to find the transient current and the steady state current.
Highly patient teacher.Very meticulous in steps, drawings and explanations.
100, 100 and 100, thanks dear Professor Biezen, the best explanation I have had ever.
You are welcome!
Great explanation mister, I am getting ready for my next week exam and your videos are very helpful. Greetings from Mexico
Welcome to the channel. Good luck on your exam. Keep studying.
Such a helpful and well-explained video! Thank you very much!
you said at around 2:53 minutes non of the current that comes from the source passes through the conductor... i dont understand how after this you calculate that the current through the conductor equals 2*e^-t.... i dont get this part.... espicially the part where you take the equivalent circuit and calculate from it the time constant ....
at 4:10, when you redraw the equivalent circuit, why did you skip the 2 ohm resistor?
We are only drawing the circuit to the right of the short. No current will flow through that part of the circuit (after passing through the 2 ohm resistor) coming from the voltage supply. So we drew the right part of the circuit separately.
You could also use current division at the end to find I6(t) and know that the inductor is acting as a source for polarity.
Current division is a handy technique.
Great series on RC and RL. One thing: in this example, the switch is closed at t=0, so the current starts increasing eventually becoming steady at 5tao. When using the equation i(t)=Io*e^-tR/L (as used above to find i(t) and also used in the previous example where the switch is opened at t=0) this would put the current near zero at 5 time constants which doesn't make sense. Shouldn't we be using the equation i(t) = Io(1-(e^-tRL)) for this example? This would result in having the same answer and sign for i4ohmR(t) whether you do the above method or use a current divider. Otherwise these two answers give opposite signs for current across the same resistor which doesn't make sense to me
I take that back, I got confused with the switch placement in the circuit. When would you use the above equation I mentioned?
when the switch is closed at t
never mind I just saw that the direction of the switch swing actually dictates the before and after .
Thanks so much sir I am jee advanced aspirant this has helped me better to understand well the trick to solve the problem thanks love from india
Most welcome
As always, very helpful. I feel like it would have been good to show one more example (before going to #15), or even just briefly explain here the “what if” scenario, how it would be different if there were a 4th resistor between the 6Ω and 2H, to show how the 6Ω would NOT be short. Some people might realize this already from doing your previous example (#13) but I think it would be clearer for some who are still trying to figure out exactly what’s going on, because for me, it’s little things like that, that I get stuck on when it’s something very basic but just slipped my mind and has me second guessing myself.
Why do you only include the 3 ohm and 6 ohm resistors but not the 2 ohm resistor when creating RL equivalent circuit?
Due to s. c line, and current flows always low resistance path
Great vid! However, if all the current will take the short path shouldn't the 3 ohm resistor be discluded because no current will be passing through it
No, the short path would only affect components that are parallel to it.
Does the sign of the current obtained matter? I used the current divider method so the current through the 6 ohm resistor = (3/3+6)*2e^-t = (2/3)*e^-t
The negative sign means that that current direction is opposite from the direction assumed.
How come when t>0 the inductor is no longer acting like a short, thus preventing current from going into the 6ohm resistor? I.e, why doesn't the current just go through the inductor, then the 3ohm resistor, then back to the inductor ignoring the 6ohm? Is it because the inductor is no longer in a steady-state, thus opposing current movement?
First question: An inductor only affects the circuit when the current through the inductor changes (during the transient state). Second question: Current will flow through every possible path. If there is a path through the 3 ohm resistor and another path through the 6 ohm resistor, current will flow through both paths. (but more through the 3 ohm resistor).
Thank you.
Sir im a little bit confused. In the previous video (video #13) when your t is greater than 0 the switch was open. But here when your t is greater than 0 the switch is closed, am i missing something?
There are both types of problems. Sometimes the switch opens up at t = 0 and sometimes the switch closes at t = 0
@@MichelvanBiezen ahhh ok. Thank you Sir!!
So please how will you know because I’m facing the same problem
Based on the arrow, the arrow represent what will happen AFTER time 0
@ t>0, how did you know that current only flow across 10V and 2 Ohms ? How did you know that there is no current flowing across the 3 and 6 ohms ? Just curious.. Ty for reply.
After the switch closes, there branch with the switch offers no resistance (essentially a short circuit), so all of the current will flow through than branch and none of the current will flow through the branches with resistance.
Sir, I have a little question here with the behaviors of the switch, in the previous video the switch was closed at t
Just a different example with a different initial condition.
Just check the diagram and the direction of arrow on the switch, the head of the arrow shows the final condition
So any situation that tells me t
It depends on the situation, what other components are there, if the switch opens or closes, etc. It is better to think about it like this: "If there is a steady state current flowing through an inductor, the reactance is zero and the inductor acts as a short.
at around 318 he said that there would still be current throught the resistive path after the short circuit. can somone explain why any current would pass through anywhere beside the path of least resistance namely the short
Inductors oppose a change in current. Therefore the current will continue to flow for a little while.
That's how it's done:). Great! Thank you!!
Good Day sir! What reference book do you use for this discussion? I have watched your video discussions about RL circuit and I Aced my test yesterday. Thank you very much
alexander and sadiku
When t
Never mind I see it.
So under DC condition the L acted as short circuit. Then it shorted the R=6(ohms).
thank you
What will be voltage across 3 ohm resistor??and which book are you following??
Great resource for my reference!
When t
The inductor essentially become the current source (for a short while) and the current can either flow through the 6 ohm resistor OR the 3 ohm resistor. That is why they are in parallel. If the current must flow through both to get back to the inductor, then they would be in series.
when t
Great vid as always! Just tell me why at @3:10 the 2 ohm resistor and the 10 V source weren't included.
After the switch closed, all of the current driven by the 10 V source will take the path along the "short", and the energy in the inductor will drive the current temporarily down the "short" path as well until the current through the inductor reaches zero. The 2 ohm resistor does not affect the inductor after the switch closes.
Your question is a fundamental part for solving RL circuits so if anyone else sees this, and has the same question, and doesn’t understand why he did this (especially after his reply) please go back and either watch his videos and/or read the book or your notes from class. Learning RC and RL took me many practice problems and TH-cam videos (hints why I’m here) to finally understand but it will click. Just give it time.
Why doesn't the inductor short out?
its rule
At steady state, (current doesn't change), the inductor does not oppose the current. (only when the current is changing). Therefore at steady state, the inductor looks like a wire with no resistance.
Why doesnt any current go the the 6 ohm resistor when t is less than zero?
Because in a steady state condition, the inductor acts like a short (inductors only oppose a CHANGE in current) and thus all of the current will flow through the inductor and none through the 6 ohm resistor.
@@MichelvanBiezen love your response time. Thank you sir.
why is none of the current going through the 6omh resistor
Once the switch is closed, there is a path with no resistance, which is the preferred path of the current. The inductor will have a current for a little while longer until steady state is reached.
Thanks sir 🙏🏻🙏🏻🇮🇳
You are welcome. Glad it was helpful. 🙂
Thank you! Sir
You are welcome. Glad the video was helpful.
when the switch is closed at t
Correct, as shown on the sketch to the right.
Is he following sadiku and alexander?
sir, why there is no current will go the 3 ohm resitance when the switch is closed? i hope you noticed me. thx
Not at the moment the switch is closed, because it takes a finite amount of time for the current to start flowing through the circuit
@@MichelvanBiezen thx sir, what is reference book in teaching this dc transient?
why the current @t > 0 is flowing through 10V and 2 ohms resistor only?
When you close the switch, that branch is now a short (zero resistance) and the current follows the path of least resistance.
ok thank you
Sir what does it mean when you're only required to find the voltage for t>0 but there is no initial stored energy?
In the problem given, the switch closes at t = 0 sec. That means there will be a transient period, before the current settles into a steady state condition. You will have to find the transient current and the steady state current.
Super sir great explenation
:) your videos are legendary
Thank you for sharing. And welcome to the channel!
Good video
thank you
You didn't explain the RC and why you mentioned in the description RC
Did you look at the other videos in the playlist?
you are graet
💜💜💜💜💜💜💜💜💜💜💜💜