Figure 13.15

(13.35) Thus, the mutual inductance cannot be greater than the geometric mean of the self-inductances of the coils. The e xtent to which the mutual inductance *M* approaches the upper limit is specified by the *coefficient of coupling k*, given by

k = \frac{M}{\sqrt{L_1 L_2}}\tag{13.36}

M = k\sqrt{L_1 L_2} \tag{13.37}

where 0 ≤ *k* ≤ 1 or equivalently 0 ≤ *M* ≤  √ \_\_\_\_ *L*1*L*2 . The coupling coefficient is the fraction of the total flux emanating from one coil that links the other coil. For example, in Fig. 13.2,

k = \frac{\phi_{12}}{\phi_1} = \frac{\phi_{12}}{\phi_{11} + \phi_{12}}

(13.38) or or and in Fig. 13.3, # **Figure 13.15** Windings: (a) loosely coupled, (b) tightly coupled; cutaway view demonstrates both windings.

k = \frac{\phi_{21}}{\phi_2} = \frac{\phi_{21}}{\phi_{21} + \phi_{22}}

(13.39) If the entire flux produced by one coil links another coil, then *k* = 1 and we have 100 percent coupling, or the coils are said to be *perfectly coupled*. For *k* < 0.5, coils are said to be *loosely coupled*; and for *k* > 0.5, they are said to be *tightly coupled*. Thus, The coupling coefficient k is a measure of the magnetic coupling between two coils; 0 \leq k \leq 1. We expect *k* to depend on the closeness of the two coils, their core, their orientation, and their windings. Figure 13.15 sho ws loosely coupled windings and tightly coupled windings. The air -core trans formers used in radio frequenc y circuits are loosely coupled, whereas iron-core transformers used in po wer systems are tightly coupled. The linear transformers discussed in Section 3.4 are mostly air -core; the ideal transformers discussed in Sections 13.5 and 13.6 are principally iron-core. **Figure 13.16** For Example 13.3. Consider the circuit in Fig. 13.16. Determine the coupling coef ficient. Calculate the ener gy stored in the coupled inductors at time *t* = 1 s if *v* = 60 cos(4*t* + 30°) V. # **Solution:** The coupling coefficient is

k = \frac{M}{\sqrt{L_1 L_2}} = \frac{2.5}{\sqrt{20}} = 0.56

indicating that the inductors are tightly coupled. To find the energy stored, we need to calculate the current. To find the current, we need to obtain the frequency-domain equivalent of the circuit. 60 cos(4*t* + 30°)

\Rightarrow

60/30°, $\omega = 4$ rad/s \n5 H $\Rightarrow$ $j\omega L_1 = j20 \Omega$ \n2.5 H $\Rightarrow$ $j\omega M = j10 \Omega$ \n4 H $\Rightarrow$ $j\omega L_2 = j16 \Omega$ \n $\frac{1}{16}F \Rightarrow \frac{1}{j\omega C} = -j4 \Omega$ The frequency-domain equivalent is shown in Fig. 13.17. We now apply mesh analysis. For mesh 1,

(10 + j20)\mathbf{I}_1 + j10\mathbf{I}_2 = 60/30^{\circ}

(13.3.1) For mesh 2,

j10\mathbf{I}_1 + (j16 - j4)\mathbf{I}_2 = 0

\overline{a}

I_1 = -1.2I_2 \tag{13.3.2}

<span id="page-587-0"></span>Substituting this into Eq. (13.3.1) yields

I_2(-12 - j14) = 60/30^{\circ} \qquad \Rightarrow \qquad I_2 = 3.254/160.6^{\circ} \text{ A}

and

I_1 = -1.2I_2 = 3.905 \div 19.4^{\circ}

A In the time-domain,

i_1 = 3.905 \cos(4t - 19.4^\circ),

$i_2 = 3.254 \cos(4t + 160.6^\circ)$ At time *t* = 1 s, 4*t* = 4 rad = 229.2°, and

i_1 = 3.905 \cos(229.2^\circ - 19.4^\circ) = -3.389 \text{ A}

i_2 = 3.254 \cos(229.2^\circ + 160.6^\circ) = 2.824 \text{ A}

The total energy stored in the coupled inductors is

w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 + Mi_1i_2

= $\frac{1}{2}(5)(-3.389)^2 + \frac{1}{2}(4)(2.824)^2 + 2.5(-3.389)(2.824) = 20.73 \text{ J}$ **Figure 13.17** Frequency-domain equivalent of the circuit in Fig. 13.16. For the circuit in Fig. 13.18, determine the coupling coefficient and the energy stored in the coupled inductors at *t* = 1.5 s. Practice Problem 13.3 **Answer:** 0.7071, 246.2 J.