You can just add up the values. Call the inductances of the individual components L1, L2,and L3, and the total inductance L. Then LL1L2L3404040120µH.186 Inductance10-3Inductors in series.Problem 10-2Suppose there are three inductors, with no mutual inductance, and their values are 20.0mH, 55.0 µH, and 400 nH. What is the total inductance of these components if they areconnected in series as shown in Fig. 10-3?First, convert the inductances to the same units. You might use microhenrys, be-cause that’s the “middle-sized” unit here. Call L1 20.0 mH 20,000µH; L 2 55.0µH;L3 400 nH 0.400 uH. Then the total inductance is L 20,00055.00.400 uH 20,055.4µH. The values of the original separate components were each given to threesignificant figures, so you should round the final figure off to 20,100 µH.Note that subscripts are now used in designators. An example is L2 (rather thanL2). Some engineers like subscripts, while others don’t want to bother with them. Youshould get used to seeing them both ways. They’re both alright. If there are several inductors in series, and one of them has a value much larger than the values of the others, then the total inductance will be only a little bit more than the value of thelargest inductor.Inductors in parallelIf there is no mutual inductance among two or more parallel-connected inductors, theirvalues 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 in-ductance, 1/L, using the following formula:1/L 1/L1 1/L2 1/L3… 1/LnThe 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 unitshave to agree. Don’t mix microhenrys with millihenrys, or henrys with nanohenrys. Theunits you use for the individual component values will be the units you get for the finalanswer.