Two Inductors in Parallel

Put two inductors across the same pair of nodes and they share the same voltage. Each carries the current that voltage drives into it, the currents add, and since inductance is voltage per rate-of-change-of-current, the reciprocals add: $1/L_t = 1/L_1 + 1/L_2$. For two parts that rearranges to product over sum, and it is the identical arithmetic used for resistors in parallel, springs in series and pipes in parallel. A 100 mH beside a 400 mH gives $(0.1 \times 0.4)/0.5 = 80\ \text{mH}$.

The result is always smaller than the smaller member, and a large inductor in parallel with a small one is very nearly the small one alone — the low-inductance path takes the current, just as the low-resistance path does. That is worth having as a sanity check: if a parallel combination comes out larger than either part, the wrong rule has been applied.

Paralleling is done for current, not for inductance. Two chokes side by side carry twice the current before saturating and dissipate half the copper loss of one, at the cost of halving the inductance — so designers parallel two of double the value to get back where they started with twice the headroom. It is also how a designer reaches an awkward value: 60 mH from a 100 mH and a 150 mH.

The coupling caveat is sharper here than in the series case. Coupled parallel inductors follow $(L_1L_2 - M^2)/(L_1 + L_2 \mp 2M)$, and the difference is not a correction — at high $k$ it changes the answer by a large factor and can even produce a circulating current between the two windings that dissipates real power while contributing nothing. Two further practical points. Current does not divide evenly between mismatched parts: it divides in inverse proportion to inductance at low frequency and in inverse proportion to DC resistance at true DC, so a parallel pair should be identical parts if the current sharing is meant to be equal. And as with the series case, the ratings do not follow the arithmetic — saturating one of the pair dumps its share of the current onto the other, which then saturates too.