If you have read the previous post about what is an inductor, let us proceed to the next level, what is the inductance formula circuits.

Now that the inductor has been added to our list of passive elements, it is necessary to extend the powerful tool of series-parallel combination. We need to know how to find the equivalent inductance of a series-connected or parallel-connected set of inductors found in practical circuits.

## Series Inductors

Consider a series connection of *N *inductors, as shown in Figure.(1a), with the equivalent circuit shown in Figure.(1b).

The inductors have the same current through them. Applying KVL to the loop,

Substituting *v*_{k}*L*_{k }*di*/*dt *results in

where

Thus,

The

equivalent inductanceof series-connected inductors is the sum of the individual inductances.

Inductors in series are combined in exactly the same way as resistors in series.

## Parallel Inductors

We now consider a parallel connection of *N *inductors, as shown in Figure.(2a), with the equivalent circuit in Figure.(2b). The inductors have the same voltage across them. Using KCL,

But

hence,

where

The initial current *i*(*t*_{0}) through *L*_{eq} at *t**t*_{0} is expected by KCL to be the *t*_{0}. Thus, according to Equation.(5),

According to Equation.(6),

The

equivalent inductanceof parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.

Note that the inductors in parallel are combined in the same way as resistors in parallel.

For two inductors in parallel (*N *= 2), Equation.(6) becomes

## Inductance Formula Circuits Examples

1. Find the equivalent inductance of the circuit shown in Figure.(3).

**Solution:**

The 10-H, 12-H, and 20-H inductors are in series; thus, combining them gives a 42-H inductance. This 42-H inductor is in parallel with the 7-H inductor so that they are combined, to give

This 6-H inductor is in series with the 4-H and 8-H inductors. Hence,

2. For the circuit in Figure.(4),*i*(*t*) = 4(2 − *e*^{−10t}) mA. If *i*_{2}(0) = −1 mA, find:

(a) *i*1(0);

(b) *v*(*t*), *v*_{1}(*t*), and *v*_{2}(*t*);

(c) *i*_{1}(*t*) and *i*_{2}(*t*).

**Solution:**

(a) From *i*(*t*) = 4(2 − *e*^{−10t}) mA, *i*(0) = 4(2 − 1) = 4 mA.

Since *i *= *i*_{1} + *i*_{2},

(b) The equivalent inductance is

Thus,

and

Since *v *= *v*_{1} + *v*_{2},

(c) The current *i*_{1} is obtained as

Similarly,

Note that *i*_{1}(*t*) + *i*_{2}(*t*) = *i*(*t*).