Alternatively,

← Back to Fundamentals of Electric Circuits Overview v_1 = N_1 \frac{d\phi_1}{di_1} \frac{di_1}{dt} = L_1 \frac{di_1}{dt}

$$(13.8) where *L*1 =  *N*1 *dϕ*1∕*di*1 is the self-inductance of coil 1. Similarly, Eq. (13.7) can be written as$$

v_2 = N_2 \frac{d\phi_{12}}{di_1} \frac{di_1}{dt} = M_{21} \frac{di_1}{dt}

$$(13.9) where$$

M_{21} = N_2 \frac{d\phi_{12}}{di_1}

$$(13.10) *M*21 is known as the *mutual inductance* of coil 2 with respect to coil 1. Subscript 21 indicates that the inductance *M*21 relates the voltage induced in coil 2 to the current in coil 1. Thus, the open-circuit *mutual voltage* (or induced voltage) across coil 2 is$$

v_2 = M_{21} \frac{di_1}{dt}

$$(13.11) Suppose we now let current *i*2 flow in coil 2, while coil 1 carries no current (Fig. 13.3). The magnetic flux *ϕ*2 emanating from coil 2 comprises flux *ϕ*22 that links only coil 2 and flux *ϕ*21 that links both coils. Hence,$$

\phi_2 = \phi_{21} + \phi_{22} \tag{13.12}

$$The entire flux *ϕ*2 links coil 2, so the voltage induced in coil 2 is$$

v_2 = N_2 \frac{d\phi_2}{dt} = N_2 \frac{d\phi_2}{di_2} \frac{di_2}{dt} = L_2 \frac{di_2}{dt}

$$(13.13) where *L*² = *N*² *dϕ*2∕*di*2 is the self-inductance of coil 2. Since only flux *ϕ*21 links coil 1, the voltage induced in coil 1 is$$

v_1 = N_1 \frac{d\phi_{21}}{dt} = N_1 \frac{d\phi_{21}}{di_2} \frac{di_2}{dt} = M_{12} \frac{di_2}{dt}

$$(13.14) where$$

M_{12} = N_1 \frac{d\phi_{21}}{di_2}

$$(13.15) which is the *mutual inductance* of coil 1 with respect to coil 2. Thus, the open-circuit *mutual voltage* across coil 1 is$$

v_1 = M_{12} \frac{di_2}{dt}

$$(13.16) We will see in the next section that *M*12 and *M*21 are equal; that is,$$

M_{12} = M_{21} = M \tag{13.17}

and we refer to *M* as the mutual inductance between the tw o coils. Like self-inductance *L*, mutual inductance *M* is measured in henrys (H). Keep in mind that mutual coupling only exists when the inductors or coils are in close proximity, and the circuits are dri ven by time-varying sources. We recall that inductors act like short circuits to dc. From the two cases in Figs. 13.2 and 13.3, we conclude that mutual inductance results if a v oltage is induced by a time-v arying current in another circuit. It is the property of an inductor to produce a v oltage in reaction to a time-varying current in another inductor near it. Thus, Mutual inductance *M*12 of coil 1 with respect to coil 2. Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor, measured in henrys (H). Although mutual inductance *M* is al ways a positi ve quantity, the mutual voltage *M di*∕*dt* may be negative or positive, just like the self-induced voltage *L di*∕*dt*. However, unlike the self-induced *L di*∕*dt*, whose polarity is determined by the reference direction of the current and the reference polarity of the v oltage (according to the passi ve sign convention), the polarity of mutual v oltage *M di*∕*dt* is not easy to determine, because four terminals are involved. The choice of the correct polarity for *M di*∕*dt* is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz's law in conjunction with the right-hand rule. Since it is inconvenient to show the construction details of coils on a circuit schematic, we apply the *dot convention* in circuit analysis. By this convention, a dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside the coils so that we need not bother about how to place them. The dots are used along with the dot con vention to determine the polarity of the mutual voltage. The dot convention is stated as follows: If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil. # Alternatively, If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil. Thus, the reference polarity of the mutual v oltage depends on the ref erence direction of the inducing current and the dots on the coupled coils. Application of the dot con vention is illustrated in the four pairs of mutually coupled coils in Fig. 13.5. Fo r the coupled coils in Fig. 13.5(a), the sign of the mutual voltage *v*2 is determined by the reference polarity for *v*2 and the direction of *i*1. Since *i*1 enters the dotted terminal of coil 1 and *v*2 is positive at the dotted terminal of coil 2, the mutual voltage is +*M di*1∕*dt*. For the coils in Fig. 13.5(b), the current *i*1 enters Illustration of the dot convention. **Figure 13.5** Examples illustrating how to apply the dot convention. the dotted terminal of coil 1 and *v*2 is ne gative at the dotted terminal of coil 2. Hence, the mutual v oltage is −*M di*1∕*dt*. The same reasoning applies to the coils in Figs. 13.5(c) and 13.5(d). Figure 13.6 shows the dot convention for coupled coils in series. For the coils in Fig. 13.6(a), the total inductance is

L = L_1 + L_2 + 2M

$$(Series-aiding connection) (13.18) For the coils in Fig. 13.6(b),$$

L = L_1 + L_2 - 2M

$$(Series-opposing connection) (13.19) Now that we know how to determine the polarity of the mutual voltage, we are prepared to analyze circuits involving mutual inductance. As the first example, consider the circuit in Fig. 13.7(a). Applying KVL to coil 1 gives$$

v_1 = i_1 R_1 + L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}

$$(13.20a) For coil 2, KVL gives$$

v_2 = i_2 R_2 + L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}

$$(13.20b) We can write Eq. (13.20) in the frequency domain as$$

\mathbf{V}_1 = (R_1 + j\omega L_1)\mathbf{I}_1 + j\omega M \mathbf{I}_2 \tag{13.21a}

$$$$

\mathbf{V}_2 = j\omega M \mathbf{I}_1 + (R_2 + j\omega L_2) \mathbf{I}_2 \tag{13.21b}

$$As a second e xample, consider the circuit in Fig. 13.7(b). We analyze this in the frequency domain. Applying KVL to coil 1, we get$$

\mathbf{V} = (\mathbf{Z}_1 + j\omega L_1)\mathbf{I}_1 - j\omega M\mathbf{I}_2

$$(13.22a) For coil 2, KVL yields$$

0 = -j\omega M I_1 + (Z_L + j\omega L_2)I_2 \tag{13.22b}

Equations (13.21) and (13.22) are solv ed in the usual manner to deter mine the currents. One of the most important things in making sure one solv es problems accurately is to be able to check each step during the solution pro cess and to mak e sure assumptions can be v erified. Too often, solving mutually coupled circuits requires the problem solv er to track tw o or more steps made at once re garding the sign and v alues of the mutually induced voltages. # **Figure 13.7** Time-domain analysis of a circuit containing coupled coils (a) and frequency-domain analysis of a circuit containing coupled coils (b). # **Figure 13.6** Dot convention for coils in series; the sign indicates the polarity of the mutual voltage: (a) seriesaiding connection, (b) seriesopposing connection. Model that makes analysis of mutually coupled easier to solve. Experience has sho wn that if we break the problem into steps of solving for the value and the sign into separate steps, the decisions made are easier to track. We suggest that model (Figure 13.8 (b)) be used when analyzing circuits containing a mutually c oupled circuit shown in Figure 13.8(a): Notice that we have not included the signs in the model. The reason for that is that we first determine the value of the induced voltages and then we determine the appropriate signs. Clearly, I1 induces a voltage within the second coil represented by the value *jω*I1 and I2 induces a voltage of *jω*I2 in the first coil. Once we have the values we next use both circuits to find the correct signs for the dependent sources as shown in Figure 13.8(c). Since I1 enters *L*1 at the dotted end, it induces a voltage in *L*2 that tries to force a current out of the dotted end of *L*2 which means that the source must have a plus on top and a minus on the bottom as sho wn in Figure 13.8(c). I2 leaves the dotted end of *L*2 which means that it induces a voltage in *L*1 which tries to force a current into the dotted end of *L*¹ requiring a dependent source that has a plus on the bottom and a minus on top as shown in Figure 13.8(c). No w all we have to do is to analyze a circuit with two dependent sources. This process allows you to check each of your assumptions. At this introductory level we are not concerned with the determination of the mutual inductances of the coils and their dot placements. Lik e *R*, *L*, and *C*, calculation of *M* would involve applying the theory of elect romagnetics to the actual ph ysical properties of the coils. In this te xt, we assume that the mutual inductance and the placement of the dots are the "gi vens'' of the circuit problem, like the circuit components *R*, *L*, and *C*. For Example 13.1.

-12 + (-j4 + j5)\mathbf{I}_1 - j3\mathbf{I}_2 = 0

$$$$

jI_1 - j3I_2 = 12 \tag{13.1.1}

$$For loop 2, KVL gives$$

-j3\mathbf{I}_1 + (12 + j6)\mathbf{I}_2 = 0

$$or$$

\mathbf{I}_1 = \frac{(12 + j6)\mathbf{I}_2}{j3} = (2 - j4)\mathbf{I}_2

$$(13.1.2) Substituting this in Eq. (13.1.1), we get ( *j*2 + 4 − *j*3)**I**² = (4 − *j*)**I**² = 12 or$$

\mathbf{I}_2 = \frac{12}{4 - j} = 2.91 \underline{ / 14.04^{\circ}} A \tag{13.1.3}

$$From Eqs. (13.1.2) and (13.1.3),$$

\mathbf{I}_1 = (2 - j4)\mathbf{I}_2 = (4.472 \angle -63.43^\circ)(2.91 \angle 14.04^\circ)

= 13.01 \angle -49.39^\circ A Practice Problem 13.1 Determine the voltage **V***o* in the circuit of Fig. 13.10. **Figure 13.10** For Practice Prob. 13.1. **Answer:** 12⧸ −45° V. Example 13.2 Calculate the mesh currents in the circuit of Fig. 13.11. **Figure 13.11** For Example 13.2. # **Solution:** The key to analyzing a magnetically coupled circuit is knowing the polarity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11, suppose coil 1 is the one whose reactance is 6 Ω, and coil 2 is the one whose reactance is 8 Ω. To figure out the polarity of the mutual voltage in coil 1 due to current **I**2, we observe that **I**2 leaves the dotted terminal of coil 2. Since we are applying KVL in the clockwise direction, it implies that the mutual voltage is negative, that is, −*j*2**I**2. Alternatively, it might be best to figure out the mutual voltage by redrawing the rele vant portion of the circuit, as sho wn in Fig. 13.12, where it becomes clear that the mutual voltage is **V**¹ = −2*j* **I**2. Thus, for mesh 1 in Fig. 13.11, KVL gives

-100 + \mathbf{I}_1(4 - j3 + j6) - j6\mathbf{I}_2 - j2\mathbf{I}_2 = 0

$$or$$

100 = (4+j3)\mathbf{I}_1 - j8\mathbf{I}_2 \tag{13.2.1}

$$Similarly, to figure out the mutual voltage in coil 2 due to current **I**1, consider the relevant portion of the circuit, as shown in Fig. 13.12. Applying the dot convention gives the mutual voltage as **V**² = −2*j***I**1. Also, current **I**2 sees the two coupled coils in series in Fig. 13.11; since it leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore, for mesh 2 in Fig. 13.11, KVL gives$$

0 = -2jI_1 - j6I_1 + (j6 + j8 + j2 \times 2 + 5)I_2

$$or$$

0 = -j8I_1 + (5+j18)I_2 \tag{13.2.2}

$$Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get$$

\begin{bmatrix} 100 \ 0 \end{bmatrix} = \begin{bmatrix} 4+j3 & -j8 \ -j8 & 5+j18 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \ \mathbf{I}_2 \end{bmatrix}

$$The determinants are$$

\Delta = \begin{vmatrix} 4+j3 & -j8 \ -j8 & 5+j18 \end{vmatrix} = 30 + j87

$$\n$$

\Delta_1 = \begin{vmatrix} 100 & -j8 \ 0 & 5+j18 \end{vmatrix} = 100(5+j18)

$$\n$$

\Delta_2 = \begin{vmatrix} 4+j3 & 100 \ -j8 & 0 \end{vmatrix} = j800

$$Thus, we obtain the mesh currents as Thus, we obtain the mesh currents as \n$$

\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{100(5 + j18)}{30 + j87} = \frac{1,868.2/74.5^{\circ}}{92.03/71^{\circ}} = 20.3/3.5^{\circ} \text{ A}

$$\n$$

\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{j800}{30 + j87} = \frac{800/90^{\circ}}{92.03/71^{\circ}} = 8.693/19^{\circ} \text{ A}

Determine the phasor currents **I**1 and **I**2 in the circuit of Fig. 13.13. Practice Problem 13.2 **Figure 13.13** For Practice Prob. 13.2. **Answer:** I1 = 17.889⧸86.57° A, I2 = 26.83⧸86.57° A. **Figure 13.12** Model for Example 13.2 showing the polarity of the induced voltages. # **13.3** Energy in a Coupled Circuit In Chapter 6, we saw that the energy stored in an inductor is given by

w = \frac{1}{2} L i^2

$$(13.23) We now want to determine the energy stored in magnetically coupled coils. Consider the circuit in Fig. 13.14. We assume that currents *i*1 and *i*² are zero initially, so that the ener gy stored in the coils is zero. If we let *i*¹ increase from zero to *I*1 while maintaining *i*2 = 0, the power in coil 1 is$$

p_1(t) = v_1 i_1 = i_1 L_1 \frac{di_1}{dt}

$$(13.24) and the energy stored in the circuit is$$

w_1 = \int p_1 dt = L_1 \int_0^{I_1} i_1 dt_1 = \frac{1}{2} L_1 I_1^2

$$(13.25) If we now maintain *i*1 = *I*1 and increase *i*2 from zero to *I*2, the mutual voltage induced in coil 1 is *M*¹² *di*2∕*dt*, while the mutual v oltage induced in coil 2 is zero, since *i*1 does not change. The power in the coils is now$$

p_2(t) = i_1 M_{12} \frac{di_2}{dt} + i_2 v_2 = I_1 M_{12} \frac{di_2}{dt} + i_2 L_2 \frac{di_2}{dt}

$$(13.26) and the energy stored in the circuit is$$

w_2 = \int p_2 dt = M_{12} I_1 \int_0^{I_2} di_2 + L_2 \int_0^{I_2} i_2 di_2

$$$$

= M_{12} I_1 I_2 + \frac{1}{2} L_2 I_2^2

$$(13.27) The total ener gy stored in the coils when both *i*1 and *i*2 have reached constant values is$$

w = w_1 + w_2 = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M_{12} I_1 I_2

$$(13.28) If we reverse the order by which the currents reach their final values, that is, if we first increase *i*2 from zero to *I*2 and later increase *i*1 from zero to *I*1, the total energy stored in the coils is$$

w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + M_{21}I_1I_2

$$\n(13.29) Because the total energy stored should be the same regardless of how we reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us to conclude that$$

M_{12} = M_{21} = M \tag{13.30a}

$$and$$

w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2

$$(13.30b) This equation was derived based on the assumption that the coil currents both entered the dotted terminals. If one current enters one dotted <span id="page-584-0"></span>**Figure 13.14** The circuit for deriving energy stored in a coupled circuit. terminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy *MI*1*I*2 is also negative. In that case,$$

w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 - MI_1I_2

$$(13.31) Also, because *I*1 and *I*2 are arbitrary v alues, they may be replaced by *i*1 and *i*2, which gives the instantaneous ener gy stored in the circuit the general expression$$

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

$$(13.32) The positive sign is selected for the mutual term if both currents enter or leave the dotted terminals of the coils; the ne gative sign is selected otherwise. We will now establish an upper limit for the mutual inductance M. The energy stored in the circuit cannot be negative because the circuit is passive. This means that the quantity 1∕2*L*1 *i* 1 ² + 1∕2*L*2 *i* 2 ² − *Mi*1*i*2 must be greater than or equal to zero:$$

\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 - Mi_1i_2 \ge 0

$$\n(13.33) To complete the square, we both add and subtract the term *i*1*i*2( √ *L*1*L*2 on the right-hand side of Eq. (13.33) and obtain$$

\frac{1}{2}(i_1\sqrt{L_1} - i_2\sqrt{L_2})^2 + i_1i_2(\sqrt{L_1L_2} - M) \ge 0

$$(13.34) The squared term is ne ver negative; at its least it is zero. Therefore, the sec ond term on the right-hand side of Eq. (13.34) must be greater than zero; that is, √$$

\overline{L_1 L_2} - M \ge 0

$$$$

M \le \sqrt{L_1 L_2}