How this instrument works
Two inductors in parallel share the same pair of terminals, so at every instant they carry the same voltage across them. Faraday's law ties that shared voltage to the rate of change of current in each branch: v = L di/dt, or turned around, di/dt = v/L. Each branch changes its own current in inverse proportion to its own inductance, so the branch with the smaller inductance ramps its current fastest. Add the two branch currents together and the combined pair behaves like one coil whose reciprocal inductance is the sum of the two separate reciprocals — 1/L_total = 1/L₁ + 1/L₂ — which algebra alone collapses to product over sum for exactly two coils.
That reciprocal-sum pattern is the same one conductances follow in a parallel resistor network, because 1/L plays a bookkeeping role identical to 1/R: both measure how readily a branch accepts a shared driving signal, just governed by a different law of physics. The henry itself honors Joseph Henry, who discovered self-induction independently of Michael Faraday around 1830 and whose name the unit received in 1893 at the International Electrical Congress in Chicago. One henry is the inductance that induces one volt when current through it changes at one ampere per second — a definition written directly into the formula above.
The formula assumes the two coils are magnetically isolated from each other — no shared flux, no mutual inductance. Wind two chokes on separate cores or keep them well apart on a board and that assumption holds comfortably. Mount them close together, stack them on a shared bobbin, or let one sit inside a transformer's stray field, and a mutual term creeps into the true total, either raising or lowering it depending on whether the two windings' fields add or cancel. Real coils also carry winding resistance and core losses this idealized formula ignores entirely, and a saturating core can make L itself fall as current climbs, so the simple total is a starting point for a design, not a guarantee once current gets large.
- Enter your first coil's value into Inductance 1, in henries — switch the unit menu to mH or µH if that's how the part is printed.
- Enter your second coil's value into Inductance 2. Which coil goes in which box makes no difference to the result.
- Read Total inductance. It should land below whichever of L1 or L2 is the smaller value — that inequality is your quickest sanity check.
- Keep the two coils physically separated and their windings angled apart as you wire them — the formula assumes no shared magnetic flux between them.
- For three or more coils, combine them two at a stretch: feed each running total back into Inductance 1 alongside your next coil in Inductance 2.
Worked example — building a 120 mH choke from spares
An EMI filter design calls for a 120 mH differential-mode choke ahead of a switching supply, but the parts drawer holds no 120 mH part — only two toroidal chokes salvaged from old equipment, rated 200 mH and 300 mH. Wired in parallel rather than shelved, Inductance 1 takes 0.2 H, Inductance 2 takes 0.3 H, and Total inductance returns (0.2 × 0.3) ⁄ (0.2 + 0.3) = 0.06 ⁄ 0.5 = 0.12 H — 120 mH, exactly the filter's target, out of parts that were otherwise headed for the bin.
Mounting matters here as much as the arithmetic. The two toroids sit several centimetres apart on the board and their windings run at right angles to each other, which keeps their magnetic fields from coupling — satisfy that condition and 0.12 H is what a bench inductance meter will actually read. Current splits between the pair in inverse proportion to inductance, so the 200 mH core carries the larger share, 300⁄500 or 60 percent of the ripple current, while the 300 mH core carries the remaining 40 percent, worth checking against each part's rated saturation current before trusting the pair under full load.
Questions
Why do inductors in parallel use the same reciprocal-sum rule as resistors?
Because both quantities measure how readily a branch accepts change under one shared driving signal — conductance 1/R for a shared voltage across resistive branches, and 1/L for a shared voltage that drives changing current through inductive ones. Add the branch responses and invert the sum for a total. That collapses to product over sum only with exactly two branches; a third requires the full form, 1/L_total = 1/L₁ + 1/L₂ + 1/L₃.
Does the formula still work if the two coils are wound close together?
Not reliably. Product over sum assumes zero mutual inductance between the branches — no shared magnetic flux. Mount two chokes on one core, stack them tightly, or let one sit inside another's stray field, and a mutual term enters the true total, pushing it above or below the plain calculation depending on whether the windings' fields reinforce or oppose. Separate the coils, or angle their windings apart, to keep the simple formula trustworthy.
Why is the combined inductance always smaller than either coil alone?
Because paralleling gives current a second path, not a bigger obstacle. Whatever a smaller coil alone would resist, a second coil beside it now shares, so the pair together opposes changing current less than either one by itself. Enter 0.2 H beside 0.3 H and any result above 0.2 H signals an arithmetic slip; the correct combined value, 0.12 H, sits below both inputs, exactly as reciprocal addition guarantees.
How do I combine three or more inductors in parallel?
Combine them in pairs, or add reciprocals directly: 1/L_total = 1/L₁ + 1/L₂ + 1/L₃ + ⋯. Feeding three inductances straight into one product-over-sum expression is a common error — that shortcut only holds for exactly two branches. This instrument accepts two values at once; take the total it returns for your first pair, put that back into Inductance 1, and enter your remaining coil into Inductance 2.
What units does this calculator accept?
Henries by default, with each field's unit menu also taking millihenries and microhenries, the ranges most real chokes and small inductors are actually printed in. Every entry converts to henries before arithmetic runs, so a 200 mH part and a 0.2 H part are the identical input typed two different ways; the readout can likewise be switched to whichever prefix suits your bench notes.
Do inductors in parallel split current the same way resistors do?
The shape is identical but the roles trade places. A resistor current divider places the branch you are not solving for on top of its fraction, so a smaller resistor still ends up carrying the bigger share of current. Swap L in for R in that same fraction and the outcome for inductors matches: the branch with the smaller inductance carries the larger current share, because it resists a changing current the least — mirroring how the smaller resistor conducts the most steady current.