Tutorial 17 — Magnetically Coupled Circuits
When two coils share a common magnetic flux, a current in one induces a voltage in the other through the mutual inductance \(M = k\sqrt{L_1 L_2}\) , where the coupling coefficient \(0 \le k \le 1\) . The dot convention fixes the sign of the mutual term. These problems cover induced voltage and stored energy, phasor analysis of coupled coils, mesh analysis with mutual terms, equivalent and reflected impedance, and series aiding/opposing connections.
AC Analysis · Electric Circuit Analysis · 13 solved problems
Let \(L_1 = 0.4\ \text{H}\), \(L_2 = 2.5\ \text{H}\), \(k = 0.6\), and \(i_1 = 4i_2 = 20\cos(500t-20^\circ)\ \text{mA}\). Determine \(v_1(0)\) and the total energy stored in the system at \(t=0\).
Solution
The mutual inductance is
\[ M = k\sqrt{L_1 L_2} = 0.6\sqrt{(0.4)(2.5)} = 0.6\ \text{H} \]
With \(i_1 = 20\cos(500t-20^\circ)\) mA and \(i_2 = \tfrac14 i_1 = 5\cos(500t-20^\circ)\) mA, the primary voltage is
\[ v_1(t) = L_1\frac{di_1}{dt} + M\frac{di_2}{dt} \]
Evaluating the derivatives at \(t=0\):
\[ v_1(0) = 0.4\big[-10\sin(-20^\circ)\big] + 0.6\big[-2.5\sin(-20^\circ)\big] = 1.881\ \text{V} \]
The instantaneous currents at \(t=0\):
\[ i_1(0) = 20\cos(-20^\circ) = 18.79\ \text{mA},\qquad i_2(0) = 0.25\,i_1(0) = 4.698\ \text{mA} \]
The total energy stored in the coupled system (currents both enter the dots, so the mutual term adds):
\[ w = \tfrac12 L_1 i_1^2 + \tfrac12 L_2 i_2^2 + M\,i_1 i_2 = 151.2\ \mu\text{J} \]
Answer \(v_1(0) = 1.881\ \text{V},\quad w = 151.2\ \mu\text{J}\)
The coils have \(L_1 = 40\) mH, \(L_2 = 5\) mH, and coupling coefficient \(k = 0.6\). Find \(i_1(t)\) and \(v_2(t)\), given that \(v_1(t) = 10\cos\omega t\) and \(i_2(t) = 2\sin\omega t\), with \(\omega = 2000\) rad/s.
Solution
The mutual inductance and the branch reactances at \(\omega = 2000\) rad/s:
\[ \begin{aligned}
M &= k\sqrt{L_1 L_2} = 0.6\sqrt{40\times 5} = 8.485\ \text{mH}\\
j\omega L_1 &= j80,\quad j\omega L_2 = j10,\quad j\omega M = j16.97
\end{aligned} \]
From the dot markings (one current enters a dot, the other the undotted end) the mutual term is subtractive, giving the symmetric set
\[ \begin{aligned}
V_1 &= j80\,I_1 - j16.97\,I_2\\
V_2 &= -j16.97\,I_1 + j10\,I_2
\end{aligned} \]
With the given \(V_1 = 10\angle 0^\circ\) and \(I_2 = 2\angle{-90^\circ} = -j2\), the first equation gives
\[ 10 = j80\,I_1 - j16.97(-j2) = j80\,I_1 - 33.94 \;\Rightarrow\; I_1 = 0.5493\angle{-90^\circ} \]
Substituting into the second equation:
\[ V_2 = -j16.97(-j0.5493) + j10(-j2) = -9.32 + 20 = 10.68\angle 0^\circ\ \text{V} \]
Answer \(i_1(t) = 0.5493\sin 2000t\ \text{A},\quad v_2(t) = 10.68\cos 2000t\ \text{V}\)
Find \(V_0\) marked in the circuit.
Solution
Writing the two mesh equations (the mutual reactance is \(j1\,\Omega\)):
\[ \begin{aligned}
\text{Mesh-1:}\quad 12 &= (2+j6)I_1 + jI_2\\
\text{Mesh-2:}\quad 0 &= jI_1 + (2 - j1 + j4)I_2
\end{aligned} \]
In matrix form:
\[ \begin{bmatrix} 12 \\ 0 \end{bmatrix} =
\begin{bmatrix} 2+j6 & j \\ j & 2+j3 \end{bmatrix}
\begin{bmatrix} I_1 \\ I_2 \end{bmatrix} \]
Solving for the second mesh current:
\[ I_2 = -0.4381 + j0.3164 \]
The output voltage across the \(1\ \Omega\) resistor:
\[ V_0 = I_2 \times 1 = 540.5\angle 144.16^\circ\ \text{mV} \]
Answer \(V_0 = 540.5\angle 144.16^\circ\ \text{mV}\)
Find \(V_x\) marked in the network.
Solution
After a source transformation (the \(2\angle 0^\circ\) A source with its \(-j1\ \Omega\) capacitor becomes a \(-j2\) V source in series with \(-j1\ \Omega\)), apply mesh analysis with mutual reactance \(j1\ \Omega\):
\[ \begin{aligned}
\text{Mesh-1:}\quad 8\angle 30^\circ &= (2+j4)I_1 - jI_2\\
\text{Mesh-2:}\quad (2 - j1 + j4)I_2 - jI_1 &= j2
\end{aligned} \]
Solving the pair:
\[ I_2 = 1.037\angle 21.12^\circ\ \text{A} \]
The voltage across the \(2\ \Omega\) resistor:
\[ V_x = 2I_2 = 2.074\angle 21.12^\circ\ \text{V} \]
Answer \(V_x = 2.074\angle 21.12^\circ\ \text{V}\)
Determine \(L_{eq}\) for the coupled network.
Solution
Take \(\omega = 1\) rad/s and drive the port with a \(1\) V test source. The two loop equations are
\[ \begin{aligned}
1 &= j8\,I_1 + j4\,I_2\\
0 &= j4\,I_1 + j18\,I_2
\end{aligned} \]
Solving for the input-loop current:
The input impedance is purely inductive, \(Z = 1/I_1 = jL_{eq}\), so
\[ L_{eq} = \frac{1}{jI_1} = 7.111\ \text{H} \]
Answer \(L_{eq} = 7.111\ \text{H}\)
If the load is a 15 mH inductor of impedance \(Z_L = j40\ \Omega\), determine \(Z_{in}\) when \(k = 0.6\).
Solution
The load \(Z_L = j40\ \Omega\) is a 15 mH inductor, fixing the operating frequency:
\[ j\omega L = j40 \;\Rightarrow\; \omega = \frac{40}{15\times 10^{-3}} = 2667\ \text{rad/s} \]
The mutual inductance and the reactances of each element at this frequency:
\[ \begin{aligned}
M &= k\sqrt{L_1 L_2} = 0.6\sqrt{12\times 10^{-3}\cdot 30\times 10^{-3}} = 11.384\ \text{mH}\\[2pt]
12\ \text{mH} &\Rightarrow j32,\quad 30\ \text{mH} \Rightarrow j80,\quad 11.384\ \text{mH} \Rightarrow j30.36
\end{aligned} \]
The input impedance is the primary self-impedance plus the reflected impedance \((\omega M)^2/Z_{22}\):
\[ \begin{aligned}
Z_{in} &= 10 + j32 + \frac{(\omega M)^2}{j80 + 60 + j40}
= 10 + j32 + \frac{(30.36)^2}{60 + j120}\\
&= 13.073 + j25.86\ \Omega
\end{aligned} \]
Answer \(Z_{in} = 13.073 + j25.86\ \Omega\)
Find \(I_0\).
Solution
After a source transformation (\(4\angle 60^\circ\) A with \(50\ \Omega \Rightarrow 200\angle 60^\circ\) V), the coupling reactance is
\[ \omega M = k\sqrt{\omega L_1\,\omega L_2} = 0.6\sqrt{20\times 40} = 17 \]
The mesh equations are
\[ \begin{aligned}
\text{Mesh-1:}\quad 200\angle 60^\circ &= (50 - j30 + j20)I_1 - j17\,I_2\\
\text{Mesh-2:}\quad 0 &= (10 + j40)I_2 - j17\,I_1
\end{aligned} \]
Solving simultaneously:
\[ I_2 = I_0 = 1.516\angle 92.04^\circ\ \text{A} \]
Answer \(I_0 = 1.516\angle 92.04^\circ\ \text{A}\)
Find the input impedance \(Z\).
Solution
Drive the port with a \(1\angle 0^\circ\) V source and write the mesh equations:
\[ \begin{aligned}
\text{Mesh-1:}\quad 1 &= (1 + j10)I_1 - j4\,I_2\\
\text{Mesh-2:}\quad 0 &= -jI_1 + (2 + j3)I_2
\end{aligned} \]
Solving for the input current:
\[ I_1 = 0.019 - j0.1068 \]
The input impedance:
\[ Z = \frac{1}{I_1} = 9.219\angle 79.91^\circ\ \Omega \]
Answer \(Z = 9.219\angle 79.91^\circ\ \Omega\)
Determine \(I_1,\ I_2,\ I_3\) and the energy stored in the coupled coils at \(t = 2\ \text{ms}\), taking \(\omega = 1000\) rad/s.
Solution
After a source transformation (\(3\angle 90^\circ\) A with \(4\ \Omega \Rightarrow j12\) V), the coupling reactance is
\[ \omega M = k\sqrt{\omega L_1\,\omega L_2} = 0.5\times 10 = 5 \]
The mesh equations are
\[ \begin{aligned}
\text{Mesh-1:}\quad j12 &= (4 + j5)I_1 + j10\,I_2\\
\text{Mesh-2:}\quad -20 &= j10\,I_1 + (8 + j5)I_2
\end{aligned} \]
Solving gives the three branch currents:
\[ \begin{aligned}
I_1 &= 2.462\angle 72.18^\circ\ \text{A}\\
I_2 &= 0.878\angle{-97.48^\circ}\ \text{A}\\
I_3 &= I_1 - I_2 = 3.329\angle 74.89^\circ\ \text{A}
\end{aligned} \]
Converting to the time domain and evaluating at \(t = 2\) ms, where \(1000t = 2\ \text{rad} = 114.6^\circ\):
\[ \begin{aligned}
i_1(t) &= 2.462\cos(1000t + 72.18^\circ)\ \text{A}, &\; i_1(2\,\text{ms}) &= 2.462\cos(186.78^\circ) = -2.445\ \text{A}\\
i_2(t) &= 0.878\cos(1000t - 97.48^\circ)\ \text{A}, &\; i_2(2\,\text{ms}) &= 0.878\cos(17.12^\circ) = +0.839\ \text{A}
\end{aligned} \]
With \(\omega L_1 = 10\) and \(\omega = 1000\), the coils are \(L_1 = L_2 = 10\) mH and \(M = 0.5L_1 = 5\) mH. The energy stored at \(t = 2\) ms is
\[ W = \tfrac12 L_1 i_1^2 + \tfrac12 L_2 i_2^2 - M\,i_1 i_2 = 43.67\ \text{mJ} \]
Answer \(I_1 = 2.462\angle 72.18^\circ,\ I_2 = 0.878\angle{-97.48^\circ},\ I_3 = 3.329\angle 74.89^\circ\ \text{A};\ W = 43.67\ \text{mJ}\)
Calculate the total inductance of the three coupled coils.
Solution
Summing the self term and the mutual contributions for each coil (sign set by the dot markings):
\[ \begin{aligned}
\text{Coil-1:}\quad & L_1 - M_{12} + M_{13} = 6 - 4 + 2 = 4\\
\text{Coil-2:}\quad & L_2 - M_{21} - M_{23} = 8 - 4 - 5 = -1\\
\text{Coil-3:}\quad & L_3 + M_{31} - M_{32} = 10 + 2 - 5 = 7\\
& L_T = 4 - 1 + 7 = 10\ \text{H}
\end{aligned} \]
Equivalently, using the series formula for three coupled coils:
\[ \begin{aligned}
L_T &= L_1 + L_2 + L_3 - 2M_{12} - 2M_{23} + 2M_{13}\\
&= 24 - 8 - 10 + 4 = 10\ \text{H}
\end{aligned} \]
Answer \(L_T = 10\ \text{H}\)
Calculate \(i_2(t)\) if \(v_1(t) = 8\sin 720t\).
Solution
At \(\omega = 720\) rad/s, replacing each element by its reactance and writing the three mesh equations leads to the system
\[ \begin{bmatrix}
1.8 + j0.72 & -j0.72036 & j3.6\times 10^{-4}\\
-j0.72036 & j0.77183 & -j1.44036\\
j3.6\times 10^{-4} & -j1.44036 & 2 + j1.44
\end{bmatrix}
\begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix}
= \begin{bmatrix} 8 \\ 0 \\ 0 \end{bmatrix} \]
Solving the system for the middle mesh current:
\[ I_2 = 1.6958 + j2.6335 = 3.132\angle 57.22^\circ\ \text{A} \]
Returning to the time domain:
\[ i_2(t) = 3.132\sin(720t + 57.22^\circ)\ \text{A} \]
Answer \(i_2(t) = 3.132\sin(720t + 57.22^\circ)\ \text{A}\)
Additional Practice Problems
Two self-contained worked examples with schematics: the series aiding/opposing connection of two coupled coils, and finding the coupling coefficient together with the magnetic energy stored.
Two coupled coils with \(L_1 = 4\ \text{H}\), \(L_2 = 6\ \text{H}\) and \(M = 2\ \text{H}\) are connected in series. Find the equivalent inductance for the series-aiding and the series-opposing connections.
a
L₁ = 4 H
M = 2 H
L₂ = 6 H
b
Solution
For two coils in series the equivalent inductance is \(L_{eq} = L_1 + L_2 \pm 2M\). With both dots on the same (entering) side, the fluxes aid and the mutual term adds:
\[ L_{eq,\text{aiding}} = L_1 + L_2 + 2M = 4 + 6 + 2(2) = 14\ \text{H} \]
If one coil is reversed so the dots are on opposite ends, the fluxes oppose and the mutual term subtracts:
\[ L_{eq,\text{opposing}} = L_1 + L_2 - 2M = 4 + 6 - 2(2) = 6\ \text{H} \]
Answer \(L_{eq,\text{aiding}} = 14\ \text{H},\quad L_{eq,\text{opposing}} = 6\ \text{H}\)
Two coupled coils \(L_1 = 2\ \text{H}\) and \(L_2 = 8\ \text{H}\) have a coupling coefficient \(k = 0.5\). They carry steady currents \(i_1 = 4\ \text{A}\) and \(i_2 = 2\ \text{A}\), both entering the dotted terminals. Find the mutual inductance \(M\) and the total magnetic energy stored.
L₁ = 2 H
i₁ = 4 A
L₂ = 8 H
i₂ = 2 A
k = 0.5
Solution
The mutual inductance follows from the coupling coefficient:
\[ M = k\sqrt{L_1 L_2} = 0.5\sqrt{(2)(8)} = 0.5\times 4 = 2\ \text{H} \]
Both currents enter the dotted terminals, so the fluxes aid and the mutual energy term is positive:
\[ \begin{aligned}
W &= \tfrac12 L_1 i_1^2 + \tfrac12 L_2 i_2^2 + M\,i_1 i_2\\
&= \tfrac12(2)(4)^2 + \tfrac12(8)(2)^2 + (2)(4)(2)\\
&= 16 + 16 + 16 = 48\ \text{J}
\end{aligned} \]
Answer \(M = 2\ \text{H},\quad W = 48\ \text{J}\)
