DC Current in Parallel Inductors?

For extra bonus points, also consider what effect, if any, there would be to your final answer if there was a current flowing in a loop through the two inductors at time zero. (Ideal parallel inductors will happily support a circulating current without loss.)

Would the troll like a cookie? Nice troll, come here, gootch gootchy goo

SPLAT

y,

Thanks Dan, but how would I calculate such a thing? And I'm still not sure whether two different value inductors in parallel will share the mainline DC current unequally or equally after reaching steady state. (I feel, but half my class does not, that after steady state is reached that both parallel inductors could simply be replaced by a zero ohm piece of wire...).

Thanks!

-Desiree

Instead of looking at the steady state, you need to look at how you get there.

The remainder of your homework problem is left to the student.

Has anyone got a Digkey part number for some of these inductors? Sounds like they would be really useful.

Right, this is a theoretical question about ideal inductors, so they never actually reach "steady state," meaning constant current in this case. For a circuit made of ideal components, you would have to define steady state: in the limit as t increases without bound.

It doesn't. An ideal inductor has zero resistance of course, and you have two of these lovely devices in parallel. But ask yourself how long does it take to reach steady state in an ideal circuit? Tell your teacher you'll give him the answer after "the big crunch".

Dave.

I guess we assume no initial currents before we switch on the supply.

Put the two inductors, in parallel, into a black box. Now you have 10 volts through 1K ohms driving a 0.909 uH inductor.

Calculate the voltage versus time across the black box.

Now consider what would happen if that voltage profile were applied to the 1 uH inductor, and separately to the 10 uH inductor.

The issue isn't so much what the circuit looks like "after it reaches its steady state" but the path it took to get there. An inductor integrates voltage into current, so it remembers everything that ever happened to it.

What did Spice say?

John

Yeah, room temperature superconductors will be mighty handy. Finite-Q filter designs are a nuisance.

John

Given the level of the question steady state means after all the math is done. Parallel devices have equal voltages.

Tom

Thanks guys. I'm trying to put all your answers together to clearly figure this all out, but its tough! John, here is a clearer explanation, and what I am seeing in Spice: In a circuit with a (10V) DC power supply, and a series current limiting (100 Ohm) resistor, and two ideal* inductors (with no mutual coupling) that are in parallel with each other -- one being 1uH and the other 10uH -- why do the DC currents take >>5xL/R to reach equality in each branch? Why should an ideal inductor of ANY value have ANY effect whatsoever on the DC current *after* it reaches its steady state? My Spice simulator shows that it takes a HUGE amount of time (25ms) to reach equal current of 50mA in each branch, and until then the current in the 10uH branch is 9.1mA, and the current in the 1uH branch is 91mA. Since 25ms is WAY past five time constants, why does it take so darn long to even-out the currents in each leg?

(* Rser=3D0.001 to make Spice happy.)

Confused,

-Desiree

• C:\Program Files\LTC\LTspiceIV\Draft-INDS.asc L1 N001 N002 1=B5 Rser=3D0.001 R1 N002 0 100 V1 N001 0 10 L2 N001 N002 10=B5 Rser=3D0.001 .TRAN 500us 0.05 UIC .PLOT TRAN I(L1) I(L2) .backanno .end

Thanks guys. I'm trying to put all your answers together to clearly figure this all out, but its tough! John, here is a clearer explanation, and what I am seeing in Spice: In a circuit with a (10V) DC power supply, and a series current limiting (100 Ohm) resistor, and two ideal* inductors (with no mutual coupling) that are in parallel with each other -- one being 1uH and the other 10uH -- why do the DC currents take >>5xL/R to reach equality in each branch? Why should an ideal inductor of ANY value have ANY effect whatsoever on the DC current *after* it reaches its steady state? My Spice simulator shows that it takes a HUGE amount of time (25ms) to reach equal current of 50mA in each branch, and until then the current in the 10uH branch is 9.1mA, and the current in the 1uH branch is 91mA. Since 25ms is WAY past five time constants, why does it take so darn long to even-out the currents in each leg?

=============================

(* Rser=0.001 to make Spice happy.)

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and

Thanks Mike. So at least I'm not doing anything wrong with the Spice simulator! But I'll not rest until I find out exactly *why* this occurs. The mechanism behind it has me completely baffled, since it is not an LC tank circuit, so energy is not being exchanged back and forth between an inductor and capacitor. It is merely two inductors in parallel (I always assumed that such a circuit would simply act like single, lower value, inductor). So strange, but none of my (many) school books seems to cover any of this, they only say that an ideal inductor is a "short" to DC.

Thanks again,

-Desiree

Perhaps one of the things your instructor wanted to do was to wean you away from using Spice -- which is a fine tool for some things -- for everything.

Try doing it using Laplace domain analysis; John Larkin's suggestion of finding the voltage and then the current is to the point if you want to simplify the math.

Remember that Spice is a real-world tool, and an ideal inductor is not a real-world device. So using Spice and expecting it to cancel out infinities is inappropriate. OTOH, this is a fairly simple problem with linear circuit elements -- hence my suggestion of using Laplace analysis.

Thanks Mike. So at least I'm not doing anything wrong with the Spice simulator! But I'll not rest until I find out exactly *why* this occurs. The mechanism behind it has me completely baffled, since it is not an LC tank circuit, so energy is not being exchanged back and forth between an inductor and capacitor. It is merely two inductors in parallel (I always assumed that such a circuit would simply act like single, lower value, inductor). So strange, but none of my (many) school books seems to cover any of this, they only say that an ideal inductor is a "short" to DC.

Thanks again,

-Desiree

Are you pursuing an engineering degree?

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Yes Mike, but I am only in my first year of electronics.

Okay. Now we need to know a little more about what you have learned.

Have you had calculus? Do you know what this equation means and how to use it:

v = L di/dt

Your intuition might work better with capacitors rather than inductors.

What would you expect if you had a resistor, small cap, and big cap in series?

Yes Mike, but I am only in my first year of electronics.

Good. I'm going to show you a simple solution. It amounts to a fairly rigorous demonstration that you can show your instructor. Assume the inductors have zero current in them at t=0, which is when you apply the voltage. Dan Coby gave you formula v=L*(di/dt) I'm going to call the inductances L1 and L2. Simillarly, currents in the inductors I1 and I2. I1 and I2 are variables, functions of t (time). Now, v is always equal for both inductors becuase they are in parallel. Therefore L1 (dI1/dt) = L2 (dI2/dt) Say L2 is the "bigger" inductor. L2 = 10 L1 implies dI1/dt = 10 dI2/dt which we can integrate over time: I1 = 10 I2 + C (I hope you've had some basic calculus) now, C has to be zero because at t=0, I1 = I2 = 0 so I1 = 10 I2 and that's always true, for any value of t