5.454545
mH
0.005454545
H
0.183333333
1/mH
10
mH
5.454545
mH
0.005454545
H
0.183333333
1/mH
10
mH
The Inductors in Parallel Calculator determines the equivalent inductance when two or three inductors share the same voltage across their terminals. In a parallel configuration, voltage is common but current divides among branches, resulting in a total inductance that is always less than the smallest individual inductor.
Parallel inductor arrangements appear in current-sharing networks, multi-winding transformer designs, EMI filter topologies, and situations where a specific small inductance value is needed from larger available components.
For inductors connected in parallel with no mutual coupling, the reciprocal of the total inductance equals the sum of the reciprocals:
$$\frac{1}{L_{total}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}$$
This is derived from the fact that all parallel inductors share the same voltage $$V$$, while the total current is the sum of branch currents. Since $$I_k = \frac{1}{L_k}\int V\,dt$$, the total current is:
$$I_{total} = \left(\frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}\right)\int V\,dt = \frac{1}{L_{total}}\int V\,dt$$
For two inductors, this simplifies to the product-over-sum formula:
$$L_{total} = \frac{L_1 \cdot L_2}{L_1 + L_2}$$
The parallel combination always yields a value smaller than the smallest individual inductor in the group, analogous to resistors in parallel. This property is useful for creating precise small inductance values from standard components.
The Total Inductance shown is always less than the smallest input value. This makes parallel combinations useful for obtaining low inductance values. In frequency-selective circuits, lower inductance shifts resonant and cutoff frequencies higher. When designing current dividers, the branch with the smallest inductance carries the largest share of transient current.
Inputs
Results
1/L = 1/10 + 1/20 + 1/30 = 0.1 + 0.05 + 0.0333 = 0.1833, so L = 5.45 mH — less than the smallest (10 mH).
Inputs
Results
Two equal 100 mH inductors in parallel give approximately 50 mH. Set L₃ very large to effectively remove it.
Because adding more current paths reduces the overall opposition to current change. Each parallel branch provides an additional path for current flow, lowering the effective inductance just as parallel resistors lower total resistance.
For exactly two inductors: $$L_{total} = \frac{L_1 \cdot L_2}{L_1 + L_2}$$. This is a convenient shortcut derived from the reciprocal formula.
Set the third inductor (L₃) to a very large value (e.g., 100000 mH). Its reciprocal becomes negligible, effectively removing it from the calculation.
Yes. If the inductors are magnetically coupled, the mutual inductance $$M$$ modifies the equivalent inductance. The uncoupled formula used here assumes the magnetic fields do not interact.
Common applications include achieving a specific low inductance from available parts, increasing current-handling capacity (each inductor carries only a fraction of the total current), and in multi-winding EMI choke designs.
In an LC circuit, $$f = \frac{1}{2\pi\sqrt{LC}}$$. Lower parallel inductance increases the resonant frequency, useful for tuning to higher frequencies with existing capacitor values.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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