# What is the Current in a Circuit with a Newly Introduced Inductor?

• Mandeep Deka
In summary: L1 = 5L, L2 = 4L, and let the initial current through L1 be I1 = E/R ThenI1*L1 = I2*(L1+L2)I2 = I1*L1/(L1+L2) = (E/R)*5/9

## Homework Statement

A circuit has an inductor of self inductance 5L connected in series with a resistor of resistance R and a cell of emf E. After a long time, a new inductor of inductance 4L is suddenly connected in series to the arrangement, keeping the rest of the arrangement intact. What is the current in the circuit immediately, after the new inductor has been introduced?

-

## The Attempt at a Solution

I am not really able to figure out how to approach the problem. After a long time, in the initial arrangement, the current in the circuit will just be E/R. The moment i introduce the the 4L inductor, it will tend to provide infinite resistance to the circuit but at the same time it can't stop the flow of current as the 5L inductor won't allow the current through it to drop to zero immediately. So how will the current flow at the immediate instant. Mathematically too i haven't found anything that can really help to get through.

Mandeep Deka said:

## Homework Statement

A circuit has an inductor of self inductance 5L connected in series with a resistor of resistance R and a cell of emf E. After a long time, a new inductor of inductance 4L is suddenly connected in series to the arrangement, keeping the rest of the arrangement intact. What is the current in the circuit immediately, after the new inductor has been introduced?

-

## The Attempt at a Solution

I am not really able to figure out how to approach the problem. After a long time, in the initial arrangement, the current in the circuit will just be E/R. The moment i introduce the the 4L inductor, it will tend to provide infinite resistance to the circuit but at the same time it can't stop the flow of current as the 5L inductor won't allow the current through it to drop to zero immediately. So how will the current flow at the immediate instant. Mathematically too i haven't found anything that can really help to get through.

The question is not physical, or more accurately, you can calculate things, but nothing is going to happen "instantly". The voltages and currents depend on the parasitic capacitances and such. Trying to make "instantaneous" changes to inductor currents generates very large voltages.

When you try to "open circuit" the loop to insert the 4L inductor, the open circuit voltage goes to infinity (in reality, it goes to a voltage limited by the parasitic capacitance the that inductance resonates with). That "infinite" voltage would then try to instantaneously start up the current in the total inductance in the new configuration.

I think it's a trick question, with non-physical consequences. However, if forced to answer it, I'd approach it in two ways.

First, I'd use an energy approach, stating that the infinite voltage spike is not going to change the stored energy in the inductances. So I'd figure out the new t=0+ current from the energy stored in the one inductor at t=0-.

Second, and as a check on that first answer, I'd introduce an explicit capacitor into the circuit, so that the "infinite" voltage is limited by the LC resonance peak of the half-cycle flyback behavior of open circuiting the loop with the first inductor. Connect the 2nd inductor as the voltage falls back through where it was before the open circuit, and see what the current result is. Does it match the energy argument answer from above?

If not, I'd still look at arguments like these to try to get around the "ideal infinite voltage" discontinuity problem.

Last edited:
If you consider the mechanical analog of the situation, in a model where mass is analogous to inductance and current to velocity, then sticking the new inductor into the circuit is analogous to putting a stationary mass in the way of an already moving one.

So, assuming an inelastic collision (the inductors are stuck together afterwords)...

Let the inductors be L1 = 5L, L2 = 4L, and let the initial current through L1 be I1 = E/R Then

I1*L1 = I2*(L1 + L2)

I2 = I1*L1/(L1 + L2) = (E/R)*5/9

Which corresponds to the suggested solution (as well as a SPICE simulation -- I checked).

Like any inelastic collision, we can't expect conservation of 'kinetic' energy:

(1/2)I12L1 = (5/2)(E/R)2L

(1/2)I22(L1 + L2) = (25/18)*(E/R)2*L

No doubt the 'missing' energy was dissipated in the resistance and/or absorbed by the battery during the 'adjustment period' when L1 panicked.

gneill said:
If you consider the mechanical analog of the situation, in a model where mass is analogous to inductance and current to velocity, then sticking the new inductor into the circuit is analogous to putting a stationary mass in the way of an already moving one.

So, assuming an inelastic collision (the inductors are stuck together afterwords)...

Let the inductors be L1 = 5L, L2 = 4L, and let the initial current through L1 be I1 = E/R Then

I1*L1 = I2*(L1 + L2)

I2 = I1*L1/(L1 + L2) = (E/R)*5/9

Which corresponds to the suggested solution (as well as a SPICE simulation -- I checked).

Like any inelastic collision, we can't expect conservation of 'kinetic' energy:

(1/2)I12L1 = (5/2)(E/R)2L

(1/2)I22(L1 + L2) = (25/18)*(E/R)2*L

No doubt the 'missing' energy was dissipated in the resistance and/or absorbed by the battery during the 'adjustment period' when L1 panicked.

Weird. I don't get the analogy, and it looks like my energy approach would not give the right answer, eh?

berkeman said:
Weird. I don't get the analogy, and it looks like my energy approach would not give the right answer, eh?

There are a number of mechanical systems (including springs and masses and dampers, fluid 'circuits', temperature 'circuits', etc.) that can be modeled with electrical circuit analogs. The reverse is also true. It can be very handy to be able to bring sophisticated circuit analysis methods to bear on them.

If you look at the expressions for how energy is stored in a given system, you can usually spot the analogous components and variables. There are analogous differential equations across all sorts of physical systems.

So, in mechanics we've got energy stored in masses, (1/2)M*V2. In an inductive circuit we have (1/2)L*I2. You can do the same with capacitive circuits, but the 'velocity' becomes 'voltage'.

I just figured it was fair game to do the reverse of the usual, and model the the 'collision' in the electrical circuit in terms of a mechanical analog. Speaking of which, there are forces that become indefinite and energy that seems to vanish instantly during an 'ideal' mechanical collision, too, when we gloss over the nasty nanoseconds!

Thanks a lot, for giving a way out of this problem,
But frankly speaking, i am finding it a little difficult to accept the way the problem is solved. Its is a smart way to solve the problem. As for a mechanical system we have defined quantities like momentum and energy and laws related to them, we can deduce the analogy of mechanical momentum from the inductor energy equation and hence forth approach the problem. But i have some questions to ask:
1.Could you explain me why you have assumed an inelastic collision? (obviously you wrote the inductors 'stick' together, but i want a more physical explanation)
2.Can there be a case of sort of elastic collision, when a new inductor is introduced to the system?
3. (Just wondering) What if in the same initial system, i, instead of adding a new inductor, replace the initial inductor with it? What mechanical analogy would you apply to solve the problem?

The fact that i am trying to highlight is, we have got laws of physics to apply in mechanical systems like conservation of momentum, conservation of energy, etc,
but i for one haven't found any law, related to the quantity L.I you used as the momentum of the electrical circuit. How would i assure myself that an analogy used for an undefined (sort of) quantity, would give me a correct answer??

Hi everyone!

Isn't this a simple application of good ol' Kirchhoff's rules (adjusted with an "opposing" emf of LdI/dt across any inductor)?
Mandeep Deka said:
A circuit has an inductor of self inductance 5L connected in series with a resistor of resistance R and a cell of emf E. After a long time, a new inductor of inductance 4L is suddenly connected in series to the arrangement, keeping the rest of the arrangement intact. What is the current in the circuit immediately, after the new inductor has been introduced?

Initially, IR + (dI/dt)L1 = E.

but dI/dt = 0, so I = E/R.

Immediately after L2 is "impulsively" added into the circuit (presumably it was there all along, in parallel to a bit of wire of almost zero resistance, but in series with a huge resistance … then the sizes of the resistances are suddenly changed so that virtually the whole current goes through it), I is "impulsively" decreased to J in time ∆t, so ∆I for L1 is ∆I1 = J - I, and ∆I for L2 is ∆I2 = J (because virtually no current was originally flowing through it) …

since now IR∆t + ∆I1L1 + ∆I2L2 = E∆t,

we can ignore ∆t as being infinitesimal, and write

∆I1L1 + ∆I2L2 = 0,

ie (I - J)L1 = JL2,

or J = IL1/(L1 + L2) = (5/9)E/R.

Mandeep Deka said:
... But i have some questions to ask:
1.Could you explain me why you have assumed an inelastic collision? (obviously you wrote the inductors 'stick' together, but i want a more physical explanation)
Since the new inductor remains in the circuit sharing the same current after it is inserted, and thenceforward the two behave like a single inductance L1 + L2, it is analogous to the two masses M1 + M2 of the inelastic collision sharing the same velocity.

2.Can there be a case of sort of elastic collision, when a new inductor is introduced to the system?

You could switch the inductor into and then out of the circuit. The switching time would have to be very brief in order to mimic the short interaction time of the ideal elastic collision.

3. (Just wondering) What if in the same initial system, i, instead of adding a new inductor, replace the initial inductor with it? What mechanical analogy would you apply to solve the problem?
That would be like removing the moving mass along with its momentum and replacing it with a new mass at rest.

Here's a more elaborate mechanical analogy for the circuit. Five linked railroad cars of mass L each are moving on a long circular track. There's a constant friction R acting on the cars, and the energy lost is compensated for by the application of a constant force E to the train of cars. At some point in time, someone moves four more cars from a siding onto the track ahead. The new cars are at rest. When the two trains of cars collide, they link together. What's the new velocity of the combined cars?

The fact that i am trying to highlight is, we have got laws of physics to apply in mechanical systems like conservation of momentum, conservation of energy, etc,
but i for one haven't found any law, related to the quantity L.I you used as the momentum of the electrical circuit. How would i assure myself that an analogy used for an undefined (sort of) quantity, would give me a correct answer??

Because the underlying differential equations are the same for all sorts of physical systems, the various 'laws' that are derived from them will also be found across all of them with the same mathematical structure. http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Analogs/ElectricalMechanicalAnalogs.html" [Broken] to a page where someone has gone into a bit more detail about the various parallels.

Last edited by a moderator:
Thanks for the explanation.
Both the solutions to the problems are quite satisfactory to me now!

Thanks a lot!

## 1. What is self-inductance?

Self-inductance is a property of a circuit or electrical component that describes its ability to generate an electromotive force (EMF) in response to a change in the current flowing through it. It is measured in henrys (H) and is represented by the symbol L.

## 2. How does self-inductance affect electronic devices?

Self-inductance can cause electromagnetic interference (EMI) in electronic devices. This interference can disrupt the normal operation of the device and cause malfunctions. It is important for electronic devices to be designed with proper shielding and grounding to minimize the effects of self-inductance.

## 3. What factors affect the self-inductance of a circuit?

The self-inductance of a circuit is affected by the number of turns in a conductor, the shape and size of the conductor, and the material it is made of. It also depends on the permeability of the surrounding medium and the presence of any nearby conductive materials.

## 4. Can self-inductance be beneficial in electronic circuits?

Yes, self-inductance can be beneficial in electronic circuits. It can be used to store energy in a magnetic field, which can then be released when the current changes. This property is used in devices such as transformers and inductors, which are essential components in many electronic circuits.

## 5. How can self-inductance be reduced or controlled?

To reduce the effects of self-inductance, electronic devices can be designed with proper grounding and shielding to minimize the interference. The use of materials with lower permeability can also help to control self-inductance. Additionally, inductors and capacitors can be used together in a circuit to cancel out the effects of self-inductance.

Replies
7
Views
612
Replies
16
Views
2K
Replies
5
Views
1K
Replies
4
Views
128
Replies
42
Views
551
Replies
52
Views
11K
Replies
3
Views
121
Replies
34
Views
5K
Replies
1
Views
152
Replies
18
Views
1K