Inductors in Parallel

If there is no mutual inductance among two or more parallel-connected inductors, their values add up like the values of resistors in parallel. Suppose you have inductances L1, L2, L3, . . . , Ln all connected in parallel. Then you can find the reciprocal of the total inductance, 1/L, using the following formula:

1/L = 1/L1 + 1/L2 + 1/L3 + . . . + 1/Ln

The total inductance, L, is found by taking the reciprocal of the number you get for 1/L. Again, as with inductances in series, it’s important to remember that all the units have to agree during the calculation process. Once you have completed the calculation, you can convert the result to any inductance unit.


Problem 1

Suppose there are three inductors, each with a value of 40 μH, connected in parallel with no mutual inductance, as shown in above Figure. What is the net inductance of the combination?
Let’s call the inductances L1 = 40 μH, L2 = 40 μH, and L3 = 40 μH. Use the preceding formula to obtain 1/L = 1/40 + 1/40 + 1/40 = 3/40 = 0.075. Then L = 1/0.075 = 13.333 μH. This should be rounded off to 13 μH, because the original inductances are specified to only two significant digits.

Problem 2

Imagine four inductors in parallel, with no mutual inductance and values of L1 = 75.0 mH, L2 = 40.0 mH, L3 = 333 μH, and L4 = 7.00 H. What is the net inductance of this combination?
You can use henrys, millihenrys, or microhenrys as the standard units in this problem. Suppose you decide to use henrys. Then L1 = 0.0750 H, L2 = 0.0400 H, L3 = 0.000333 H, and L4 = 7.00 H. Use the preceding formula to obtain 1/L = 13.33 + 25.0 + 3003 + 0.143 = 3041.473. The reciprocal of this is the inductance L = 0.00032879 H = 328.79 μH. This should be rounded off to 329 μH. This is only a little less than the value of the 333 μH inductor alone.

If there are several inductors in parallel, and one of them has a value that is much smaller than the values of all the others, then the total inductance is a little smaller than the value of the smallest inductor.