Inductors in Series and Parallel

Electrical Circuit Analysis > Inductors

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Inductors in series

Inductors, like resistors and capacitors, can be placed in series or parallel. Increasing levels of inductance can be obtained by placing inductors in series, while decreasing levels can be obtained by placing inductors in parallel. For inductors in series, the total inductance is found in the same manner as the total resistance of resistors in series. Consider a series connection of N inductors, as shown in [Fig. 1(a)], with the equivalent circuit shown in[ Fig. 1(b)].

Fig. 1: (a) A series connection of N inductors, (b) equivalent circuit for the series inductors.

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

$$ v = v_{1}+v_{2}+v_{3}+...+ v_{N} \tag{1}$$

Substituting $v_k = L_k di/dt$ results in

$$ \begin{split} v &= L_1 {di \over dt}+L_2 {di \over dt}+L_3 {di \over dt}+...+ L_N {di \over dt}\\ &= (L_1 + L_2 + L_3+...+L_N)( {di \over dt})\\ &= (L_{eq})( {di \over dt})\\ \end{split} $$

where

$$\bbox[10px,border:1px solid grey]{L_{eq}= L_1 + L_2 + L_3+...+L_N} \tag{2} $$

Thus,

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

Inductors in parallel

We now consider a parallel connection of N inductors, as shown in [Fig. 2(a)], with the equivalent circuit in [Fig. 2(b)].

Fig. 2: (a) A parallel connection of N inductors, (b) equivalent circuit for the parallel inductors.

The inductors have the same voltage across them. Using KCL,

$$i = i_1 + i_2 + i_3 +...+ i_N$$

But

$$i_k= {1 \over L_k} \int_{t_0}^{t}v \,dt$$

hence,

$$ \begin{split} i &={1 \over L_1} \int_{t_0}^{t}v \, dt +{1 \over L_2} \int_{t_0}^{t}v \, dt+{1 \over L_3} \int_{t_0}^{t}v \, dt+...+{1 \over L_N} \int_{t_0}^{t}v \, dt \\ &=({1 \over L_1}+{1 \over L_2}+{1 \over L_3}+...+{1 \over L_N})\int_{t_0}^{t}v \, dt \\ &=({1 \over L_{eq}})\int_{t_0}^{t}v \, dt \\ \end{split} $$

where

$$ \bbox[10px,border:1px solid grey]{{1 \over L_{eq}} = {1 \over L_1}+{1 \over L_2}+{1 \over L_3}+...+{1 \over L_N}} \tag{3}$$

According to Eq. (3),

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

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 1].

Table 1.

Example 1: For the circuit in [Fig. 3], $i(t) = 4(2 - e^{-10t}) mA$. 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 $i_2(t)$.

Fig. 3:

Solution:
(a) From

$$i(t) = 4(2 - e^{-10t}) mA$$

$$i(0) = 4(2 - 1) = 4 mA$$

Since

$$i = i_1 + i_2$$

$$i_1(0) = i(0) - i_2(0) = 4 - (-1) = 5 mA$$

$$\bbox[10px,border:1px solid grey]{i_1(0) = 5 mA}$$

(b) The equivalent inductance is

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

$$\bbox[10px,border:1px solid grey]{L_{eq} = 5 H}$$

Thus,

$$ \begin{split} v(t) &= L_{eq} {di \over dt} = 5(4){d(2 - e^{-10t}) \over dt}\\ &= 20(0 - e^{-10t} (-10)) = 200 e^{-10t}mV \end{split}$$

$$\bbox[10px,border:1px solid grey]{v(t)= 200 e^{-10t}mV}$$

and

$$\begin{split} v_1(t) &= 2 {di \over dt} = 2(4){d(2 - e^{-10t}) \over dt}\\ &= 8(0 - e^{-10t}(-10)) = 80e^{-10t}\\ \end{split}$$

$$ \bbox[10px,border:1px solid grey]{v_1(t)= 80e^{-10t}}$$

Since

$$v(t) = v_1(t) + v_2(t)$$

$$v_2(t) = v(t) - v_1(t) $$ $$=200e^{-10t}mV -80e^{-10t}mV$$

$$\bbox[10px,border:1px solid grey]{v_2(t) =120e^{-10t}mV}$$

(c) The current $i_1$ is obtained as

$$ \begin{split} i_1(t) &= {1 \over 4} \int_0^t v_2(t)dt + i_1(0)\\ &= {1 \over 4} \int_0^t 120e^{-10t}dt + 5mA\\ &=30 e^{-10t} {1 \over -10} |_0^t+ 5mA\\ &= -3e^{-10t}|_0^t +5mA\\ &= (-3e^{-10t}) - (-3e^{-10(0)})+5mA \\ &= -3e^{-10t}+3(1)+5mA\\ &= -3e^{-10t} +8mA = 8mA -3e^{-10t}\\ \end{split}$$

$$\bbox[10px,border:1px solid grey]{i_1(t) = 8mA -3e^{-10t} }$$

Similarly,

$$\begin{split} i_2(t) &= {1 \over 12} \int_0^t v_2(t)dt + i_2(0)\\ &= {1 \over 12} \int_0^t 120e^{-10t}dt - 1mA\\ &=10 e^{-10t} {1 \over -10} |_0^t-1mA\\ &= -1e^{-10t}|_0^t -1mA\\ &= (-e^{-10t}) - (-e^{-10(0)})-1mA \\ &= -e^{-10t}+(1)-1mA\\ \end{split} $$

$$\bbox[10px,border:1px solid grey]{i_2(t) = -e^{-10t} }$$

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