6.5 Series and Parallel Inductors

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6.5 Series and Parallel Inductors

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.

Consider a series connection of N inductors, as shown in Fig. 6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors have the same current through them. Applying KVL to the loop,

$$v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.25}$$

Substituting vk=Lk di∕dt results in

$$v = L_1 \frac{di}{dt} + L_2 \frac{di}{dt} + L_3 \frac{di}{dt} + \dots + L_N \frac{di}{dt}$$

= $(L_1 + L_2 + L_3 + \dots + L_N) \frac{di}{dt}$
= $\left(\sum_{k=1}^{N} L_k\right) \frac{di}{dt} = L_{eq} \frac{di}{dt}$ (6.26)

where

Thus,

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

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

We now consider a parallel connection of N inductors, as shown in Fig. 6.30(a), with the equi valent circuit in Fig. 6.30(b). The inductors have the same voltage across them. Using KCL,

$$i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.28}$$

But ik = __1 Lk ∫ t0 t v dt + ik(t0); hence, i = __1 L1 ∫ t0 t v dt + i1(t0) + __1 L2 ∫ t0 t v dt + i2(t0) + ⋯ + ___1 LN ∫ t0 t v dt + iN (t0) ⁼ ( __1 L1 ⁺ __1 L2 ^{+ ⋯ +} ___1 LN)^∫ t0 t v dt + i1(t0) + i2(t0) + ⋯ + iN (t0) N t N t

$$= \left(\sum_{k=1}^{N} \frac{1}{L_k}\right) \int_{t_0}^{t} v \, dt + \sum_{k=1}^{N} i_k(t_0) = \frac{1}{L_{\text{eq}}} \int_{t_0}^{t} v \, dt + i(t_0) \tag{6.29}$$

where

$$\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \dots + \frac{1}{L_N}$$

(6.30)

The initial current i(t0) through Leq at t=t0 is expected by KCL to be the sum of the inductor currents at t0. Thus, according to Eq. (6.29),

$$i(t_0) = i_1(t_0) + i_2(t_0) + \dots + i_N(t_0)$$

According to Eq. (6.30),

The equivalent inductance of 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), Eq. (6.30) becomes

$$\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} \qquad \text{or} \qquad L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2} \tag{6.31}$$

As long as all the elements are of the same type, the ∆-Y transformations for resistors discussed in Section 2.7 can be e xtended to capacitors and inductors.

Figure 6.30

(a) A parallel connection of N inductors, (b) equivalent circuit for the parallel inductors.

TABLE 6.1

Important characteristics of the basic elements.†

Relation	Resistor (R)	Capacitor (C)	Inductor (L)
v-i:	v = iR	t __1 v = C ∫ i(τ) dτ + v(t0) t0	v = L __di dt
i-v:	i = v∕R	i = C ___ dv dt	t __1 i = ∫ v(τ) dτ + i(t0) L t0
p or w:	v2 __ 2 p = i R = R	__1 Cv2 w = 2	__1 Li2 w = 2
Series:	Req = R1 + R2	_______ C1C2 Ceq = C1 + C2	Leq = L1 + L2
Parallel:	_______ R1R2 Req = R1 + R2	Ceq = C1 + C2	_______ L1L2 Leq = L1 + L2
At dc:	Same	Open circuit	Short circuit
Circuit variable that cannot
change abruptly:	Not applicable	v	i

† Passive sign convention is assumed.

It is appropriate at this point to summarize the most important characteristics of the three basic circuit elements we have studied. The summary is given in Table 6.1.

Example 6.11 Find the equivalent inductance of the circuit shown in Fig. 6.31.

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

$$\frac{7 \times 42}{7 + 42} = 6$$

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

$$L_{\text{eq}} = 4 + 6 + 8 = 18 \text{ H}$$

Practice Problem 6.11 Calculate the equivalent inductance for the inductive ladder network in Fig. 6.32.

For the circuit in Fig. 6.33, i(t) = 4(2 ^{− e} Example 6.12 ⁻10*^t* ) mA. If i2(0) = −1 mA, find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t) and i2(t).

Solution:

(a) From i(t) = 4(2 − e⁻10*^t* ) mA, i(0) = 4(2 − 1) = 4 mA. Since i = i¹ + i2,

$$i_1(0) = i(0) - i_2(0) = 4 - (-1) = 5 \text{ mA}$$

(b) The equivalent inductance is

$$L_{\text{eq}} = 2 + 4 || 12 = 2 + 3 = 5 \text{ H}$$

Thus,

$$v(t) = L_{eq} \frac{di}{dt} = 5(4)(-1)(-10)e^{-10t} \text{ mV} = 200e^{-10t} \text{ mV}$$

and

$$v_1(t) = 2\frac{di}{dt} = 2(-4)(-10)e^{-10t}$$

mV = 80 $e^{-10t}$ mV

Since v = v¹ + v2,

$$v_2(t) = v(t) - v_1(t) = 120e^{-10t} \text{ mV}$$

(c) The current i1 is obtained as

$$i_1(t) = \frac{1}{4} \int_0^t v_2 dt + i_1(0) = \frac{120}{4} \int_0^t e^{-10t} dt + 5 \text{ mA}$$

= $-3e^{-10t} \Big|_0^t + 5 \text{ mA} = -3e^{-10t} + 3 + 5 = 8 - 3e^{-10t} \text{ mA}$

Similarly,

$$i_2(t) = \frac{1}{12} \int_0^t v_2 dt + i_2(0) = \frac{120}{12} \int_0^t e^{-10t} dt - 1 \text{ mA}$$ $$= -e^{-10t} \Big|_0^t - 1 \text{ mA} = -e^{-10t} + 1 - 1 = -e^{-10t} \text{ mA}$$

Note that i1(t) + i2(t) = i(t).

In the circuit of Fig. 6.34, i1(t) = 3e Practice Problem 6.12 ⁻2*^t* A. If i(0) = 7 A, find: (a) i2(0); (b) i2(t) and i(t); (c) v1(t), v2(t), and v(t).

Answer: (a) 4 A, (b)

$$(-2 + 6e^{-2t})$$

A, $(-2 + 9e^{-2t})$ A, (c) $-36e^{-2t}$ V, $-144e^{-2t}$ V, $-180e^{-2t}$ V.

Figure 6.34 For Practice Prob. 6.12.