It is really entertaining watching your videos unlike other IIT professors because of your unique comic nature. You create great interactive environment.I seriously learned a lot by this lecture that too with a continuous smile on my face. It is very rare (haha).This triggers me too watch the whole series even though some of topics are out of syllabus and i'm sure i'll learn all of them smiling.Thank you so much sir. At 52:06 "I can afford to do that but you can't!" (hahaha).
At 33:29 when both the capacitor and inductor have same voltage then how the capacitor starts dicharging??. potential is same that means no current will flow between capacitor and inductor.. Anybody answer?
Okay, when current stops flowing, the capacitor hold the charges, the electrons... But the inductor when pumped all charges in the capacitor now has no magnetic energy and hence it short, or no voltage appearance due to change in magnetic field... So the stored charges now starts to flow current, as the voltage accros capacitor is same to inductor... Its analogous to spring mass oscillation system, where spring is capacitor, and mass having kinetic energy is inductor.
To me D Roy Chaudhary seems to be undermining his students potential. They were correct to ask why current is divided half in inductor and closed switch. "At minus infinity when battery is plugged into the circuit, current finds two path, one is having inductor and other is short circuit switch. Since initially relaxed inductor opposes any current (essentially act like open circuit). So current will follow short circuit switch path and inductor path will never get any magnetic flux to establish any current across it." He did not justify this argument. Instead diverted topic to Indian history and never answered that question clearly. By the way in Time domain you can not forget all the history. In fact you are not allowed to do so.
It takes time for the inductor to conduct, depends upon potential difference at its terminals, and how large the inductor is... Eventually the current will be forced to flow due to potential difference, and will be stored in magnetic field... Giving rise to half current of total. ALSO THE INDUCTOR L1 WOULDN'T HAD ALLOWED SUDDEN RISE IN VOLTAGE.... SO L2 GETS CHANCE TO SLOWLY INCREASE SHORT.
@@chanakyasinha8046 At 44:22 the assumption that the steady-state current through *L1* for all *t < 0* divides evenly between *L2* and the closed switch is arbitrary. Any other distribution of currents would also satisfy all network equations! The reason is that for *t < 0* the "DC steady-state network" has two parallel shorts, therefore it either has no solution at all or an infinite family of solutions, in this case the latter. A short analysis of the "DC steady-state network" yields: *t < 0: i_L1(t) = V / R, i_L2(t) = J (arbitrary const)* The current through the closed switch is the difference between the two, of course. ----------------------------------------------------------------------------------------------------------------------------------------------- *Discussion of **44:22** ff.:* The main argument given why the steady-state current should evenly divide between the two shorts is the idea that we can substitute both L2 and the closed switch by the SAME small resistor *r -> 0* . The fact that both have to be the same in that limiting process is arbitrary. *Counterexample:* You could also substitute *"L2 -> a * r" and "closed switch -> b * r", a, b > 0, r -> 0* Depending on *a, b* you get infinitely many different steady-state current distributions. In fact, it is even arbitrary to use such a limiting process to try and determine both currents - if you only use the information given (aka KCL, KVL and the element equations), you have a network with infinitely many solutions. Why not acknowledge that fact? Depending on your view on such networks, the problem was either under-specified or ill-formed. Of course you can always calculate a general solution still dependent on *J* . I'd argue that is the way to go, as you can learn the most! ----------------------------------------------------------------------------------------------------------------------------------------------- *Rem.:* Even if you solve the differential equations for *t < 0* with initial conditions at *t -> -oo* you get the same result: *right mesh: 0 = L2 * d / dt i_L2(t) => i_L2(t) = J (arbitrary const)* *big loop: 0 = -V + R * i_L1(t) + L1 * d / dt i_L1(t) => i_L1(t) = V / R* Of course the second solution contains a limiting process for the initial conditions at *t -> -oo* .
@@carstenmeyer7786 you elaborated it well,thanks. or we can reframe the question ,such that battery was added at t=0,and switch s was closed initially ...for case 1,resistance of both L2 and switch s =0,then at t-> infinity ,L2 remains open .and for case 2 ,for both having non zero resistances ,there will be minor voltage difference across switch s,which will ultimately force some current and energise L2 at t->infinity . I guess you are saying this!
![[TH] 2024 PMSL SEA Finals D2 | Fall | ต้องรักษาฟอร์ม อย่ายอมให้ใครแซง](http://i.ytimg.com/vi/x38QsGn7A3M/mqdefault.jpg)
It is really entertaining watching your videos unlike other IIT professors because of your unique comic nature. You create great interactive environment.I seriously learned a lot by this lecture that too with a continuous smile on my face. It is very rare (haha).This triggers me too watch the whole series even though some of topics are out of syllabus and i'm sure i'll learn all of them smiling.Thank you so much sir.
At 52:06 "I can afford to do that but you can't!" (hahaha).
☺☺☺☺
At 33:29 when both the capacitor and inductor have same voltage then how the capacitor starts dicharging??. potential is same that means no current will flow between capacitor and inductor.. Anybody answer?
Okay, when current stops flowing, the capacitor hold the charges, the electrons... But the inductor when pumped all charges in the capacitor now has no magnetic energy and hence it short, or no voltage appearance due to change in magnetic field... So the stored charges now starts to flow current, as the voltage accros capacitor is same to inductor... Its analogous to spring mass oscillation system, where spring is capacitor, and mass having kinetic energy is inductor.
To me D Roy Chaudhary seems to be undermining his students potential. They were correct to ask why current is divided half in inductor and closed switch.
"At minus infinity when battery is plugged into the circuit, current finds two path, one is having inductor and other is short circuit switch. Since initially relaxed inductor opposes any current (essentially act like open circuit). So current will follow short circuit switch path and inductor path will never get any magnetic flux to establish any current across it."
He did not justify this argument. Instead diverted topic to Indian history and never answered that question clearly.
By the way in Time domain you can not forget all the history. In fact you are not allowed to do so.
It takes time for the inductor to conduct, depends upon potential difference at its terminals, and how large the inductor is... Eventually the current will be forced to flow due to potential difference, and will be stored in magnetic field... Giving rise to half current of total.
ALSO THE INDUCTOR L1 WOULDN'T HAD ALLOWED SUDDEN RISE IN VOLTAGE.... SO L2 GETS CHANCE TO SLOWLY INCREASE SHORT.
@@chanakyasinha8046 but how can the current be half of L1 before the switch was open ?
@@arifromIITKGP at what video timing?
@@chanakyasinha8046 At 44:22 the assumption that the steady-state current through *L1* for all *t < 0* divides evenly between *L2* and the closed switch is arbitrary. Any other distribution of currents would also satisfy all network equations!
The reason is that for *t < 0* the "DC steady-state network" has two parallel shorts, therefore it either has no solution at all or an infinite family of solutions, in this case the latter. A short analysis of the "DC steady-state network" yields:
*t < 0: i_L1(t) = V / R, i_L2(t) = J (arbitrary const)*
The current through the closed switch is the difference between the two, of course.
-----------------------------------------------------------------------------------------------------------------------------------------------
*Discussion of **44:22** ff.:* The main argument given why the steady-state current should evenly divide between the two shorts is the idea that we can substitute both L2 and the closed switch by the SAME small resistor *r -> 0* . The fact that both have to be the same in that limiting process is arbitrary.
*Counterexample:* You could also substitute
*"L2 -> a * r" and "closed switch -> b * r", a, b > 0, r -> 0*
Depending on *a, b* you get infinitely many different steady-state current distributions. In fact, it is even arbitrary to use such a limiting process to try and determine both currents - if you only use the information given (aka KCL, KVL and the element equations), you have a network with infinitely many solutions. Why not acknowledge that fact?
Depending on your view on such networks, the problem was either under-specified or ill-formed. Of course you can always calculate a general solution still dependent on *J* . I'd argue that is the way to go, as you can learn the most!
-----------------------------------------------------------------------------------------------------------------------------------------------
*Rem.:* Even if you solve the differential equations for *t < 0* with initial conditions at *t -> -oo* you get the same result:
*right mesh: 0 = L2 * d / dt i_L2(t) => i_L2(t) = J (arbitrary const)*
*big loop: 0 = -V + R * i_L1(t) + L1 * d / dt i_L1(t) => i_L1(t) = V / R*
Of course the second solution contains a limiting process for the initial conditions at *t -> -oo* .
@@carstenmeyer7786 you elaborated it well,thanks.
or we can reframe the question ,such that battery was added at t=0,and switch s was closed initially ...for case 1,resistance of both L2 and switch s =0,then at t-> infinity ,L2 remains open .and for case 2 ,for both having non zero resistances ,there will be minor voltage difference across switch s,which will ultimately force some current and energise L2 at t->infinity .
I guess you are saying this!
1,2,3 vedios are not opening please solve that issue
You can download them from somewhere else buddy.