ECE 211 · Lecture 17: Mutual Inductance & the Dot Convention

§1 · Chapter 13 opens

Two coils, one magnetic field

Chapter 13 studies circuits in which two coils share a magnetic field. Two consequences are developed. The first is mutual inductance: the magnetic flux caused by a current in one coil induces a voltage in a second coil, even though no conducting path joins them. The second is the ideal transformer, a four-terminal device comprising two magnetically coupled coils, which is used to step a voltage up or down and to match impedances.

mutual inductance m (H) both coils in the same field step up or step down voltage impedance matching

Review: where self-inductance comes from

The inductor law used since Chapter 6 is worth re-deriving, because mutual inductance is obtained by repeating the derivation across two coils. Faraday's law states that the voltage across a coil of $N$ turns is proportional to the rate of change of the flux it links,

\tm{sf.v}{v_L} = N\,\frac{d\tm{sf.flux}{\phi}}{dt}

A coil of N turns carrying $i_L$ . The current produces the flux $\phi$ , and a change in that flux produces the voltage $v_L$ across the coil.

The flux itself is produced by the current, so a change in flux is due to a change in current. The chain rule therefore converts the flux derivative into a current derivative:

v_L = N\,\frac{d\phi}{di}\cdot\frac{di}{dt} \;\Longrightarrow\; \boxed{\;v_L = L\,\frac{di}{dt}\;} \qquad \text{where}\quad L = N\,\frac{d\phi}{di}

The bracketed group $N\,d\phi/di$ is a property of the coil geometry alone, and it is given the name inductance. Nothing in this derivation required the flux to link only the coil that produced it, which is precisely the opening that mutual inductance exploits.

§2 · The dot convention

A second coil in the same flux

When a second coil is placed near the first, part of the flux produced by $i_1$ links it as well, and the derivation above is repeated with the turns of the second coil. Two voltages are therefore present:

self-induced: v₁ = L₁ di₁/dt mutual: v₂ = ± m di₁/dt

The mutual inductance $m$ is measured in henrys, exactly as $L$ is. The essential observation is that $v_2$ is controlled by a current in a different branch, so the second coil behaves as a current-controlled voltage source and is handled with the dependent-source machinery of Lecture 5.

Only the sign remains open, and it is settled by the dots printed next to the coils.

The dot convention

A current entering the dot side (before the coil) of one coil induces a voltage across the second coil with the plus on the dot side.

Two coils near each other

A current $i_1$ is driven through $L_1$ . Its flux reaches $L_2$ , which carries no source of its own and is left open. The strength of the shared flux path is the mutual inductance $m$ . No conducting path joins the two loops.

The rule, read in one direction

The current enters the dot of coil 1. The induced voltage across coil 2 therefore carries its plus mark on the dot side, which here is the upper terminal, and its value is

v_2 = +\,m\,\frac{di_1}{dt}

Move the dot, and the plus mark follows

Nothing about the current has changed, but the dot of coil 2 is now printed at the lower terminal, so the plus mark of the induced voltage moves there as well. Measured against a reference that is held fixed at the upper terminal, the induced voltage has reversed:

v_2 = -\,m\,\frac{di_1}{dt}

The dots are therefore not decoration. They record how the two coils are wound relative to each other, which the schematic cannot otherwise show.

The phasor form

For a sinusoidal steady state at frequency $\omega$ , differentiation becomes multiplication by $j\omega$ , so the mutual relation becomes an impedance-like statement in the phasor domain:

v_2 = m\,\frac{di_1}{dt} \;\Longrightarrow\; \boxed{\;\tilde{V}_2 = j\omega m\,\tilde{I}_1\;}

The quantity $\omega m$ carries units of ohms, but it is not the impedance of the second coil; it is the transfer coefficient from the first coil's current to the second coil's voltage.

The three cases are collected below. Only the relative sense of the dot and the current matters, so reversing the current has exactly the effect that reversing one dot has.

Current at coil 1	Dot of coil 2	Induced voltage, plus mark held at the upper terminal
enters the dot	upper terminal	+ m di₁/dt
enters the dot	lower terminal	− m di₁/dt
leaves the dot	upper terminal	− m di₁/dt

§3 · The process

Every coupled pair becomes two dependent sources

The board notes reduce magnetically coupled circuits to a mechanical procedure. Once it has been applied, no new analysis technique is required: the circuit that results is an ordinary phasor circuit with dependent sources, and mesh analysis, nodal analysis, or plain KVL finishes it.

The process

A dependent voltage source is added in series with each coil.
The circuit is converted to the phasor (frequency) domain.
$\tilde{I}_1$ and $\tilde{I}_2$ are defined into the dot before the coil on each coil.
A plus is placed on the dot side of the coil on each dependent source.
The value of each dependent source is found: $j\omega m$ times the other coil's current.
The circuit is solved.
The result is carried back to the time domain.

The model that steps 1 through 5 produce. Each coil contributes its own impedance, $j\omega L_1$ and $j\omega L_2$ , driven by its own dot-entering current; the coupling contributes the two dependent sources $j\omega m \tilde{I}_2$ and $j\omega m \tilde{I}_1$ , each with its plus mark on the dot side of its own coil.

Written out, the two branch voltages, each measured with the plus mark at the dotted terminal, are

\tm{md.v1}{\tilde{V}_1} = \tm{md.l1}{j\omega L_1 \tilde{I}_1} + \tm{md.d1}{j\omega m \tilde{I}_2} \qquad\qquad \tm{md.v2}{\tilde{V}_2} = \tm{md.l2}{j\omega L_2 \tilde{I}_2} + \tm{md.d2}{j\omega m \tilde{I}_1}

The first term of each expression is the self-induced voltage, which would be present with the other coil removed; the second term is the mutual voltage, which vanishes when the coupling is broken. The sliders below carry the coil values of Section 4 ( $L_1 = 1/2$ H, $L_2 = 1/4$ H) and retune all three reactances at once.

ω (rad/s) 100

m (H) 0.2

ωL₁ = 50.0 Ω ωL₂ = 25.0 Ω ωm = 20.0 Ω (coupling k = m/√(L₁L₂) = 0.566)

The mutual inductance is bounded

A coil cannot induce more flux in its neighbor than it produces, so $m \le \sqrt{L_1 L_2}$ always holds, and the slider stops at that ceiling for the coils above. The ratio $k = m/\sqrt{L_1 L_2}$ is the coupling coefficient: $k = 1$ describes perfect coupling and $k = 0$ describes coils that ignore each other.

Notation

The board writes phasors with a tilde, as $\tilde{V}$ and $\tilde{I}$ , and that notation is kept in the equations on this page. The schematics drop the accent for legibility, but every voltage and current symbol printed in a figure of Sections 3 to 6 is a phasor.

§4 · Example A: one loop, two coupled coils

A current source in series with the pair

The problem

The source $i_s = 2\cos(100t)$ A drives a 1/2 H coil and a 1/4 H coil connected in one loop. The coils are coupled with $m = 1/5$ H. Find the voltage $v(t)$ across the current source.

Step 2: the impedances at ω = 100 rad/s

j\omega L_1 = j(100)\tfrac{1}{2} = \tm{a.l1}{j50\ \Omega} \qquad j\omega L_2 = j(100)\tfrac{1}{4} = \tm{a.l2}{j25\ \Omega} \qquad j\omega m = j(100)\tfrac{1}{5} = j20\ \Omega

The source phasor is $\tilde{I}_s = 2\angle 0^\circ$ A.

Step 3: the currents defined into the dots

The dot of the first coil is at its left end, and the source current enters there, so $\tilde{I}_1 = 2\angle 0^\circ$ A. The dot of the second coil is at its right end, and the same physical current leaves by that end, so the dot-entering current of the second coil is the negative of it:

\tm{a.i1}{\tilde{I}_1 = 2\angle 0^\circ\ \text{A}} \qquad\qquad \tm{a.i2}{\tilde{I}_2 = -2\angle 0^\circ\ \text{A}}

This sign is the entire content of the dot convention in this problem, and getting it wrong changes the answer.

Steps 4 and 5: the two dependent sources

Each source takes $j\omega m$ times the other coil's dot-entering current, with its plus mark on the dot side of its own coil:

\tm{a.d1}{j20\,\tilde{I}_2 = j20(-2) = -j40\ \text{V}} \qquad\qquad \tm{a.d2}{j20\,\tilde{I}_1 = j20(2) = +j40\ \text{V}}

The first source has its plus mark at the left (the dot side of coil 1); the second has its plus mark at the right (the dot side of coil 2).

Step 6: one KVL around the loop

Travelling clockwise, each coil contributes its impedance times the current in the direction of travel, and each dependent source contributes a drop or a rise according to its plus mark:

\tm{a.l1}{j50(2\angle 0^\circ)} - \tm{a.d1}{j20(2\angle 0^\circ)} - \tm{a.d2}{j20(2\angle 0^\circ)} + \tm{a.l2}{j25(2\angle 0^\circ)} - \tm{a.v}{\tilde{V}} = 0

\tilde{V} = j100 - j40 - j40 + j50 \;\Longrightarrow\; \boxed{\tilde{V} = j70 = 70\angle 90^\circ\ \text{V}}

Step 7: back to the time domain

v(t) = 70\cos(100t + 90^\circ)\ \text{V}

The voltage leads the source current by a quarter cycle, as it must: every element in the loop is an inductance, and the coupling only changes how much reactance the loop presents.

Check: the same answer from the two coil voltages

The branch voltages may also be assembled one coil at a time, each measured with its plus mark at the dotted terminal:

\tilde{V}_1 = j50(2) + j20(-2) = j60\ \text{V} \qquad \tilde{V}_2 = j25(-2) + j20(2) = -j10\ \text{V}

$\tilde{V}_1$ is oriented with its plus at the left of coil 1, so travelling left to right it is a drop of $j60$ . $\tilde{V}_2$ is oriented with its plus at the right of coil 2, so travelling left to right it is a rise of $-j10$ , that is, a drop of $+j10$ . The total drop across the top branch is $j60 + j10 = j70$ , which the source must hold: $\tilde{V} = j70$ V ✓

It is also worth noting the effective reactance seen by the source: $j70 / 2\angle 0^\circ = j35\ \Omega$ , whereas the two coils alone would give $j50 + j25 = j75\ \Omega$ . The coupling as connected is opposing, and it has removed $2\omega m = j40\ \Omega$ from the loop.

§5 · Example B: the pair feeds a capacitor

Two loops, two unknowns

The problem

The source $12\cos(100t + 60^\circ)$ V drives a 5 Ω resistor and a 1/20 H coil, which is coupled with $m = 1/50$ H to a 1/10 H coil. A 1/400 F capacitor closes the circuit. Find $v_o(t)$ , the capacitor voltage.

Step 2: every element in ohms, at ω = 100 rad/s

j\omega L_1 = j(100)\tfrac{1}{20} = \tm{b.l1}{j5\ \Omega} \qquad j\omega L_2 = j(100)\tfrac{1}{10} = \tm{b.l2}{j10\ \Omega}

j\omega m = j(100)\tfrac{1}{50} = j2\ \Omega \qquad \frac{1}{j\omega C} = \frac{-j}{100\,(1/400)} = \tm{b.c}{-j4\ \Omega}

Step 3: the two mesh currents, and what the dots see

Two branch currents are named: $\tilde{I}_1$ in the source branch and $\tilde{I}_o$ in the capacitor branch. KCL then fixes the middle branch at $\tilde{I}_o - \tilde{I}_1$ upward.

The dot of coil 1 is at its left end, which $\tilde{I}_1$ enters. The dot of coil 2 is at its bottom end, which the upward middle current enters. Hence the dot-entering currents are $\tilde{I}_1$ and $\tilde{I}_o - \tilde{I}_1$ , with no sign flip on either coil.

Steps 4 and 5: the dependent sources

\tm{b.d1}{j2\,(\tilde{I}_o - \tilde{I}_1)} \quad\text{plus mark at the left, the dot side of coil 1}

\tm{b.d2}{j2\,\tilde{I}_1} \quad\text{plus mark at the bottom, the dot side of coil 2}

Step 6a: KVL around the left loop

-12\angle 60^\circ + \tm{b.r5}{5\tilde{I}_1} + \tm{b.l1}{j5\tilde{I}_1} + \tm{b.d1}{j2(\tilde{I}_o - \tilde{I}_1)} - \tm{b.d2}{j2\tilde{I}_1} - \tm{b.l2}{j10(\tilde{I}_o - \tilde{I}_1)} = 0

Collecting terms gives the first equation:

(5 + j11)\,\tilde{I}_1 - j8\,\tilde{I}_o = 12\angle 60^\circ

Step 6b: KVL around the right loop

\tm{b.l2}{j10(\tilde{I}_o - \tilde{I}_1)} + \tm{b.d2}{j2\tilde{I}_1} + \tm{b.c}{(-j4)\tilde{I}_o} = 0 \;\Longrightarrow\; j6\,\tilde{I}_o - j8\,\tilde{I}_1 = 0

The right loop alone therefore ties the two currents together: $\tilde{I}_1 = \tfrac{3}{4}\tilde{I}_o$ .

Step 6c and step 7: solve, then return to time

Substituting $\tilde{I}_1 = 0.75\,\tilde{I}_o$ into the left-loop equation:

\left[0.75(5 + j11) - j8\right]\tilde{I}_o = (3.75 + j0.25)\,\tilde{I}_o = 12\angle 60^\circ

\tm{b.io}{\tilde{I}_o = \frac{12\angle 60^\circ}{3.758\angle 3.81^\circ} = 3.19\angle 56.19^\circ\ \text{A}}

\tm{b.vo}{\tilde{V}_o = (-j4)\tilde{I}_o = 12.77\angle{-33.81^\circ}\ \text{V}} \;\Longrightarrow\; \boxed{v_o(t) = 12.77\cos(100t - 33.81^\circ)\ \text{V}}

Check: back-substitution into both loop equations

With $\tilde{I}_o = 3.193\angle 56.19^\circ$ A and $\tilde{I}_1 = 2.395\angle 56.19^\circ$ A, the right-loop equation reads $j(6)(3.193) - j(8)(2.395) = j19.16 - j19.16 = 0$ ✓, since both currents carry the same angle. The left-loop equation reads

(5 + j11)(2.395\angle 56.19^\circ) - j8(3.193\angle 56.19^\circ) = 6.00 + j10.40 = 12.0\angle 60.0^\circ \ \checkmark

Both equations balance, which confirms the solution independently of how it was obtained. (The board notes carry $2.9\angle 85^\circ$ A at this point; that value does not satisfy either loop equation, and the corrected result is printed above.)

§6 · Example C: a Thevenin equivalent with coupling

Coupled coils are dependent sources, so a test source is required

Once the coupled pair has been replaced by its two dependent sources, the network at the port contains dependent sources, and Lecture 5 applies without change: the impedance cannot be read off by series and parallel combination, because the dependent sources remain active when the independent sources are turned off. The port impedance is therefore measured with a test source.

Z_T from a 1 A test source V_T from the open port I_N = V_T / Z_T

Experiment 1: the port impedance

Turn the independent source off, then drive the port

The 10∠0° V source is replaced by a short; both dependent sources remain, because they belong to the coils rather than to any generator. A 1∠0° A test source is connected at the terminals a and b, and the resulting port voltage $\tilde{V}_x$ is sought, because $Z_T = \tilde{V}_x / 1\angle 0^\circ$ .

The dot decides the sign of the first current

The test current enters terminal a and flows to the right through the 20 Ω resistor and the j4 Ω coil. The dot of that coil is printed at its right end, so the current defined into the dot runs against the test current:

\tm{z.i1}{\tilde{I}_1 = -1\angle 0^\circ\ \text{A}}

KVL around the right loop

By KCL at the upper node, the capacitor carries $1 + \tilde{I}_2$ downward. Travelling up the middle branch, across the shorted source, and down through the capacitor:

\tm{z.l2}{j3\tilde{I}_2} + \tm{z.d2}{j2\tilde{I}_1} + \tm{z.c}{(-j5)(1 + \tilde{I}_2)} = 0

j3\tilde{I}_2 - j2 - j5 - j5\tilde{I}_2 = 0 \;\Longrightarrow\; \tm{z.i2}{\tilde{I}_2 = -3.5\angle 0^\circ\ \text{A}}

KVL around the left loop delivers the port voltage

-\tm{z.vx}{\tilde{V}_x} + \tm{z.r20}{20(1)} + \tm{z.l1}{j4(1)} - \tm{z.d1}{j2\tilde{I}_2} - \tm{z.d2}{j2\tilde{I}_1} - \tm{z.l2}{j3\tilde{I}_2} = 0

\tilde{V}_x = 20 + j4 + j7 + j2 + j10.5 \;\Longrightarrow\; \tilde{V}_x = 20 + j23.5\ \text{V}

The impedance

Z_T = \frac{\tilde{V}_x}{1\angle 0^\circ} \;\Longrightarrow\; \boxed{Z_T = 20 + j23.5\ \Omega = 30.86\angle 49.59^\circ\ \Omega}

The port therefore behaves as 20 Ω in series with an inductive reactance of 23.5 Ω. Note that neither number could have been obtained by combining the 20 Ω resistor with the coil impedances: the mutual terms contribute $j7 + j2$ of the $j23.5$ , and the capacitor influences the result only through $\tilde{I}_2$ .

Experiment 2: the open-circuit voltage

Open the port and restore the source

The test source is removed, the 10∠0° V source is restored, and the terminals are left open. No current can enter terminal a, so the top branch is dead and $\tilde{I}_1 = 0$ . The dependent source in the middle branch, whose value is $j2\tilde{I}_1$ , therefore vanishes as well.

KVL around the right loop

\tm{t.l2}{j3\tilde{I}_2} + \tm{t.d2}{j2\tilde{I}_1} + \tm{t.vs}{10} + \tm{t.c}{(-j5)\tilde{I}_2} = 0 \;\Longrightarrow\; -j2\tilde{I}_2 + 10 = 0

\tm{t.i2}{\tilde{I}_2 = \frac{10}{j2} = -j5 = 5\angle{-90^\circ}\ \text{A}}

The open-circuit voltage

With no current in the top branch, the 20 Ω resistor and the self-impedance of the first coil hold no voltage, and only the mutual terms survive:

-\tm{t.vt}{\tilde{V}_T} - \tm{t.d1}{j2\tilde{I}_2} - \tm{t.d2}{j2\tilde{I}_1} - \tm{t.l2}{j3\tilde{I}_2} = 0

\tilde{V}_T = -j2(-j5) - j3(-j5) = -10 - 15 \;\Longrightarrow\; \boxed{\tilde{V}_T = -25 = 25\angle 180^\circ\ \text{V}}

In the time domain, $v_T(t) = 25\cos(\omega t + 180^\circ)$ V. The angle of 180° records that the open-circuit voltage opposes the assumed polarity.

The equivalent

The Thevenin equivalent seen at a-b: 25∠180° V in series with 20 Ω and j23.5 Ω. Any load connected at the terminals sees exactly this circuit.

The Norton form follows without further analysis:

\tilde{I}_N = \frac{\tilde{V}_T}{Z_T} = \frac{25\angle 180^\circ}{30.86\angle 49.59^\circ} = 0.810\angle 130.41^\circ\ \text{A}

Check: the short-circuit current, computed from scratch

Shorting a to b makes the port voltage zero and leaves two unknowns, $\tilde{I}_1$ (defined into the dot of the first coil, and therefore running from the network out through terminal a) and $\tilde{I}_2$ . The left loop gives $(20 + j6)\tilde{I}_1 + j5\tilde{I}_2 = 0$ , and the right loop gives $j2\tilde{I}_2 - j7\tilde{I}_1 = 10$ . Eliminating $\tilde{I}_2$ :

\tilde{I}_2 = (-1.2 + j4)\,\tilde{I}_1 \;\Longrightarrow\; (-8 - j9.4)\,\tilde{I}_1 = 10 \;\Longrightarrow\; \tilde{I}_1 = 0.810\angle 130.41^\circ\ \text{A}

That current flows out of terminal a and through the short, so it is the Norton current, and it agrees with $\tilde{V}_T / Z_T$ to every digit carried ✓. The redundant experiment confirms both parameters at once.

The impedance is frequency-specific

$Z_T = 20 + j23.5\ \Omega$ holds at the one frequency at which the element impedances were given. A Thevenin equivalent in the phasor domain is valid for a single $\omega$ ; a different source frequency requires the whole calculation again.

§7 · Your turn

Practice: series-aiding and series-opposing coils

The circuit below places two coupled coils in series in a single loop, which is the arrangement in which the effect of the dots is easiest to feel. The problem should be attempted on paper before the hints are opened, in order.

The source $v_s = 10\cos(50t)$ V drives 20 Ω, $L_1 = 0.4$ H, and $L_2 = 0.2$ H in series. The coils are coupled with $m = 0.1$ H, and both dots are printed at the left ends. Find $i(t)$ . Then repeat the calculation with the dot of the second coil moved to its right end.

Hint 1: what the dots say about this loop

One current circulates. It enters the dot of the first coil and, because the second dot is also at a left end, it enters the dot of the second coil as well. Both dot-entering currents are therefore the same phasor $\tilde{I}$ , with no sign flip anywhere.

Hint 2: assemble the branch voltages

At $\omega = 50$ rad/s the reactances are $\omega L_1 = 20\ \Omega$ , $\omega L_2 = 10\ \Omega$ , and $\omega m = 5\ \Omega$ . Each coil holds its own self term plus a mutual term of $j5\tilde{I}$ , and every one of the four terms is traversed as a drop.

Hint 3: the KVL equation

-10\angle 0^\circ + 20\tilde{I} + \left[j20\tilde{I} + j5\tilde{I}\right] + \left[j10\tilde{I} + j5\tilde{I}\right] = 0

Solution

The bracketed groups are the two coil branches, each written as its self term plus its mutual term. Collecting:

(20 + j40)\,\tilde{I} = 10\angle 0^\circ \;\Longrightarrow\; \tilde{I} = \frac{10\angle 0^\circ}{44.72\angle 63.43^\circ} = 0.2236\angle{-63.43^\circ}\ \text{A}

\boxed{i(t) = 0.224\cos(50t - 63.43^\circ)\ \text{A}}

The check by equivalent inductance. Because one current passes through both coils, the loop may be collapsed to a single inductance. With both currents entering their dots, the mutual terms add:

L_{eq} = L_1 + L_2 + 2m = 0.4 + 0.2 + 0.2 = 0.8\ \text{H} \;\Longrightarrow\; j\omega L_{eq} = j40\ \Omega \ \checkmark

This is the series-aiding connection.

The second connection. Moving the dot of the second coil to its right end means the current leaves that dot, so both mutual terms reverse and the coils are series-opposing:

L_{eq} = L_1 + L_2 - 2m = 0.4\ \text{H} \;\Longrightarrow\; Z = 20 + j20\ \Omega \;\Longrightarrow\; \tilde{I} = 0.3536\angle{-45^\circ}\ \text{A}

i(t) = 0.354\cos(50t - 45^\circ)\ \text{A}

Reversing one dot has increased the current by 58 % and halved the effective inductance, without a single element value being changed. Note also that $m = 0.1$ H satisfies $m \le \sqrt{L_1 L_2} = 0.283$ H, so the pair is physically realizable, with a coupling coefficient of $k = 0.354$ .