I have a doubt regarding the circuit in the figure. It is assumed that the switch opens at t=0; At t=0- (i.e. just before opening the switch)we can determine the current thru 1H inductor as 2.5 A and current thru 2H inductor as 1.25 A. But what is the value of current at t=0+ ie the instant immediately after the switch is opened?

 

I have a doubt regarding the circuit in the figure. It is assumed that the switch opens at t=0; At t=0- (i.e. just before opening the switch)we can determine the current thru 1H inductor as 2.5 A and current thru 2H inductor as 1.25 A. But what is the value of current at t=0+ ie the instant immediately after the switch is opened?

If the switch was closed (= shorted) until t=0, the initial current in the 2H inductor would be zero, wouldn't it? There is no coupling apparent between the coils, so where does the 1.25A come from?

 

I guess the exercise wants us to look at the 2H inductor as a pure short-circuit and the current through two 0Ω branches is ...uh... divided. Silly if you ask me.

 

Good point Georacer.

Even teachers / professors who pose these problems sometimes miss the 'obvious' (?) anomalies. Heaven help their students with piles of crumpled up solutions in their WPB, wringing their hands, tearing out their hair, tossing and turning on their beds, etc.

 

common guys there is nothing silly in this problem as it appears. Obviously as someone said before we have to consider the 2H inductor equivalent to a short circuit and thus the current through it would be 1.25 A and thru the closed switch would also be 1.25 A before the switch is opened. If you still cannot grasp the obvious let me explain. Let us assume at t= - (minus) infinity we short the 2H inductor. We know that for dc the inductor acts like a short circuit at steady state. At t=0 already infinite time has elapsed since at t=-infinity the switch is closed. So obviously the inductor must have reached a steady state ie became equivalent to a short. So there is nothing wrong in the current in the 2H inductor to be 1.25A at t=0-. So my question is friends, how to find the current through 2H and 1H inductors at t=0+ ie time just after the switch has been opened.

 

If the current in the 2H was 2.5A at t=0- then it would normally be 2.5A at t=0+.

If the current in the 1H was 1.25A at t=0- then it would normally be 1.25A at t=0+.

Why?: Current in an inductor cannot change instantaneously. Or so we are told.

The problem with this exercise is that it produces an indeterminate condition at t=0+ because of the anomaly of having only a single series path with two inductors having different currents. Even if we assume the current in the 1H was 0A at t=0- the anomaly still exists.

Notionally, each inductor would produce an infinite emf at t=0+. But that doesn't give us a sensible answer.

Perhaps another approach is to do an energy based solution.

The 2H inductor has 6.25J and the 1H inductor 0.78125J or 7.03125J total stored energy.

Their equivalent series value is 3H. To have 7.03125J in 3H requires 2.165A.

But I'm suspicious of this approach since this situation is similar to the duality of instantaneous dumping of charge between two parallel capacitors having unequal initial voltages.

So you could go for the Q=CV equivalent - presumably with inductor "charge" being L*I. Total L*I at t=0- is 2*2.5+1*1.25=6.25 units {??}.

We'd probably denote the equivalent of "charge" in the inductors as their internal self "flux linkages".

The equivalent inductance of 3H would then have an equivalent "charge" of 6.25 units and hence a current of 6.25/3 or 2.0833A. This allows for an energy loss during the instantaneous transition - which would probably be accounted for by an emitted electromagnetic pulse.

That's my take on this problem. Like Georacer, I don't regard this as a very useful question.

 

thank u very much for your explanation tnk. i am more inclined to go by the inductor "charge" method as you say so. But why cant the inductor "charge" change instantaneously ie (from t=0- to t=0+). Another doubt: can capacitor charge change from t=0- to t=0+ ie instantaneously? If so that would mean infinite current na? or more so likely an impulse current?

 

An inductor current cannot really change instantaneously because this implies infinite voltage (E=-Ldi/dt), in the same way that changing a capacitor voltage instantaneously implies infinite current. For a theoretical problem of this kind, you could assume that the situation changes abruptly with a voltage or current impulse.

In any similar practical situation, the conditions would be modified by secondary effects such as parasitic capacitances, inductances and resistances, and insulation breakdown. The transient voltage or current amplitude would not be infinite, and the redistribution of charge or flux linkages would take time.

 

The obvious way to solve the problem is to force the model into a 'real world' equivalent by simply shunting the 1H inductor with a resistance of suitable value - let's say 10kΩ. In this way the circuit may be analyzed or simulated without significant loss of conditions approached by the 'impossible' model.

I have simulated the circuit with the proposed 10kΩ shunt resistor for the case of 1.25A in the 1H at t=0-.

Surprised or not - the initial current in the 2H at t=0+ is 2.0834A - not far from what I suggested in may last post for the un-shunted case.

If the current in the 1H at t=0- were assumed to be 0A, then the current at t=0+ would be 1.666A - also verified by the simulation.

 

Interesting approach t_n_k. I have never thought of it in that way. Well, I never needed to, to be honest, all my exercises were well founded in the Linear Circuits lesson.

I suppose that 10k resistor will go parallel to the 2H inductor, right?

Also, can you explain the difference between the approach from the energy scope, and the approach from the scope of the L*I quantity?
Where did you base your line of thoughts?

 

Interesting approach t_n_k. I have never thought of it in that way. Well, I never needed to, to be honest, all my exercises were well founded in the Linear Circuits lesson.

I suppose that 10k resistor will go parallel to the 2H inductor, right?

Also, can you explain the difference between the approach from the energy scope, and the approach from the scope of the L*I quantity?
Where did you base your line of thoughts?

Yes it would be the 2H - did I say the 1H?? My mistake. The one with the switch across it is the problem.

We seem to be comfortable with the conservation of charge (Q=CV) in capacitors but the conservation of flux linkages (λ=LI) in inductors is a little more unfamiliar territory. But that's the way it works.

 

Actually Georacer you picked up on the fact that I had inadvertently transposed the 2H & 1H values in my analysis & subsequent simulations.

So the 1H has 2.5A at t=0- and the 2H has 1.25A at t=0+.

Giving a total λ=1*2.5 + 2*1.25=5 Wb(Turns?)

The t=0+ series current would therefore be 5/3 or 1.666A.

If the 2H had an initial current (at t=0-) of 0A then the final outcome at t=0+ would be 2.5/3 or 0.833A.

Apologies for any confusion that may have arisen from my oversight.

 

Webber is the SI unit for magnetic flux. http://en.wikipedia.org/wiki/Weber_(unit). Just for our daily positive knowledge balance.

I understand that we work on a basis of flux here, but on what ground will we abolish the current before = current after principle?

 

If you are referring to the the current in the inductor from time t=0- to 0+ being the same ....

I guess the valid criterion for that assumption, would be that there is a unique [sensible?] transition, in which we don't end up with infinite or indeterminate rates of change of current in inductors - or voltage across capacitors, for that matter.

This problem belongs to a class in which we are faced with just that improbable (or perhaps impossible) scenario.

Another criterion may be that we can account for any energy difference between the initial and final states of the circuit as being entirely attributable to lossy elements in the circuit - such as any resistance(s) present. Otherwise we have to account for the energy loss by some more esoteric proposition such as the emitting of an electromagnetic impulse - my old 'favorite'. Not necessarily an agreed idea with anyone else.

In a nutshell - if the solution can't simultaneously satisfy both the conservation of energy and conservation of flux linkage (or charge) requirements then the problem lacks a valid solution.

Most informed people skilled in the art of electronics would probably ridicule these problems as being entirely impractical, physically unrealizable and therefore of no importance to real 'sparkies'.

 

It's food for thought. Maybe I 'll find a century-old professor in the uni and tease him about it a bit. Thanks again for your time!