This is actually a series connecting of the two inductors after the switches are set to the B position. At t=0, it would be like connecting two current sources in series, creating a high voltage across the air between the contact point and the common ground. Each inductor will continue to sink/source its current into the virtual high impedance load at the connection, while energy is redistributed until both indictors will reach the same current values (at same or opposing direction, depending on the initial direction of each current). If a load was connected between the B bus and ground, it will be a parallel connection and the total current can be calculated by equating energies before and after the switch.
Not reading other comments, so here my take - inductors are in essence current sources, without any other parasitic values to work with - the answer is - undefined (voltages will go to infinity as pushing current into infinite resistance raises the voltage and nothing flows nowhere, the same way with capacitors, but in that case the currents will be infinite).
Strictly speaking, this is an ill posed problem. Tis a violation of KCL or KVL (in the case of shorting two different capacitors of different charges together, the analog problem). Explanation on ill posed: Recall that if you believe currents and voltages are bounded, then the inductor current must be continuous (can use Fundamental Theorem of Calculus here) and so therefore you have a KCL situation where Current from inductor 1 = Current from inductor 2, an impossibility as posed. (I'm making some other assumptions, like the inductances aren't coupled, switches are instantaneous, etc.) Still, a "solution" can be observed as follows. (like you can insert the right "real" element into ckt to allow a transient to final solution). Since I'm lazy I'll do the capacitor version first. Capacitors: Conserve Initial Charge = Final Charge, distribute charge to the parallel capacitors Q initial = Q1.C1 + Q2.C2 = Q final =Q1final+Q2final C final = C1 + C2. Since C1 Q1final = V final = C2 Q2final then solve accordingly. Inductors. Conserve Initial Flux = Final Flux, distribute flux to the series inductors. Phi initial = L1.I1 + L2.I2 = Phi final = Phi1final+Phi2final Lfinal = L1 + L2. Since L1.Phi1final = I final = L2.Phi2final then solve accordingly. i'm sure i screwed up something so i'll edit in a few hours, but the idea of conserving charge (for caps in parallel) or flux (for inductors in series) is the essence of the equilibrium solution sought
Did you consider both the left hand and right hand schematic diagrams? The initial current in L2 in the LHS case is reversed relative to the RHS case. I suggest the outcomes would be different.
Strickly speaking your observation lacks engineering insight, There is no inductance without resistance and even a 1f (femto) Ohm resistor solves the singularity issue.
@@sambenyaakov hey professor, please don't chew me out, I think I fully admitted that strict adherence to bounded state variables and zero resistances is a mathematical construct. It was meant to try to generate insight into the "paradox"
If we assume that the current direction is as shown in the right figure (where the current can be either positive or negative), the current after connection is the same for both inductors and should be I = (L1*I1 + L2*I2)/(L1 + L2). The energy loss is simply then 0.5(L1*I1^2 + L2*I2^2 - (L1 + L2)*I^2). This value of the current after connection can be obtained by solving a simple differential equation (it is not necessary to add any parasitic element, Laplace transform can deal with the inductor current discontinuity). Other way around would be to use the Maxwell equation div B = 0, about the flux conservation, which gives the result in one step actually.
@@sambenyaakov I will be happy to see other approach, since I am quute convinced that my approach is completely correct from the point of view of circuit theory.
Another idea is to remember the theorem about the superconducting ring: the flux through it will never change. Assuming the coils are not mutually connected and are ideal, initial flux referencing direction into the page: Ф1=I1*L1-I2*L2. After the reconnection current is the same in them: Ф2=I(L1+L2). Thus the final current I=(I1*L1-I2*L2)/(L1+L2).
Thank you for your video. I had to solve the differential equation assuming that we connect a small lossy capacitor in parallel with inductors. And the answer for steady-state is that the current in inductances (I1*L1-I2*L2)/(L1+L2).
I can only guess as if i can handle the inductors in this case the same way as capacitors, because they "feel" more like intuitive understood. I think something like I1*L1 - I2*L2 = the current between them after connecting them. Which current the thus must also go through both inductors. But i can hardly hope that could even be close because now i think, that wouldn't even be right with capacitors. After thinking about such, thanks to these riddles, i am even more curious for the real solution!
Like the classic two capacitor problem, we are left with the question as to where the lost energy "went". In the absence of any inductor winding resistance and assumed ideal switching, we have to discount the 'obvious' choices of Joule heating and arcing losses. One option is to postulate that the energy is 'lost' as electromagnetic radiation. This is difficult to explain when we have only idealised lumped parameter circuit models to work with.
Back of the envelope. We can use the resistive circuit analog. The final current magnitude would be the same in each inductor and given by the absolute value of [I1 * L1 - I2 * L2]/(L1 + L2). The lost energy would be [L1 * L2 * (I1 + I2)^2]/[2 * (L1 + L2)]. Will recheck to ensure the result is correct.
@@sambenyaakov Happy to join in. Depending on the inductor current directions the answer will presumably differ. I analysed for the left hand side circuit case with both currents directed upwards. The right hand side circuit case would give different results.
This is similar to the problem of connecting 2 capacitors in //. With capacitors you'll have a very large current spike to equalize the capacitor's voltage. The final voltage can be calculated using conservation of charge ( C∙V ) Here with inductors we'll have a very large voltage spike to equalize the inductor's currents. The final current can be calculated using conservation of magnetic flux ( L∙I ) The final inductors current will be: ⅠL = ( L₁∙ⅠL₁ + L₂∙ⅠL₂ ) / ( L₁ + L₂ ) The lost energy is: E = 1/2 [ L₁∙(ⅠL₁)² + L₂∙(ⅠL₂)² - ( L₁ + L₂ )∙(ⅠL)² ] ⇒ E = 1/2 [ L₁∙(ⅠL₁)² + L₂∙(ⅠL₂)² - ( L₁∙ⅠL₁ + L₂∙ⅠL₂ )²/( L₁ + L₂ ) ] We can see that for instance, one may have no energy lost if ⅠL₁ = ⅠL₂ to start with. But one can also loose all the energy if for instance L₁ = L₂ and ⅠL₂ = - ⅠL₁.
If the inductors are lossless, the circuit will oscillate due to the unavoidable parasitic capacitance. It's the same for connecting 2 lossless capacitors with different voltages, it will oscillate due to the unavoidable parasitic inductance. No energy is ever 'lost', it just keeps bouncing from one part of the circuit to another.
@@sambenyaakov I was not sure whether you were claiming it didn't matter whether both currents were directed upwards or in opposite directions. I took both to be directed upwards. I think the answer would be different if one current was upwards and the other downwards.
The key equation to derive energy lost will be L1 |dI1/dt| = L2 |dI2/dt |. Assume I1 > I2 in case-1, the current I1 will decrease and current I2 will increase such that eventually, the current in the circuit become I. The value of I can be derived in terms of L1, L2, I1 and I2. Subsequently, energy lost can be estimated.
@@sambenyaakov Thanks for the reply. Assuming I1 > I2, the final series current in the inductor L1, L2 can be derived to be I = I2+ L1(I1-I2)/(L1+L2). It is easy to model the circuit in LTspice with the initial currents and verify the derivation. For a special case when L1 = L2 and I2 = 0, it can be shown that half of the initial inductor energy will be lost. Elost = 1/4*L1*I1^2.
@@sambenyaakov The voltage where switch is connected will tend to infinity and that’s how energy will be lost and energy conservation will be maintained. I derived those equations assuming final current in both inductor will be eventually same and change in the current is approximated to be linear.
All energy is lost in both cases since no other storage for the energy is available - no capacitances have been specified and as soon as the contacts open , the energy is lost to ionisation of the ambient gas .
I'm not sure about the correct answer, but my guess is the energy is not conserved in this inductor connection, like in the parallel connection of charged capacitors. Charged capacitors when puted in parallel conserves "electrical charge", not "electrical energy". Am I right?
The schematic looks like a relay? So it's line one and line 2 220 volts? If so then one of the coils on the relay operate with half the voltage? And this would cause the oil not to pull in all the way and burn up.
