RL 10-14-2020

Let the current in the left inductor be $I_1$ and that through the right inductor be $I_2$ . The circuit equations read:

$L\dot{I}_1 + R(I_1+I_2) = V_1$ $L\dot{I}_2 + R(I_1+I_2) = V_2$

Adding the above equations:

$L(\dot{I}_1 +\dot{I}_2)+2R(I_1+I_2)=V_1+V_2$

Let $\omega = 2\pi$ . Here, $I=I_1+I_2$ is the resistor current. It can also be seen that $I(0)=0$ . The final differential equation for resistor current is:

$L\dot{I}+ 2IR = V_1+V_2$

The above equation has a null solution and a particular solution. The null and particular solutions are:

$I_n = I_o \mathrm{e}^{-2t/3}$ $I_p = \frac{10(2 + 3\omega)\sin(\omega t) + 10(2 - 3\omega)\cos(\omega t)}{9\omega^2+4}$

Adding up both solutions and solving for $I_o$ gives:

$I(t) = \frac{10(2 + 3\omega)\sin(\omega t) + 10(2 - 3\omega)\cos(\omega t)}{9\omega^2+4} -\frac{10(2 - 3\omega)}{9\omega^2+4}\mathrm{e}^{-2t/3}$

The steps for solving have been left out. I used the $D$ operator method to solve this equation. Other techniques such as the method of integrating factors, or Laplace transformation will work just as well here. Plotting the above function:

The inductor currents can be solved for by integrating the following equations:

$I_1 = \int_{0}^{t} \frac{V_1(\tau)-RI(\tau)}{L} \ d\tau$ $I_2 = \int_{0}^{t} \frac{V_2(\tau)-RI(\tau)}{L} \ d\tau$

I do not observe anything unexpected. The currents are a little higher initially due to the presence of the decaying exponential which goes away with time. A phase difference in the steady state solution between inductor currents is also not unexpected. I think I am unable to latch onto any interesting behaviour at the moment.