# Question on inductors with initial current

#### triggetrac

Joined Oct 4, 2014
1
Question: Initially the switch position is aa', 3 H is connected across a 3 A supply and 6 H is connected across 6 A supply. After some time the make before break switch is actuated to the new position bb'. Determine the steady state current, fluxes associated with each inductor in the new position. Assume there are no switching losses. Fig 1 : Initial condition (switch in aa') Fig 2: circuit while operating the make before break switch Fig 3 : switch in bb'

Pre switching period is fine, each inductor having a current of 3 A and 6 A respectively. Inductors don't allow a sudden change in current when disconnected from supply as in fig 3, their magnetic field collapses and allow a decreasing current to flow in the same direction as previous and their voltages are reversed. But how does it behave when two inductors having different initial currents are switched together? It shouldn't be 9 A otherwise total energy associated with the series inductors is more than pre switching violating law of conservation of energy. Is this case similar to connecting two current sources in series which is not allowed?

#### MisterBill2

Joined Jan 23, 2018
3,998
This look a whole lot like a homework problem, or perhaps a test. Unless the inductors have zero resistance, that will also affect the current. But as soon as the current sources are disconnected the only source of voltage comes from the energy in that collapsing magnetic field. If the switches are lossless and instant, then the voltage on each inductor is L(di/dt), which is very large as dt approaches zero.
What practical experience tells me is that the circuit is going to oscillate for a while, until the losses turn all of the stored energy into heat.

#### crutschow

Joined Mar 14, 2008
23,494
That's not an experiment you would want to do with a real circuit.

If this is an ideal circuit with no stray capacitance, then the voltage at the junctions of the two inductors will approach infinity when the switches change to connect them in series.

In real life the stray capacitances would charge until the capacitive energy equals the excess energy in the 6H inductor (or the voltage gets so high that it creates an arc-over).
Then the circuit will oscillate with the excess energy transferring back and forth between the inductors and capacitance.

#### MrAl

Joined Jun 17, 2014
6,608
Question: Initially the switch position is aa', 3 H is connected across a 3 A supply and 6 H is connected across 6 A supply. After some time the make before break switch is actuated to the new position bb'. Determine the steady state current, fluxes associated with each inductor in the new View attachment 156963 View attachment 156963 position. Assume there are no switching losses. Fig 1 : Initial condition (switch in aa') Fig 2: circuit while operating the make before break switch Fig 3 : switch in bb'

Pre switching period is fine, each inductor having a current of 3 A and 6 A respectively. Inductors don't allow a sudden change in current when disconnected from supply as in fig 3, their magnetic field collapses and allow a decreasing current to flow in the same direction as previous and their voltages are reversed. But how does it behave when two inductors having different initial currents are switched together? It shouldn't be 9 A otherwise total energy associated with the series inductors is more than pre switching violating law of conservation of energy. Is this case similar to connecting two current sources in series which is not allowed?
Hello,

Is this really homework or just a somewhat interesting theoretical question?

As others have alluded to, there must be at least one other component in the circuit to make this a practical question. That's because for the first instant the two (ideal) inductors would act like two current sources of different currents in series, and then the question that comes up is which one prevails.
If we have two ideal current sources in series one 3 amps and one 6 amps which one takes over? Does the 6 amps somehow force 6 amps through the 3 amp one, or does the 3 amp one somehow force the 6 amp one to somehow reduce it's current?
Well since the 3 amp one can only pass 3 amps and the 6 amp one can only pass 6 amps and we have the circuital law "the current in a series circuit is the same in each element" then we see that this law would not hold so there must be another law that also comes into play.
This other law would have to include at least one other element. You have your choice of elements like a resistor, capacitor, or even just a wire. A wire would radiate, while an ideal inductor would not radiate, so a wire would loose some energy in the circuit.
Of course we might also want to assume some coupling because all inductors radiate some energy, but ideal inductors do not so we're back to introducing another component. If we do allow some coupling though then we can see some kind of flux interaction which changes the whole circuit again anyway.
So two ideal inductors in series with different initial currents would break every law unless we add some other element such as a resistor, capacitor, wire, mutual inductance between the two, etc. With each type of added element, we would see different types of responses but those responses would be those of a type that we could actually see in real life.
Note that the placement of the added element is also important. A resistor in series may not change anything because it does not change anything about what each current source has to do in order to act like a true current source. A parallel resistor on the other hand across either element would allow the circuit to work normally.

#### MisterBill2

Joined Jan 23, 2018
3,998
Mister Al has made not considered that the switches change instantly and in a lossless manner. That takes us to the voltage, because the current has stopped instantly. At this point, excuse the lack of correct symbols, but V=L di/dt, translating to voltage equals inductance multiplied by the rate of current change. But since both the switches and the inductors are perfect, both voltages tend towards infinity, or at least a very high number. But if the circuit were assembled from real world parts the results would be less exciting.

#### MrAl

Joined Jun 17, 2014
6,608
Mister Al has made not considered that the switches change instantly and in a lossless manner. That takes us to the voltage, because the current has stopped instantly. At this point, excuse the lack of correct symbols, but V=L di/dt, translating to voltage equals inductance multiplied by the rate of current change. But since both the switches and the inductors are perfect, both voltages tend towards infinity, or at least a very high number. But if the circuit were assembled from real world parts the results would be less exciting.
Hello there,

Not only have i considered that the switches change instantly and in a lossless manner, i insist that they do and take the next step by eliminating them entirely by stating that the switches serve nothing more than to show that the two current sources are just initial current generators for the two inductors. What this does is eliminate the switche(s) and turns the circuit into a regular circuit with initial conditions. So far nothing new at all.

This shows the problem right off because we can not determine what the intial current is because we can only have a valid initial current in a circuit that allows that initial current.

For some examples, take an inductor with current source (as in the schematic) of 1 amp in parallel with a resistor of 10 ohms. What is the equivalent initial current? It is 1 amp.
Similarly, with a 2 amp source we have initial current of 2 amps. Nothing difficult there either.

Before we can solve a circuit like these we have to establish the initial conditions and that is usually very simple because the REST of the circuit allows that current to flow with no issues.

Now what about an inductor with initial current of 1 amp and a parallel resistance that is infinite (open circuit). The voltage we might say goes to infinity. But is that really a solution. Theoretically maybe it is, but it's not of much use and an infinite voltage is not possible in a real circuit.

Try to solve for the voltage between the two inductors see what you get.
Use the long connecting line as ground.
Also keep in mind that even if the voltage does theoretically go to infinity that does not mean that it STAYS at infinity for a finite amount of time. It could be just an impulse.

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#### MisterBill2

Joined Jan 23, 2018
3,998
It may be that since I was very tired my previous response was not totally correct.
This is a magical circuit and so in the instant prior to switching the magical switches there are the specified currents flowing through the perfect inductors. Then, when the perfect inductors are instantly placed in series, we see be that the current must change because they are suddenly in parallel with the same voltage across each one, and since V=L x di/dt, the answer is only valid in the same magic world of homework land, which is in some cases far away from the real world.

#### MrAl

Joined Jun 17, 2014
6,608
It may be that since I was very tired my previous response was not totally correct.
This is a magical circuit and so in the instant prior to switching the magical switches there are the specified currents flowing through the perfect inductors. Then, when the perfect inductors are instantly placed in series, we see be that the current must change because they are suddenly in parallel with the same voltage across each one, and since V=L x di/dt, the answer is only valid in the same magic world of homework land, which is in some cases far away from the real world.
Hi again,

Very well put I found the time response to the same circuit with a resistor (R1) connected from the junction of the two inductors to 'ground' where i made ground the long wire on the bottom in the schematic. Here is what i got:
v=(I1-I2)*R1*e^(-(t*(L2+L1)*R1)/(L1*L2))

where v is the voltage across the resistor.

The response is interesting as we allow R1 to go higher and higher in value (thus eventually taking it out of the circuit completely in the limit). As R1 goes higher, we see the response initial amplitude get higher but at the same time the duration of the time that it stays higher becomes shorter. Interestingly, in the limit as R1 goes infinite we get v=0 volts, and that is not dependent on time anymore.

Another interesting question though if it has not been asked before is:
What is the initial current in the wire? (the wire here is the long wire on the bottom of the schematic and given again ideal elements).
That's something to think about Also, did anyone try this in a simulator yet? I have a feeling it may be harder to get good results that way.
I just tried it with the two inductors and didnt even add the two current sources yet and guess what happened. MicroCap rejects the circuit on the grounds that there is an inductor/voltage loop and will not do the analysis.
Looking in HELP, the text reads as follows:
"A loop containing only voltage sources or inductors was found. Such loops usually violate Kirchoff's Voltage Law and are not allowed."

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• nsaspook

#### MrAl

Joined Jun 17, 2014
6,608
Hi again,

I did one more thing with this by looking at the dual of this circuit, which is a circuit with two caps in parallel with different initial voltages.

It's easy to see if one of the voltages is zero too. The cap with the voltage causes an impulse current to flow into the other cap. I suppose if the first cap had a certain energy E and the caps where the same value, half the energy would end up in both caps. With some resistance though we'd loose some energy as heat.
So this question partly depends on what we want to believe about the practicality of the circuits we choose to represent the reality impossible circuit.