DC Current in Parallel Inductors?

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?

John Larkin wrote in news: snipped-for-privacy@4ax.com:

So there's nothing wrong with Greg's statement. In this special case it just means you have to take their zero resistance into account.

Right, just as you have to take their zero Leprechaun content into account.

John

John Larkin wrote in news: snipped-for-privacy@4ax.com:

Only if general calculations expected to have to handle leprechauns.

The general equations that define the behavior of an inductor don't include resistance.

I = Io + time_integral(E/L)

So why include a resistance term and then make an effort to remove its effects?

Plugging into cookbook equations is a risky way to really understand things, which was maybe one of the points that Desiree's instructor was making.

John

The even split assumption is certainly questionable since superconductors are involved. In fact, a little thought should reveal that the only thing that will determine the final current in each inductor will be the history of the current flows through each.

At steady state both inductors will have a constant current and zero voltage drop. Without considering how the final currents obtain, that condition (zero voltage drop, constant current) can be met by any aritrary currents that add up to the required total. There could even be a very large circulating current going around the superconducting ring formed by the two inductors quite separate from the current passing through the pair from the voltage source and resistor.

With superconductors, current has a sort of inertia.

Here's an example to consider. Suppose you have two superconducting circuits in the form of squares. They are side by side. The one on the left has a 100A current circulating counterclockwise in it, while the one on the right has a 10A current circulating clockwise.

Now the two nearest sides of the squares are brought into contact in such a way that they merge into a single conductor. After the merge, what is the steady state current in each part of the circuit?

. .---------------. . | 100 A | | 10 A | . | | | | . | | | | . | | | | . | | | | . | | | | . '----------------' '----------------'

Ouch. You'd have to assume that they were very close (the gap you drew was very small) before the merger; otherwise they will change currents (and the system energy will change) as they move towards one another, since their magnetic fields will interact.

Given that, I'd guess that the currents won't change. That's the easiest guess that conserves energy.

I have at least a 52% probability of being right.

John

Is that a technical term? I don't see it in many data sheets.

:)

it's

So, what's your final answer to the OP's question? How many mA in each ideal inductor?

John

True, it results in an overdefined matrix, which it can't solve.However,specifying a very small series resistance, say 10^-18 ohms, together with zero parallel resistance and capacitance makes the solver happy, and is near enough to ideal inductors to give results near to what you get using calculus.

LTSpice inserts a default (1 milliohm, IIRC) series resistance in its inductor model if you don't specify one. That's too big for the inductor to be near-ideal.

Try this, it's your posted circuit, modified using 1e-18 ohm series resistance in the inductor models and the series resistors deleted. It gives a 10:1 share of current, just like the differential equation says it should.

Version 4 SHEET 1 880 680 WIRE 320 128 128 128 WIRE 448 128 400 128 WIRE -16 192 -96 192 WIRE 128 192 128 128 WIRE 128 192 64 192 WIRE 448 192 448 128 WIRE 528 192 448 192 WIRE -96 240 -96 192 WIRE 128 256 128 192 WIRE 320 256 128 256 WIRE 448 256 448 192 WIRE 448 256 400 256 WIRE -96 352 -96 320 WIRE 528 352 528 192 WIRE 528 352 -96 352 WIRE -96 400 -96 352 FLAG -96 400 0 SYMBOL ind 304 144 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L1 SYMATTR Value 10e-6 SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0 SYMBOL ind 304 272 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L2 SYMATTR Value 1e-6 SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0 SYMBOL res 80 176 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 1000 SYMBOL voltage -96 224 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 10 .1 1e-7) TEXT -128 424 Left 0 !.tran 10 uic

"Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock)

=A0 =A0 =A0 =A0 =A0(Stephen Leacock)

Thanks Fred, It's good to know that spice can be made to cough up the right answer.

George H.

Unless you really want a 1:1 current split and are willing to wait until it happens.

John

it's

:)