Answers and Discussions for
Problem 1, part a
Voltage Calculation
The voltage equation for these five identical, magnetically coupled inductors may be written down in matrix form (resistance is neglected).
$ \begin{align} \vec{v}_{12345s} &= \cancelto{\mathbf{0}}{\mathbf{r}_s} \vec{i}_{12345s} + p \vec{\lambda}_{12345s} \\ \vec{v}_{12345s} &= p \vec{\lambda}_{12345s} \end{align} $
The flux linkage equation needs to be written (linear magnetics are assumed).
$ \begin{align} \vec{\lambda}_{12345s} &= \mathbf{L}_{ss} \vec{i}_{12345s} \\ \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \\ \lambda_5 \end{bmatrix} &= \begin{bmatrix} L_s & M_{ns} & M_{fs} & M_{fs} & M_{ns} \\ M_{ns} & L_s & M_{ns} & M_{fs} & M_{fs} \\ M_{fs} & M_{ns} & L_s & M_{ns} & M_{fs} \\ M_{fs} & M_{fs} & M_{ns} & L_s & M_{ns} \\ M_{ns} & M_{fs} & M_{fs} & M_{ns} & L_s \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \end{bmatrix} \end{align} $
Because the inductors are identical to each other, only three unique, constant entries exist in the inductance matrix $ \mathbf{L}_{ss} $. The self-inductance $ L_s > 0 $ combines the effects of leakage inductance and magnetizing inductance. The mutual inductance between inductors near each other, $ \frac{2\pi}{5} \, \text{rad} $ apart in other words, is denoted $ M_{ns} $. The mutual inductance between inductors far from each other, $ \frac{4\pi}{5} \, \text{rad} $ apart in other words, is denoted $ M_{fs} $. Because inductors with spacing $ -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2} $ produce a positive value of cosine, the mutual inductance of near conductors is positive $ M_{ns} > 0 $. The inductors with spacing $ -\pi < \Delta\theta < -\frac{\pi}{2} $ or $ +\frac{\pi}{2} < \Delta\theta < \pi $ produce a negative value of cosine, leading to a negative mutual inductance $ M_{fs} < 0 $.
Solving Imbalanced Conditions
The current through the circuit is given knowing that four the the five inductors are open-circuited and carry no current.
$ \begin{equation} \begin{bmatrix} i_1(t) \\ i_2(t) \\ i_3(t) \\ i_4(t) \\ i_5(t) \end{bmatrix} = \begin{bmatrix} 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t\right) \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \text{A} \end{equation} $
The following, partially complete information is measured from the circuit.
$ \begin{equation} \begin{bmatrix} v_1(t) \\ v_2(t) \\ v_3(t) \\ v_4(t) \\ v_5(t) \end{bmatrix} = \begin{bmatrix} 40 \cos(\omega_1 t + \phi_1) \\ 10 \cos(\omega_2 t + \phi_2) \\ v_3(t) \\ 1 \cos(\omega_4 t + \phi_4) \\ v_5(t) \end{bmatrix} \, \text{V} \end{equation} $
Because of the linear magnetics, the frequency of the voltage response must match the frequency of the current excitation: $ \omega_j = 100 \, \frac{\text{rad}}{\text{s}} \, \forall j \in \{1, 2, 3, 4, 5\} $.
$ \begin{equation} \boxed{\omega_1 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $
$ \begin{equation} \boxed{\omega_2 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $
$ \begin{equation} \boxed{\omega_4 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $
By analyzing the voltage equation and flux linkage equation, the phase shift of the voltage of each coil becomes known. The first coil is analyzed.
$ \begin{align} v_1(t) &= p \lambda_1(t) \\ v_1(t) &= p \left[L_s i_1(t)\right] \\ v_1(t) &= L_s \frac{d}{dt} \left[10 \cos\left(\omega_1 t\right) \, \text{A}\right] \\ v_1(t) &= L_s \left[-10 \omega_1 \sin\left(\omega_1 t\right) \, \text{A}\right] \\ v_1(t) &= 10^3 L_s \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_2(t) &= 10^3 M_{ns} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_4(t) &= 10^3 M_{fs} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_4(t) &= 10^3 |M_{fs}| \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \end{align} $
Similar calculations unfold for $ v_2(t) $ and $ v_4(t) $, just with different amplitudes. The voltages all have a phase shift $ \phi_j = -\frac{\pi}{2} \, \text{rad} \, j \in \{1, 2, 5\} $ and $ \phi_j = +\frac{\pi}{2} \, \text{rad} \, j \in \{3, 4\} $. In other words, the current lags the voltage with quadrature phase shift for inductors shifted $ -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2} $, and the current leads the voltage with quadrature phase shift for the remaining inductors.
$ \begin{equation} \boxed{\phi_1 = -\frac{\pi}{2} \, \text{rad}} \end{equation} $
$ \begin{equation} \boxed{\phi_2 = -\frac{\pi}{2} \, \text{rad}} \end{equation} $
$ \begin{equation} \boxed{\phi_4 = +\frac{\pi}{2} \, \text{rad}} \end{equation} $
